# How Far Can Fairness Constraints Help Recover From Biased Data?

Mohit Sharma1 Work done during internship at Microsoft Research India. The
author is supported by Microsoft Research India Joint PhD fellowship. Amit
Deshpande2

(1IIIT Delhi
2Microsoft Research India
)

###### Abstract

Blum & Stangl (2019) propose a data bias model to simulate under-
representation and label bias in underprivileged population. For a stylized
data distribution with i.i.d. label noise, under certain simple conditions on
the bias parameters, they show that fair classification with equal opportunity
constraints even on extremely biased distribution can recover an optimally
accurate and fair classifier on the original distribution. Although their
distribution is stylized, their result is interesting because it demonstrates
that fairness constraints can implicitly rectify data bias and simultaneously
overcome a perceived fairness-accuracy trade-off. In this paper, we give an
alternate proof of their result using threshold-based characterization of
optimal fair classifiers. Moreover, we show that their conditions on the bias
parameters are both necessary and sufficient for their recovery result. Our
technique is arguably more flexible, as it readily extends to more general
distributions, e.g., when the labels in the original distribution have Massart
noise instead of i.i.d. noise. Finally, we prove that for any data
distribution, if the optimally accurate classifier in a hypothesis class is
fair and _robust_ , then it can be recovered through fair classification on
the biased distribution, whenever the bias parameters satisfy certain simple
conditions.

##  1 Introduction

Fairness in machine learning has been an important topic of both qualitative
and quantitative research for more than a decade [1, 7]. Several quantitative
metrics of fairness and various bias mitigation techniques developed over the
years are available to practitioners as open-source fairness toolkits [2].
Fairness-accuracy trade-offs seem often inevitable and bias mitigation
guarantees can fail when there is a mismatch between train and test data
distributions; it is a double whammy when we get neither the desired accuracy
nor the desired fairness on deployment. Hence, it is important to understand
the role of fair classification and fairness constraints when the training
data has systematic biases [3, 5, 10].

Blum & Stangl [3] propose a model for data bias that simulates systematic
under-representation and label bias in underprivileged population. Their model
induces bias in a given data distribution as follows: The positively labeled
and negatively labeled underprivileged population is systematically under-
sampled, and the positive labeled underprivileged population undergoes a
random label flip with some probability. Applying this data bias model to a
stylized data distribution with i.i.d. label noise, they prove a recovery
result that makes the following high-level point: careful choice of fairness
constraints can implicitly rectify even extreme data bias and can also
overcome a perceived fairness-accuracy trade-off. Formally, for their stylized
data distribution with i.i.d. label noise, they prove that the optimal fair
classifier satisfying equal opportunity on a biased (or bias-induced) data
distribution coincides with the Bayes optimal classifier (that also happens to
be perfectly fair) on the original data distribution, if the bias parameters
satisfy certain simple conditions. Their proof is tailored to equal
opportunity constraints, and uses a careful case analysis that is not easily
amenable to other distributions. They also show an example distribution where
a similar result fails to hold if we use demographic parity constraints
instead of equal opportunity constraints.

Our approach is significantly different from that of Blum & Stangl [3] in that
we first show that the class label probabilities conditioned on the input
features undergo a linear fractional transformation when we induce under-
representation and label bias in the data distribution. This plays an
important role in our proof because the optimal group-fair classifiers can be
mathematically characterized by thresholds on the above probabilities [8]. We
show that, under the conditions on bias parameters given by Blum & Stangl [3],
the fairness-corrected thresholds applied to a linear fractional
transformation recover the Bayes optimal classifier on the original
distribution that has i.i.d. label noise. Our proof makes it easier to observe
that these conditions on the bias parameters are both necessary and
sufficient. Moreover, unlike the proof by Blum & Stangl [3], our proof readily
extends to the case when the original distribution has Massart label noise
instead of i.i.d. label noise. Finally, we define _robustness_ of a classifier
using a natural definition of stability w.r.t. small changes to the
distribution. Using our proof technique, we can show that for _any_ data
distribution (and not just the stylized distribution used in Blum & Stangl
[3]), if the optimally accurate classifier in a hypothesis class is fair and
_robust_ , then it can be recovered through fair classification using equal
opportunity constraints on the biased distribution, whenever the bias
parameters satisfy certain simple conditions.

The rest of our paper is organized as follows.

  * •

In section 2, we describe the theoretical setup. In Subsection 2.1, we
describe the data bias model of Blum & Stangl [3] for under-representation and
label bias in any given data distribution. Proposition 1 shows that under the
above data bias model, the class probability conditioned on the input
undergoes a linear fractional transform. We combine this observation with a
threshold-based characterization of optimal fair classifiers [8], which helps
us characterize the optimal fair classifiers for equal opportunity and
demographic parity on the biased data distribution (Propositions 5 and 6).
Interestingly, we note that the hypothesis class of fair classifiers
satisfying equal opportunity or demographic parity remains unchanged under the
data bias model of Blum & Stangl (Propositions 3 and 4). Finally, in
Subsection 2.2, we rephrase the construction of a stylized data distribution
used in the recovery result of Blum & Stangl [3].

  * •

In section 3, we use Proposition 5 to prove a slightly stronger version of the
data bias recovery guarantees from Theorem $4.1$ in Blum & Stangl [3] (Theorem
7). Our proof gives necessary and sufficient conditions for recovery, whereas
the earlier theorem only provided sufficient conditions.

  * •

In section 4, we use our proof strategy from Theorem 7 to prove the same set
of recovery guarantees for a slightly more general data distribution with
Massart Noise [6].

  * •

Finally, in section 5, assuming $\epsilon$-robustness on the optimal fair
classifier from a hypothesis class, we obtain data bias recovery guarantees
for any arbitrary data distribution and hypothesis class.

##  2 Theoretical Setup for Recovery from Data Bias

We divide the theoretical setup of Blum & Stangl result on recovery from data
bias [3] into two parts. The first part describes the data bias model that
takes any data distribution $D$ and introduces under-representation and label
bias for the unprivileged population or group within it to get a biased
distribution $\tilde{D}$. The second part describes a particular distribution
$D$ used by Blum & Stangl to prove their main result on recovery under data
bias.

###  2.1 Data Bias Model for under-representation and Label Bias

Let $(X,A,Y)$ be a random data point from the joint distribution $D$ over
$\mathcal{X}\times\mathcal{A}\times\mathcal{Y}$, where $\mathcal{X}$ denotes
the set of features, $\mathcal{A}$ denotes the set of sensitive attributes,
and $\mathcal{Y}$ denotes the set of class labels. We consider group-aware
classifiers $f:\mathcal{X}\times\mathcal{A}\rightarrow\mathcal{Y}$, and we
work with the simple case of binary sensitive attributes
$\mathcal{A}=\\{0,1\\}$ and binary class labels $\mathcal{Y}=\\{0,1\\}$
throughout this paper. It is well-known that the classifier of maximum
accuracy $h^{*}=\mathrm{argmax}_{f}\mathrm{Pr}\left(f(X,A)=Y\right)$, also
known as the Bayes optimal classifier, is given by
$h^{*}(x,a)=\mathbb{I}\left(\eta(x,a)\geq 1/2\right)$, where
$\eta(x,a)=\mathrm{Pr}\left(Y=1|X=x,A=a\right)$.

Consider a biased distribution $\tilde{D}$ obtained from the original
distribution $D$ by the following process defined by bias parameters
$\beta_{p},\beta_{n},\nu\in(0,1)$ [3]. A random data point $(X,A,Y)$ from $D$
with $A=1$ remains unchanged, the points with $A=0,Y=1$ have an independent
survival probability of $\beta_{p}$, and the points with $A=0,Y=0$ have an
independent survival probability of $\beta_{n}$. Finally, the survived points
with $A=0,Y=1$ keep their class label $1$ with probability $1-\nu$, and it
gets flipped to $0$ with probability $\nu$. The above transformation keeps the
joint distribution $(X,A)$ unchanged since the survival probabilities and
label flips are applied uniformly. Therefore, we use $(X,A,\tilde{Y})$ to
denote a random data point from the biased distribution $\tilde{D}$.

######  Proposition 1.

Let $(X,A,Y)$ denote a random data point from any given distribution $D$ and
let $(X,A,\tilde{Y})$ be a random data point from its corresponding biased
distribution $\tilde{D}$ as described above. Let
$\eta(x,a)=\mathrm{Pr}\left(Y=1|X=x,A=a\right)$ and
$\tilde{\eta}(x,a)=\mathrm{Pr}\left(\tilde{Y}=1|X=x,A=a\right)$. Then

$\tilde{\eta}(x,a)=\begin{cases}\eta(x,1),\qquad\qquad\qquad\quad\text{for
$a=1$}\\\ \dfrac{(1-\nu)\eta(x,0)}{(1-c)\eta(x,0)+c},\qquad\text{for $a=0$,
where $c=\dfrac{\beta_{n}}{\beta_{p}}$}.\end{cases}$

###### Proof.

The data bias applies only to the unprivileged group $a=0$, so for the
privileged group $a=1$, we have $\tilde{\eta}(x,1)=\eta(x,1)$, for all
$x\in\mathcal{X}$, whereas for the underprivileged group $A=0$ we have

$\displaystyle\tilde{\eta}(x,0)$
$\displaystyle=\mathrm{Pr}\left(\tilde{Y}=1|X=x,A=0\right)$
$\displaystyle=\frac{(1-\nu)\beta_{p}\eta(x,0)}{\beta_{p}\eta(x,0)+\beta_{n}(1-\eta(x,0))}$
$\displaystyle=\frac{(1-\nu)\beta_{p}\eta(x,0)}{(\beta_{p}-\beta_{n})\eta(x,0)+\beta_{n}}$
$\displaystyle=\frac{(1-\nu)\eta(x,0)}{(1-c)\eta(x,0)+c},\qquad\text{where
$c=\frac{\beta_{n}}{\beta_{p}}$}.$

∎

The fact that $\tilde{\eta}(x,a)$ can be written as a linear fractional
transformation of $\eta(x,a)$ is interesting because, as we will see, it plays
an important role when we consider certain threshold classifiers that apply a
threshold to $\tilde{\eta}(x,a)$ or $\eta(x,a)$. Another advantage is that a
composition of two linear fractional transformations is another linear
fractional transformation, which lends itself nicely to repeated data bias in
pipelines and long-term fairness.

We make a few observations about fair classifiers on the original distribution
$D$ and the biased distribution $\tilde{D}$. Given any classifier
$f:\mathcal{X}\times\mathcal{A}\rightarrow\mathcal{Y}$, let $TPR_{a}(f)$ and
$\widetilde{TPR}_{a}(f)$ denote its true positive rates conditioned on $A=a$
for distributions $D$ and $\tilde{D}$, respectively. Then

$\widetilde{TPR}_{1}(f)=TPR_{1}(f)=\frac{\mathrm{Pr}\left(f(X,A)=1,A=1,Y=1\right)}{\mathrm{Pr}\left(Y=1,A=1\right)}=\frac{\mathrm{E}_{X|A=1}\left[f(X,1)\eta(X,1)\right]}{\mathrm{E}_{X|A=1}\left[\eta(X,1)\right]},$

and

$\displaystyle TPR_{0}(f)$
$\displaystyle=\frac{\mathrm{Pr}\left(f(X,A)=1,A=0,Y=1\right)}{\mathrm{Pr}\left(Y=1,A=0\right)}=\frac{\mathrm{E}_{X|A=0}\left[f(X,0)\eta(X,0)\right]}{\mathrm{E}_{X|A=0}\left[\eta(X,0)\right]},$
$\displaystyle\widetilde{TPR}_{0}(f)$
$\displaystyle=\frac{\mathrm{Pr}\left(f(X,A)=1,A=0,\tilde{Y}=1\right)}{\mathrm{Pr}\left(\tilde{Y}=1,A=0\right)}=\frac{\mathrm{E}_{X|A=0}\left[f(X,0)\tilde{\eta}(X,0)\right]}{\mathrm{E}_{X|A=0}\left[\tilde{\eta}(X,0)\right]}.$

