# Optimal investment in illiquid market
with search frictions and transaction costs

Jin Hyuk Choi, Tae Ung Gang

###### Abstract.

We consider an optimal investment problem to maximize expected utility of the
terminal wealth, in an illiquid market with search frictions and transaction
costs. In the market model, an investor’s attempt of transaction is successful
only at arrival times of a Poisson process, and the investor pays proportional
transaction costs when the transaction is successful. We characterize the no-
trade region describing the optimal trading strategy. We provide asymptotic
expansions of the boundaries of the no-trade region and the value function,
for small transaction costs. The asymptotic analysis implies that the effects
of the transaction costs are more pronounced in the market with less search
frictions.

Department of mathematics, Ulsan National Institute of Science and Technology;
<EMAIL_ADDRESS><EMAIL_ADDRESS>

Keywords: stochastic control, optimal investment, illiquidity, transaction
costs, search frictions

##  1\. Introduction

Understanding the effects of liquidity on optimal investment is one of the
main topics in mathematical finance and financial economics. According to
[53], stated simply, liquidity is the ease of trading a security. Various
sources of illiquidity include exogenous transaction costs, search frictions
such as the difficulty of locating a counterparty with whom to trade, and
price impacts due to private information.111For instance, [37, 4, 21, 13]
study how asymmetric information effects price impact and optimal trading
strategy in equilibrium. [2, 24, 46, 48] consider optimal order execution
problems with exogenously given price impacts. This paper studies an optimal
investment problem in the market model with two different types of
illiquidity: search frictions and transaction costs.

Under the assumption of the perfect liquidity,222Here, perfect liquidity
assumption means that assets can be traded in any quantity and at any moment
in time, without any transaction costs. the pioneering papers [41, 42]
formulate the optimal investment problem (so-called Merton’s portfolio
problem) with a geometric Brownian motion and a CRRA (constant relative risk
aversion) investor, and show that the optimal solution is to keep the constant
fraction of wealth invested in the risky asset. With the same assumption of
perfect liquidity, more general stochastic processes and utility functions
have been considered to obtain more general characterizations of the optimal
investment strategies (e.g., [30, 31, 34, 33, 28]).

In the optimal investment problems, the assumption of perfect liquidity can be
relaxed by considering search frictions in the market. As Table 1 in [3]
shows, many financial assets are illiquid in the sense that it is difficult to
find a counterparty who is willing to trade. One way to incorporate this type
of illiquidity, search frictions, into the optimal investment problem is to
impose some restrictions on trade times. In the classical Merton framework,
[49] considers an investor who is allowed to change portfolio only at times
which are multiple of a constant $h>0$, and [50, 40, 3] assume that an
illiquid asset can only be traded on the arrival of a randomly occurring
trading opportunity that is represented by jump times of a Poisson process.
[44, 14] consider an optimal investment/consumption problem under the
assumption that the asset price is observed only at the random trade times.
[23] complicates the model by using random intensity of trade times, regime-
switching, and liquidity shocks. In [18], a risky asset can be traded only on
deterministic time intervals and the investor pays proportional transaction
costs.

Transaction costs (such as order processing fees or transaction taxes) are
another source of market illiquidity, and the optimal investment problems with
transaction costs have been extensively studied in mathematical finance
community. [39, 20, 52] study the model in [42] with the assumption that
proportional transaction costs are levied on each transaction, and show (with
different level of mathematical rigorosity) that it is optimal to keep the
fraction of wealth invested in the risky asset in an interval so called no-
trade region. The boundaries of the no-trade region are characterized in terms
of the free-boundaries determined by the HJB (Hamilton-Jacobi-Bellman)
equation of the control problem. The models with transaction costs and
multiple risky assets have been studied (e.g., [1, 38, 43, 10] for costs on
all assets and [17, 8, 27, 12] for costs on only one assets) to characterize
the value function or no-trade region. More general stochastic processes have
been also considered in the framework of optimal investment with transaction
costs (e.g., [15, 16, 6]).

