MS-0001-1922.65

Human-AI Interactions and Societal Pitfalls

Francisco Castro, Jian Gao Anderson School of Management, UC Los Angeles, Los
Angeles, CA 90024<EMAIL_ADDRESS><EMAIL_ADDRESS>Sébastien Martin Kellogg School of Management,
Northwestern University, Evanston, IL 60208,
<EMAIL_ADDRESS>
When working with generative artificial intelligence (AI), users may see
productivity gains, but the AI-generated content may not match their
preferences exactly. To study this effect, we introduce a Bayesian framework
in which heterogeneous users choose how much information to share with the AI,
facing a trade-off between output fidelity and communication cost. We show
that the interplay between these individual-level decisions and AI training
may lead to societal challenges. Outputs may become more homogenized,
especially when the AI is trained on AI-generated content. And any AI bias may
become societal bias. A solution to the homogenization and bias issues is to
improve human-AI interactions, enabling personalized outputs without
sacrificing productivity.

##  1 Introduction

Generative artificial intelligence (AI) systems, particularly large language
models (LLMs), have improved at a rapid pace. For example, ChatGPT recently
showcased its advanced capacity to perform complex tasks and human-like
behaviors (OpenAI 2023b), reaching 100 million users within two months of its
2022 launch (Hu 2023). This progress is not limited to text generation, as
demonstrated by other recent generative AI systems such as Midjourney
(Midjourney 2023) (a text-to-image generative AI) and GitHub Copilot (Github
2023) (an AI pair programmer that can autocomplete code). Eloundou et al.
(2023) estimated that about 80% of the U.S. workforce could be affected by the
introduction of LLMs, and 19% of the workers may have at least 50% of their
tasks impacted. In particular, AI can make users more productive by generating
complex content in seconds, while users can simply communicate their
preferences. For example, Noy and Zhang (2023) highlighted that ChatGPT can
substantially improve productivity in writing tasks, and GitHub claims that
Copilot increases developer productivity by up to 55% (Kalliamvakou 2023).

However, content generated with the help of AI is not exactly the same as
content generated without AI. The boost in productivity may come at the
expense of users’ idiosyncrasies, such as personal style and tastes,
preferences we would naturally express without AI. To let users express their
preferences, many AI systems let users edit their prompt (e.g., Midjourney) or
allow more natural interactions (e.g., ChatGPT), and users can always review
and edit the AI-generated output themselves (Vaithilingam et al. 2022).
However, aligning a user’s intentions with an AI’s output can take time and
may not always be worth it if the AI’s first or default output “does the job.”
Consider a simple example where we use Copilot to code a Python function that
calculates the sum of numbers in a nested list. Figure 1 shows that Copilot’s
default output (the first to the left) was correct and functional. However, it
did not correspond to our own way of writing the same function given enough
time (at the bottom of the figure). To push Copilot to better match our style,
we could provide more information by articulating a more detailed prompt.
However, the figure shows this may require many steps, which goes against the
goal of being more productive. Similarly, Lingard (2023) described guiding
ChatGPT through incremental prompting. In essence, users’ time and effort to
convey information about their desired outcome to an AI can enhance the
output’s alignment with their preferences, albeit at the expense of an
increased communication cost. In short, users face a trade-off between AI
_output fidelity_ and _communication cost_.111In fact, the importance and the
associated costs of communicating with AIs have given rise to a new profession
called prompt engineering (Mok 2023), and spurred the creation of novel
marketplaces like PromptBase (URL: https://promptbase.com/).

Successive Prompts Function to return the sum of numbers in a nested list Make
it shorter by using a generator expression Make sure you check if the element
is a list by using isinstance() Copilot’s Output Our solution without AI:

Figure 1: The incremental information provided to align GitHub Copilot’s
Python code output with our preference. While the initial output in Attempt 1
is functional, it significantly differs from our solution without AI. Bridging
the gap requires several iterations.

Different users may respond to this trade-off differently, but those who value
productivity more than fidelity will rely on AI more and willingly let go of
their own preferences. We are interested in the potential societal
consequences of these choices. First, working with AI may be more beneficial
for some users than others: in the Copilot example, users who prefer the
default output would not even need to communicate with the AI to have high
fidelity to their preferences. Second, as users do not share complete
preferences with the AI and let it “choose” for them, the produced content may
be, on average, _homogenized_ towards the AI’s default choices. For example,
ChatGPT has been trained with reinforcement learning from human feedback
(RLHF) (Kinsella 2023) to have a specific tone and language. If students use
ChatGPT’s help for their homework, their writing style may be influenced by
ChatGPT’s. More generally, AIs are built by a few but used by many, and there
is a risk any AI bias could turn into a _societal bias_. The AI training
process may involve censoring (e.g., the choice of the dataset) and human
input (e.g., RLHF), which could intentionally or unintentionally lead to bias.
For example, some studies discuss ChatGPT’s inclination towards left-leaning
political stances (Hartmann et al. 2023, Rozado 2023, Motoki et al. 2023). All
in all, because of the benefits of increased productivity and the balance
between output fidelity and communication costs, users could willingly produce
less diverse content that is vulnerable to potential AI biases.

We propose a Bayesian model to study the societal consequences of human-AI
interactions. For a given task, rational users can exchange information with
the AI to align its output with their heterogeneous preferences. The AI has a
knowledge of the distribution of preferences in the population and uses a
Bayesian update to create the optimal output with maximal expected fidelity
given the information shared by the user. Users choose the amount of
information they share to maximize their utility, balancing the cost of
communication with the fidelity of the output. In this setting, we aim to
formalize and evaluate the societal risks of homogenization and AI bias and
how they could be mitigated.

When solving for each user’s optimal decision, we find that their use of AI
depends on how “unique” they are. Users with more common preferences simply
accept the default output, avoiding any communication costs at the expense of
a small fidelity mismatch. In contrast, users with more unique preferences
share information with the AI to reduce fidelity errors, albeit with higher
communication costs. And the most unique users do not benefit from the AI and
simply perform the task themselves. Interestingly, we find that for less
common users, their fidelity improves with the uniqueness of their
preferences. In other words, when we compare two users with relatively unique
preferences (neither the most common nor the most unique), the user with the
most unique preferences between the two will experience better fidelity.

To establish the homogenization effect, we prove that any output resulting
from human-AI interactions is less unique than what a user would have done
without AI. This is confirmed at the population level, where the AI-generated
output distribution has a lower variance than the users’ preference
distribution. This phenomenon is exacerbated when AI-generated content is used
to train the next generation of AI: we show numerically that the users’
rational decisions and the AI’s training process can mutually reinforce each
other, leading to a homogenization “death spiral.” However, we also
demonstrate numerically that facilitating human-AI interaction by offering
improved means for users to express their preferences can serve as a valuable
tool in mitigating this effect and preserving population diversity. For
example, Open AI has experimented with custom instructions (OpenAI 2023a) and
voiced-based interactions (OpenAI 2023c), while Jina AI offers tools that
optimize prompts.222E.g., automatic prompt optimization such as Prompt Perfect
(URL: https://promptperfect.jina.ai/).

We also study the effects of AI bias, identifying who benefits or loses when
using an AI model that does not accurately reflect the population preference
distribution. At the population level, the censoring type of bias (e.g.,
biasing against the more unique preferences) negatively impacts the population
utility as whole, especially users with uncommon preferences who rely on AI
interactivity the most. This may seem counter-intuitive as we might assume
that the majority with common preferences would benefit from censorship. Yet,
our findings reveal that the benefits for this majority are marginal, while
the harm to the minority with unique preferences is substantial, leading to an
overall loss in the population utility. On the other hand, directional biases
(e.g., a slightly left-leaning AI) are not as harmful in terms of utility, but
any directional bias will influence the users’ chosen output, leading to a
societal bias. On the positive side, the user interactions with the AI
partially counter the effects of AI bias, highlighting the need to consider
human decisions to fully understand the impact of generative AI.

We show that tasks that are either hard to do without AI (e.g., image
generation) or for which speed is particularly important (e.g., grammar
correction) are especially sensitive to the risks of homogenization and bias.
However, our research demonstrates that creating models that facilitate human-
AI interactions can significantly limit these risks and preserve the
population preference diversity.

##  2 Literature review

##### Related studies on human-AI interactions.

Our paper is related to some recent modeling studies about human-AI
interaction in operations management (e.g., Agrawal et al. (2018), Ibrahim et
al. (2021), Bastani et al. (2022), Chen et al. (2022), Dai and Singh (2023),
Boyacı et al. (2023), Mclaughlin and Spiess (2023), de Véricourt and Gurkan
(2023)). Essentially, their primary focus lies in examining the potential
impact of the coexistence of humans and an AI on performance, such as
accuracy, and exploring how the predictive performance can be enhanced or
hindered compared to decisions made solely by humans or AI. For example, de
Véricourt and Gurkan (2023) consider the human-AI interactions in which a
human agent supervises an AI to make some high-stakes decisions. They show
that the agent may be subject to a verification bias and hesitates forever
whether the AI performs better than the agent because the agent can overrule
the AI before observing the correctness of the AI’s predictions. Ibrahim et
al. (2021) build a stylized model to analyze how human judgments can improve
AI predictions. In the paper of Boyacı et al. (2023), the authors consider a
situation in which a human agent has to spend a cognitive cost collecting
information in a decision process, whereas an AI can provide him with some
additional information without cognitive cost. They show that the AI input can
improve the overall accuracy of human decisions but may incur a higher
propensity for certain types of errors. These papers primarily focus on
decision-making when human and AI options exist separately. In contrast, our
paper considers a more interactive setting, where users can provide a
generative AI with more information to improve the AI’s outputs but have to
spend a communication cost.

##### Related studies on generative AIs.

With the popularity of ChatGPT, many scholars have engaged in research on its
impact on people’s lives and in their respective fields, such as labor markets
(Eloundou et al. 2023), marketing (Brand et al. 2023), healthcare (Sallam
2023), and so on. Most of the research uses empirical analysis to investigate
whether generative AI, represented by ChatGPT, can truly bring us more
benefits and conveniences. For instance, Noy and Zhang (2023) show that
ChatGPT can substantially improve productivity in mid-level professional
writing tasks. Binz and Schulz (2023) tested GPT-3 with some experiments from
the cognitive psychology literature. They find that GPT-3 can solve many of
those tasks well and even sometimes outperform humans’ performance. Our study
approaches this question from a different angle. Through a modeling method, we
attempt to foreshadow how our lives may change under the widespread
application of generative AIs due to people’s rational decision-making when
interacting with AIs. To the best of our knowledge, our paper is the first
modeling study that employs a human-centric perspective to understand the
societal consequences of generative AIs.

We assume that the output of AIs depends on the information provided by users.
In fact, many empirical studies have observed that AIs are quite sensitive to
users’ inputs (Liu et al. 2023, Brand et al. 2023, Binz and Schulz 2023). For
example, Brand et al. (2023), who adopted ChatGPT to conduct marketing
research, found that GPT is sensitive to the phrasing of queries in their
empirical work. When querying GPT with a list of options, they found that GPT
is more likely to choose the first option. Denny et al. (2023) also indicated
that sending proper prompts is critical for the performance of Copilot.

##### Related studies on the issues of homogenization and bias.

A few studies have a focus related to the homogenization issue (Saatci and
Wilson 2017, Chaney et al. 2018, Shumailov et al. 2023). They mainly focus on
how the technical algorithm design of an AI may reduce the diversity of AI-
generated content. For example, Shumailov et al. (2023) observed that the
tails of the original content distribution disappear when AIs are successively
trained from AI-generated content (they call it model collapse). They analyze
how the functional approximation error and sampling error can contribute to
the process of model collapse. In addition, a similar homogenization issue has
also been found in the literature of recommendation systems. For instance, by
using a simulation, Chaney et al. (2018) demonstrate that a feedback loop,
where a recommendation system is trained on data from users already exposed to
AI recommendations, may homogenize user behavior. On the other hand, the bias
issue of AI has also been shown (Hartmann et al. 2023, Rozado 2023, Motoki et
al. 2023). For example, Rozado (2023) implemented 15 different political
orientation tests to ChatGPT. The author found that ChatGPT’s answers
manifested a preference for left-leaning opinions in 14 of the 15 tests.

This paper considers these societal issues from a human-centric perspective.
Instead of analyzing AI’s inherent technical design or training process, we
concentrate on how the issues of homogenization and bias may be influenced by
human-AI interactions. In our setting, the issues would disappear if every
user is able to provide infinite information. Our results highlight the
importance of improving the usability of AIs to mitigate the issues of
homogenization and bias.

##### Related studies on the modeling approach.

The way we model the human-AI interaction shares similarities with the
frameworks of information design (Kamenica and Gentzkow 2011), costly
persuasion (Gentzkow and Kamenica 2014), the theory of rational inattention
(Sims 2003), as well as the interpretation of LLMs with Bayesian inference
(Wei et al. 2021, Xie et al. 2022). The user’s decision is modeled similarly
to an information design process (Alizamir et al. 2020, de Véricourt et al.
2021). The sender (i.e., the user) sends a signal to the receiver (i.e., the
AI) to inform the receiver about a true state (i.e., the user’s preference).
The utility of the sender is determined by the receiver’s decision (i.e., the
AI’s output). Additionally, we employ the framework of costly persuasion
(Gentzkow and Kamenica 2014) and the theory of rational inattention (Sims
2003) to model the user’s communication cost when sending the signal. In
particular, we follow the standard way in the literature to model the cost of
information as the expected reduction in entropy. This assumption can also be
found in other modeling papers, such as the cognitive cost defined in Boyacı
et al. (2023). Note that we assume the reduction in entropy is relative to the
population distribution of users’ preferences (see Section 3) instead of AI’s
prior (defined in Section 3). As Gentzkow and Kamenica (2014) suggested, the
reduction in entropy can be defined relative to any proper fixed reference
belief. So we use the population distribution of users’ preferences as the
fixed reference belief to indicate that the communication cost is independent
of AI’s prior but relevant to the difficulty of distinguishing a user’s
preference from the others. Furthermore, we model the AI’s behavior as a
Bayesian inference (Wei et al. 2021, Xie et al. 2022). For instance, Xie et
al. (2022) interpret that the in-context learning of an LLM can be viewed as
an implicit Bayesian inference. The prior of the LLM is formulated during
training. Conditional on a prompt, the LLM characterizes a posterior
distribution to make an output.