######  Proposition 2.

Let $D$ and $\tilde{D}$ be the original distribution and its biased version,
respectively, defined as above. Let $h$ be any hypothesis, and let
$TPR_{0}(h)$ and $\widetilde{TPR}_{0}(h)$ be its true positive rates according
to $D$ and $\tilde{D}$, respectively, conditioned on the underprivileged group
$A=0$. Let $TNR_{a}(h)$ and $\widetilde{TNR}_{a}(h)$ be the true negative
rates defined similarly. Then $\widetilde{TPR}_{0}(h)=TPR_{0}(h)$ and

$\widetilde{TNR}_{0}(h)=\mathrm{Pr}\left(Y=0|\tilde{Y}=0,A=0\right)~{}TNR_{0}(h)+\mathrm{Pr}\left(Y=1|\tilde{Y}=0,A=0\right)~{}FNR_{0}(h).$

The proof of Proposition 2 is given in Appendix A.1. From Proposition 2 we
have that $\widetilde{TPR}_{0}(h)=TPR_{0}(h)$. Since the data bias model
leaves the group $a=1$ unchanged, $\widetilde{TPR}_{1}(h)=TPR_{1}(h)$. Thus,
$TPR_{0}(h)=TPR_{1}(h)$ iff $\widetilde{TPR}_{0}(h)=\widetilde{TPR}_{1}(h)$.
Hence, we get the following interesting consequence stated in Proposition 3
stated below. Given any hypothesis class $\mathcal{H}$, let
$\mathcal{H}_{\text{fair, EO}}$ be its subset consisting of classifiers that
satisfy equal opportunity on the original distribution $D$, i.e.,
$\mathcal{H}_{\text{fair,
EO}}=\\{f\in\mathcal{H}\;:\;TPR_{0}(f)=TPR_{1}(f)\\}$. Similarly, let
$\widetilde{\mathcal{H}}_{\text{fair, EO}}$ be the subset of classifiers in
$\mathcal{H}$ that satisfy equal opportunity on the biased distribution
$\tilde{D}$, i.e., $\widetilde{\mathcal{H}}_{\text{fair,
EO}}=\\{f\in\mathcal{H}\;:\;\widetilde{TPR}_{0}(f)=\widetilde{TPR}_{1}(f)\\}$.

######  Proposition 3.

For any data distribution $D$ and its biased version $\tilde{D}$ described as
above and any hypothesis class $\mathcal{H}$, the subsets of classifiers in
$\mathcal{H}$ that satisfy equal opportunity on $D$ also satisfy equal
opportunity on $\tilde{D}$ and vice versa, i.e., $\mathcal{H}_{\text{fair,
EO}}=\widetilde{\mathcal{H}}_{\text{fair, EO}}$.

Similarly, we can define $\mathcal{H}_{\text{fair, DP}}$ and
$\widetilde{\mathcal{H}}_{\text{fair, DP}}$ to be the subset of classifiers in
$\mathcal{H}$ that satisfy demographic parity on the distribution $D$ and
$\tilde{D}$, respectively. In other words, $\mathcal{H}_{\text{fair,
DP}}=\\{f\in\mathcal{H}\;:\;\mathrm{Pr}\left(f(X,A)=1|A=0\right)=\mathrm{Pr}\left(f(X,A)=1|A=1\right)\\}$.
An analog of Proposition 3 holds trivially for demographic parity because the
demographic parity constraint depends on neither $Y$ nor $\tilde{Y}$.

######  Proposition 4.

For any data distribution $D$ and its biased version $\tilde{D}$ described as
above and any hypothesis class $\mathcal{H}$, the subsets of classifiers in
$\mathcal{H}$ that satisfy demographic parity on $D$ also satisfy demographic
parity on $\tilde{D}$ and vice versa, i.e., $\mathcal{H}_{\text{fair,
DP}}=\widetilde{\mathcal{H}}_{\text{fair, DP}}$.

######  Remark.

Proposition 3 and 4 show that the hypothesis class of classifiers that satisfy
equal opportunity and demographic parity remains unchanged the under-
representation and label bias model. It is easy to see that the same holds
also for _approximately_ fair classifiers that satisfy these fairness
constraints upto a small additive or multiplicative error. However, similar to
Blum & Stangl [3], we focus only on exact fairness in this paper.

Proposition 1 combined with known characterizations of optimal fair
classifiers [8, 4] allows us to express the optimal fair classifiers on the
biased distribution $\tilde{D}$ as threshold classifiers on $\eta(x,a)$.

######  Proposition 5.

For any distribution $D$ and its biased version $\tilde{D}$ described above,
let $\tilde{h}_{EO}$ be a classifier of the maximum accuracy among all binary
classifiers that satisfy equal opportunity on $\tilde{D}$. Then there exists
$\lambda^{*}\in\mathbb{R}$ such that

$\tilde{h}_{EO}(x,a)=\begin{cases}\mathbb{I}\left(\eta(x,0)\geq\dfrac{1}{1+\dfrac{1-2\nu}{c}+\dfrac{\lambda^{*}}{\beta_{n}\mathrm{Pr}\left(Y=1,A=0\right)}}\right),\quad\text{for
$a=0$}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2-\dfrac{\lambda^{*}}{\mathrm{Pr}\left(Y=1,A=1\right)}}\right),\quad\text{for
$a=1$}.\end{cases}.$

The Proof of Proposition is given in Appendix A.2. Combining Proposition 1
with a similar characterization of optimal fair classifiers for demographic
parity [8, 4] gives us the following proposition.

######  Proposition 6.

For any distribution $D$ and its biased version $\tilde{D}$ described above,
let $\tilde{h}_{DP}$ be a classifier of the maximum accuracy among all binary
classifiers that satisfy demographic parity on $\tilde{D}$. Then there exists
$\lambda^{*}\in\mathbb{R}$ such that

$\tilde{h}_{DP}(x,a)=\begin{cases}\mathbb{I}\left(\eta(x,0)\geq\dfrac{c(r-\lambda^{*})}{2r(1-\nu)-(1-c)(r-\lambda^{*})}\right),\quad\text{for
$a=0$}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2}+\dfrac{\lambda^{*}}{2(1-r)}\right),\quad\text{for
$a=1$}\end{cases}$

The Proof of Proposition 6 is given in Appendix A.3. Note that unlike
Proposition 5, the threshold expressions in Proposition 6 depend only on the
bias parameters and $\lambda^{*}$.

###  2.2 The Original Data Distribution in Blum & Stangl Recovery Result [3]

Now we describe the specific distribution $D$ on which Blum & Stangl [3] show
their recovery result. We rephrase its construction below in a way consistent
with our notation. Take any joint distribution on
$\mathcal{X}\times\mathcal{A}$ and denote a random data point from it as
$(X,A)$. Let $\mathrm{Pr}\left(A=0\right)=r$ and
$\mathrm{Pr}\left(A=1\right)=1-r$, for some $0<r<1$. Now take a group-aware
classifier $h:\mathcal{X}\times\mathcal{A}\rightarrow\mathcal{Y}$ that has
equal base rates on the two groups, i.e.,
$\mathrm{Pr}\left(h(X,0)=1\right)=\mathrm{Pr}\left(h(X,1)=1\right)$, and
construct a joint distribution $D$ on
$\mathcal{X}\times\mathcal{A}\times\mathcal{Y}$ as

$Y=\begin{cases}h(X,A),\quad\text{with probability $1-\delta$},\\\ \neg
h(X,A),\quad\text{with probability $\delta$},\end{cases}$

for some $0<\delta<1/2$. In what follows, we consider the hypothesis class
$\mathcal{H}$ to be the set of all binary classifiers on
$\mathcal{X}\times\mathcal{A}$. It is well-known that the Bayes optimal
classifier $h^{*}:\mathcal{X}\times\mathcal{A}\rightarrow\mathcal{Y}$ that
maximizes accuracy among all classifiers in $\mathcal{H}$ is given by
$h^{*}(x,a)=\mathbb{I}\left(\eta(x,a)\geq 1/2\right)$, where
$\eta(x,a)=\mathrm{Pr}\left(Y=1|X=x,A=a\right)$. Let $h_{EO}$ and $h_{DP}$ be
the classifiers of maximum accuracy among the equal opportunity classifiers
$\mathcal{H}_{\text{fair, EO}}$ and the demographic parity classifiers
$\mathcal{H}_{\text{fair, DP}}$, respectively. Observe that in the above
distribution $D$ used by Blum & Stangl [3], we have $h\equiv h^{*}\equiv
h_{EO}\equiv h_{DP}$, since the labeling function $h$ has equal base rates.

Blum & Stangl [3] then construct the biased distribution $\tilde{D}$ from $D$
using bias parameters $\beta_{p},\beta_{n}$ and $\nu$ in the same way as what
is described in the previous subsection (see Subsection 2.1). Let
$\tilde{h}_{EO}$ and $\tilde{h}_{DP}$ be the optimal fair classifiers on
$\tilde{D}$ under equal opportunity and demographic parity constraints,
respectively. In other words, let $\tilde{h}_{EO}$ and $\tilde{h}_{DP}$ be the
classifiers of maximum accuracy from $\widetilde{\mathcal{H}}_{\text{fair,
EO}}$ and $\widetilde{\mathcal{H}}_{\text{fair, DP}}$, respectively.

An interesting thing about the above distribution $D$ is that
$\eta(x,a)=\mathrm{Pr}\left(Y=1|X=x,A=a\right)$ takes only two possible
values, $\delta$ and $1-\delta$. It essentially divides the population into
two clear points as low-risk and high-risk populations, respectively. As a
consequence of that and Proposition 1, $\tilde{\eta}(x,a)$ takes four possible
values: $\delta$ and $1-\delta$ for the privileged group $a=1$, and two other
values that are linear fractional transformation of $\delta$ and $1-\delta$
for the unprivileged group $a=0$. Note that the Bayes optimal classifier on
$\tilde{D}$ that gives $\mathbb{I}\left(\tilde{\eta}(x,a)\geq 1/2\right)$ can
fail to recover $h^{*}$ because of the data bias and its corresponding linear
fractional transformation between $\eta(x,0)$ and $\tilde{\eta}(x,0)$ shown in
Proposition 1. However, if we can find a classifier that uses group-wise
thresholds to separate the high-risk and low-risk populations for each group,
separately, then we can recover $h^{*}$. This is exactly what we show and our
approach based on threshold classifiers is completely different compared to
the careful case analysis of Blum & Stangl [3].

##  3 An Alternate Proof of Blum & Stangl Recovery Result [3]

We now restate a slightly stronger version of the main theorem of Blum &
Stangl on recovery from data bias [3]. We show that the sufficient conditions
used in their Theorem 4.1 [3] are actually both necessary and sufficient to
recover $h^{*}$ using optimal equal opportunity classifier on the biased data
distribution $\tilde{D}$.

######  Theorem 7.

(a slightly stronger version of Theorem 4.1 in [3]) For the data distribution
$D$ described in Subsection 2.2 and its biased version $\tilde{D}$ described
in Subsection 2.1, the data distribution and bias parameters satisfy

$(1-r)(1-2\delta)+r((1-\delta)\beta_{p}(1-2\nu)-\delta\beta_{n})>0\quad\text{and}\quad(1-r)(1-2\delta)+r((1-\delta)\beta_{n}-(1-2\nu)\delta\beta_{p})>0,$

if and only if the optimal equal opportunity classifier on $\tilde{D}$
recovers the Bayes optimal classifier on $D$, i.e., $\tilde{h}_{EO}\equiv
h_{EO}\equiv h^{*}$.

Later we show a stronger Theorem 9 that subsumes Theorem 7, and its complete
proof is given in Appendix A.4. Here we outline the main idea for the reader
anyway. Proposition 5 characterizes $\tilde{h}_{EO}$ as
$\tilde{h}_{EO}(x,a)=\mathbb{I}\left(\eta(x,a)\geq t_{a}\right)$ that applies
group-dependent thresholds $t_{a}$ on $\eta(x,a)$, where the threshold $t_{a}$
is actually a function of $\lambda^{*}$, the data distribution parameters, and
the bias parameters. As long as these thresholds applied to $\eta(x,a)$ for
both the groups $a=0$ and $a=1$ lie within the interval $(\delta,1-\delta)$,
the decisions of the corresponding classifier coincide with
$\tilde{h}_{EO}\equiv h^{*}$. We show that the characterization of optimal
equal opportunity classifier in Proposition 5 narrows the choice of candidate
classifiers down to only three, namely, $h^{*}$ and the trivial all-zero and
all-one classifiers. We prove that the conditions in Theorem 7 are necessary
and sufficient for the optimal $\lambda^{*}$ parameter in Proposition 5 to
satisfy that the group-dependent thresholds $t_{0}$ and $t_{1}$ lie in the
interval $(\delta,1-\delta)$, and equivalently, $\tilde{h}_{EO}\equiv h^{*}$.