In this paper, we merge the aforementioned frameworks and analyze an optimal
investment problem in a market model with both search frictions and
transaction costs. We consider the classical Merton’s portfolio problem with a
log-utility investor, whose goal is to maximize the expected utility of wealth
at the terminal time $T>0$. We assume that the investor’s attempt of trading
is successful only when a Poisson process with intensity $\lambda$ jumps (as
in [50, 40]), and the investor needs to pay proportional transaction costs for
successful trading (as in [20, 52]). We show that there exists a unique
classical solution of the HJB equation and provide the standard verification
argument. As in the aforementioned models for transaction costs, the optimal
trading strategy is characterized by a no-trade region: if the investor can
trade at time $t\in[0,T)$, then there are two constants
$0\leq\underline{y}(t)\leq\bar{y}(t)\leq 1$ such that the investor should
minimally trade to keep the fraction of wealth invested in the risky asset
inside of the interval $[\underline{y}(t),\bar{y}(t)]$. Strict concavity of
the value function uniquely determines the boundaries $\underline{y}(t)$ and
$\bar{y}(t)$ of the no-trade region.

We provide asymptotic expansions of the value function and the no-trade
boundaries for small transaction costs. For the transaction cost parameter
$\epsilon$, in the similar model set up without the search frictions, it is
well known that the width of the no-trade region is the order of
$\epsilon^{\frac{1}{3}}$ and the decrease of the value function is the order
of $\epsilon^{\frac{2}{3}}$ (e.g., see [52, 29, 25, 11, 9, 45, 8, 12]). In our
model with the search frictions, we show that both the width of the no-trade
region and the decrease of the value function are the order of $\epsilon$, not
$\epsilon^{\frac{1}{3}}$ nor $\epsilon^{\frac{2}{3}}$. The search frictions in
our model induce the discrete trading times, and as in the discrete time
portfolio selection problems (e.g., see [5, 47]), we have the $\epsilon$ order
in the asymptotics.333In contrast to [5, 47], the trading times are random in
our model.

The previous paragraph suggests that the illiquidity due to the search
fractions makes the effect of the transaction costs less severe. To clarify
this effect more, we differentiate the coefficients of the first-order terms
with respect to the search friction parameter $\lambda$, and examine the signs
of these quantities as $\lambda\to\infty$. This analysis implies that the
effects of the transaction costs are more pronounced (more widening effect of
the no-trade region and more diminishing effect of the value function) in the
market with less search frictions.444To be more specific, let
$0<\lambda_{1}<\lambda_{2}$ and consider two markets $\mathcal{M}_{1}$ and
$\mathcal{M}_{2}$ with search friction parameters $\lambda_{1}$ and
$\lambda_{2}$, respectively. In words, $\mathcal{M}_{2}$ has less search
frictions than $\mathcal{M}_{1}$. Our result implies that if we increase the
transaction costs, the widening speed of the no-trade region in
$\mathcal{M}_{2}$ is faster than that in $\mathcal{M}_{1}$. Similarly, if we
increase the transaction costs, the diminishing speed of the optimal value in
$\mathcal{M}_{2}$ is faster than that in $\mathcal{M}_{1}$.

Our modeling assumption of the Poisson arrivals of the successful trading
times has a similar taste to the assumptions for the liquidity provision in
some models on limit order markets. [22, 51] derive equilibrium order
placement strategies in a limit order market model, where patient/impatient
agents arrive at the market according to a Poisson process. [26] solves an
optimal market-making model with execution/inventory risks, where the limit
order book is modeled by a Markov chain that jumps according to a stochastic
clock described by a Poisson process. [36] studies Merton’s portfolio problem
where the investor has the choice between market orders (immediate
transactions) and limit orders (Poisson arrivals of execution times). In the
model, trading with limit orders has a flavor of negative proportional
transaction costs.

The remainder of the paper is organized as follows. Section 2 describes the
model. In Section 3, we provide the verification argument and the strict
concavity of the value function. In Section 4, we characterize the optimal
trading strategy in terms of the no-trade region, and present some properties
of the no-trade region. In Section 5, we provide asymptotic analysis for small
transaction costs. Section 6 summarizes this paper. Proof of the technical
lemmas can be found in Appendix.