##  3 Model Setup

We develop a Bayesian model to represent the process of working with
generative AI on a given task. Users have preferences on how to complete the
task, and the AI knows the population’s distribution of these preferences
(through its training). Each user can also interact with the AI to share
information about her specific preferences. This interaction will help the AI
produce an output closer to what a user would have done without AI, leading to
a better output _fidelity_. However, sharing information requires effort,
which entails a _communication cost_. Users must choose how much information
they share to balance the benefits of fidelity and the cost of communication.

##### User preferences and Fidelity

We use $\theta\in\mathbb{R}$ to denote a user’s specific preference, and we
assume that $\theta$ is normally distributed in the population: $\theta\sim
N(\mu_{\theta},\sigma_{\theta}^{2})$. Here $\mu_{\theta}$ represents the
average population preference and $\sigma_{\theta}$ the diversity of the
population’s preferences. In practice, the users’ preferences should be
represented by a high-dimensional space, but we will interpret $\theta$ as a
specific _feature_ of a user’s preferences, as illustrated in the following
example.

######  Example 3.1 (News Article)

A journalist would like to write an article about a piece of breaking news and
wants to use ChatGPT to write faster. $\theta$ measures the political
orientation of the article this journalist would have written without AI. If
$\theta>\mu_{\theta}$, the journalist is more right-leaning than the average
journalist. When using AI, the journalist may be able to write faster but may
not meet her exact political orientation $\theta$ (low fidelity).

We will refer to a “user $\theta$” to describe a user with preference
$\theta$, and to an “output $\theta_{A}$” to refer to an AI’s output matching
some preference $\theta_{A}$. We define the output’s _fidelity loss_ as
$(\theta-\theta_{A})^{2}$, and we interpret it as a loss in utility for a user
$\theta$ receiving an output $\theta_{A}$ (users prefer an output matching
their preference). For example, if $\theta_{A}=\theta-1$, the AI of Example
3.1 outputs a more left-leaning article than the journalist’s preferred
political orientation, and the fidelity loss is 1.

##### AI Bayesian Inference

We model the interaction between a user and AI as an exchange of information
about $\theta$. The AI has a prior belief of the population distribution of
$\theta$. This belief corresponds to a normal distribution
$N(\mu_{A},\sigma_{A}^{2})$ with density $\pi_{A}(\cdot)$. To capture that the
AI has been trained on a representative dataset, we assume that the AI’s prior
is exactly the population distribution, $\mu_{A}=\mu_{\theta}$ and
$\sigma_{A}=\sigma_{\theta}$ (this assumption is relaxed in Section 5 to study
the effects of a biased AI). We model the exchange of information between a
user $\theta$ and the AI as a Bayesian update with normal distributions: the
user shares a noisy signal $q=\theta+\epsilon_{q}$ where $\epsilon_{q}\sim
N(0,\sigma_{q}^{2})$, and the AI refines its belief using Bayes’ rule:
$\theta|q\sim\pi_{A}(\cdot|q)$. It then returns the optimal output with the
maximum expected fidelity:

$\theta_{A}\triangleq\argmin_{\hat{\theta}}E[(\hat{\theta}-\theta)^{2}|q]=E[\theta|q]=\dfrac{\sigma_{A}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}\cdot
q+\dfrac{\sigma_{q}^{2}}{\sigma_{A}^{2}+\sigma_{q}^{2}}\cdot\mu_{A}.$ (1)

Note that $\theta_{A}$ is a weighted average between $q$ and the prior mean
(Berger 1985), which is a random variable since $q$ is noisy. Additionally, a
lower value of $\sigma_{q}$ corresponds to more information shared with the
AI: if $\sigma_{q}=0$, then the user $\theta$ shares her exact preference, and
the AI returns $\theta_{A}=\theta$. In the limit
$\sigma_{q}\rightarrow+\infty$, the signal is uninformative, and the AI
outputs the mean of its prior, $\theta_{A}=\mu_{A}$.

The fidelity error of a user $\theta$ given $\sigma_{q}$ is the expected
output fidelity, denoted by:

$e(\theta,\sigma_{q})\triangleq E[(\theta_{A}-\theta)^{2}|\theta].$ (2)

We can decompose it into two terms,

$e(\theta,\sigma_{q})=Var(\theta_{A}|\theta)+[E(\theta_{A}|\theta)-\theta]^{2}.$

For a user $\theta$, the first term corresponds to the variability in the AI’s
output that stems from the information exchange, while the second term is the
impact of the bias in the AI response, which will be a focus of the paper. In
the context of Example 3.1, the bias is high if the AI-written article
consistently leans more to the left than the journalist orientation.

##### Information and Communication Cost

Given an exchange of information parametrized by $\sigma_{q}$, we measure the
“communication cost” of the user to share this information with the AI.
Following standard assumptions in the rational inattention (Sims 2003) and
costly persuasion (Gentzkow and Kamenica 2014) literature, we assume the
communication cost to be proportional to the expected reduction in the AI
uncertainty of $\theta$ relative to the population distribution of $\theta$
given $\sigma_{q}$:

$\gamma
I(\sigma_{q})\triangleq\gamma[H(\theta)-E[H(\theta|q)]]=\gamma\left[\ln(\sigma_{\theta}\sqrt{2\pi
e})-\ln\left(\sqrt{\dfrac{\sigma_{\theta}^{2}\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}}\sqrt{2\pi
e}\right)\right]=-\dfrac{\gamma}{2}\ln\left(\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}\right),$

where $\gamma>0$ is the marginal cost of communication, $I(\sigma_{q})$ is the
mutual information, and $H(\cdot)$ denotes the differential entropy.
Intuitively, $I(\sigma_{q})$ corresponds to the “amount of information” the
user shares about her preference $\theta$. Note that sharing the exact value
of $\theta$ ($\sigma_{q}=0$) requires an _infinite_ amount of information
$I=+\infty$. Conversely, providing an uninformative signal about $\theta$
($\sigma_{q}\rightarrow+\infty$) requires no information $I=0$. To interpret
this situation, remember that our model assumes that the AI knows the task
description and that the human-AI interaction is about sharing user
preferences. Without preference information, the AI uses its knowledge of the
preference distribution to return a “default” output for the task,
$\mu_{A}=\mu_{\theta}$ (e.g., the first answer of ChatGPT) . In Example 3.1,
ChatGPT would write an article that expresses an “average” political
orientation that does not necessarily reflect the journalist’s views. And, in
the GitHub Copilot coding example of Figure 1, the initial Copilot’s output is
a default function that does not consider the user’s specific preference. In
both cases, the AI requires more information to deliver better fidelity.

##### User’s decision

Each user $\theta$ chooses the information $I$ they share with the AI
(parametrized by $\sigma_{q}$) to minimize their utility loss $l$ given by the
sum of the fidelity error and the communication cost

$l(\theta,\sigma_{q})\triangleq e(\theta,\sigma_{q})+\gamma I(\sigma_{q}).$
(3)

That is, a user $\theta$ chooses an optimal $\sigma_{q}^{\star}(\theta)$ that
minimizes her utility loss,

$\sigma_{q}^{\star}(\theta)\triangleq\argmin_{\sigma_{q}\geq
0}l(\theta,\sigma_{q}).$ (4)

Importantly, $\gamma$ controls the tradeoff between fidelity error and
communication cost, and we will refer to it as the _cost of human-AI
interactions_. A task has a low $\gamma$ if it is particularly easy to
interact with the AI and share preferences (e.g., a chat interface like
ChatGPT or voice-based interactions (OpenAI 2023c)) and/or if users care a lot
about fidelity. $\gamma$ will be high if users care more about minimizing
effort than matching their specific preferences. Because the task of Example
3.1 is about breaking news, the journalist may be in a hurry and have a high
$\gamma$.

##### Choosing to work with AI

If the cost of human-AI interaction is high and fidelity is important, a user
might be better off not using the AI and doing the work herself. In this case,
the output would have no fidelity error by definition. However, manual work
takes time, which we model as a fixed utility cost $\Gamma>0$ that depends on
the task but is the same for everyone.

If $\Gamma$ is smaller than the expected utility loss
$l(\theta,\sigma_{q}^{\star})$, then a user $\theta$ will not use the AI. We
define the output $\theta^{\star}$ chosen by a user $\theta$ and the
corresponding expected utility loss $l^{*}$ as:

$\theta^{\star}\triangleq\begin{cases}\theta_{A}|(\theta,\sigma_{q}^{\star})&\text{if
}l(\theta,\sigma_{q}^{\star})\leq\Gamma\\\
\theta&\text{otherwise}\end{cases},\qquad
l^{*}\triangleq\min(l(\theta,\sigma_{q}^{\star}),\Gamma).$ (5)

##  4 Human-AI Interactions and Homogenization

A consequence of our model is that different users may interact with the AI
differently, sharing varying amounts of information about their preferences or
even choosing not to use the AI. We first describe these individual-level
choices and then study their implied societal consequences and how to mitigate
them.

###  4.1 Individual Level: Heterogeneous Use of AI

Analyzing the optimal decision of each user $\theta$ requires solving Problem
(4), and the results are presented in Proposition 4.1. Users’ choices depend
on their _uniqueness_ , the distance of their preference $\theta$ to the
population mean $\mu_{\theta}$, $d(\theta)\triangleq|\theta-\mu_{\theta}|$. We
note that the derivation of $\sigma_{q}^{\star}(\theta)$, presented in Section
7, is non-trivial as Equation 3 is neither concave nor convex.333We present
the proofs for all the statements in Section 8.

###### Proposition 4.1

Optimal user strategies solving Equation 5 have the following characteristics:

  1. 1.

More unique users have a higher utility loss: $l^{\star}$ increases444All
references to “increasing” or “decreasing” functions are meant in a weak sense
(i.e., “non-decreasing”). in $d(\theta)$.

  2. 2.

More unique users interact more with the AI (if they choose to use it):
$\gamma I^{\star}$ increases in $d(\theta)$.

  3. 3.

Users use the AI if they are below a uniqueness threshold $\tau_{a}>0$:
$d(\theta)\leq\tau_{a}\iff l(\theta,\sigma_{q})\leq\Gamma$.

  4. 4.

Users that use AI are characterized by another uniqueness threshold $\tau_{d}$
such that:

    1. (a)

If $d(\theta)\leq\tau_{d}$, users choose the default AI output ($I^{\star}=0$)
and their fidelity error $e(\theta,\sigma_{q}^{\star})$ increases with their
uniqueness $d(\theta)$.

    2. (b)

If $d(\theta)>\tau_{d}$, users interact with AI ($I^{\star}>0$) and their
fidelity error _decreases_ with their uniqueness.

$-5$$-4$$-3$$-2$$-1$$0$$1$$2$$3$$4$$5$$0$$0.5$$1$$1.5$user preference $\theta$
(a)fidelity error $e^{\star}$communication cost $\gamma I^{\star}$utility loss
$l^{\star}$

$-5$$-4$$-3$$-2$$-1$$0$$1$$2$$3$$4$$5$$-4$$-2$$0$$2$$4$user preference
$\theta$ (b)user preference $\theta$AI’s default output $\mu_{\theta}$expected
output $E[\theta^{\star}|\theta]$

Figure 2: The black dashed vertical lines are at $d(\theta)=\tau_{d}$, and the
black dotted vertical lines are at $d(\theta)=\tau_{a}$. The white region
indicates the users who choose the default output; the yellow region indicates
those who send information to the AI; the red region indicates those not using
AI. We use $\mu_{\theta}=0,\sigma_{\theta}=1,\gamma=1,\Gamma=1.4$.

The main takeaway from Proposition 4.1 is that users with more “common”
preferences have a utility advantage (Footnote 4) and choose to interact less
with the AI (Item 2). The fundamental driver of this, crucial throughout the
paper, is that more common users can have a small fidelity error with limited
shared information. There are three types of users: users who use AI but do
not share information, those who share information, and users who do not use
AI. The most common users, with $d(\theta)\leq\tau_{d}$ as described in Item
4a, accept the default output of the AI, zero communication cost but rapidly
increasing fidelity error as these users become more unique, and the default
output $\theta_{A}=\mu_{\theta}$ becomes worse (see the region in the center
of Figure 2. Users with $\theta>\tau_{d}$ then choose to interact with the AI
(Item 4b), which will reduce their fidelity error at the expense of
communication cost (Item 2). Indeed, as illustrated in Figure 2 (a), while the
fidelity error (green curve) dominates the utility loss of users with common
preferences, more unique users prefer to pay an increasing communication cost
(red curve) that dominates a decreasing fidelity error.555A more detailed
description of effects at play can be found in Section 10. Interacting with
the AI eventually reaches such high communication costs for the most unique
users ($d(\theta)>\tau_{a}$) that the no-AI option becomes preferable (Item 3)
as shown in the red area of Figure 2 (a).

Many users have a positive fidelity error, so the AI’s output may not always
align perfectly with a user’s preference. The next proposition shows that this
output is misaligned in a specific way: on average, a user’s chosen output
$\theta^{\star}$ tends to revert toward the population’s mean preference.

###### Proposition 4.2

The expected chosen output $E[\theta^{\star}|\theta]$ of any user $\theta$ is
closer to the population’s mean than their actual preference:
$|\mathbb{E}[\theta^{\star}|\theta]-\mu_{\theta}|\leq|\theta-\mu_{\theta}|$.
Moreover, the inequality is strict for almost all users that use the AI: when
$d(\theta)<\tau_{a}$ and $\theta\neq\mu_{\theta}$.

We illustrate this result in Figure 2 (b). The most common users (with
$d(\theta)\leq\tau_{d}$) provide an uninformative signal to the AI
($I^{\star}=0$) and accept the AI’s default output $\theta_{A}=\mu_{\theta}$,
which is a direct revert to the mean. As users become more unique, they
interact with the AI ($I^{\star}>0$), which mitigates the mean reversion in
the AI’s output. However, it doesn’t completely vanish due to the high
communication cost. The mean reversion disappears only for those very unique
users who choose to work themselves and not to use the AI. In Example 3.1, a
journalist with $\theta>\mu_{\theta}$ would write a (slightly) more left-
leaning article than her preference. As discussed in the next section, this
can be an issue at the population level.