Similarly, we obtain necessary and sufficient conditions for when the optimal
demographic parity classifier on $\tilde{D}$ recovers $h^{*}$. Blum & Stangl
[3] only give a specific example where such a recovery is impossible via
demographic parity constraints (see Subsection 3.1 of [3]) but they do not
give any recovery conditions for demographic parity similar to Theorem 7 (see
Table 1 in [3]).

######  Theorem 8.

For the data distribution $D$ described in Subsection 2.2 and its biased
version $\tilde{D}$ described in Subsection 2.1, the data distribution and
bias parameters satisfy

$\beta_{p}(1-\delta)(1-2\delta-2r(\nu-\delta))+\delta\beta_{n}(1-2\delta-2r(1-\delta))>0\text{
and }\\\
\beta_{p}\delta(1-2r(1-\nu)-2\delta(1-r))+(1-\delta)\beta_{n}(1-2\delta(1-r))>0$

if and only if the optimal demographic parity classifier on $\tilde{D}$
recovers the Bayes optimal classifier on $D$, i.e., $\tilde{h}_{DP}\equiv
h_{DP}\equiv h^{*}$.

The overall proof idea is very similar to what is described above, and the
complete proof of Theorem 8 is included in Appendix A.5.

Finally, a few words about the original proof of Blum & Stangl recovery result
[3]. They start with any classifier that satisfies equal opportunity on
$\tilde{D}$, and propose a clever way to iteratively edit its decisions to
increase accuracy while not violating the equal opportunity constraints. They
show that this iterative editing stops when it reaches one of the three
possible candidates, namely, $h^{*}$ and the trivial all-zero and all-one
classifier. They use the conditions to rule out the two trivial candidates for
having lower accuracy on $\tilde{D}$. Their proof technique of iterative edits
is specific to their distribution $D$ and is arguably difficult to generalize
when $D$ is any general distribution. Secondly, it is not easy to perform
iterative edits for a data bias pipeline, whereas our techniques (especially
Propositions 1, 5, and 6) lend themselves easily to data bias pipelines and
long-term fairness problems where data bias model gets applied repeatedly
across multiple stages.

##  4 Recovery Result under Data Bias for Massart Noise

We now assume any arbitrary data distribution $D$ on
$\mathcal{X}\times\mathcal{A}$. Let
$h:\mathcal{X}\times\mathcal{A}\rightarrow\mathcal{Y}$ be any hypothesis that
satisfies
$\mathrm{Pr}\left(h(X,A)=1|A=0\right)=\mathrm{Pr}\left(h(X,A)=1|A=1\right)$.
Let $\delta<1/2$, and extend the distribution to
$\mathcal{X}\times\mathcal{A}\times\mathcal{Y}$ as follows. $Y|X=x,A=a$ takes
value $h(x,a)$ with probability $1-\delta(x,a)$, and $\neg h(x,a)$ with
probability $\delta(x,a)$, for some $\delta(x,a)\leq\delta$. Let $D$ be the
resulting distribution on $\mathcal{X}\times\mathcal{A}\times\mathcal{Y}$.
This type of bounded noise in class label is popularly known as Massart
noise111Equivalently known as malicious classification noise in previous work
[9, 11]. in literature, based on a noise model proposed by Massart and
Nédélec[6]. We assume that the Massart noise is added in a way that equalizes
the base rates on the two protected groups, i.e., $Pr(Y=1|A=0)=Pr(Y=1|A=1)=q$.
Since $\delta<1/2$, the Bayes optimal classifier $h^{*}$ on the distribution
$D$ coincides with $h$ and satisfies Equal Opportunity.

We prove the following stronger theorem for the above distribution $D$ with
Massart noise that extends and subsumes Theorem 7.

######  Theorem 9.

For any distribution $D$ defined as above and data bias parameters
$\beta_{p},\beta_{n},\nu$ that satisfy

$(1-r)(1-2\delta)+r\left((1-\delta)\beta_{p}(1-2\nu)-\delta\beta_{n}\right)>0\quad\text{and}\quad(1-r)(1-2\delta)+r\left((1-\delta)\beta_{n}-\delta\beta_{p}(1-2\nu)\right)>0,$

the optimal equal opportunity classifier on the biased distribution
$\tilde{D}$ recovers the Bayes optimal classifier on the original distribution
$D$, i.e., $\tilde{h}_{EO}\equiv h^{*}$. _Note that the above conditions are
the same as the ones in Theorem 4.1 of Blum-Stangl._

The proof of Theorem 9 is given in Appendix A.4. The same proof also works for
group-dependent Massart noise, i.e., there exist $\delta_{0},\delta_{1}<1/2$
such that $\delta(x,a)\leq\delta_{a}$, for all
$(x,a)\in\mathcal{X}\times\mathcal{A}$.

##  5 Recovering Robust Hypothesis under Data Bias on Arbitrary Data
Distributions and Hypothesis Classes

In the previous sections, we assumed $\mathcal{H}$ to be the class of all
group-aware classifiers, i.e., all functions
$h:\mathcal{X}\times\mathcal{A}\rightarrow\mathcal{Y}$ and $D$ to be a
stylized distribution coming from i.i.d. or Massart label noise added to an
underlying deterministic _ground truth_ classifier. In this section, we remove
both these restrictions and work with arbitrary data distributions and
hypothesis classes. As a result, we need to make an assumption that the most
accurate and fair classifier that we want to recover under data bias must be
somewhat _stable_ or _robust_. We describe the precise definition of
_robustness_ below and then proceed to prove our recovery result for arbitrary
data distributions and hypothesis classes in Theorem 10.

Let $\mathcal{H}$ be any given hypothesis class. Let $D$ be any distribution
on $\mathcal{X}\times\mathcal{A}\times\mathcal{Y}$. Let
$\mathrm{Pr}\left(A=0\right)=r$ and $\mathrm{Pr}\left(A=1\right)=1-r$. Let
$\mathrm{Pr}\left(Y=1|A=0\right)=\mathrm{Pr}\left(Y=1|A=1\right)=q$. Let
$h^{*}$ be the most accurate hypothesis from $\mathcal{H}$, i.e.,

$h^{*}=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\mathrm{Pr}\left(h(X,A)=Y\right)=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\mathrm{E}_{(X,A)}\left[h(X,A)(2\eta(X,A)-1)\right].$

Assume that $h^{*}$ satisfies equal opportunity, i.e.,
$\mathrm{Pr}\left(h^{*}(X,A)=1|Y=1,A=0\right)=\mathrm{Pr}\left(h^{*}(X,A)=1|Y=1,A=1\right)$.

###  5.1 Definition of $\epsilon$-robustness

######  Definition.

We define $h^{*}\in\mathcal{H}$ to be _$\epsilon$ -robust_ if, for any
distribution $D^{\prime}$ with random data points $(X,A,Y^{\prime})$ such that

$\eta^{\prime}(x,a)=\mathrm{Pr}\left(Y^{\prime}=1|X=x,A=a\right)=\frac{P_{a}\eta(x,a)+Q_{a}}{R_{a}\eta(x,a)+S_{a}},$

with $S_{a}=1,|R_{a}|\leq\epsilon,|Q_{a}|\leq\epsilon,1-\epsilon\leq P_{a}\leq
1+\epsilon$, and $P_{a}S_{a}-Q_{a}R_{a}\geq 0$, the optimal classifier $h^{*}$
remains unchanged, i.e.,
$h^{*}=h^{\prime}=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\mathrm{Pr}\left(h(X,A)=Y^{\prime}\right)$.
This is a scale-invariant proxy to say that the linear fractional
transformation defined by $P_{a},Q_{a},R_{a},S_{a}$ is order-preserving and
when appropriately scaled to make $S_{a}=1$, it is $\epsilon$-close to the
identity transformation.

The above definition of $\epsilon$-robustness of $h^{*}$ can be rewritten as

$\displaystyle h^{*}$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\mathrm{Pr}\left(h(X,A)=Y^{\prime}\right)$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\mathrm{E}_{(X,A)}\left[h(X,A)(2\eta^{\prime}(X,A)-1)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)\mathrm{E}_{X|A=a}\left[h(X,a)(2\eta^{\prime}(X,a)-1)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)\mathrm{E}_{X|A=a}\left[h(X,a)\left(2~{}\frac{P_{a}\eta(X,a)+Q_{a}}{R_{a}\eta(X,a)+S_{a}}-1\right)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)\mathrm{E}_{X|A=a}\left[h(X,a)~{}\frac{(2P_{a}-R_{a})\eta(X,a)+(2Q_{a}-S_{a})}{R_{a}\eta(X,a)+S_{a}}\right],$

for any $S_{a}=1,|R_{a}|\leq\epsilon,|Q_{a}|\leq\epsilon,1-\epsilon\leq
P_{a}\leq 1+\epsilon$, and $P_{a}S_{a}-Q_{a}R_{a}\geq 0$.

###  5.2 Recovery theorem from biased data when $h^{*}$ is $\epsilon$-robust

Now we state our result for recovering robust hypothesis under data bias on
arbitrary data distributions and hypothesis classes.

######  Theorem 10.

For any distribution $D$ and any hypothesis class $\mathcal{H}$, if the
optimal classifier $h^{*}\in\mathcal{H}$ is $\epsilon$-robust and the bias
parameters $\beta_{p},\beta_{n},\nu$ satisfy
$(1-\epsilon)\beta_{n}\leq\beta_{p}\leq(1+\epsilon)\beta_{n}$,

$r\left((1-\nu)\beta_{p}-(1-\epsilon)\beta_{n}\right)+\epsilon(1-r)\geq
0\quad\text{and}\quad
r\left((1+\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)+\epsilon(1-r)\geq 0,$

then the optimal equal opportunity classifier from $\mathcal{H}$ on the biased
distribution $\tilde{D}$ recovers the optimal classifier from $\mathcal{H}$ on
the original distribution $D$, i.e., $\tilde{h}_{EO}\equiv h^{*}$.

Its proof can be found in Appendix A.6. We reuse the basic characterization of
optimal fair classifiers using Langrange duality, and although our proof uses
the class probabilities $\eta(x,a)$’s in a crucial way, we circumvent the need
for threshold-based arguments seen in the proofs from previous sections.

$0.2$$0.4$$0.6$$0.8$$1$$0.2$$0.4$$0.6$$0.8$$1$$\beta_{p}$$\beta_{n}$$0.9\beta_{p}-0.95\beta_{n}+0.2\geq
0$$1.05\beta_{n}-0.9\beta_{p}+0.2\geq 0$ Figure 1: Recovery region for
$\beta_{p}\geq 0$ and $\beta_{n}\geq 0$ given by the constraints
$r\left((1-\nu)\beta_{p}-(1-\epsilon)\beta_{n}\right)+\epsilon(1-r)\geq 0$ and
$r\left((1+\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)+\epsilon(1-r)\geq 0$ in
Theorem 10, when $r=0.2$, $\nu=0.1$, and $\epsilon=0.05$. The region close to
the origin $(0,0)$ shows that it is possible to recover optimal and fair
classifiers even from extreme under-representation in data by applying just
equal opportunity constraints.

##  6 Conclusion, Ongoing and Future Work

In this work, we investigate the recovery of optimal and fair classifiers in
the presence of under-representation and label bias in data. This problem
setting was proposed by Blum & Stangl [3]. They showed a recovery result on a
stylized data distribution to emphasize that fairness constraints on biased
data can even help improve accuracy. We reprove and extend their result to
more general data distributions. We intend to extend our results to
unrestricted optimal hypotheses and even more general data distributions.

## References

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##  Appendix A Proofs

###  A.1 Proof of Proposition 2

###### Proof.

We show that for any classifier $h$ and any $a\in\\{0,1\\}$, we have
$\widetilde{TPR}_{a}(h)=TPR_{a}(h)$.

$\displaystyle\widetilde{TPR}_{a}(h)$
$\displaystyle=\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=a\right)$
$\displaystyle=\frac{\mathrm{Pr}\left(h(X,A)=1,A=a,\tilde{Y}=1\right)}{\mathrm{Pr}\left(\tilde{Y}=1,A=a\right)}$
$\displaystyle=\frac{\mathrm{Pr}\left(h(X,A)=1,A=a,\tilde{Y}=1,Y=0\right)+\mathrm{Pr}\left(h(X,A)=1,A=a,\tilde{Y}=1,Y=1\right)}{\mathrm{Pr}\left(A=a,\tilde{Y}=1,Y=0\right)+\mathrm{Pr}\left(A=a,\tilde{Y}=1,Y=1\right)}$
$\displaystyle=\frac{\mathrm{Pr}\left(h(X,A)=1,A=a,\tilde{Y}=1,Y=1\right)}{\mathrm{Pr}\left(A=a,\tilde{Y}=1,Y=1\right)}\qquad\text{because
$\mathrm{Pr}\left(\tilde{Y}=1,Y=0\right)=0$.}$
$\displaystyle=\frac{\mathrm{Pr}\left(h(X,A)=1,A=a,Y=1\right)\mathrm{Pr}\left(\tilde{Y}=1|h(X,A)=1,A=a,Y=1\right)}{\mathrm{Pr}\left(A=a,Y=1\right)\mathrm{Pr}\left(\tilde{Y}=1|A=a,Y=1\right)}$
$\displaystyle=\frac{\mathrm{Pr}\left(h(X,A)=1,A=a,Y=1\right)\mathrm{Pr}\left(\tilde{Y}=1|A=a,Y=1\right)}{\mathrm{Pr}\left(A=a,Y=1\right)\mathrm{Pr}\left(\tilde{Y}=1|A=a,Y=1\right)}$
because $\tilde{Y}$ depends only on $A$ and $Y$
$\displaystyle=\frac{\mathrm{Pr}\left(h(X,A)=1,A=a,Y=1\right)}{\mathrm{Pr}\left(A=a,Y=1\right)}$
$\displaystyle=\mathrm{Pr}\left(h(X,A)=1|Y=1,A=a\right)$
$\displaystyle=TPR_{a}(h).$

Similarly, we show that $\widetilde{TNR}_{0}(h)$ can be written as a linear
combination of $TNR_{0}(h)$ and $FNR_{0}(h)$.

$\displaystyle\widetilde{TNR}_{0}(h)$
$\displaystyle=\mathrm{Pr}\left(h(X,A)=0|\tilde{Y}=0,A=0\right)$
$\displaystyle=\frac{\mathrm{Pr}\left(h(X,A)=0,\tilde{Y}=0|A=0\right)}{\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)}$
$\displaystyle=\frac{\mathrm{Pr}\left(Y=0|A=0\right)\mathrm{Pr}\left(h(X,A)=0,\tilde{Y}=0|Y=0,A=0\right)+\mathrm{Pr}\left(Y=1|A=0\right)\mathrm{Pr}\left(h(X,A)=0,\tilde{Y}=0|Y=1,A=0\right)}{\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)}$
$\displaystyle=\frac{\mathrm{Pr}\left(Y=0|A=0\right)\mathrm{Pr}\left(\tilde{Y}=0|h(X,A)=0,Y=0,A=0\right)\mathrm{Pr}\left(h(X,A)=0|Y=0,A=0\right)}{\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)}$
$\displaystyle\qquad\qquad+\frac{\mathrm{Pr}\left(Y=1|A=0\right)\mathrm{Pr}\left(\tilde{Y}=0|h(X,A)=0,Y=1,A=0\right)\mathrm{Pr}\left(h(X,A)=0|Y=1,A=0\right)}{\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)}$
$\displaystyle=\frac{\mathrm{Pr}\left(Y=0|A=0\right)\mathrm{Pr}\left(\tilde{Y}=0|Y=0,A=0\right)\mathrm{Pr}\left(h(X,A)=0|Y=0,A=0\right)}{\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)}$
$\displaystyle\qquad\qquad+\frac{\mathrm{Pr}\left(Y=1|A=0\right)\mathrm{Pr}\left(\tilde{Y}=0|Y=1,A=0\right)\mathrm{Pr}\left(h(X,A)=0|Y=1,A=0\right)}{\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)}$
because $\tilde{Y}$ depends only on $A$ and $Y$
$\displaystyle=\frac{\mathrm{Pr}\left(\tilde{Y}=0,Y=0|A=0\right)}{\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)}~{}TNR_{0}(h)+\frac{\mathrm{Pr}\left(\tilde{Y}=0,Y=1|A=0\right)}{\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)}~{}FNR_{0}(h)$
$\displaystyle=\mathrm{Pr}\left(Y=0|\tilde{Y}=0,A=0\right)~{}TNR_{0}(h)+\mathrm{Pr}\left(Y=1|\tilde{Y}=0,A=0\right)~{}FNR_{0}(h).$

∎

###  A.2 Proof of Proposition 5

###### Proof.

We begin with a known techniques that uses Lagrange duality to give threshold-
based characterization of optimal equal opportunity classifiers [8, 4]. We
define $\tilde{h}_{EO}=\underset{h\in\widetilde{\mathcal{H}}_{\text{fair,
EO}}}{\mathrm{argmax}}~{}\mathrm{Pr}\left(h(X,A)=\tilde{Y}\right)$. By
Lagrange duality, we can write

$\displaystyle\tilde{h}_{EO}$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\mathrm{Pr}\left(h(X,A)=\tilde{Y}\right)+\lambda~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=0\right)-\lambda~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=1\right)$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)~{}\mathrm{E}_{X|A=a}\left[h(X,a)(2\tilde{\eta}(X,a)-1)\right]+\frac{(-1)^{\mathbb{I}\left(a=1\right)}~{}\lambda}{\mathrm{E}_{X|A=a}\left[\tilde{\eta}{(X,a)}\right]}~{}\mathrm{E}_{X|A=a}\left[h(X,a)\tilde{\eta}(X,a)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)~{}\mathrm{E}_{X|A=a}\left[h(X,a)\left(\left(2+\dfrac{(-1)^{\mathbb{I}\left(a=1\right)}\lambda}{\mathrm{Pr}\left(\tilde{Y}=1,A=a\right)}\right)\tilde{\eta}(X,a)-1\right)\right].$

Thus, there exists $\lambda^{*}\in\mathbb{R}$ such that

$\displaystyle\tilde{h}_{EO}(x,a)$
$\displaystyle=\mathbb{I}\left(\tilde{\eta}(x,a)\geq\dfrac{1}{2+\dfrac{(-1)^{\mathbb{I}\left(a=1\right)}\lambda^{*}}{\mathrm{Pr}\left(\tilde{Y}=1,A=a\right)}}\right)$
$\displaystyle=\begin{cases}\mathbb{I}\left(\dfrac{(1-\nu)\eta(x,0)}{(1-c)\eta(x,0)+c}\geq\dfrac{1}{2+\dfrac{\lambda^{*}}{\mathrm{Pr}\left(\tilde{Y}=1,A=0\right)}}\right),\quad\text{for
$a=0$}\qquad\text{(Using Proposition \ref{prop:eta-tilde})}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2-\dfrac{\lambda^{*}}{\mathrm{Pr}\left(Y=1,A=1\right)}}\right),\quad\text{for
$a=1$}\end{cases}$
$\displaystyle=\begin{cases}\mathbb{I}\left(\eta(x,0)\geq\dfrac{c}{1-2\nu+c+\dfrac{\lambda^{*}(1-\nu)}{\mathrm{Pr}\left(\tilde{Y}=1,A=0\right)}}\right),\quad\text{for
$a=0$}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2-\dfrac{\lambda^{*}}{\mathrm{Pr}\left(Y=1,A=1\right)}}\right),\quad\text{for
$a=1$}\end{cases}$

Let $\mathrm{Pr}\left(A=0\right)=r$ and $\mathrm{Pr}\left(A=1\right)=1-r$, and
let $\mathrm{Pr}\left(Y=1|A=0\right)=\mathrm{Pr}\left(Y=1|A=1\right)=q$. Then
$\mathrm{Pr}\left(Y=1,A=1\right)=\mathrm{Pr}\left(A=1\right)\mathrm{Pr}\left(Y=1|A=1\right)=(1-r)q$
and

$\displaystyle\mathrm{Pr}\left(\tilde{Y}=1,A=0\right)$
$\displaystyle=\mathrm{Pr}\left(A=0\right)\mathrm{Pr}\left(Y=1|A=0\right)\mathrm{Pr}\left(\tilde{Y}=1|Y=1,A=0\right)$
$\displaystyle\qquad\qquad+\mathrm{Pr}\left(A=0\right)\mathrm{Pr}\left(Y=0|A=0\right)\mathrm{Pr}\left(\tilde{Y}=1|Y=0,A=0\right)$
$\displaystyle=rq\beta_{p}(1-\nu).$

Therefore,

$\displaystyle\tilde{h}_{EO}(x,a)$
$\displaystyle=\begin{cases}\mathbb{I}\left(\eta(x,0)\geq\dfrac{c}{1-2\nu+c+\dfrac{\lambda^{*}}{\beta_{p}rq}}\right),\quad\text{for
$a=0$}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2-\dfrac{\lambda^{*}}{(1-r)q}}\right),\quad\text{for
$a=1$}\end{cases}.$

Equivalently, we can also write

$\displaystyle\tilde{h}_{EO}(x,a)$
$\displaystyle=\begin{cases}\mathbb{I}\left(\eta(x,0)\geq\dfrac{1}{1+\dfrac{1-2\nu}{c}+\dfrac{\lambda^{*}}{\beta_{n}rq}}\right),\quad\text{for
$a=0$}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2-\dfrac{\lambda^{*}}{(1-r)q}}\right),\quad\text{for
$a=1$}.\end{cases}.$

∎

###  A.3 Proof of Proposition 6

###### Proof.

Similar to previous work that gives a threshold-based characterization of
optimal demographic parity classifiers using Lagrange duality [8], we use the
same idea on the biased distribution.
$\tilde{h}_{DP}=\underset{h\in\widetilde{\mathcal{H}}_{\text{fair,
DP}}}{\mathrm{argmax}}~{}\mathrm{Pr}\left(h(X,A)=\tilde{Y}\right)$. By
Lagrange duality, we can write

$\displaystyle\tilde{h}_{DP}$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\mathrm{Pr}\left(h(X,A)=\tilde{Y}\right)+\lambda~{}\mathrm{Pr}\left(h(X,A)=1|A=0\right)-\lambda~{}\mathrm{Pr}\left(h(X,A)=1|A=1\right)$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)~{}\mathrm{E}_{X|A=a}\left[h(X,a)(2\tilde{\eta}(X,a)-1)\right]+(-1)^{\mathbb{I}\left(a=1\right)}~{}\lambda~{}\mathrm{E}_{X|A=a}\left[h(X,a)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)~{}\mathrm{E}_{X|A=a}\left[h(X,a)\left(2\tilde{\eta}(X,a)-1+\dfrac{(-1)^{\mathbb{I}\left(a=1\right)}~{}\lambda}{\mathrm{Pr}\left(A=a\right)}\right)\right].$

Thus, there exists some optimal $\lambda^{*}\in\mathbb{R}$ such that

$\displaystyle\tilde{h}_{DP}(x,a)$
$\displaystyle=\mathbb{I}\left(\tilde{\eta}(x,a)\geq\frac{1}{2}-\frac{(-1)^{\mathbb{I}\left(a=1\right)}\lambda^{*}}{2\mathrm{Pr}\left(A=a\right)}\right)$
$\displaystyle=\begin{cases}\mathbb{I}\left(\dfrac{(1-\nu)\eta(x,0)}{(1-c)\eta(x,0)+c}\geq\dfrac{1}{2}-\dfrac{\lambda^{*}}{2\mathrm{Pr}\left(A=0\right)}\right),\quad\text{for
$a=0$}\quad\text{(Using Proposition \ref{prop:eta-tilde})}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2}+\dfrac{\lambda^{*}}{2\mathrm{Pr}\left(A=1\right)}\right),\quad\text{for
$a=1$}\end{cases}$
$\displaystyle=\begin{cases}\mathbb{I}\left(\eta(x,0)\geq\dfrac{c(r-\lambda^{*})}{2r(1-\nu)-(1-c)(r-\lambda^{*})}\right),\quad\text{for
$a=0$}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2}+\dfrac{\lambda^{*}}{2(1-r)}\right),\quad\text{for
$a=1$}\end{cases}$

∎

###  A.4 Proof of Theorem 9 (that subsumes Theorem 7)

Here we give a complete proof of Theorem 9. Note that Theorem 9 implies
Theorem 7 as i.i.d. noise is a special case of Massart noise when we have
$\delta(x,a)=\delta$, for all $(x,a)\in\mathcal{X}\times\mathcal{A}$. Thus,
Theorem 9 is a stronger statement than Theorem 7 (the original recovery
theorem of Blum & Stangl [3]) while having the same necessary and sufficient
conditions on the data and bias parameters.