###  4.2 Societal Level: Homogenization

In a world without AI, the distribution of people’s output would exactly match
the distribution of their preference $\theta\sim
N(\mu_{\theta},\sigma_{\theta}^{2})$. However, with AI, the output is
$\theta^{\star}$, which does not have the same distribution, as we saw that
users of AI tend to choose an output closer to the mean $\mu_{\theta}$. At the
population level, this leads to _homogenization_ , where the output
distribution has a lower variance than the distribution of preferences.

###### Theorem 4.3

When everyone uses AI ($\Gamma\rightarrow+\infty$), the variance of the
population output is lower than the variance of the population preferences,
$Var(\theta^{\star})<Var(\theta)$, and strictly decreases in the cost of
human-AI interactions $\gamma$. In the general case ($\Gamma<+\infty$),
$\lim_{\gamma\rightarrow 0}Var(\theta^{\star})=Var(\theta)$ and
$\lim_{\gamma\rightarrow+\infty}Var(\theta^{\star})<Var(\theta)$.

$0$$0.5$$1$$1.5$$2$$2.5$$3$$0.2$$0.4$$0.6$$0.8$$1$cost of human-AI
interactions $\gamma$diversity of AI
outputs$Var(\theta^{\star})$$\Gamma=1$$\Gamma=1.5$$\Gamma=2$$\Gamma=\infty$

Figure 3: We use $\mu_{\theta}=0$, $\sigma_{\theta}=1$.

Theorem 4.3 formalizes the risk of homogenization and points to its solution.
When everyone uses AI ($\Gamma\rightarrow+\infty$), reducing the cost of
human-AI interactions $\gamma$ encourages users to interact more with the AI
and share their specific preferences more accurately, limiting homogenization
and helping to preserve the population’s diversity. When $\gamma\rightarrow
0$, there is no more homogenization as users can share their precise
preference for free. The case $\Gamma<+\infty$ is more involved, as some users
renounce the AI when the cost of human-AI interactions is high, partially
improving the chosen output’s diversity. We illustrate it in Figure 3 and
present a more in-depth analysis in Section 11. An interesting special case is
when $\Gamma<+\infty$ and $\gamma\rightarrow+\infty$. Only two types of users
remain: those who complete the task themselves and those who accept the
default AI output, leading to homogenization on average. In all cases, Theorem
4.3 underscores that enhancing the interactivity of AI tools (e.g., through
better interfaces, multi-modal inputs, or real-time feedback mechanisms) to
achieve a sufficiently low $\gamma$ is an effective strategy to encourage
users toward higher fidelity, reduce homogenization, and ultimately, preserve
population preference diversity.

###  4.3 AI-generated content and the “Death Spiral” of Homogenization

While homogenization can easily be perceived as a negative societal outcome,
we argue it may also have long-term consequences. As more and more content
becomes AI-generated, this content could be used to train the next generation
of AI. Because of the homogenization issue, this would lead to an incorrect AI
distribution of human preference (the AI’s prior). The next AI generation
would be even more likely to return homogenized outputs, resulting in a “death
spiral” of homogenization, a dreadful outcome for human preference diversity.
We study this phenomenon within our model, considering a _self-training loop_
where the population’s output distribution at time $t$ becomes the AI prior at
time $t+1$ (as illustrated in Figure 4).

User $\theta$AI ModelOutput $\theta^{\star}$AI updates its prior
$\pi_{A}(\cdot)$decides new optimal $\sigma_{q}^{\star}$ (step 1)user learnsAI
behavior (step 4)generates (step 2)becomes training data(step 3)

Figure 4: Steps in each iteration of the self-training loop.

(a)

(b)

(c)

Figure 5: The iterative convergence of the distribution of
$\theta_{A}^{\star}$. We use $\mu_{\theta}=0$, $\sigma_{\theta}=1$,
$\Gamma=+\infty$. A full simulation description is provided in Section 9.

This model is not tractable, as our analysis does not apply when the AI has a
non-normal prior, which happens after the first iteration of the self-training
loop. However, we use simulations, as shown in Figure 5, to illustrate its
consequences for homogenization. For simplicity, we choose a setting with
$\Gamma=+\infty$, and everyone uses AI. Note that while the human distribution
of preference $\theta$ (in blue) remains unchanged, the distribution of the
outputs $\theta^{\star}=\theta_{A}$ (in orange) becomes more and more
homogeneous. As the AI’s prior becomes increasingly erroneous, the
communication cost necessary to reduce the fidelity error becomes
unmanageable, and more and more users start to accept the default output until
we converge to a completely homogenized world.

These findings offer a different perspective than the technical explanation
for the homogenization problem in Shumailov et al. (2023) (they call it model
collapse), where the authors suggest the problem is caused by sampling and
approximation errors. Our model also indicates that human and technical
factors may mutually reinforce each other, potentially leading to an
exacerbated homogenization problem. In our model, the homogenization loop is
due to technical factors, the misspecified AI prior, and human behavior, who
maximize their utility and are willing to let go of their specificity to limit
the communication costs. We also offer a solution: creating models
facilitating human-AI interactions (i.e., low $\gamma$) can significantly slow
down the homogenization process (see Figure 5 (c)).

##  5 Human-AI Interactions and AI Bias

The homogenization phenomenon shows that the use of AI “influences” the user
outputs, in the sense that $\theta^{\star}\neq\theta$ for many users $\theta$.
This is potentially concerning, as any choices made in the AI training, any
bias it might have, would then influence the users’ choice of output. Indeed,
generative AIs are not necessarily trained to reflect the population’s
preferences exactly. For example, the AI’s training data may be censored to
avoid illegal or dangerous behavior (Thompson 2023). Moreover, the training of
LLMs uses Reinforcement Learning from Human Feedback (Ziegler et al. 2020), in
which a small group of humans “teach” the model what output is preferable.
These training choices of a few can then influence the output of the entire
population interacting with AI. We model this potential AI “bias” via an AI
prior that does not exactly reflect the population’s preference distribution
(i.e., $\mu_{A}\neq\mu_{\theta}$ or $\sigma_{A}\neq\sigma_{\theta}$), leaving
the true user preference distribution and the rest of the Bayesian inference
unchanged. We refer to $\mu_{A}\neq\mu_{\theta}$ as a _directional bias_ and
to $\sigma_{A}<\sigma_{\theta}$ as a _censoring bias_. In Example 3.1, the AI
may have a slight bias towards a political side (directional bias), or it may
avoid extreme political views (censoring bias). We first discuss how the two
types of AI bias affect users and the effectiveness of human-AI interactions.
We then evaluate how much influence a biased AI can have on society and ways
to mitigate this influence.

###  5.1 AI Bias and User Utility

A biased AI may be less useful for some users but may also help others, as
summarized below.

###### Proposition 5.1

The utility loss $l^{\star}$ of a user $\theta$ changes when with a biased AI
as follows:

  1. 1.

the directional bias favors users the AI is biased towards: $l^{\star}$
increases with $|\mu_{A}-\theta|$;

  2. 2.

the censoring bias benefits users with common preferences: $l^{\star}$ is
increasing in $\sigma_{A}$ when $\sigma_{A}\geq|\mu_{A}-\theta|$, and
decreasing in $\sigma_{A}$ when $\sigma_{A}<|\mu_{A}-\theta|$.

Item 1 of the proposition states that directional bias is detrimental to users
of the “opposite” direction. In Example 3.1, if the AI is slightly right-
leaning, a left-leaning journalist may need more communication cost to obtain
an article they will be satisfied with. However, a right-leaning journalist
could be directly satisfied with the default output. The ideal case for user
$\theta$ is $\mu_{A}=\theta$, as the default AI output would correspond to a
perfect utility $l^{\star}=0$. Item 2 states a similar result for the
censoring bias. To clarify it, suppose $\mu_{A}=\mu_{\theta}$, and consider a
user with “common” preferences less than a standard deviation away from the
mean, i.e., $|\mu_{\theta}-\theta|<\sigma_{\theta}$. Then she would be better
off if a slight censoring is used, with $\sigma_{A}$ such that
$|\mu_{\theta}-\theta|<\sigma_{A}<\sigma_{\theta}$. When reducing
$\sigma_{A}$, the AI is more likely to return outputs closer to the mean,
benefiting this user. However, this hurts users with more unique preferences,
who will need more communication costs to maintain a reasonable fidelity or
will stop using the AI. Therefore, both types of bias can increase some users’
utility loss and decrease others’. The next results consider the aggregate-
level consequences of bias and its effect on the population utility, defined
as the expected utility loss $E[l^{\star}]$ taken across the users $\theta$.

###### Proposition 5.2

Directional and censoring bias have contrasting effects on the population
utility:

  1. 1.

A small directional bias has a limited negative effect on the population
utility: $\dfrac{\partial
E[l^{\star}]}{\partial\mu_{A}}\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}=0$
and $E[l^{\star}]$ is minimized at $\mu_{A}=\mu_{\theta}$.

  2. 2.

A small censoring bias can have a stronger negative impact: for example,
$\dfrac{\partial
E[l^{\star}]}{\partial\sigma_{A}}\Big{\lvert}_{\mu_{A}=\mu_{\theta},\sigma_{A}=\sigma_{\theta}}<0$
when $\gamma\geq 2\sigma_{\theta}^{2}$ and $\Gamma\rightarrow\infty$.

The proposition first shows that, while any directional bias hurts the
population utility, a small directional bias has a negligible effect.
Intuitively, if $\mu_{A}=\mu_{\theta}+\varepsilon$ for $\varepsilon>0$ small,
slightly less than half of the users (above $\mu_{A}+\epsilon/2$) benefit from
the bias because they have a lower communication cost for the same fidelity,
while the other half (below $\mu_{\theta}$) is hurt because of an increased
communication cost for the same fidelity. These two populations balance each
other, which limits the total loss of utility.

$0$$1$$2$$3$$4$$5$$6$$0.2$$0.4$$0.6$$0.8$AI’s prior variance $\sigma_{A}^{2}$
(a)population utility loss,
$E[l^{\star}]$$\gamma=0.1$$\gamma=0.3$$\gamma=0.5$$\gamma=1.0$$\gamma=1.5$

$-10$$-7.5$$-5$$-2.5$$0$$2.5$$5$$7.5$$10$$-0.6$$-0.4$$-0.2$$0$$0.2$user
preference $\theta$ (b)difference in utility loss $l^{\star}$with increased
$\sigma_{A}$better off usersworse off usersindifferent users

Figure 6: (a) the black circles indicate the AI prior variance
$\sigma_{A}^{2}$ that would minimize the population utility loss. (b) the
utility loss $l^{\star}$ with $\sigma_{A}^{2}=2$ minus those with
$\sigma_{A}^{2}=1$, when $\gamma=1$. In both panels, we use
$\mu_{A}=\mu_{\theta}=0,\sigma_{\theta}^{2}=1,\Gamma=+\infty$.

The case of censoring bias (Item 2 of Proposition 5.2) is maybe more
surprising. Unlike the effect directional bias, setting
$\sigma_{A}=\sigma_{\theta}$ (an unbiased prior) does not generally minimize
the population utility loss $E[l^{\star}]$. Both the proposition and Figure 6
(a) show that when $\Gamma\rightarrow+\infty$, it is preferable to have
$\sigma_{A}>\sigma_{\theta}$ (the opposite of censoring). Remember from
Section 4.1 that there are two types of user behavior when
$\Gamma\rightarrow+\infty$: using the default AI output or interacting with
the AI, and the choice of $\sigma_{A}$ only influences the utility of the
interacting users. Therefore, the AI Bayesian update is more accurate when the
choice of $\sigma_{A}$ better represents the interacting users, who are the
ones with more unique preferences (Proposition 4.1). This is why choosing
$\sigma_{A}>\sigma_{\theta}$ improves the population utility. This effect is
illustrated in Figure 6 (b): when increasing $\sigma_{A}$, common-preference
users do not lose utility, but more unique users see a large improvement in
utility loss. While this result may have implications for the design of
interactive AI, it also warns against the potential negative effects of
censoring bias. Decreasing $\sigma_{A}$ is particularly hurtful to the most
unique users, _who rely on human-AI interactions the most_. While censoring
can be useful in preventing dangerous or illegal uses of AI, our results also
highlight the importance of training AI on datasets that reflect a wide range
of preferences.

###  5.2 AI Bias Becomes Societal Bias

Another interpretation of Item 1 of Proposition 5.2 is that a small
directional bias $|\mu_{A}-\mu_{\theta}|>0$ (referred to as _AI bias_ in this
section) may be hard to detect in practice, as it does not strongly affect the
population’s utility. However, it may still significantly influence the user
output $\theta^{\star}$. For example, users who accept the default output
($I^{\star}=0$) have $\theta^{\star}=\mu_{A}$, directly inheriting the AI
bias. On the other hand, users may choose to share more information to correct
this bias and maintain a high-fidelity output. To study which effect
dominates, we analyze the consequences of the AI bias on the _societal bias_ ,
defined as the bias of the output distribution:
$|E[\theta^{\star}]-\mu_{\theta}|$.

###### Theorem 5.3

Given the AI bias $|\mu_{A}-\mu_{\theta}|$ and the societal bias
$|E[\theta^{\star}]-\mu_{\theta}|$,

  1. 1.

the societal bias is lower than the AI bias,

  2. 2.

the societal bias is minimized when $\gamma\rightarrow 0$ _or_
$\Gamma\rightarrow 0$: $|E[\theta^{\star}]-\mu_{\theta}|=0$,

  3. 3.

the societal bias is maximized when $\gamma\rightarrow+\infty$ _and_
$\Gamma\rightarrow+\infty$:
$|E[\theta^{\star}]-\mu_{\theta}|=|\mu_{A}-\mu_{\theta}|$,

  4. 4.

if everyone uses AI ($\Gamma\rightarrow+\infty$), the societal bias increases
with the cost of human-AI interactions $\gamma$.

$0$$0.5$$1$$1.5$$2$$2.5$$3$$0$$0.2$$0.4$$0.6$cost of human-AI interactions
$\gamma$societal
bias$|E[\theta^{\star}]-\mu_{\theta}|$$\Gamma=1$$\Gamma=1.5$$\Gamma=2$$\Gamma=\infty$

Figure 7: We use $\mu_{\theta}=0$, $\mu_{A}=1$, $\sigma_{\theta}=\sigma_{A}=1$
(e.g., the AI bias is $|\mu_{A}-\mu_{\theta}|=1$).

This theorem is illustrated in Figure 7 and shows an encouraging result:
human-AI interactions can partially prevent AI bias from becoming societal
bias. For example, a left-wing journalist in Example 3.1 may increase their
interactions with the AI to correct the output if the AI is biased to the
right. This is particularly true when either the cost of human-AI
interactions, $\gamma$, or the cost of not using AI, $\Gamma$, is low. It is
much easier for users to correct bias if they can easily interact or simply
stop using AI. However, Theorem 5.3 also comes with a warning. For larger
tasks that are painful to do without AI (high $\Gamma$), and if the human-AI
interactions are not efficient (high $\gamma$), rational users will simply
accept the AI bias, which will be fully converted into a societal bias. For
example, generative AI systems that favor speed over interactivity (e.g., the
AI writing assistant Grammarly) or tackle complex tasks with limited
interactivity (e.g., the image generator Midjourney) may fall into this
category. Any bias they introduce may have a stronger influence on societal
output than systems designed for interactivity (e.g., ChatGPT).

##  6 Conclusions

The widespread introduction of generative AI enables significant productivity
gains. However, we show that the power of these tools may lead users to accept
homogenized or biased outputs and abandon their idiosyncrasies, even when
given the possibility to communicate their preferences. At the societal level,
this can lead to homogenization (reinforced by training loop effects) and the
potential influence of AI training choices on the societal output. These risks
are particularly strong for labor-intensive tasks (e.g., image/sound
generation) or with AI tools that favor speed over preference-sharing (e.g.,
grammar correction). Nonetheless, we also show that enabling easier human-AI
communication and training the AI on diverse data can significantly limit
these negative effects, allowing the best of both worlds: high productivity
and human diversity.

The topic studied in this work combines technical and behavioral complexity,
as we need to capture how the AI tool works and how users interact with it.
Our Bayesian framework is a simplified representation of this interaction that
still enables nontrivial insights, but there are effects we do not capture.
For example, it is a simplification to assume that a one-dimensional normal
distribution can represent the vast space of human preferences and outputs and
that the complexity of human-AI communication can be represented as a simple
normal signal and Bayesian inference. We also assume all users have the same
no-AI utility loss $\Gamma$, and the same human-AI interaction cost $\gamma$
for a given task. Nonetheless, we believe our framework is versatile enough to
study deeper variants and is a first step towards understanding the societal
consequences of human-AI interactions.

Recent empirical studies examine the multifaceted implications of generative
AIs across various domains, such as education (Baidoo-Anu and Owusu Ansah
2023), labor markets (Eloundou et al. 2023), and marketing (Brand et al.
2023). Understanding the general effects of user behaviors while interfacing
with an AI remains an open question that is difficult to study empirically. We
hope our analytical approach highlights the importance of adopting a human-
centric perspective rather than solely focusing on AI technology. Indeed,
while AIs could surpass human abilities in various aspects (Binz and Schulz
2023, Webb et al. 2023, Chen et al. 2023), their impact may largely depend on
how we employ them. The interaction with AIs could offer a novel medium for
production and creation but also introduce an extra risk: AIs may filter and
even replace our original preferences, styles, and tastes, thereby leading to
a world dictated by the AI creators’ perspective — a potentially homogenized
and biased world. Improving human-AI interactions and encouraging users to
authentically voice their unique views is crucial to avoid these societal
pitfalls.

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Appendix

##  7 Characterization of Optimal Decision

In this section, we characterize the user’s optimal decision. We first show
the closed form of the fidelity error $e(\theta,\sigma_{q})$ and illustrate
how the user’s decision may impact the error. Then, the optimal solution to
Problem (4) is derived. As in Section 4, we assume $\mu_{A}=\mu_{\theta}$ and
$\sigma_{A}=\sigma_{\theta}$.

Let’s first explore how the fidelity error $e(\theta,\sigma_{q})$ varies with
respect to $1/\sigma_{q}$.

###### Proposition 7.1

For any $\theta,\sigma_{q}$, the fidelity error is

$e(\theta,\sigma_{q})=\dfrac{\sigma_{q}^{2}(\sigma_{\theta}^{4}+\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{2}}$
(6)

Furthermore,

  * •

$e(\theta,\sigma_{q})$ increase in $d(\theta)$.

  * •

$\lim_{\sigma_{q}^{2}\to 0}e(\theta,\sigma_{q})=0$,
$\lim_{\sigma_{q}^{2}\to\infty}e(\theta,\sigma_{q})=d(\theta)^{2}$

  * •

If $d(\theta)\geq\sigma_{\theta}/\sqrt{2}$, $e(\theta,\sigma_{q})$ is
monotonically increasing in $\sigma_{q}$; if
$d(\theta)<\sigma_{\theta}/\sqrt{2}$, there exists a threshold $t>0$ such that
$e(\theta,\sigma_{q})$ increases in $1/\sigma_{q}\in(0,t)$ and decreases in
$1/\sigma_{q}\in(t,\infty)$.

Proposition 7.1 reveals that for any given $\sigma_{q}$, the uniqueness of a
user’s preference results in a larger error. And if the user provides no
information and simply accepts the default output, the fidelity error
increases the uniqueness metric, $d(\theta)$. Conversely, if the user offers
substantial information, the fidelity error approaches zero. More
intriguingly, the third part of Proposition 7.1 suggests that the fidelity
error is monotonically decreasing in $1/\sigma_{q}$ only when the uniqueness
$d(\theta)$ is sufficiently large (as demonstrated in the left panel of Figure
8). In other words, if a user’s preference significantly deviates from the
average, any additional information to articulate their preference can yield a
reduction in the AI’s fidelity error. However, when a user’s preference aligns
closely with the majority (i.e., $d(\theta)<\sigma_{\theta}/\sqrt{2}$), there
exists a threshold, $t$ such that if the user is reluctant to provide
sufficient information such that $1/\sigma_{q}>t$, sending little information
may backfire, causing the user to be worse off than if they provided no
information.

$0$$0.25$$0.5$$0.75$$1$$1.25$$1.5$$1.75$$2$$0$$2$$4$$6$$8$$1/\sigma_{q}^{2}$fidelity
error $e_{A}$$\theta=3$$\theta=1$

$0$$0.25$$0.5$$0.75$$1$$1.25$$1.5$$1.75$$2$$0$$0.5$$1$$1.5$$2$$2.5$$1/\sigma_{q}^{2}$fidelity
error $e$fidelity uncertainty $e_{u}$fidelity bias $e_{b}$

Figure 8: Left panel: the fidelity error with respect to $1/\sigma_{q}^{2}$.
Right panel: the decomposition of fidelity error for $\theta=1$. In both
panels, We use $\mu_{\theta}=0$ and $\sigma_{\theta}^{2}=9$.

To further investigate the cause of the non-monotonic fidelity error, as
introduced in Section 3, we can decompose the fidelity error into a bias and a
variance term,

$e(\theta,\sigma_{q})=Var(\theta_{A}|\theta)+[E(\theta_{A}|\theta)-\theta]^{2},$

We call $Var(\theta_{A}|\theta)$ the fidelity uncertainty error denoted by
$e_{u}(\theta,\sigma_{q})$, and $[E(\theta_{A}|\theta)-\theta]^{2}$ the
fidelity bias error denoted by $e_{b}(\theta,\sigma_{q})$. This decomposition
is depicted in the right panel of Figure 8, highlighting that the non-
monotonic fidelity error is primarily driven by the variance component.
Intuitively, when a user knows that the AI’s default output $\mu_{\theta}$ is
closely aligned with their preference without the need for further
information, any vague information could introduce ambiguity and cause the AI
to deviate from the user’s true preference. For example, in Example 3.1, users
with a neutral opinion may find it advantageous to accept the AI’s default
output (suppose $\mu_{\theta}=0$). If they were to loosely explain their
reasoning without detailing specifics, they risk introducing noisy information
and receiving a less desirable result. Hence, if you’re not inclined to invest
enough effort in providing precise information and you’re aware your
preference aligns closely with the AI’s default output, it may be beneficial
to exert less effort or allow the AI to make decisions on your behalf. In
other words, offering nothing may be preferable to providing ambiguous
information.

By Proposition 7.1, we can solve Problem (4) and derive the following Lemma
7.2.

###### Lemma 7.2

The optimal solution to Problem (4) is

$\sigma_{q}^{\star}=\begin{cases}\sqrt{\dfrac{w^{\star}\sigma_{\theta}^{2}}{1-w^{\star}}}&d(\theta)\geq\tau_{d}\\\
\infty&\text{otherwise}\end{cases}$ (7)

where
$w^{\star}=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\theta-\mu_{\theta})^{2}-\sigma_{\theta}^{2})}}{4((\theta-\mu_{\theta})^{2}-\sigma_{\theta}^{2})}$,
and $\tau_{d}>0$ is a threshold that strictly increases in $\gamma$ and is not
less than
$\sqrt{\max\\{0,\sigma_{\theta}^{2}-\dfrac{\sigma_{\theta}^{4}}{4\gamma}\\}}$.
In particular, $\tau_{d}=\dfrac{1}{2}\sigma_{\theta}^{2}+\dfrac{1}{4}\gamma$
when $\gamma>\sigma_{\theta}^{2}$.

It is not trivial to solve Problem (4), since the objective function is
neither concave nor convex when $d(\theta)$ is small. This non-convexity
emerges from the non-monotonicity of the fidelity error, as outlined in
Proposition 7.1. Lemma 7.2 implies that the users with common preferences are
best suited to send no information. As discussed previously, these users may
find it advantageous to rely on the AI’s default output instead of introducing
ambiguity.

###  7.1 Proof of the Results in Section 7.

###### Proof 7.3

Proof of Proposition 7.1.

By the definition of $e(\theta,\sigma_{q})$,

$\displaystyle
e(\theta,\sigma_{q})=E[(\theta_{A}(\sigma_{q}^{2})-\theta)^{2}|\theta]$
$\displaystyle=E\left[\left(\dfrac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}(\theta+\epsilon_{q})+\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}\cdot\mu_{\theta}-\theta\right)^{2}\Big{|}\theta\right]$
$\displaystyle=E\left[\left(\dfrac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}\epsilon_{q}+\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}(\mu_{\theta}-\theta)\right)^{2}\Big{|}\theta\right]$
$\displaystyle=\left(\dfrac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}\right)^{2}E[\epsilon_{q}^{2}]+\left(\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}(\mu_{\theta}-\theta)\right)^{2}$
$\displaystyle=\left(\dfrac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}\right)^{2}\sigma_{q}^{2}+\left(\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}(\mu_{\theta}-\theta)\right)^{2}$
$\displaystyle=\dfrac{\sigma_{q}^{2}(\sigma_{\theta}^{4}+\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{2}}$

  * •

$\lim_{\sigma_{q}^{2}\to 0}e(\theta,\sigma_{q})=\lim_{\sigma_{q}^{2}\to
0}\dfrac{\sigma_{q}^{2}(\sigma_{\theta}^{4}+\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{2}}=0$

And by L’Hôpital’s rule:

$\lim_{\sigma_{q}^{2}\to\infty}e(\theta,\sigma_{q})=\lim_{\sigma_{q}^{2}\to\infty}\dfrac{\sigma_{q}^{2}(\sigma_{\theta}^{4}+\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{2}}=\lim_{\sigma_{q}^{2}\to\infty}\dfrac{\sigma_{\theta}^{4}+2\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2}}{2(\sigma_{\theta}^{2}+\sigma_{q}^{2})}=\lim_{\sigma_{q}^{2}\to\infty}\dfrac{2(\mu_{\theta}-\theta)^{2}}{2}=(\mu_{\theta}-\theta)^{2}$

  * •

Take the derivative of $e(\theta,\sigma_{q})$ with respect to
$\sigma_{q}^{2}$:

$\displaystyle\dfrac{\partial e(\theta,\sigma_{q})}{\partial\sigma_{q}^{2}}$
$\displaystyle=\dfrac{\partial\dfrac{\sigma_{q}^{2}(\sigma_{\theta}^{4}+\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{2}}}{\partial\sigma_{q}^{2}}$
$\displaystyle=\dfrac{(\sigma_{\theta}^{4}+2\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{2}-2(\sigma_{\theta}^{2}+\sigma_{q}^{2})\sigma_{q}^{2}(\sigma_{\theta}^{4}+\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{4}}$
$\displaystyle=\dfrac{(\sigma_{\theta}^{4}+2\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})(\sigma_{\theta}^{2}+\sigma_{q}^{2})-2\sigma_{q}^{2}(\sigma_{\theta}^{4}+\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{3}}$
$\displaystyle=\dfrac{\sigma_{\theta}^{6}+\sigma_{\theta}^{4}\sigma_{q}^{2}+2\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2}\sigma_{\theta}^{2}+2\sigma_{q}^{4}(\mu_{\theta}-\theta)^{2}-2\sigma_{q}^{2}\sigma_{\theta}^{4}-2\sigma_{q}^{4}(\mu_{\theta}-\theta)^{2}}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{3}}$
$\displaystyle=\dfrac{\sigma_{\theta}^{6}-\sigma_{\theta}^{4}\sigma_{q}^{2}+2\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2}\sigma_{\theta}^{2}}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{3}}$
$\displaystyle=\dfrac{\sigma_{\theta}^{2}(\sigma_{\theta}^{4}-\sigma_{\theta}^{2}\sigma_{q}^{2}+2\sigma_{q}^{2}(\mu_{\theta}-\theta)^{2})}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{3}}$
$\displaystyle=\dfrac{\sigma_{\theta}^{2}(\sigma_{\theta}^{4}+\sigma_{q}^{2}(2(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{(\sigma_{\theta}^{2}+\sigma_{q}^{2})^{3}}$

which is non-negative for all $\sigma_{q}\geq 0$ if and only if
$(\mu_{\theta}-\theta)\geq\sigma_{\theta}/\sqrt{2}$. And when
$(\mu_{\theta}-\theta)<\sigma_{\theta}/\sqrt{2}$, $e(\theta,\sigma_{q})$
increases for
$\sigma_{q}\in\left(0,\sqrt{\dfrac{\sigma_{\theta}^{4}}{\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}}}\right)$,
and decreases for
$\sigma_{q}\in\left(\sqrt{\dfrac{\sigma_{\theta}^{4}}{\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}}},\infty\right)$,
so
$t=\sqrt{\dfrac{\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}}{\sigma_{\theta}^{4}}}$.