We first state the following simple lemma about the flexibility of thresholds
for recovering $h^{*}$ before proving Theorem 9. Let $D$ be any distribution
$\mathcal{X}\times\mathcal{A}$.

######  Lemma 11.

Let $h:\mathcal{X}\times\mathcal{A}\rightarrow\\{0,1\\}$ be a deterministic
classifier. Let $Y|X=x,A=a$ be a random variable that takes value $h(x,a)$
with probability $1-\delta(x,a)$ and $\not h(x,a)$ with probability
$\delta(x,a)$, for some $\delta(x,a)\leq\delta<1/2$. Let
$\eta(x,a)=\mathrm{Pr}\left(Y=1|X=x,A=a\right)$ and let $g(x,a)$ be a
threshold classifier given by $g(x,a)=\mathbb{I}\left(\eta(x,a)\geq
t_{a}\right)$, using group-dependent thresholds $t_{0}$ and $t_{1}$,
respectively. If $t_{0},t_{1}\in(\delta,1-\delta)$ then $g\equiv h\equiv
h^{*}$, where $h^{*}$ is the Bayes optimal classifier for the joint
distribution on $\mathcal{X}\times\mathcal{A}\times\mathcal{Y}$.

###### Proof.

It is folklore that the (group-aware) Bayes optimal classifier is given by
$h^{*}(x,a)=\mathbb{I}\left(\eta(x,a)\geq 1/2\right)$. Consider the following
two cases for $h(x,a)$. If $h(x,a)=0$, then $Y|X=x,A=a$ takes value $0$ with
probability $\delta(x,a)\leq\delta$, and hence, $\mathbb{I}\left(\eta(x,a)\geq
t_{a}\right)=\mathbb{I}\left(\delta(x,a)\geq
t_{a}\right)=\mathbb{I}\left(\delta(x,a)\geq 1/2\right)=0$, for
$\delta(x,a)\leq\delta<1/2$ and $t_{a}\in(\delta,1-\delta)$. On the other
hand, if $h(x,a)=1$, then $Y|X=x,A=a$ takes value $1$ with probability
$1-\delta(x,a)\geq 1-\delta$, and hence, $\mathbb{I}\left(\eta(x,a)\geq
t_{a}\right)=\mathbb{I}\left(1-\delta(x,a)\geq
t_{a}\right)=\mathbb{I}\left(1-\delta(x,a)\geq 1/2\right)=1$, for
$\delta(x,a)\leq\delta<1/2$ and $t_{a}\in(\delta,1-\delta)$. Therefore,
$g\equiv h\equiv h^{*}$. ∎

Now we proceed to the proof of Theorem 9 (that subsumes Theorem 7).

###### Proof.

(Proof of Theorem 9) Let

$\displaystyle\tilde{h}_{\lambda}(x,a)$
$\displaystyle=\begin{cases}\mathbb{I}\left(\eta(x,0)\geq\dfrac{1}{1+\dfrac{1-2\nu}{c}+\dfrac{\lambda}{\beta_{n}rq}}\right),\quad\text{for
$a=0$}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2-\dfrac{\lambda}{(1-r)q}}\right),\quad\text{for
$a=1$}.\end{cases}$

From the characterization of optimal equal opportunity classifier shown
earlier, there exists some optimal $\lambda^{*}\in\mathbb{R}$ such that
$\tilde{h}_{EO}=h_{\lambda^{*}}$. We will show that if the conditions in
Theorem 4.1 of Blum-Stangl hold, then the only possible choices left for
$\lambda^{*}\in\mathbb{R}$ are the ones that give $h_{\lambda^{*}}=h^{*}$.

Step 1: If the given conditions on the bias parameters hold, then there exists
a $\lambda\in\mathbb{R}$ such that $\tilde{h}_{\lambda}=h^{*}$. The threshold
on $\eta(x,1)$ lies in the interval $(\delta,1-\delta)$ if and only if
$\dfrac{1}{1-\delta}<2-\dfrac{\lambda}{(1-r)q}<\dfrac{1}{\delta}$, or
equivalently,
$\dfrac{-(1-2\delta)(1-r)q}{\delta}<\lambda<\dfrac{(1-2\delta)(1-r)q}{1-\delta}$.
Similarly, the threshold on $\eta(x,0)$ lies in the interval
$(\delta,1-\delta)$ if and only if
$\dfrac{1}{1-\delta}<1+\dfrac{1-2\nu}{c}+\dfrac{\lambda}{\beta_{n}rq}<\dfrac{1}{\delta}$,
or equivalently,

$\beta_{n}rq\left(\frac{\delta}{1-\delta}-\frac{1-2\nu}{c}\right)<\lambda<\beta_{n}rq\left(\frac{1-\delta}{\delta}-\frac{1-2\nu}{c}\right).$

There exists $\lambda$ that simultaneously satisfies both sets of constraints
given above if and only if

$\displaystyle\beta_{n}rq\left(\frac{\delta}{1-\delta}-\frac{1-2\nu}{c}\right)$
$\displaystyle<\dfrac{(1-2\delta)(1-r)q}{1-\delta}$
$\displaystyle\dfrac{-(1-2\delta)(1-r)q}{\delta}$
$\displaystyle<\beta_{n}rq\left(\frac{1-\delta}{\delta}-\frac{1-2\nu}{c}\right).$

Using $c=\beta_{n}/\beta_{p}$, the above conditions can be rewritten as

$\displaystyle(1-r)(1-2\delta)+r\left((1-\delta)\beta_{p}(1-2\nu)-\delta\beta_{n}\right)$
$\displaystyle>0\quad\text{and}$
$\displaystyle(1-r)(1-2\delta)+r\left((1-\delta)\beta_{n}-\delta\beta_{p}(1-2\nu)\right)$
$\displaystyle>0.$

Step 2: If the given conditions on the bias parameters hold, then there cannot
exist any $\lambda\in\mathbb{R}$ such that both the group-wise thresholds
applied to $\eta(x,a)$ in $h_{\lambda}$ are at most $\delta$, or both the
thresholds are at least $1-\delta$. (Thresholds lying on the same side) The
threshold on $\eta(x,1)$ is at most $\delta$ if and only if
$2-\dfrac{\lambda}{(1-r)q}\geq\dfrac{1}{\delta}$, or equivalently,
$\dfrac{-(1-2\delta)(1-r)q}{\delta}\geq\lambda$. Similarly, the threshold on
$\eta(x,0)$ is at most $\delta$ if and only if
$1+\dfrac{1-2\nu}{c}+\dfrac{\lambda}{\beta_{n}rq}\geq\dfrac{1}{\delta}$, or
equivalently,
$\lambda\geq\beta_{n}rq\left(\dfrac{1-\delta}{\delta}-\dfrac{1-2\nu}{c}\right)$.
If both were to hold simultaneously, then
$\dfrac{-(1-2\delta)(1-r)q}{\delta}\geq\beta_{n}rq\left(\dfrac{1-\delta}{\delta}-\dfrac{1-2\nu}{c}\right)$,
or equivalently,
$(1-r)(1-2\delta)+r\left((1-\delta)\beta_{n}-\delta\beta_{p}(1-2\nu)\right)\leq
0$, which would violate the second condition in Theorem 4.1 of Blum-Stangl.

On the other hand, the threshold on $\eta(x,1)$ is at least interval
$1-\delta$ if and only if $\dfrac{1}{1-\delta}\geq 2-\dfrac{\lambda}{(1-r)q}$,
or equivalently, $\lambda\geq\dfrac{(1-2\delta)(1-r)q}{1-\delta}$. Similarly,
the threshold on $\eta(x,0)$ is at least $1-\delta$ if and only if
$\dfrac{1}{1-\delta}\geq 1+\dfrac{1-2\nu}{c}+\dfrac{\lambda}{\beta_{n}rq}$, or
equivalently,
$\beta_{n}rq\left(\dfrac{\delta}{1-\delta}-\dfrac{1-2\nu}{c}\right)\geq\lambda$.
If both were to hold simultaneously, then
$\beta_{n}rq\left(\dfrac{\delta}{1-\delta}-\dfrac{1-2\nu}{c}\right)\geq\dfrac{(1-2\delta)(1-r)q}{1-\delta}$,
or equivalently,
$(1-r)(1-2\delta)+r\left((1-\delta)\beta_{p}(1-2\nu)-\delta\beta_{n}\right)\leq
0$, which would violate the first condition in Theorem 4.1 of Blum-Stangl.

Step 3: For any $\lambda$, if the group-wise thresholds on $\eta(x,a)$ in
$\tilde{h}_{\lambda}$ have one of them at most $\delta$ and another at least
$1-\delta$, then $\tilde{h}_{\lambda}$ cannot satisfy equal opportunity unless
$\tilde{h}_{\lambda}=h^{*}$. (Thresholds lying on opposite sides) We know that
$h^{*}$ satisfies equal opportunity. If $h_{\lambda}$ has its threshold on
$\eta(x,0)$ at most $\delta$ and its threshold on $\eta(x,1)$ at least
$1-\delta$, then $TPR_{0}(\tilde{h}_{\lambda})\geq
TPR_{0}(h^{*})=TPR_{1}(h^{*})\geq TPR_{1}(\tilde{h}_{\lambda})$. Hence, if
$\tilde{h}_{\lambda}$ satisfies equal opportunity on $\tilde{D}$, then it must
satisfy equal opportunity on $D$ too. So the above inequality must actually be
an equality $TPR_{0}(\tilde{h}_{\lambda})\geq
TPR_{0}(h^{*})=TPR_{1}(h^{*})\geq TPR_{1}(\tilde{h}_{\lambda})$, which is only
possible if $\tilde{h}_{\lambda}=h^{*}$ (as both are threshold classifiers).
The other case where $\tilde{h}_{\lambda}$ has its threshold on $\eta(x,0)$ at
least $1-\delta$ and its threshold on $\eta(x,1)$ is at most $\delta$ can be
argued similarly.