###### Proof 7.4

Proof of Lemma 7.2. Let
$w\triangleq\dfrac{\sigma_{q}^{2}}{\sigma_{q}^{2}+\sigma_{\theta}^{2}}$, and
by Proposition 7.1, we can rewrite Problem (4) as:

$w^{\star}(\theta)\triangleq\argmin_{w\in[0,1]}w(1-w)\sigma_{\theta}^{2}+w^{2}(\mu_{\theta}-\theta)^{2}-\dfrac{\gamma}{2}\ln
w$ (8)

Let $l(w)\triangleq
w(1-w)\sigma_{\theta}^{2}+w^{2}(\mu_{\theta}-\theta)^{2}-\dfrac{\gamma}{2}\ln
w$. On the boundary, we have $l(0)=\infty$ and
$l(1)=(\mu_{\theta}-\theta)^{2}$.

Take the first-order condition:

$l^{\prime}(w)=\dfrac{\partial l}{\partial
w}(w)=2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})w+\sigma_{\theta}^{2}-\dfrac{\gamma}{2w}=0$

we can get the roots of the above equation are

$w_{1}=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})},\
w_{2}=\dfrac{-\sigma_{\theta}^{2}-\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$

Moreover, we have to make sure $w^{\star}(\theta)\in[0,1]$ and
$l(w^{\star}(\theta))\leq(\mu_{\theta}-\theta)^{2}$ because Problem (4) is
non-convex.

  1. 1.

$(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}$

Because $w^{\star}(\theta)\geq 0$ but $w_{2}<0$, it is only possible to have
$w^{\star}(\theta)=w_{1}$. Also,

$\displaystyle w_{1}\leq 1$ $\displaystyle\iff$
$\displaystyle-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\leq
4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$ $\displaystyle\iff$
$\displaystyle\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\leq(4(\mu_{\theta}-\theta)^{2}-3\sigma_{\theta}^{2})^{2}$
$\displaystyle\iff$ $\displaystyle
4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\leq(4(\mu_{\theta}-\theta)^{2}-4\sigma_{\theta}^{2})(4(\mu_{\theta}-\theta)^{2}-2\sigma_{\theta}^{2})$
$\displaystyle\iff$ $\displaystyle\gamma\leq
2(2(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$ $\displaystyle\iff$
$\displaystyle(\mu_{\theta}-\theta)^{2}\geq\dfrac{1}{2}\sigma_{\theta}^{2}+\dfrac{1}{4}\gamma$

Additionally, when $(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}$,
$\dfrac{\partial l}{\partial w}$ is negative for $w<w_{1}$ and positive for
$w>w_{1}$, so $l(w^{\star}(\theta))<l(1)=(\mu_{\theta}-\theta)^{2}$.

Therefore, when $(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}$,
$w^{\star}(\theta)=w_{1}$ is optimal if
$(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}/2+\gamma/4$; otherwise,
$w^{\star}(\theta)=1$ is optimal.

  2. 2.

$(\mu_{\theta}-\theta)^{2}<\sigma_{\theta}^{2}$

In what follows, let us discuss when we have the optimal solution
$w^{\star}(\theta)<1$.

First, to make sure $l^{\prime}(w)=0$ has a real root (otherwise,
$w^{\star}(\theta)=1$ is optimal), we need
$\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\geq
0$. That is,
$(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)$.

In addition, we can see that $l^{\prime}(w)$ is negative for $w<w_{1}$ or
$w>w_{2}$. And $l^{\prime}(w)$ is positive for $w\in(w_{1},w_{2})$ Thus, the
local minimum is at $w=w_{1}$, and the local maximum is at $w=w_{2}$. This
means $w=w_{2}$ is never optimal.

Feasibility of $w=w_{1}$:

First, because $(\mu_{\theta}-\theta)^{2}<\sigma_{\theta}^{2}$, we must have
$w_{1}>0$.

Second, we want to find the conditions such that $w_{1}\leq 1$

$\displaystyle w_{1}=$
$\displaystyle\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\leq
1$ $\displaystyle\iff$
$\displaystyle\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\geq
4(\mu_{\theta}-\theta)^{2}-3\sigma_{\theta}^{2}$

The above inequality always holds if $(\mu_{\theta}-\theta)^{2}\leq
3\sigma_{\theta}^{2}/4$, otherwise

$\displaystyle\iff$
$\displaystyle\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\geq(4(\mu_{\theta}-\theta)^{2}-3\sigma_{\theta}^{2})^{2}$
$\displaystyle\iff$ $\displaystyle\gamma\leq
2(2(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$ $\displaystyle\iff$
$\displaystyle(\mu_{\theta}-\theta)^{2}\geq\dfrac{1}{2}\sigma_{\theta}^{2}+\dfrac{1}{4}\gamma$

This implies that if $\gamma\leq\sigma_{\theta}^{2}$, we always have
$w_{1}\leq 1$.

Thus, if $\gamma\leq\sigma_{\theta}^{2}$, we only need
$(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)$
such that $w=w_{1}\in[0,1]$. Otherwise, $w^{\star}(\theta)=1$ is optimal.

If $\gamma>\sigma_{\theta}^{2}$, we need
$(\mu_{\theta}-\theta)^{2}\geq\max\\{\sigma_{\theta}^{2}/2+\gamma/4,\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)\\}$.
However, notice that
$\sigma_{\theta}^{2}/2+\gamma/4\geq\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)$
because
$\sigma_{\theta}^{2}/2+\gamma/4-[\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)]=(\gamma-\sigma_{\theta}^{2})^{2}/(4\gamma)\geq
0$. So we need $(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}/2+\gamma/4$
such that $w=w_{1}\in[0,1]$. Otherwise, $w^{\star}(\theta)=1$ is optimal.

Optimality of $w=w_{1}$,

Now we need to show the conditions when $w^{\star}(\theta)=w_{1}$ is optimal.

Notice that $w^{\star}(\theta)=w_{1}$ is the global minimum if $w_{2}\geq 1$,
since $l^{\prime}(w)$ is negative for $w<w_{1}$ and positive for
$w\in(w_{1},w_{2})$. And,

$\displaystyle w_{2}=$
$\displaystyle\dfrac{-\sigma_{\theta}^{2}-\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\geq
1$ $\displaystyle\iff$
$\displaystyle-\sigma_{\theta}^{2}-\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\leq
4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\quad\text{since
$(\mu_{\theta}-\theta)^{2}<\sigma_{\theta}^{2}$}$ $\displaystyle\iff$
$\displaystyle-\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\leq
4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}$

The above inequality always holds if $(\mu_{\theta}-\theta)^{2}\geq
3\sigma_{\theta}^{2}/4$, otherwise

$\displaystyle\iff$
$\displaystyle\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\geq(4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2})^{2}$
$\displaystyle\iff$
$\displaystyle\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\geq
16((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}+8\sigma_{\theta}^{2}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{4}$
$\displaystyle\iff$
$\displaystyle\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\geq
4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}+2\sigma_{\theta}^{2}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$
$\displaystyle\iff$ $\displaystyle\gamma\leq
4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+2\sigma_{\theta}^{2}\quad\text{since
$(\mu_{\theta}-\theta)^{2}<\sigma_{\theta}^{2}$}$ $\displaystyle\iff$
$\displaystyle\gamma\leq 4(\mu_{\theta}-\theta)^{2}-2\sigma_{\theta}^{2}$
$\displaystyle\iff$
$\displaystyle(\mu_{\theta}-\theta)^{2}\geq\dfrac{1}{2}\sigma_{\theta}^{2}+\dfrac{1}{4}\gamma$

Thus, if
$\sigma_{\theta}^{2}/2+\min\\{\sigma_{\theta}^{2},\gamma\\}/4\leq(\mu_{\theta}-\theta)^{2}\leq\sigma_{\theta}^{2}$,
$w^{\star}(\theta)=w_{1}$ is optimal.

  3. 3.

Let’s see what we have shown now. We have shown that if
$(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}/2+\gamma/4$,
$w^{\star}(\theta)=w_{1}$ is feasible and optimal; if
$(\mu_{\theta}-\theta)^{2}<\sigma_{\theta}^{2}/2+\gamma/4$ and
$\gamma>\sigma_{\theta}^{2}$, $w=w_{1}$ is not feasible and
$w^{\star}(\theta)=1$ is optimal; if $\gamma\leq\sigma_{\theta}^{2}$ and
$(\mu_{\theta}-\theta)^{2}\leq\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)$,
$w=w_{1}$ is not feasible and $w^{\star}(\theta)=1$ is optimal. In addition,
we have shown that if $\gamma\leq\sigma_{\theta}^{2}$ and
$(\mu_{\theta}-\theta)^{2}\in[\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma),\
\sigma_{\theta}^{2}/2+\gamma/4]$, $w=w_{1}\in[0,1]$ is feasible, but we need
to show whether it is optimal since $w=1$ is another local minimum.

We want to show that if $\gamma\leq\sigma_{\theta}^{2}$, there exists a
threshold $\eta\geq\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)$ such
that when $(\mu_{\theta}-\theta)^{2}>\eta$, $w^{\star}(\theta)=w_{1}$ is
optimal; otherwise, $w^{\star}(\theta)=1$ is optimal. Because we have seen
that $w^{\star}(\theta)=w_{1}$ is optimal whenever
$(\mu_{\theta}-\theta)^{2}\geq\sigma_{\theta}^{2}/2+\gamma/4$, we only need to
consider the case when
$(\mu_{\theta}-\theta)^{2}\in[\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma),\sigma_{\theta}^{2}/2+\gamma/4]$.
That is, we want to show if $\gamma\leq\sigma_{\theta}^{2}$ and
$(\mu_{\theta}-\theta)^{2}\in[\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma),\sigma_{\theta}^{2}/2+\gamma/4]$,

$g((\mu_{\theta}-\theta)^{2})\triangleq
l(1)-l(w_{1})=(\mu_{\theta}-\theta)^{2}-l\left(\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\right)$

has at most one zero point. In particular, we want to show that
$g((\mu_{\theta}-\theta)^{2})$ is monotonically increasing for any
$(\mu_{\theta}-\theta)^{2}\in[\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma),\sigma_{\theta}^{2}/2+\gamma/4]$.

By Lemma 8.7,

$\displaystyle\dfrac{\partial g}{\partial(\mu_{\theta}-\theta)^{2}}$
$\displaystyle=1-\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})}{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$
$\displaystyle=\dfrac{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}-\sigma_{\theta}^{2}\sqrt{\Delta}+\sigma_{\theta}^{4}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})}{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$

Let
$h(\gamma)=8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}-\sigma_{\theta}^{2}\sqrt{\Delta}+\sigma_{\theta}^{4}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})$
represents the numerator of $\dfrac{\partial
g}{\partial(\mu_{\theta}-\theta)^{2}}$. We have
$h(0)=8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}\geq 0$ and

$\displaystyle h(\gamma=\sigma_{\theta}^{2})$
$\displaystyle=8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}+\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\sigma_{\theta}^{2}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})$
Since $\displaystyle\text{$\gamma\leq\sigma_{\theta}^{2}$ and
$(\mu_{\theta}-\theta)^{2}\leq\sigma_{\theta}^{2}/2+\gamma/4\implies
8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}\geq
2(\gamma-2\sigma_{\theta}^{2})((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})$},$
$\displaystyle h(\gamma=\sigma_{\theta}^{2})$ $\displaystyle\geq
2(\sigma_{\theta}^{2}-2\sigma_{\theta}^{2})((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})+\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\sigma_{\theta}^{2}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})$
$\displaystyle=\sigma_{\theta}^{2}\sqrt{\Delta}-\Delta$
$\displaystyle=\sqrt{\Delta}(\sigma_{\theta}^{2}-\sqrt{\Delta})$
$\displaystyle\geq 0\quad\text{since $\gamma\leq\sigma_{\theta}^{2}$ and
$(\mu_{\theta}-\theta)^{2}\leq\dfrac{1}{2}\sigma_{\theta}^{2}+\dfrac{1}{4}\gamma\implies(\mu_{\theta}-\theta)^{2}\leq\sigma_{\theta}^{2}$}$

In addition,

$\displaystyle\dfrac{\partial h}{\partial\gamma}$
$\displaystyle=\dfrac{\sigma_{\theta}^{2}}{2\sqrt{\Delta}}4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})-2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})$
$\displaystyle=2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})(\dfrac{\sigma_{\theta}^{2}}{\sqrt{\Delta}}-1)\leq
0\quad\text{since $\gamma\leq\sigma_{\theta}^{2}$ and
$(\mu_{\theta}-\theta)^{2}\leq\dfrac{1}{2}\sigma_{\theta}^{2}+\dfrac{1}{4}\gamma\implies(\mu_{\theta}-\theta)^{2}\leq\sigma_{\theta}^{2}$}$

This implies $h(\gamma)\geq h(\gamma=\sigma_{\theta}^{2})\geq 0$ for any
$\gamma\leq\sigma_{\theta}^{2}$, which further implies that $\dfrac{\partial
g}{\partial(\mu_{\theta}-\theta)^{2}}\geq 0$.

Therefore, if $\gamma\leq\sigma_{\theta}^{2}$, $g((\mu_{\theta}-\theta)^{2})$
is monotonically increasing for any
$(\mu_{\theta}-\theta)^{2}\in(\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma),\
\sigma_{\theta}^{2}/2+\gamma/4)$. This implies that if
$\gamma\leq\sigma_{\theta}^{2}$, there exists a threshold
$\eta\geq\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)$ such that when
$(\mu_{\theta}-\theta)^{2}>\eta$, $w^{\star}(\theta)=w_{1}$ is optimal.

In summary, when $\gamma>\sigma_{\theta}^{2}$, then
$\tau_{d}(\gamma)\triangleq\sqrt{\sigma_{\theta}^{2}/2+\gamma/4}$ is a
threshold such that $w^{\star}(\theta)=w_{1}$ is optimal if and only if
$|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)$; and when
$\gamma\leq\sigma_{\theta}^{2}$, then $\tau_{d}(\gamma)\triangleq\sqrt{\eta}$
is a threshold such that $w^{\star}(\theta)=w_{1}$ is optimal if and only if
$|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)$.

Additionally, it is clear that $\sigma_{\theta}^{2}/2+\gamma/4$ strictly
increases in $\gamma$; and by Equation 21,

$\dfrac{\partial
l(w_{1})}{\partial\gamma}=\dfrac{3\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{2\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}-\dfrac{1}{2}\ln
w_{1}=\dfrac{3\gamma}{8\sqrt{\Delta}w_{1}}-\dfrac{1}{2}\ln w_{1}>0$

which implies $g((\mu_{\theta}-\theta)^{2})$ strictly decreases in $\gamma$.
Because we have shown $\dfrac{\partial
g}{\partial(\mu_{\theta}-\theta)^{2}}\geq 0$, then must have $\eta$ strictly
increases in $\gamma$. These imply that $\tau_{d}(\gamma)$ strictly increases
in $\gamma$.

Hence, the optimal solution to Problem (4) is

$\sigma_{q}^{\star}=\begin{cases}\sqrt{\dfrac{w^{\star}(\theta)\sigma_{\theta}^{2}}{1-w^{\star}(\theta)}}&|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)\\\
\infty&\text{otherwise}\end{cases}$ (7)

where
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$,
and $\tau_{d}(\gamma)>0$ is a threshold that increases in $\gamma$ and is not
less than
$\sqrt{\max\\{0,\sigma_{\theta}^{2}-\dfrac{\sigma_{\theta}^{4}}{4\gamma}\\}}$.

##  8 Proof of the Main Results

###  8.1 Proof of the Results in Section 4.

####  8.1.1 Auxiliary lemmas

###### Lemma 8.1

Let
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$,
then

  1. 1.

$\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}=\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
(9)

  2. 2.

$\dfrac{\partial w^{\star}(\theta)}{\partial\gamma}=\dfrac{1}{2\sqrt{\Delta}}$
(10)

  3. 3.

$\dfrac{\partial
w^{\star}(\theta)}{\partial\sigma_{\theta}^{2}}=\dfrac{(\sigma_{\theta}^{2}-\sqrt{\Delta})(\mu_{\theta}-\theta)^{2}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
(11)

where
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$.

###### Proof 8.2

Proof of Lemma 8.1. Let
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$.

Since
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\Delta}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$,

  1. 1.

$\displaystyle\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}$
$\displaystyle=\dfrac{\dfrac{1}{2\sqrt{\Delta}}4\gamma\cdot
4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-4(-\sigma_{\theta}^{2}+\sqrt{\Delta})}{16((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=\dfrac{2\gamma\cdot((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}\sqrt{\Delta}-\Delta}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$

  2. 2.

$\displaystyle\dfrac{\partial w^{\star}(\theta)}{\partial\gamma}$
$\displaystyle=\dfrac{\dfrac{1}{2\sqrt{\Delta}}\cdot
4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$
$\displaystyle=\dfrac{1}{2\sqrt{\Delta}}$

  3. 3.

$\displaystyle\dfrac{\partial w^{\star}(\theta)}{\partial\sigma_{\theta}^{2}}$
$\displaystyle=\dfrac{\left(-1+\dfrac{2\sigma_{\theta}^{2}-4\gamma}{2\sqrt{\Delta}}\right)4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+4(-\sigma_{\theta}^{2}+\sqrt{\Delta})}{16((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=\dfrac{(-\sqrt{\Delta}+\sigma_{\theta}^{2}-2\gamma)((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-\sigma_{\theta}^{2}\sqrt{\Delta}+\Delta}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=\dfrac{(\sigma_{\theta}^{2}-\sqrt{\Delta})((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-\sigma_{\theta}^{2}\sqrt{\Delta}+\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=\dfrac{(\sigma_{\theta}^{2}-\sqrt{\Delta})((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}(\sigma_{\theta}^{2}-\sqrt{\Delta})+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=\dfrac{(\sigma_{\theta}^{2}-\sqrt{\Delta})(\mu_{\theta}-\theta)^{2}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$

###### Lemma 8.3

Let $w=\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}$, then we
can rewrite $e(\theta,\sigma_{q})$ as

$e(\theta,w)=w(1-w)\sigma_{\theta}^{2}+w^{2}(\mu_{\theta}-\theta)^{2}$ (12)

In addition, if
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$,

  1. 1.

$\dfrac{\partial
e(\theta,w^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=\dfrac{\sigma_{\theta}^{2}(\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$
(13)

  2. 2.

$\dfrac{\partial
e(\theta,w^{\star}(\theta))}{\partial\gamma}=\dfrac{\sigma_{\theta}^{2}+\sqrt{\Delta}}{4\sqrt{\Delta}}$
(14)

  3. 3.

$\dfrac{\partial
e(\theta,w^{\star}(\theta))}{\partial\sigma_{\theta}^{2}}=\dfrac{[2(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}](-\sigma_{\theta}^{2}\sqrt{\Delta}+\sigma_{\theta}^{4}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
(15)

where
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$.

###### Proof 8.4

Proof of Lemma 8.3. Let
$w=\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}$, then
$\sigma_{q}^{2}=\dfrac{w\sigma_{\theta}^{2}}{1-w}$. Substitute
$\sigma_{q}^{2\star}=\dfrac{w\sigma_{\theta}^{2}}{1-w}$ into Equation 6, then
we have Equation 12.

Let
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$,

  1. 1.

$\displaystyle\dfrac{\partial
e(\theta,w^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}$
$\displaystyle=$
$\displaystyle\sigma_{\theta}^{2}(1-2w^{\star}(\theta))\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}+2(\mu_{\theta}-\theta)^{2}w^{\star}(\theta)\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}+w^{\star 2}$ Substitute
Equation 9 into the above equation $\displaystyle=$
$\displaystyle\sigma_{\theta}^{2}\cdot\dfrac{\sigma_{\theta}^{2}+2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-\sqrt{\Delta}}{2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$
$\displaystyle+$ $\displaystyle
2(\mu_{\theta}-\theta)^{2}\cdot\dfrac{-\sigma_{\theta}^{2}+\sqrt{\Delta}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$
$\displaystyle+$
$\displaystyle\dfrac{\sigma_{\theta}^{4}-\sigma_{\theta}^{2}\sqrt{\Delta}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{\sigma_{\theta}^{4}+2\sigma_{\theta}^{2}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-\sigma_{\theta}^{2}\sqrt{\Delta}-(\mu_{\theta}-\theta)^{2}\sigma_{\theta}^{2}+(\mu_{\theta}-\theta)^{2}\sqrt{\Delta}}{2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$
$\displaystyle+$
$\displaystyle\dfrac{\sigma_{\theta}^{4}-\sigma_{\theta}^{2}\sqrt{\Delta}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{\sigma_{\theta}^{2}+\sqrt{\Delta}}{2}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}+\dfrac{\sigma_{\theta}^{4}-\sigma_{\theta}^{2}\sqrt{\Delta}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{\sigma_{\theta}^{2}(\Delta-\sigma_{\theta}^{4})-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})(\sigma_{\theta}^{2}+\sqrt{\Delta})}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}+\dfrac{\sigma_{\theta}^{4}\sqrt{\Delta}-\sigma_{\theta}^{2}\Delta+2\gamma\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{\sigma_{\theta}^{2}(\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$

  2. 2.

$\displaystyle\dfrac{\partial e(\theta,w^{\star}(\theta))}{\partial\gamma}$
$\displaystyle=$
$\displaystyle\sigma_{\theta}^{2}(1-2w^{\star}(\theta))\dfrac{\partial
w^{\star}(\theta)}{\partial\gamma}+2(\mu_{\theta}-\theta)^{2}w^{\star}(\theta)\dfrac{\partial
w^{\star}(\theta)}{\partial\gamma}$ Substitute Equation 10 into the above
equation $\displaystyle=$
$\displaystyle\sigma_{\theta}^{2}\cdot\dfrac{\sigma_{\theta}^{2}+2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-\sqrt{\Delta}}{2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\cdot\dfrac{1}{2\sqrt{\Delta}}$
$\displaystyle+$ $\displaystyle
2(\mu_{\theta}-\theta)^{2}\cdot\dfrac{-\sigma_{\theta}^{2}+\sqrt{\Delta}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\cdot\dfrac{1}{2\sqrt{\Delta}}$
$\displaystyle=$
$\displaystyle\dfrac{\sigma_{\theta}^{4}+2\sigma_{\theta}^{2}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-\sigma_{\theta}^{2}\sqrt{\Delta}-(\mu_{\theta}-\theta)^{2}\sigma_{\theta}^{2}+(\mu_{\theta}-\theta)^{2}\sqrt{\Delta}}{2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\cdot\dfrac{1}{2\sqrt{\Delta}}$
$\displaystyle=$
$\displaystyle\dfrac{\sigma_{\theta}^{2}+\sqrt{\Delta}}{4\sqrt{\Delta}}$

  3. 3.

$\displaystyle\dfrac{\partial
e(\theta,w^{\star}(\theta))}{\partial\sigma_{\theta}^{2}}$ $\displaystyle=$
$\displaystyle
w^{\star}(\theta)(1-w^{\star}(\theta))+\sigma_{\theta}^{2}(1-2w^{\star}(\theta))\dfrac{\partial
w^{\star}(\theta)}{\partial\sigma_{\theta}^{2}}+2(\mu_{\theta}-\theta)^{2}w^{\star}(\theta)\dfrac{\partial
w^{\star}(\theta)}{\partial\sigma_{\theta}^{2}}$ Substitute Equation 11 into
the above equation $\displaystyle=$
$\displaystyle\dfrac{-\sigma_{\theta}^{2}+\sqrt{\Delta}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}\cdot\dfrac{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}-\sqrt{\Delta}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}+\dfrac{\sigma_{\theta}^{2}+\sqrt{\Delta}}{2}\cdot\dfrac{(\sigma_{\theta}^{2}-\sqrt{\Delta})(\mu_{\theta}-\theta)^{2}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})(-\sigma_{\theta}^{2}+\sqrt{\Delta})-(\sigma_{\theta}^{2}-\sqrt{\Delta})^{2}}{16((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle+$
$\displaystyle\dfrac{-4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})(\mu_{\theta}-\theta)^{2}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})(\sigma_{\theta}^{2}+\sqrt{\Delta})}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})(-\sigma_{\theta}^{2}+\sqrt{\Delta})\sqrt{\Delta}-\sqrt{\Delta}(\sigma_{\theta}^{4}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})-\sigma_{\theta}^{2}\sqrt{\Delta})}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle+$
$\displaystyle\dfrac{-4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})(\mu_{\theta}-\theta)^{2}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})(\sigma_{\theta}^{2}+\sqrt{\Delta})}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})[(-\sigma_{\theta}^{2}+\sqrt{\Delta})\sqrt{\Delta}-\gamma\sqrt{\Delta}-2\gamma(\mu_{\theta}-\theta)^{2}+\gamma(\sigma_{\theta}^{2}+\sqrt{\Delta})]-\sqrt{\Delta}\sigma_{\theta}^{2}(\sigma_{\theta}^{2}-\sqrt{\Delta})}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})-2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\gamma[2(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}]+\sqrt{\Delta}\sigma_{\theta}^{2}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{[2(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}]\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})-2((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\gamma[2(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}]}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{[2(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}](-\sigma_{\theta}^{2}\sqrt{\Delta}+\sigma_{\theta}^{4}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})^{2}}$

###### Lemma 8.5

Let $w=\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}$, then we
can rewrite $I(\sigma_{q})$ as

$I(w)=-\dfrac{1}{2}\ln w$ (16)

In addition, if
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$,

  1. 1.

$\dfrac{\partial
I(w^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=-\dfrac{1}{2}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$
(17)

  2. 2.

$\dfrac{\partial
I(w^{\star}(\theta))}{\partial\gamma}=-\dfrac{(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}}{\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$
(18)

where
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$.

###### Proof 8.6

Proof of Lemma 8.5. Let
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$,

  1. 1.

$\displaystyle\dfrac{\partial
I(w^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}$
$\displaystyle=-\dfrac{1}{2}\cdot\dfrac{1}{w^{\star}(\theta)}\cdot\dfrac{\partial
w^{\star}(\theta)}{\partial(\mu_{\theta}-\theta)^{2}}$ Substitute Equation 9
into the above equation
$\displaystyle=-\dfrac{1}{2}\cdot\dfrac{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{-\sigma_{\theta}^{2}+\sqrt{\Delta}}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$
$\displaystyle=-\dfrac{1}{2}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$

  2. 2.

$\displaystyle\dfrac{\partial I(w^{\star}(\theta))}{\partial\gamma}$
$\displaystyle=-\dfrac{1}{2}\cdot\dfrac{1}{w^{\star}(\theta)}\cdot\dfrac{\partial
w^{\star}(\theta)}{\partial\gamma}$ Substitute Equation 10 into the above
equation
$\displaystyle=-\dfrac{1}{2}\cdot\dfrac{1}{w^{\star}(\theta)}\cdot\dfrac{1}{2\sqrt{\Delta}}$
$\displaystyle=-\dfrac{(\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}}{\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$

###### Lemma 8.7

Let $w=\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}$, then we
can rewrite Equation 3 as

$l(\theta,w)=w(1-w)\sigma_{\theta}^{2}+w^{2}(\mu_{\theta}-\theta)^{2}-\dfrac{\gamma}{2}\ln
w$ (19)

In addition, if
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$,

  1. 1.

$\dfrac{\partial
l(\theta,w^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=-\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$
(20)

  2. 2.

$\dfrac{\partial
l(\theta,w^{\star}(\theta))}{\partial\gamma}=\dfrac{3\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{2\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}-\dfrac{1}{2}\ln
w^{\star}(\theta)$ (21)

where
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$.