Combing Steps 1, 2, and 3 to complete the proof: Finally, from Steps 1, 2, and
3 together, it is clear that if the conditions in Theorem 4.1 of Blum-Stangl
are met then the only possible classifiers $\tilde{h}_{\lambda}$ are the ones
for which $\tilde{h}_{\lambda}=h^{*}$, and they satisfy equal opportunity on
$\tilde{D}$. So under these conditions, the optimal $\lambda^{*}$ (whatever it
may be) gives $\tilde{h}_{EO}=\tilde{h}_{\lambda^{*}}=h^{*}$.
∎

###  A.5 Proof of Theorem 8

###### Proof.

We use the characterization optimal demographic parity classifier obtained in
Proposition 6 for $\tilde{D}$, i.e., there exists $\lambda^{*}\in\mathbb{R}$
such that

$\tilde{h}_{DP}(x,a)=\begin{cases}\mathbb{I}\left(\eta(x,0)\geq\dfrac{c(r-\lambda^{*})}{2r(1-\nu)-(1-c)(r-\lambda^{*})}\right),\quad\text{for
$a=0$}\\\
\mathbb{I}\left(\eta(x,1)\geq\dfrac{1}{2}+\dfrac{\lambda^{*}}{2(1-r)}\right),\quad\text{for
$a=1$}.\end{cases}$

Step 1: If the given conditions on the bias parameters hold, then there exists
a $\lambda\in\mathbb{R}$ such that $h_{\lambda}=h^{*}$. From Lemma 11, we know
that the threshold on $\eta(x,1)$ lies in the interval $(\delta,1-\delta)$ if
and only if $\delta<\dfrac{1}{2}+\dfrac{\lambda^{*}}{2(1-r)}<1-\delta$, or
equivalently, $(2\delta-1)(1-r)<\lambda^{*}<(1-2\delta)(1-r)$. Similarly, the
threshold on $\eta(x,0)$ lies in the interval $(\delta,1-\delta)$ if and only
if $\delta<\dfrac{c(r-\lambda^{*})}{2r(1-\nu)-(1-c)(r-\lambda^{*})}<1-\delta$,
or equivalently,

$r\left(1-\dfrac{2(1-\delta)(1-\nu)}{c+(1-c)(1-\delta)}\right)<\lambda^{*}<r\left(1-\dfrac{2\delta(1-\nu)}{\delta(1-c)+c}\right).$

There exists $\lambda^{*}$ that simultaneously satisfies both sets of
constraints given above if and only if

$\displaystyle
r\left(1-\dfrac{2(1-\delta)(1-\nu)}{c+(1-c)(1-\delta)}\right)<(1-2\delta)(1-r)\text{
and }(2\delta-1)(1-r)<r\left(1-\dfrac{2\delta(1-\nu)}{\delta(1-c)+c}\right).$

Using $c=\beta_{n}/\beta_{p}$, the above conditions can be simplified as

$\displaystyle\beta_{p}(1-\delta)(1-2\delta-2r(\nu-\delta))+\delta\beta_{n}(1-2\delta-2r(1-\delta))>0\text{
and }$
$\displaystyle\beta_{p}\delta(1-2r(1-\nu)-2\delta(1-r))+(1-\delta)\beta_{n}(1-2\delta(1-r))>0$

Step 2: If the given conditions on the bias parameters hold, then there cannot
exist any $\lambda\in\mathbb{R}$ such that both the group-wise thresholds
applied to $\eta(x,a)$ in $h_{\lambda}$ are at most $\delta$, or both the
thresholds are at least $1-\delta$. (Thresholds lying on the same side) The
threshold on $\eta(x,1)$ is at most $\delta$ if and only if
$\dfrac{1}{2}+\dfrac{\lambda}{2(1-r)}\leq\delta$, or equivalently,
$\lambda\leq(2\delta-1)(1-r)$. Similarly, the threshold on $\eta(x,0)$ is at
most $\delta$ if and only if
$\dfrac{c(r-\lambda)}{2r(1-\nu)-(1-c)(r-\lambda)}\leq\delta$, or equivalently,
$\lambda\geq
r\left(1-\dfrac{2\delta\beta_{p}(1-\nu)}{\beta_{n}(1-\delta)+\delta\beta_{p}}\right)$.
If both were to hold simultaneously, then
$r\left(1-\dfrac{2\delta\beta_{p}(1-\nu)}{\beta_{n}(1-\delta)+\delta\beta_{p}}\right)\leq(2\delta-1)(1-r)$,
or equivalently,
$\beta_{p}\delta(1-2r(1-\nu)-2\delta(1-r))+(1-\delta)\beta_{n}(1-2\delta(1-r))\leq
0$, which would violate the second condition in our theorem.

On the other hand, the threshold on $\eta(x,1)$ is at least interval
$1-\delta$ if and only if $\dfrac{1}{2}+\dfrac{\lambda}{2(1-r)}\geq 1-\delta$,
or equivalently, $\lambda\geq(1-2\delta)(1-r)$. Similarly, the threshold on
$\eta(x,0)$ is at least $\dfrac{c(r-\lambda)}{2r(1-\nu)-(1-c)(r-\lambda)}\geq
1-\delta$, or equivalently, $\lambda\leq
r\left(1-\dfrac{2(1-\delta)(1-\nu)}{c+(1-c)(1-\delta)}\right)$. If both were
to hold simultaneously, then $(1-2\delta)(1-r)\leq
r\left(1-\dfrac{2(1-\delta)(1-\nu)}{c+(1-c)(1-\delta)}\right)$, or
equivalently,
$\beta_{p}(1-\delta)(1-2\delta-2r(\nu-\delta))+\delta\beta_{n}(1-2\delta-2r(1-\delta))\leq
0$, which would violate the first condition in our theorem.

Step 3: For any $\lambda$, if the group-wise thresholds on $\eta(x,a)$ in
$h_{\lambda}$ have one of them at most $\delta$ and another at least
$1-\delta$, then $h_{\lambda}$ cannot satisfy equal opportunity unless
$h_{\lambda}=h^{*}$. (Thresholds lying on opposite sides) We know that $h^{*}$
satisfies demographic parity. Let
$PR_{a}(h_{\lambda})=\mathrm{Pr}\left(h_{\lambda}=1|A=a\right)$ (group
positive rate). For demographic parity, we require $PR_{0}=PR_{1}$.

If $h_{\lambda}$ has its threshold on $\eta(x,0)$ at most $\delta$ and its
threshold on $\eta(x,1)$ at least $1-\delta$, then $PR_{0}(h_{\lambda})\geq
PR_{0}(h^{*})=PR_{1}(h^{*})\geq PR_{1}(h_{\lambda})$. Hence, if $h_{\lambda}$
satisfies demographic parity on $\tilde{D}$, it must also satisfy demographic
parity on $D$. So the above inequality must actually be an equality
$PR_{0}(h_{\lambda})\geq PR_{0}(h^{*})=PR_{1}(h^{*})\geq PR_{1}(h_{\lambda})$,
which is only possible if $h_{\lambda}=h^{*}$ (as both are threshold
classifiers). The other case where $h_{\lambda}$ has its threshold on
$\eta(x,0)$ at least $1-\delta$ and its threshold on $\eta(x,1)$ at most
$\delta$ can be argued similarly.

∎

###  A.6 Proof of Theorem 10

###### Proof.

Let
$\tilde{h}_{EO}=\underset{h\in\widetilde{\mathcal{H}}_{\text{fair}}}{\text{argmax}}~{}\mathrm{Pr}\left(h(X,A)=\tilde{Y}\right)$.
By Lagrange duality, we can write

$\displaystyle\tilde{h}_{EO}$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\mathrm{Pr}\left(h(X,A)=\tilde{Y}\right)+\lambda~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=0\right)-\lambda~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=1\right)$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)~{}\mathrm{E}_{X|A=a}\left[h(X,a)(2\tilde{\eta}(X,a)-1)\right]+\frac{(-1)^{\mathbb{I}\left(a=1\right)}~{}\lambda}{\mathrm{E}_{X|A=a}\left[\tilde{\eta}{(X,a)}\right]}~{}\mathrm{E}_{X|A=a}\left[h(X,a)\tilde{\eta}(X,a)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)~{}\mathrm{E}_{X|A=a}\left[h(X,a)(2\tilde{\eta}(X,a)-1)\right]+\frac{(-1)^{\mathbb{I}\left(a=1\right)}~{}\lambda}{\mathrm{Pr}\left(\tilde{Y}=1|A=a\right)}~{}\mathrm{E}_{X|A=a}\left[h(X,a)\tilde{\eta}(X,a)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}\sum_{a=0}^{1}\mathrm{Pr}\left(A=a\right)~{}\mathrm{E}_{X|A=a}\left[h(X,a)\left(\left(2+\dfrac{(-1)^{\mathbb{I}\left(a=1\right)}\lambda}{\mathrm{Pr}\left(\tilde{Y}=1,A=a\right)}\right)\tilde{\eta}(X,a)-1\right)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}r~{}\mathrm{E}_{X|A=0}\left[h(X,0)\left(\left(2+\frac{\lambda}{\beta_{p}(1-\nu)rq}\right)\frac{(1-\nu)\eta(X,0)}{(1-c)\eta(X,0)+c}-1\right)\right]$
$\displaystyle\qquad\qquad\qquad\qquad+(1-r)~{}\mathrm{E}_{X|A=1}\left[h(X,1)\left(\left(2-\frac{\lambda}{(1-r)q}\right)\eta(X,1)-1\right)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}r~{}\mathrm{E}_{X|A=0}\left[h(X,0)~{}\frac{\left(2(1-\nu)+\dfrac{\lambda}{\beta_{p}rq}-(1-c)\right)\eta(X,0)-c}{(1-c)\eta(X,0)+c}\right]$
$\displaystyle\qquad\qquad\qquad\qquad+(1-r)~{}\mathrm{E}_{X|A=1}\left[h(X,1)\left(\left(2-\dfrac{\lambda}{(1-r)q}\right)\eta(X,1)-1\right)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}r~{}\mathrm{E}_{X|A=0}\left[h(X,0)~{}\frac{\left(1-2\nu+c+\dfrac{\lambda}{\beta_{p}rq}\right)\eta(X,0)-c}{(1-c)\eta(X,0)+c}\right]$
$\displaystyle\qquad\qquad\qquad\qquad+(1-r)~{}\mathrm{E}_{X|A=1}\left[h(X,1)\left(\left(2-\dfrac{\lambda}{(1-r)q}\right)\eta(X,1)-1\right)\right]$
$\displaystyle=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\underset{\lambda\in\mathbb{R}}{\min}~{}r~{}\mathrm{E}_{X|A=0}\left[h(X,0)~{}\frac{\left(1+\dfrac{1-2\nu}{c}+\dfrac{\lambda}{\beta_{n}rq}\right)\eta(X,0)-1}{\dfrac{1-c}{c}\eta(X,0)+1}\right]$
$\displaystyle\qquad\qquad\qquad\qquad+(1-r)~{}\mathrm{E}_{X|A=1}\left[h(X,1)\left(\left(2-\dfrac{\lambda}{(1-r)q}\right)\eta(X,1)-1\right)\right],\qquad\text{using
$c=\dfrac{\beta_{n}}{\beta_{p}}$}.$

Now plugging in

$\displaystyle\frac{(2P_{0}-R_{0})\eta(X,0)+(2Q_{0}-S_{0})}{R_{0}\eta(X,0)+S_{0}}$
$\displaystyle=\frac{\left(1+\dfrac{1-2\nu}{c}+\dfrac{\lambda}{\beta_{n}rq}\right)\eta(X,0)-1}{\dfrac{1-c}{c}\eta(X,0)+1}\qquad\text{and}$
$\displaystyle\frac{(2P_{1}-R_{1})\eta(X,1)+(2Q_{1}-S_{1})}{R_{1}\eta(X,1)+S_{1}}$
$\displaystyle=\left(2-\frac{\lambda}{(1-r)q}\right)\eta(X,1)-1,$

we derive

$\displaystyle P_{0}$
$\displaystyle=\dfrac{1-\nu}{c}+\dfrac{\lambda}{2\beta_{n}rq},\qquad$
$\displaystyle Q_{0}=0,\qquad$ $\displaystyle R_{0}=\dfrac{1-c}{c},\qquad$
$\displaystyle S_{0}=1,$ $\displaystyle P_{1}$
$\displaystyle=1-\dfrac{\lambda}{2(1-r)q},\qquad$ $\displaystyle
Q_{1}=0,\qquad$ $\displaystyle R_{1}=0,\qquad$ $\displaystyle S_{1}=1.$

Step 1: If the given conditions on the bias parameters hold, then there exists
a $\lambda\in\mathbb{R}$ such that $1-\epsilon\leq P_{0},P_{1}\leq
1+\epsilon$, and therefore, $h_{\lambda}=h^{*}$. They satisfy our
$\epsilon$-robustness conditions if and only if