###### Proof 8.8

Proof of Lemma 8.7. By Lemma 8.3 and Lemma 8.5, it is clear that
$l(\theta,w^{\star}(\theta))=e(\theta,w^{\star}(\theta))+\gamma
I(w^{\star}(\theta))=w^{\star}(\theta)(1-w^{\star}(\theta))\sigma_{\theta}^{2}+w^{\star}(\theta)^{2}(\mu_{\theta}-\theta)^{2}-\dfrac{\gamma}{2}\ln
w^{\star}(\theta)$. In addition,

  1. 1.

$\displaystyle\dfrac{\partial
l(\theta,w^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}$
$\displaystyle=$ $\displaystyle\dfrac{\partial
e(\theta,w^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}+\gamma\cdot\dfrac{\partial
I(w^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}$ $\displaystyle=$
$\displaystyle\dfrac{\sigma_{\theta}^{2}(\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}-\dfrac{\gamma}{2}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$
$\displaystyle=$
$\displaystyle\dfrac{\sigma_{\theta}^{2}(\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))(-\sigma_{\theta}^{2}+\sqrt{\Delta})-4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})(\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$
$\displaystyle=$
$\displaystyle\dfrac{(-\sigma_{\theta}^{4}+\sigma_{\theta}^{2}\sqrt{\Delta}-4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}))(\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$
$\displaystyle=$
$\displaystyle\dfrac{\sqrt{\Delta}(-\sqrt{\Delta}+\sigma_{\theta}^{2})(\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$
$\displaystyle=$
$\displaystyle-\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$

  2. 2.

$\displaystyle\dfrac{\partial l(\theta,w^{\star}(\theta))}{\partial\gamma}$
$\displaystyle=\dfrac{\partial
e(\theta,w^{\star}(\theta))}{\partial\gamma}+\gamma\cdot\dfrac{\partial
I(w^{\star}(\theta))}{\partial\gamma}-\dfrac{1}{2}\ln w^{\star}(\theta)$
$\displaystyle=\dfrac{\sigma_{\theta}^{2}+\sqrt{\Delta}}{4\sqrt{\Delta}}+\dfrac{\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}-\dfrac{1}{2}\ln
w^{\star}(\theta)$
$\displaystyle=\dfrac{\Delta-\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{4\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}-\dfrac{1}{2}\ln
w^{\star}(\theta)$
$\displaystyle=\dfrac{3\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{2\sqrt{\Delta}(-\sigma_{\theta}^{2}+\sqrt{\Delta})}-\dfrac{1}{2}\ln
w^{\star}(\theta)$

####  8.1.2 Proof of the results.

###### Proof 8.9

Proof of Proposition 4.1. Because $d(\theta)=|\mu_{\theta}-\theta|$ by
definition and $|\mu_{\theta}-\theta|$ increases with
$(\mu_{\theta}-\theta)^{2}$, we only have to show the change of
$l(\theta,\sigma_{q}^{\star}(\theta)),I(\sigma_{q}^{\star}(\theta))$ and
$e(\theta,\sigma_{q}^{\star}(\theta))$ with respect to
$(\mu_{\theta}-\theta)^{2}$. By Lemma 7.2,

$l(\theta,\sigma_{q}^{\star}(\theta))=\begin{cases}l\left(\theta,\sqrt{\dfrac{w^{\star}(\theta)\sigma_{\theta}^{2}}{1-w^{\star}(\theta)}}\right)&|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)\\\
(\mu_{\theta}-\theta)^{2}&\text{otherwise}\end{cases}$
$I(\sigma_{q}^{\star}(\theta))=\begin{cases}I\left(\sqrt{\dfrac{w^{\star}(\theta)\sigma_{\theta}^{2}}{1-w^{\star}(\theta)}}\right)&|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)\\\
0&\text{otherwise}\end{cases}$
$e(\theta,\sigma_{q}^{\star}(\theta))=\begin{cases}e\left(\theta,\sqrt{\dfrac{w^{\star}(\theta)\sigma_{\theta}^{2}}{1-w^{\star}(\theta)}}\right)&|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)\\\
(\mu_{\theta}-\theta)^{2}&\text{otherwise}\end{cases}$

where
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$,
and $\tau_{d}(\gamma)>0$ is a threshold that increases in $\gamma$ and is not
less than $\sigma_{\theta}^{2}-\dfrac{\sigma_{\theta}^{4}}{4\gamma}$. Now,
let’s apply the results of Lemma 8.3, Lemma 8.5, and Lemma 8.7.

  1. 1.

When $|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)$, by Lemma 8.7,

$\dfrac{\partial
l(\theta,\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=-\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{8((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$

where
$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$.

We only need to show
$\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\leq
0$. When $|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)$, by the proof of Lemma
7.2, we know $\Delta\geq 0$, so
$\sigma_{\theta}^{4}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})\geq
0$. Thus,

$\displaystyle\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})\leq
0$ (22) $\displaystyle\iff\ $
$\displaystyle\sigma_{\theta}^{4}\Delta\leq[\sigma_{\theta}^{4}+2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})]^{2}$
$\displaystyle\iff\ $
$\displaystyle\sigma_{\theta}^{4}(\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))\leq\sigma_{\theta}^{8}+4\gamma\sigma_{\theta}^{4}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})+4\gamma^{2}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}$
$\displaystyle\iff\ $ $\displaystyle
4\gamma^{2}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}\geq 0$

When $|\mu_{\theta}-\theta|<\tau_{d}(\gamma)$,
$l(\theta,\sigma_{q}^{\star}(\theta))=(\mu_{\theta}-\theta)^{2}\implies\dfrac{\partial
l(\theta,\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=1$.
And $l(\theta,\sigma_{q}^{\star}(\theta))$ is continuous at
$|\mu_{\theta}-\theta|=\tau_{d}(\gamma)$. Thus,
$l(\theta,\sigma_{q}^{\star}(\theta))$ increases in $|\mu_{\theta}-\theta|$.
By definition, $l^{\star}=\min(\Gamma,\ l(\theta,\sigma_{q}^{\star}(\theta))$,
so $l^{\star}$ increases in $|\mu_{\theta}-\theta|$.

  2. 2.

When $|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)$, by Lemma 8.5,

$\dfrac{\partial
I(\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=-\dfrac{1}{2}\cdot\dfrac{\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}{\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})(-\sigma_{\theta}^{2}+\sqrt{\Delta})}$

By the proof of Lemma 7.2, we know $w^{\star}(\theta)\geq 0$ when
$|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)$, so the denominator
$((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})(-\sigma_{\theta}^{2}+\sqrt{\Delta})\geq
0$. Because of the above Inequality (22), then $\dfrac{\partial
I(\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}\geq 0$.

When $|\mu_{\theta}-\theta|<\tau_{d}(\gamma)$,
$I(\sigma_{q}^{\star}(\theta))=0\implies\dfrac{\partial
I(\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=0$.

We conclude that $I(\sigma_{q}^{\star}(\theta))$ increases in
$|\mu_{\theta}-\theta|$.

  3. 3.

Firstly, notice that $l(\theta,\sigma_{q}^{\star})=0$ for $d(\theta)=0$ and we
have shown that $l(\theta,\sigma_{q}^{\star}(\theta))$ monotonically increases
in $d(\theta)$ in part 1. In addition, we can see that $w^{\star}(\theta)\to
0$ as $d(\theta)\to\infty$, which leads to
$I(\sigma_{q}^{\star}(\theta))\to\infty$ and
$l(\theta,\sigma_{q}^{\star})\to\infty$ as $d(\theta)\to\infty$. These imply
that for any $\Gamma>0$, there must exist a threshold $\tau_{a}>0$ such that
$d(\theta)\leq\tau_{a}\iff l(\theta,\sigma_{q}^{\star})\leq\Gamma$.

  4. 4.

When $|\mu_{\theta}-\theta|<\tau_{d}(\gamma)$, by Lemma 7.2,
$\sigma_{q}^{\star}(\theta)=\infty$, thereby
$e(\theta,\sigma_{q}^{\star}(\theta))=(\mu_{\theta}-\theta)^{2}$ and
$\dfrac{\partial
e(\theta,\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=1>0$.

When $|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)$, by Lemma 8.3,

$\dfrac{\partial
e(\theta,\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}=\dfrac{\sigma_{\theta}^{2}(\sigma_{\theta}^{2}\sqrt{\Delta}-\sigma_{\theta}^{4}-2\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2}))}{8\sqrt{\Delta}((\mu_{\theta}-\theta)^{2}-\sigma_{\theta})^{2}}$

Because of Inequality (22), we have $\dfrac{\partial
e(\theta,\sigma_{q}^{\star}(\theta))}{\partial(\mu_{\theta}-\theta)^{2}}\leq
0$.

We conclude that if $|\mu_{\theta}-\theta|<\tau_{d}(\gamma)$,
$e(\theta,\sigma_{q}^{\star}(\theta))$ increases in
$(\mu_{\theta}-\theta)^{2}$; if $|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)$,
$e(\theta,\sigma_{q}^{\star}(\theta))$ decreases in $|\mu_{\theta}-\theta|$.

###### Proof 8.10

Proof of Proposition 4.2. By definition, if $d(\theta)\geq\tau_{a}$, users
will work on their own and $\theta^{\star}=\theta$, so
$|E[\theta^{\star}|\theta]-\mu_{\theta}|=|\theta-\mu_{\theta}|$.

If $d(\theta)<\tau_{a}$, $\theta^{\star}=\theta_{A}^{\star}$. By Equation 1,
we know
$E[\theta_{A}|\theta]=\dfrac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}\cdot\theta+\dfrac{\sigma_{q}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{2}}\cdot\mu_{\theta}$,
so

$|E[\theta_{A}^{\star}|\theta]-\mu_{\theta}|=\dfrac{\sigma_{\theta}^{2}}{\sigma_{\theta}^{2}+\sigma_{q}^{\star
2}(\theta)}|\theta-\mu_{\theta}|$

which is $0$ if $\theta=\mu_{\theta}$.

Additionally, since $l(\theta,\sigma_{q})\to\infty$ as $\sigma_{q}\to 0$ and
$\sigma_{q}=\infty$ is feasible, we must have $\sigma_{q}^{\star}(\theta)>0$.
Thus, $|E[\theta_{A}^{\star}|\theta]-\mu_{\theta}|<|\theta-\mu_{\theta}|$
whenever $\theta\neq\mu_{\theta}$.

###### Proof 8.11

Proof of Theorem 4.3. By Lemma 7.2, the AI’s output
$\theta_{A}(\sigma_{q}^{\star}(\sigma_{\theta}))$ is

$\theta_{A}(\sigma_{q}^{\star}(\theta))=\begin{cases}(1-w^{\star}(\theta))q+w^{\star}(\theta)\mu_{\theta}&|\mu_{\theta}-\theta|\geq\tau_{d}(\gamma)\\\
\mu_{\theta}&\text{otherwise}\end{cases}$

where
$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}$,
and $\tau_{d}(\gamma)>0$ is a threshold that increases in $\gamma$ and is not
less than $\sigma_{\theta}^{2}-\dfrac{\sigma_{\theta}^{4}}{4\gamma}$.

By definition, the unconditional variance of $\theta^{\star}$ is

$Var(\theta^{\star})=E[(\theta^{\star}-E[\theta^{\star}])^{2}]$

Let $\phi(\cdot)$ and $\Phi(\cdot)$ denote the probability density function
and the cumulative density function of $N(0,1)$, respectively. We know

$E[\theta^{\star}]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\theta^{\star}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

First, when $\tau_{d}>\tau_{a}$, we know that for any
$\theta<\tau_{a}<\tau_{d}$, $w^{\star}(\theta)=1$ and
$\theta^{\star}=\mu_{\theta}$; for any $\theta>\tau_{a}$,
$\theta^{\star}=\theta$, so

$\displaystyle E[\theta^{\star}]$
$\displaystyle=\int_{d(\theta)<\tau_{a}}\int_{-\infty}^{\infty}\mu_{\theta}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)>\tau_{a}}\int_{-\infty}^{\infty}\theta\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=\int_{d(\theta)<\tau_{a}}\mu_{\theta}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)>\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
Because
$\int_{d(\theta)>\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta=\int_{d(\theta)>\tau_{a}}(\theta-\mu_{\theta})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)>\tau_{a}}\mu_{\theta}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
and
$\int_{d(\theta)>\tau_{a}}(\theta-\mu_{\theta})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta=0$
due to the symmetry $\displaystyle=\mu_{\theta}$

When $\tau_{d}\leq\tau_{a}$,

$\displaystyle E[\theta^{\star}]$ $\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}\int_{-\infty}^{\infty}[(1-w^{\star}(\theta))q+w^{\star}(\theta)\mu_{\theta}]\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+\int_{d(\theta)<\tau_{d}}\int_{-\infty}^{\infty}\mu_{\theta}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)>\tau_{a}}\int_{-\infty}^{\infty}\theta\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}[(1-w^{\star}(\theta))\theta+w^{\star}(\theta)\mu_{\theta}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)<\tau_{d}}\mu_{\theta}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)>\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle\text{Because
$\int_{d(\theta)>\tau_{a}}\theta\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta=\int_{d(\theta)>\tau_{a}}(\theta-\mu_{\theta})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)>\tau_{a}}\mu_{\theta}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$},$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}[(1-w^{\star}(\theta))(\theta-\mu_{\theta})+\mu_{\theta}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)<\tau_{d}}\mu_{\theta}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{d(\theta)>\tau_{a}}\mu_{\theta}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}(1-w^{\star}(\theta))(\theta-\mu_{\theta})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{-\infty}^{\infty}\mu_{\theta}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
Notice that
$\int_{d(\theta)\in(\tau_{d},\tau_{a})}(1-w^{\star}(\theta))(\theta-\mu_{\theta})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta=0$,
because
$(1-w^{\star}(\theta))(\theta-\mu_{\theta})\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)$
is symmetric with respect to $\theta=\mu_{\theta}$. $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\mu_{\theta}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$ $\displaystyle\mu_{\theta}$