$\displaystyle|R_{0}|$
$\displaystyle=\left|\dfrac{1-c}{c}\right|=\left|\frac{\beta_{p}-\beta_{n}}{\beta_{n}}\right|\leq\epsilon,$
$\displaystyle 1-\epsilon\leq P_{0}$
$\displaystyle=\dfrac{1-\nu}{c}+\dfrac{\lambda}{2\beta_{n}rq}\leq 1+\epsilon$
$\displaystyle 1-\epsilon\leq P_{1}$
$\displaystyle=1-\dfrac{\lambda}{2(1-r)q}\leq 1+\epsilon.$

The above conditions also imply $P_{a}S_{a}-Q_{a}R_{a}\geq 0$, for
$a\in\\{0,1\\}$, so we do not need to include these separately. The first
inequality is equivalent to
$(1-\epsilon)\beta_{n}\leq\beta_{p}\leq(1+\epsilon)\beta_{n}$. The second and
third inequalities above are equivalent to
$2rq\left((1-\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)\leq\lambda\leq
2rq\left((1+\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)$ and
$-2\epsilon(1-r)q\leq\lambda\leq 2\epsilon(1-r)q$, respectively. There exists
a $\lambda\in\mathbb{R}$ that satisfies the above two conditions on $\lambda$
simultaneously if and only if

$2rq\left((1-\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)\leq
2\epsilon(1-r)q\quad\text{and}\quad-2\epsilon(1-r)q\leq
2rq\left((1+\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right),$

or equivalently,

$r\left((1-\nu)\beta_{p}-(1-\epsilon)\beta_{n}\right)+\epsilon(1-r)\geq
0\quad\text{and}\quad
r\left((1+\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)+\epsilon(1-r)\geq 0.$

Step 2: If the given conditions on the bias parameters hold, then there cannot
exist any $\lambda\in\mathbb{R}$ such that $\max\\{P_{0},P_{1}\\}<1-\epsilon$
or $\min\\{P_{0},P_{1}\\}>1+\epsilon$. Suppose
$\max\\{P_{0},P_{1}\\}<1-\epsilon$. Then

$\dfrac{1-\nu}{c}+\dfrac{\lambda}{2\beta_{n}rq}<1-\epsilon\quad\text{and}\quad
1-\dfrac{\lambda}{2(1-r)q}<1-\epsilon.$

Thus, $\lambda<2rq\left((1-\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)$ and
$\lambda>2\epsilon(1-r)q$. This implies
$2rq\left((1-\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)>2\epsilon(1-r)q$, or
equivalently,
$r\left((1-\nu)\beta_{p}-(1-\epsilon)\beta_{n}\right)+\epsilon(1-r)<0$,
violating one of the conditions in the theorem. The case of
$\min\\{P_{0},P_{1}\\}>1+\epsilon$ can be argued similarly.

Step 3: If the given conditions on the bias parameters hold, then an optimal
$\lambda^{*}\in\mathbb{R}$ cannot correspond to $P_{0}<1-\epsilon$ and
$P_{1}>1+\epsilon$ (or vice versa). Suppose $\lambda^{*}\in\mathbb{R}$
satisfies $P_{0}<1-\epsilon$ and $P_{1}>1+\epsilon$, then

$P_{0}=\dfrac{1-\nu}{c}+\dfrac{\lambda^{*}}{2\beta_{n}rq}<1-\epsilon\quad\text{and}\quad
P_{1}=1-\dfrac{\lambda^{*}}{2(1-r)q}>1+\epsilon,$

which implies that
$\lambda^{*}<2rq\left((1-\epsilon)\beta_{n}-(1-\nu)\beta_{p}\right)$ and
$\lambda^{*}<-2\epsilon(1-r)q$. We know that

$\tilde{h}_{EO}=\underset{h\in\mathcal{H}}{\mathrm{argmax}}~{}\mathrm{Pr}\left(h(X,A)=\tilde{Y}\right)+\lambda^{*}~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=0\right)-\lambda^{*}~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=1\right).$

Since we consider group-aware classification, when we know the optimal
$\lambda^{*}$, the optimal equal opportunity classifier $\tilde{h}_{EO}$ on
the biased distribution $\tilde{D}$ can be obtained by optimizing separately
on the two groups. Note that $\tilde{h}_{EO}$ maximizes
$\mathrm{Pr}\left(h(X,A)=\tilde{Y}\right)+\lambda^{*}~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=0\right)-\lambda^{*}~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=1\right)$
over $h\in\mathcal{H}$, for the optimal $\lambda^{*}\in\mathbb{R}$. Since we
consider group-aware classifiers, $\tilde{h}_{EO}$ maximizes the following
objective on the underprivileged group $a=0$.

$\displaystyle\mathrm{Pr}\left(h(X,A)=\tilde{Y}|A=0\right)+\frac{\lambda^{*}}{\mathrm{Pr}\left(A=0\right)}~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=0\right)$
$\displaystyle=\mathrm{Pr}\left(\tilde{Y}=1|A=0\right)~{}\widetilde{TPR}_{0}(h)+\mathrm{Pr}\left(\tilde{Y}=0|A=0\right)~{}\widetilde{TNR}_{0}(h)+\frac{\lambda^{*}}{r}~{}\widetilde{TPR}_{0}(h)$
$\displaystyle=\mathrm{Pr}\left(\tilde{Y}=1|A=0\right)~{}TPR_{0}(h)+\mathrm{Pr}\left(\tilde{Y}=0,Y=0|A=0\right)~{}TNR_{0}(h)+\mathrm{Pr}\left(\tilde{Y}=0,Y=1|A=0\right)~{}FNR_{0}(h)$
$\displaystyle\qquad\qquad\qquad\qquad+\frac{\lambda^{*}}{r}~{}TPR_{0}(h)\qquad\qquad\qquad\text{using
Proposition \ref{prop:tilde-tpr-tnr}}$
$\displaystyle=\frac{q\beta_{p}(1-\nu)}{q\beta_{p}+(1-q)\beta_{n}}~{}TPR_{0}(h)+\frac{(1-q)\beta_{n}}{q\beta_{p}+(1-q)\beta_{n}}~{}TNR_{0}(h)+\frac{q\beta_{p}\nu}{q\beta_{p}+(1-q)\beta_{n}}~{}FNR_{0}(h)+\frac{\lambda^{*}}{r}~{}TPR_{0}(h)$
$\displaystyle=\left(\frac{q\beta_{p}(1-2\nu)}{q\beta_{p}+(1-q)\beta_{n}}+\frac{\lambda^{*}}{r}\right)~{}TPR_{0}(h)+\frac{(1-q)\beta_{n}}{q\beta_{p}+(1-q)\beta_{n}}~{}TNR_{0}(h)+\frac{q\beta_{p}\nu}{q\beta_{p}+(1-q)\beta_{n}}$
$\displaystyle\hskip 256.0748pt\text{because $FNR_{0}(h)=1-TPR_{0}(h)$}.$

Thus, $\tilde{h}_{EO}$ essentially maximizes a weighted linear combination of
$TPR_{0}(h)$ and $TNR_{0}(h)$ over $h\in\mathcal{H}$, where the ratios of the
weights for $TPR_{0}(h)$ and $TNR_{0}(h)$, respectively, is

$\displaystyle\frac{q\beta_{p}(1-2\nu)}{(1-q)\beta_{n}}+\frac{\lambda^{*}(q\beta_{p}+(1-q)\beta_{n})}{r(1-q)\beta_{n}}$
$\displaystyle\leq\frac{q(1-2\nu)}{(1-q)c}+\frac{\lambda^{*}}{r(1-q)\beta_{n}}\qquad\text{using
$\beta_{p},\beta_{n}\leq 1$}$
$\displaystyle=\frac{q}{1-q}\left(\frac{1-2\nu}{c}+\frac{\lambda^{*}}{\beta_{n}rq}\right)$
$\displaystyle=\frac{q}{1-q}\left(\frac{2(1-\nu)}{c}+\frac{\lambda^{*}}{\beta_{n}rq}-\frac{1}{c}\right)$
$\displaystyle<\frac{q}{1-q}\left(2(1-\epsilon)-\frac{1}{c}\right)\qquad\text{using
$P_{0}=\dfrac{1-\nu}{c}+\dfrac{\lambda^{*}}{2\beta_{n}rq}<1-\epsilon$}$
$\displaystyle\leq\frac{q}{1-q}~{}(1-\epsilon)\qquad\qquad\text{using
$\left|\frac{1-c}{c}\right|\leq\epsilon$, and hence, $\frac{1}{c}\geq
1-\epsilon$}$

Hence, $TPR_{0}(\tilde{h}_{EO})\leq TPR_{0}(h^{*})$. Similarly,
$\tilde{h}_{EO}$ maximizes the following objective on the group $a=1$.

$\displaystyle\mathrm{Pr}\left(h(X,A)=\tilde{Y}|A=1\right)-\frac{\lambda^{*}}{\mathrm{Pr}\left(A=1\right)}~{}\mathrm{Pr}\left(h(X,A)=1|\tilde{Y}=1,A=1\right)$
$\displaystyle=\mathrm{Pr}\left(\tilde{Y}=1|A=1\right)~{}\widetilde{TPR}_{1}(h)+\mathrm{Pr}\left(\tilde{Y}=0|A=1\right)~{}\widetilde{TNR}_{1}(h)-\frac{\lambda^{*}}{1-r}~{}\widetilde{TPR}_{1}(h)$
$\displaystyle=q\left(1-\frac{\lambda^{*}}{(1-r)q}\right)~{}TPR_{1}(h)+(1-q)~{}TNR_{1}(h).$

Thus, $\tilde{h}_{EO}$ essentially maximizes a weighted linear combination of
$TPR_{1}(h)$ and $TNR_{1}(h)$ over $h\in\mathcal{H}$, where the ratio of the
weights for $TPR_{1}(h)$ and $TNR_{1}(h)$, respectively, is

$\frac{q}{1-q}\left(1-\frac{\lambda^{*}}{(1-r)q}\right)>\frac{q}{1-q}~{}(1+2\epsilon)\qquad\qquad\text{using
$P_{1}=1-\frac{\lambda^{*}}{2(1-r)q}>1+\epsilon$}.$

Hence, $TPR_{1}(\tilde{h}_{EO})\geq TPR_{1}(h^{*})$.

Combining the two observation $TPR_{0}(\tilde{h}_{EO})\leq TPR_{0}(h^{*})$ and
$TPR_{1}(\tilde{h}_{EO})\geq TPR_{1}(h^{*})$ above and that $h^{*}$ satisfies
equal opportunity, we get $TPR_{0}(\tilde{h}_{EO})\leq
TPR_{0}(h^{*})=TPR_{1}(h^{*})\leq TPR_{1}(\tilde{h}_{EO})$. However,
$\tilde{h}_{EO}\in\widetilde{\mathcal{H}}_{\text{fair,
EO}}=\mathcal{H}_{\text{fair, EO}}$ by Proposition 3, so we must have
$TPR_{0}(\tilde{h}_{EO})=TPR_{0}(h^{*})=TPR_{1}(h^{*})=TPR_{1}(\tilde{h}_{EO})$.
This means that $\tilde{h}_{EO}$ would also maximize accuracy on the original
distribution $D$, and therefore, by the $\epsilon$-robustness property (see
Subsection 5.1), we must have $\tilde{h}_{EO}\equiv h^{*}$.

The case of $P_{0}>1+\epsilon$ and $P_{1}<1-\epsilon$ can be argued similarly.
∎


# Neutron Yield Calculation from ($\alpha,n$) Reactions with SOURCES4

Vitaly A. Kudryavtsev<EMAIL_ADDRESS>Piotr Krawczun  Rayna
Bocheva  Department of Physics and Astronomy, University of Sheffield,
Sheffield, S3 7RH, UK

###### Abstract

Neutrons can induce background events in underground experiments looking for
rare processes. Neutrons in a MeV range in deep underground laboratories are
produced in spontaneous fission processes and ($\alpha,n$) reactions. A number
of computer codes are available to calculate cross-sections of ($\alpha,n$)
reactions, branching ratios to various states and neutron yields. We have used
the SOURCES4 code to calculate neutron yields and energy spectra with input
cross-sections and branching ratios taken from experimental data and
calculations based on EMPIRE2.19/3.2.3 and TALYS1.9. We report here a
comparison of SOURCES4 calculations with experimental data obtained with alpha
beams with fixed energies.

##  I Introduction

Particles produced in radioactive processes may affect sensitivity of low-
background experiments, usually constructed in underground laboratories and
looking for rare events such as dark matter or neutrino interactions. A
particular problem is caused by neutrons produced in spontaneous fission and
($\alpha,n$) reactions that can mimic, for instance, WIMP-induced events in
direct dark matter searches if a neutron scatters only once in the detector
volume and does not leave any signal in veto systems.

Several computer codes exist to calculate neutron yields and energy spectra
form these processes (see, for instance, Refs. [1, 2, 3, 4]). Spontaneous
fission (SF) is well described by the parameterisation suggested by Watt [5]
with parameters tuned to the measurements. The neutron yield from SF does not
depend on the material where the neutron emission happens, but only on the
concentration of 238U (other radioactive isotopes give negligible neutron
yield from this process). The main problem is then in accurate calculation
neutron background caused by ($\alpha,n$) reactions where alpha particles are
produced in the decay chains of radioactive isotopes of uranium and thorium.
The codes usually use as inputs cross-sections of these reactions and
transition probabilities to excited states together with energy losses of
alphas as they travel through the material until they stop. Another option is
to take neutron spectra directly from the libraries such as JENDL [6] or TENDL
[7].

A comparison between different codes to calculate neutron yields and spectra
and experimental data has been given in several papers, see for instance Refs.
[2, 3, 4, 8, 9]. In this paper, we report the new calculations of neutron
production with the SOURCES4 code that includes ‘optimised’ cross-sections and
transition probabilities. This ‘optimisation’ includes a combination of recent
experimental data for the cross-sections and a model where the data are not
available. The same model is also used in the calculation of branching ratios
to different states (transition probabilities). We focus here on ($\alpha,n$)
reactions caused by alphas up to 10 MeV. These energies are typical for alphas
from the radioactive decay chains of 238U, 235U and 232Th which are the main
contributors to neutron background in underground experiments.

In the next Section we briefly describe the SOURCES4 code and the cross-
sections used. Then we show the comparison of our calculations with
experimental data and other codes.

##  II The SOURCES4 code to calculate neutron production ($\alpha,n$)
reactions

The nuclear physics code SOURCES4 [4] has been used for a long time in a
number of applications. The code libraries contain alpha emission lines from
radioactive isotopes, cross-sections of ($\alpha,n$) reactions from
calculations and experimental data, probabilities of transitions to different
final states of the daughter nucleus and parameterisations for energy losses
of alphas in different materials. The code calculates the neutron production
rates (or yields) and energy spectra of emitted neutrons for several types of
problems. We will consider here the thick target neutron yield from
radioactive isotopes in a homogeneous material where the size of the material
sample is much bigger than the range of alphas so edge effects can be
neglected, and the neutron production by monoenergetic alphas in a beam where,
again, the size of a material sample is much bigger than the range of alphas.
The most recent version of the code is SOURCES4C [4] but for historical
reasons we are using the older version SOURCES4A [10]. Previous tests and
release notes confirmed that both versions give the same results if the same
cross-sections and transition probabilities to excited states are used in both
versions. A big advantage of the code is in its flexibility so that a user can
choose what cross-section of ($\alpha,n$) reaction is used for a particular
isotope in a material sample. The user can also add more cross-sections to the
library.

The original code calculates neutron production for alpha energies up to 6.5
MeV and is not fully suitable for calculation of neutrons from radioactive
processes that involve alphas with energies up to about 9 MeV. The original
code SOURCES4A was modified to remove the 6.5 MeV energy cut and an updated
version now allows calculations of neutron production from alphas with
energies up to 10 MeV [11, 12]. Libraries of cross-sections and branching
ratios were updated with calculated values from the EMPIRE2.19 code [13] and
extended to alpha energies up to 10 MeV [11, 12, 14, 15, 16]. Cross-sections
and branching ratios were calculated with a 0.1 MeV step in alpha energy. All
these changes have been made to the version SOURCES4A and, since no changes in
results were observed with the newer version of the code, SOURCES4A is
continued to be used in most calculations. A comparison of cross-sections from
EMPIRE2.19 with some experimental data was published in Refs. [12, 16] and the
results from the modified SOURCES4A code were used in a number of dark matter
experiments (see, for example, Refs. [17, 18, 19, 20]).

The user input to SOURCES4A includes either the energy of an alpha particle or
the $Z$ and $A$ of the radioactive isotope (or several isotopes in the case of
decay chains, for instance) with the number of atoms per unit volume. The user
should also specify material composition (where the alpha sources are located)
and isotopic composition for each element (only isotopes with cross-sections
present in the code library can be included).

The output of SOURCES4A includes several files that return the neutron yield
and spectra for the sum of the ground and all excited states, as well as
neutron spectra for individual states. The output also includes neutron
production spectra on each isotope in the material sample. In the case of
decay chains, neutron production from individual radioactive isotopes on each
isotope in the material sample is also returned by the code. The code does not
generate gammas produced from de-excitation of a nucleus in the final state
but neutron spectra are calculated for individual states.

Recently, the libraries of SOURCES4A have been updated to include some cross-
sections calculated with TALYS1.9 [7] and the newer version of EMPIRE3.2.3
[21]. The comparison of cross-sections from TALYS1.9, EMPIRE2.19, EMPIRE3.2.3
and experimental data has been presented in Ref. [22]. Ref. [22] also includes
comparison of neutron yields calculated using SOURCES4A with different cross-
sections based on calculations using TALYS1.9, EMPIRE2.19/3.2.3 and some
experimental data.

Based on the results reported in Ref. [22], the libraries of cross-sections
and branching ratios in SOURCES4A have been updated so the most reliable data
are used when available; when the data are scarce or do not exist, the model
calculations from either TALYS1.9 or EMPIRE2.19/3.2.3 are used for these
isotopes and/or alpha energies. The same model is used for branching ratios
since there are very few data on this in literature. Sometimes this approach
(a combination of most reliable data and model calculations) is called
‘evaluated’ libraries. We will refer to this approach and cross-sections as
‘optimised’.

##  III SOURCES4 results and comparison with other codes and experimental data

Here we present the results from SOURCES4A calculations using ’optimised’
cross-sections and branching ratios. Where possible the measured cross-
sections have been used in the calculations of neutron yields. If different
sets of data are not in agreement, or there are no data for a particular
isotope, the model from either TALYS1.9 or EMPIRE2.19/3.2.3 is used. Branching
ratios were calculated using TALYS1.9 or EMPIRE2.19/3.2.3. The results of
these calculations for a number of elements are compared to different data
sets that were obtained from $\alpha$-particle beams interacting with thick
targets composed of different elements/isotopes. This comparison provides
another independent test of neutron yield calculations.

The cross-section for ($\alpha,n$) reaction on 13C is plotted in Figure 2
[23]. The threshold for ($\alpha,n$) reactions on 12C is above energies of
$\alpha$ lines from radioactive decay chains of uranium and thorium so only
13C yield contributes to neutron production in carbon. Data were taken from
Refs. [24, 25, 26, 27, 28]. Calculations with TALYS1.9 and EMPIRE3.2.3 are
also shown. The cross-section marked as ‘EMPIRE2.19 + J.K. Bair et al. (1973)’
includes measurements from Ref. [26] and calculations with EMPIRE2.19 above
5.5 MeV.

Figure 2 shows the neutron yield from carbon as a function of
$\alpha$-particle energy. The measurements for natural carbon have been
reported in Ref. [29]. SOURCES4A calculations use the cross-sections for 13C
from Ref. [25] up to 8 MeV. Above this energy and for branching ratios the
model from TALYS1.9 [7] was implemented (see also Ref. [30] for the comparison
of data and models for 13C and some other isotopes). Although the cross-
section for ($\alpha,n$) reaction on 13C as input to SOURCES4A has been taken
from the measurements, the resulting neutron yield from calculation and its
trend with energy are not in good agreement with data, in particular above 6
MeV. We note, however, that statistical models and evaluated libraries [7, 13,
21, 6] do not predicted so high rise of cross-sections above 6 MeV as in Ref.
[25] so calculated neutron yield may indeed be lower and agree better with the
alpha beam measurements.

Figure 1:  Cross-section of an ($\alpha,n$) reaction on 13C [23]. Experimental
data from Refs. [24, 25, 26, 27, 28] are shown together with EMPIRE2.19/3.2.3
and TALYS1.9 models. Cross-section marked as ‘EMPIRE2.19 + J.K. Bair et al.
(1973)’ includes measurements from Ref. [26] and calculations with EMPIRE2.19
above 5.5 MeV.

Figure 2:  Neutron yield as a function of alpha energy for natural carbon.
‘Optimised’ cross-sections are used in SOURCES4A: experimental data from Ref.
[25] up to 8 MeV and the model from TALYS1.9 [7] above 8 MeV and for branching
ratios. The measurements for natural carbon have been taken from Ref. [29].

Neutron yield as a function of alpha energy from aluminium is shown in Figure
4. The cross-section from the TALYS1.9 model was used as input to SOURCES4A
due to inconsistency between available data sets and lack of the data above 6
MeV (see the discussion in Ref. [30]). The resulting neutron yield from
calculations is in good agreement with data. Neutron energy spectra from
SOURCES4A calculations for 4.5 MeV alpha beam also agree quite well with
experimental data (see Figure 4).

Figure 3:  Neutron yield as a function of alpha energy for aluminium. The
cross-sections as input to SOURCES4A were calculated with TALYS1.9.
Experimental data were from Ref. [29]

.

Figure 4:  Neutron energy spectrum from 4.5 MeV alphas stopping in aluminium.
The cross-sections as input to SOURCES4A were calculated with TALYS1.9.
Experimental data were from Ref. [31].

Figure 5 shows the neutron yield for several materials from 232Th decay chain
as calculated by different codes in comparison with experimental data. The
neutron yield is given as the number of neutrons per $10^{6}$ alphas to be
directly compared with other publications [8, 3]. The decay chain is assumed
to be in equilibrium. There are 6 alphas in the 232Th chain so to convert this
to the total neutron production rate per gram per second per ppb of thorium
the value from the figure needs to be multiplied by $6\times 4.061\times
10^{-6}/10^{6}$. SOURCES4A uses optimised cross-sections – a combination of
experimentally measured cross-sections and calculations either with TALYS1.9
or EMPIRE2.19/3.2.3. Data from the NEDIS-2.0 code were taken from Ref. [32].
USD data were obtained with the web-based toolkit developed at the University
of South Dakota (USA) [1] and reported in Ref. [8]. NeuCBOT calculations have
been reported in Ref. [2]. Yields from GEANT4 simulations using two libraries
JENDL [6] and TENDL-2017 [7] were calculated in Ref. [3]. The ‘experimental
data’ were not directly measured as neutron yields from the whole decay chain
but were evaluated from the measured neutron yields for different alpha
energies as reported in Ref. [8]. A good agreement between SOURCES4A and
evaluated data is seen for most materials.

Figure 5: Neutron yield 232Th decay chain in several materials as calculated
by different codes in comparison with measurements. Neutron yield is given as
the number of neutrons per $10^{6}$ alphas. See text for details.

Ref. [33] reported direct measurement of neutron yields from uranium and
thorium decay chains and the comparison of SOURCES4A output with these data
for thorium chain is shown in Figure 6. The calculations are again in very
good agreement with measurements.

Figure 6: Neutron yield 232Th decay chain in several materials as calculated
by SOURCES4A in comparison with measurements [33]. Neutron yield is given as
the number of neutrons per $10^{6}$ alphas.

##  IV Conclusions

We have presented the ‘optimised’ approach to the inputs to SOURCES4A when the
cross-sections are taken from available data if available and consistency
between different data sets is obvious, complemented with models from either
TALYS or EMPIRE. With this input, we have calculated neutron yields as
functions of alpha energy for a number of elements and materials and compared
them to the measurements carried out with alpha particle beams. A good
agreement is seen for most isotopes and materials tested. Calculated neutron
energy spectra also show a reasonable agreement with the measurements. We have
also presented a comparison of SOURCES4A calculations of neutron yields from
radioactive decay chains with measurements and other codes.

###### Acknowledgements.

We acknowledge support from the UKRI-STFC and the University of Sheffield.

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