Thus, when $\tau_{d}>\tau_{a}$,

$\displaystyle Var(\theta^{\star})$
$\displaystyle=\int_{d(\theta)>\tau_{a}}\int_{-\infty}^{\infty}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
(23)
$\displaystyle=\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

When $\tau_{d}\leq\tau_{a}$

$\displaystyle Var(\theta^{\star})$ $\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}\int_{-\infty}^{\infty}[\theta_{A}(\sigma_{q}^{\star}(\theta))-\mu_{\theta}]^{2}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+$
$\displaystyle\int_{d(\theta)>\tau_{a}}\int_{-\infty}^{\infty}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}\int_{-\infty}^{\infty}[(1-w^{\star}(\theta))q+w^{\star}(\theta)\mu_{\theta}-\mu_{\theta}]^{2}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+$
$\displaystyle\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}\int_{-\infty}^{\infty}(1-w^{\star}(\theta))^{2}(q-\mu_{\theta})^{2}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+$
$\displaystyle\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}\int_{-\infty}^{\infty}(1-w^{\star}(\theta))^{2}(\theta+\epsilon_{q}-\mu_{\theta})^{2}\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+$
$\displaystyle\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}\int_{-\infty}^{\infty}(1-w^{\star}(\theta))^{2}[\epsilon_{q}^{2}-2\epsilon_{q}(\mu_{\theta}-\theta)+(\mu_{\theta}-\theta)^{2}]\phi\left(\dfrac{\epsilon_{q}}{\sigma_{q}^{\star}(\theta)}\right)d\epsilon_{q}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+$
$\displaystyle\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}(1-w^{\star}(\theta))^{2}[\sigma_{q}^{\star}(\theta)^{2}+(\mu_{\theta}-\theta)^{2}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+$
$\displaystyle\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{d(\theta)\in(\tau_{d},\tau_{a})}[(1-w^{\star}(\theta))w^{\star}(\theta)\sigma_{\theta}^{2}+(1-w^{\star}(\theta))^{2}(\mu_{\theta}-\theta)^{2}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle+$
$\displaystyle\int_{d(\theta)>\tau_{a}}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

Thus,

$\displaystyle Var(\theta^{\star})$
$\displaystyle=2\Big{[}\int_{\mu_{\theta}+\tau_{d}}^{\tau_{a}}[(1-w^{\star}(\theta))w^{\star}(\theta)\sigma_{\theta}^{2}$
(24) $\displaystyle+$
$\displaystyle(1-w^{\star}(\theta))^{2}(\mu_{\theta}-\theta)^{2}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta+\int_{\mu_{\theta}+\tau_{a}}^{\infty}(\mu_{\theta}-\theta)^{2}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta\Big{]}$

  1. 1.

Now, let us first show that when $\Gamma\to\infty$, $Var(\theta^{\star})$ is
strictly decreasing in $\gamma$. In this case,

$Var(\theta^{\star})=2\int_{\mu_{\theta}+\tau_{d}}^{\infty}[(1-w^{\star}(\theta))w^{\star}(\theta)\sigma_{\theta}^{2}+(1-w^{\star}(\theta))^{2}(\mu_{\theta}-\theta)^{2}]\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

Let
$h(\theta)\triangleq[(1-w^{\star}(\theta))w^{\star}(\theta)\sigma_{\theta}^{2}+(1-w^{\star}(\theta))^{2}(\mu_{\theta}-\theta)^{2}]$,
then

$Var(\theta_{A}(\sigma_{q}^{\star}(\theta)))=2\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}h(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

By the Leibniz integral rule,

$\displaystyle\dfrac{\partial
Var(\theta_{A}(\sigma_{q}^{\star}(\theta)))}{\partial\gamma}$
$\displaystyle=-2h(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)|_{\theta=\mu_{\theta}+\tau_{d}(\gamma)}\cdot\dfrac{\partial\sqrt{\tau_{d}(\gamma)}}{\partial\gamma}$
$\displaystyle+2\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}\dfrac{\partial
h(\theta)}{\partial\gamma}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

Since $\dfrac{\partial\sqrt{\tau_{d}(\gamma)}}{\partial\gamma}>0$ by Lemma
7.2, we only need to show:

$2\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}\dfrac{\partial
h(\theta)}{\partial\gamma}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta<0$

Notice that

$\displaystyle 2\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}\dfrac{\partial
h(\theta)}{\partial\gamma}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$ $\displaystyle
2\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}\dfrac{\partial
h(\theta)}{\partial w^{\star}(\theta)}\cdot\dfrac{\partial
w^{\star}(\theta)}{\partial\gamma}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}[(1-2w^{\star}(\theta))\sigma_{\theta}^{2}+2(w^{\star}(\theta)-1)(\mu_{\theta}-\theta)^{2}]\dfrac{1}{\sqrt{\Delta}}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle\text{where}\
\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})$
$\displaystyle=$
$\displaystyle\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}[2w^{\star}(\theta)((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}]\dfrac{1}{\sqrt{\Delta}}\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$

Let
$g(\theta)\triangleq[2w^{\star}(\theta)((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}]/\sqrt{\Delta}$,
we want to show

$\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}g(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta<0$

    1. (a)

First, when $\gamma>\sigma_{\theta}^{2}/2$, we want to show $g(\theta)\leq 0$
for any $\theta\geq\mu_{\theta}+\tau_{d}(\gamma)$.

By Lemma 7.2,
$\tau_{d}(\gamma)>\sqrt{\sigma_{\theta}^{2}-\sigma_{\theta}^{4}/(4\gamma)}$,
so $\tau_{d}(\gamma)>\sigma_{\theta}/\sqrt{2}$. This implies that for any
$\theta\geq\mu_{\theta}+\tau_{d}(\gamma)$,
$(\mu_{\theta}-\theta)^{2}>\sigma_{\theta}^{2}/2$.

If $(\mu_{\theta}-\theta)^{2}>\sigma_{\theta}^{2}$,
$2w^{\star}(\theta)((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}\leq-\sigma_{\theta}^{2}<0$,
because $w^{\star}(\theta)\leq 1$. And if
$\dfrac{\sigma_{\theta}^{2}}{2}<(\mu_{\theta}-\theta)^{2}\leq\sigma_{\theta}^{2}$,
$2w^{\star}(\theta)((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}\leq\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}<0$,
because $w^{\star}(\theta)>0$. Thus,
$(\mu_{\theta}-\theta)^{2}>\dfrac{\sigma_{\theta}^{2}}{2}$ implies
$2w^{\star}(\theta)((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})+\sigma_{\theta}^{2}-2(\mu_{\theta}-\theta)^{2}<0$,
which further implies $g(\theta)<0$.

Therefore, when $\gamma>\sigma_{\theta}^{2}/2$,

$\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}g(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta<0$

    2. (b)

When $\gamma\leq\sigma_{\theta}^{2}/2$:

Let $\alpha=\gamma/\sigma_{\theta}^{2}$ (so $\gamma\leq\sigma_{\theta}^{2}/2$
implies $\alpha\leq 1/2$).

$\Delta=\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})=\sigma_{\theta}^{4}\left[1+\dfrac{4\gamma}{\sigma_{\theta}^{2}}\left(\dfrac{(\mu_{\theta}-\theta)^{2}}{\sigma_{\theta}^{2}}-1\right)\right]=\sigma_{\theta}^{4}\left[1+4\alpha\left(\left(\dfrac{\mu_{\theta}-\theta}{\sigma_{\theta}}\right)^{2}-1\right)\right]$

Similarly, we can get

$w^{\star}(\theta)=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}}{4((\mu_{\theta}-\theta)^{2}-\sigma_{\theta}^{2})}=\dfrac{-1+\sqrt{1+4\alpha\left(\left(\dfrac{\mu_{\theta}-\theta}{\sigma_{\theta}}\right)^{2}-1\right)}}{4\left[\left(\dfrac{\mu_{\theta}-\theta}{\sigma_{\theta}}\right)^{2}-1\right]}$

The substitution $x\triangleq\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}$
yields

$\displaystyle\int_{\mu_{\theta}+\tau_{d}(\gamma)}^{\infty}g(\theta)\phi\left(\dfrac{\theta-\mu_{\theta}}{\sigma_{\theta}}\right)d\theta$
$\displaystyle=$
$\displaystyle\int_{\frac{\tau_{d}(\gamma)}{\sigma_{\theta}}}^{\infty}[(1-2\hat{w}(x,\alpha))\sigma_{\theta}^{2}+2(\hat{w}(x,\alpha)-1)\sigma_{\theta}^{2}x^{2}]\dfrac{1}{\sigma_{\theta}^{2}\sqrt{\hat{\Delta}(x,\alpha)}}\phi(x)\sigma_{\theta}dx$
$\displaystyle=$
$\displaystyle\int_{\frac{\tau_{d}(\gamma)}{\sigma_{\theta}}}^{\infty}[(1-2\hat{w}(x,\alpha))+2(\hat{w}(x,\alpha)-1)x^{2}]\dfrac{1}{\sqrt{\hat{\Delta}(x,\alpha)}}\phi(x)\sigma_{\theta}dx$
$\displaystyle=$
$\displaystyle\dfrac{1}{\sqrt{2\pi}}\int_{\hat{\tau_{d}}(\alpha)}^{\infty}[(1-2\hat{w}(x,\alpha))+2(\hat{w}(x,\alpha)-1)x^{2}]\dfrac{1}{\sqrt{\hat{\Delta}(x,\alpha)}}\exp\left(-\dfrac{x^{2}}{2}\right)dx$

where $\hat{\tau_{d}}(\alpha)=\dfrac{\tau_{d}(\gamma)}{\sigma_{\theta}}$,
$\hat{w}(x,\alpha)=\dfrac{-1+\sqrt{1+4\alpha(x^{2}-1)}}{4(x^{2}-1)}$ and
$\hat{\Delta}(x,\alpha)=1+4\alpha(x^{2}-1)$.

Note that

$(1-2\hat{w}(x,\alpha))+2(\hat{w}(x,\alpha)-1)x^{2}]\dfrac{1}{\sqrt{\hat{\Delta}(x,\alpha)}}=\dfrac{1}{2}\left[1+\dfrac{1-4x^{2}}{\sqrt{1+4\alpha(x^{2}-1)}}\right]$

Define

$G(\alpha)\triangleq\displaystyle\int_{\hat{\tau_{d}}(\alpha)}^{\infty}\left[1+\dfrac{1-4x^{2}}{\sqrt{1+4\alpha(x^{2}-1)}}\right]\exp\left(-\dfrac{x^{2}}{2}\right)dx$

We want to show $\forall\alpha\in[0,1/2],\ G(\alpha)<0$.

Let’s do another change of variables: $y\triangleq x^{2}-1$, which implies
$dy=2xdx$ and $x=\sqrt{y+1}$. This yields

$G(\alpha)=\int_{\hat{\tau_{d}}^{2}(\alpha)-1}^{\infty}\left[1-\dfrac{3+4y}{\sqrt{1+4\alpha
y}}\right]\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$

Let $\omega(y,\alpha)\triangleq 1-(3+4y)/\sqrt{1+4\alpha y}$. Note that

      1. i.

If $y\geq 0$, $\omega(y,\alpha)$ is increasing $\alpha$.

      2. ii.

If $y\in[-3/4,0)$, $\omega(y,\alpha)$ is decreasing $\alpha$.

      3. iii.

If $y\in[-1,-3/4)$, $\omega(y,\alpha)$ is increasing $\alpha$.

Correspondingly,

      1. i.

Let

$G_{0}(\alpha)\triangleq\int_{0}^{\infty}\omega(y,\alpha)\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$

we have $G_{0}(\alpha)\leq G_{0}(1/2)\leq G_{0}(1)<-0.96$.

      2. ii.

$\hat{\tau_{d}}^{2}(\alpha)-1\geq-3/4\Longleftrightarrow\hat{\tau_{d}}^{2}(\alpha)\geq
1/4$

Note that $\hat{\tau_{d}}^{2}(\alpha)=\tau_{d}(\gamma)/\sigma_{\theta}$, and
by the definition of $\tau_{d}(\gamma)$ in the proof of Lemma 7.2,
$\tau_{d}(\gamma)$ solves

$(\tau_{d}^{2}(\gamma,\sigma_{\theta})-\sigma_{\theta}^{2})m^{2}+\sigma_{\theta}^{2}m-\dfrac{\gamma}{2}\ln(m)=\tau_{d}^{2}(\gamma,\sigma_{\theta})-\sigma_{\theta}^{2})$

where
$m=\dfrac{-\sigma_{\theta}^{2}+\sqrt{\sigma_{\theta}^{4}+4\gamma(\tau_{d}^{2}(\gamma,\sigma_{\theta})-\sigma_{\theta}^{2})}}{4(\tau_{d}^{2}(\gamma,\sigma_{\theta})-\sigma_{\theta}^{2})}$.
This is equivalent to that $\hat{\tau_{d}}(\alpha)$ solves

$(\hat{\tau_{d}}^{2}(\alpha)-1)m^{2}+m-\dfrac{\alpha}{2}\ln(m)=\hat{\tau_{d}}^{2}(\alpha)$

where
$m=\dfrac{-1+\sqrt{1+4\alpha(\hat{\tau_{d}}^{2}(\alpha)-1)}}{4(\hat{\tau_{d}}^{2}(\alpha)-1)}$.

Thus, we can get there exists $\alpha^{\star}$ such that
$\hat{\tau_{d}}^{2}(\alpha)\geq
1/4\Longleftrightarrow\alpha\geq\alpha^{\star}$. And we can numerically
compute $\alpha^{\star}\approx 0.13845$.

Let

$G_{1}(\alpha)\triangleq\displaystyle\int_{\hat{\tau_{d}}^{2}(\alpha)-1}^{0}\omega(y,\alpha)\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$

Since $\omega(y,\alpha)$ is decreasing in $\alpha$, we have

$G_{1}(\alpha)\leq\int_{\hat{\tau_{d}}^{2}(\alpha)-1}^{0}\omega(y,\alpha^{\star})\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy\leq\int_{-3/4}^{0}\omega(y,\alpha^{\star})\exp\left(-\dfrac{y+1}{2}\right)\dfrac{1}{2\sqrt{y+1}}dy$

We can numerically find