##  2\. The Model

Let $(\Omega,{\mathcal{F}},({\mathcal{F}}_{t})_{t\geq 0},{\mathbb{P}})$ be a
filtered probability space satisfying the usual conditions. Under the
filtration, we assume that $(B_{t})_{t\geq 0}$ is a standard Brownian motion
and $(P_{t})_{t\geq 0}$ is a Poisson process with intensity $\lambda>0$. We
assume that $B$ and $P$ are independent processes.

We consider a market that has two different types of illiquidity: (i) the
investor’s attempt of trading is successful only when the Poisson process
$(P_{t})_{t\geq 0}$ jumps,555Therefore, bigger $\lambda$ implies more frequent
trading opportunities (less search frictions), on average. and (ii) the
investor needs to pay proportional transaction costs for successful trading.

To be more specific, we consider a financial market consisting of a bond and a
stock, whose price processes $(S_{t}^{(0)})_{t\geq 0}$ and $(S_{t})_{t\geq 0}$
are given by the following stochastic differential equations (SDEs):

$\displaystyle dS_{t}^{(0)}$ $\displaystyle=S_{t}^{(0)}rdt$ (2.1)
$\displaystyle dS_{t}$ $\displaystyle=S_{t}\left(\mu dt+\sigma dB_{t}\right),$
(2.2)

where $\mu,r,\sigma,S_{0}^{(0)},S_{0}$ are constants and
$\sigma,S_{0}^{(0)},S_{0}$ are assumed to be strictly positive.

The proportional transaction costs are described by two constants
$\underline{\epsilon}\in[0,1)$ and $\bar{\epsilon}\in[0,\infty)$: the investor
gets $(1-\underline{\epsilon})S_{t}$ for selling one share of the stock, and
pays $(1+\bar{\epsilon})S_{t}$ for purchasing one share of the stock.

Let $W_{t}^{(1)}$ be the amount of wealth invested in the stock and
$W_{t}^{(0)}$ be the amount of wealth in the bond, at time $t\geq 0$. If the
investor tries to obtain the stock worth $M_{s}$ at time $s\in[0,t]$, then

$\begin{split}W_{t}^{(1)}&=w_{0}^{(1)}+\int_{0}^{t}W_{s-}^{(1)}\big{(}\mu
ds+\sigma dB_{s}\big{)}+\int_{0}^{t}M_{s}dP_{s},\\\
W_{t}^{(0)}&=w_{0}^{(0)}+\int_{0}^{t}W_{s-}^{(0)}rds+\int_{0}^{t}\left((1-\underline{\epsilon})M_{s}^{-}-(1+\bar{\epsilon})M_{s}^{+}\right)dP_{s},\end{split}$
(2.3)

where we use notation $x^{\pm}=\max{\\{0,\pm x\\}}$ for $x\in{\mathbb{R}}$,
and the constants $w_{0}^{(1)}\geq 0$ and $w_{0}^{(0)}\geq 0$ are given model
parameters that represent the initial position of the investor.

The trading strategy $(M_{t})_{t\geq 0}$ is called admissible if it is a
predictable process and the corresponding total wealth process
$W:=W^{(0)}+W^{(1)}$ is nonnegative all the time. This nonnegativity condition
is equivalent to $W^{(0)}_{t}\geq 0$ and $W^{(1)}_{t}\geq 0$ for all $t\geq
0$, because the rebalancing times are discrete.666 Indeed, for $s>t$ and
$A=\\{W^{(1)}_{t}<0,W^{(0)}_{t}>0\\}$ or $\\{W^{(1)}_{t}>0,W^{(0)}_{t}<0\\}$,
we observe that
${\mathbb{P}}(W_{s}<0\,|\,A)\geq{\mathbb{P}}(P_{t}=P_{s}\textrm{ and
}W_{s}<0\,|\,A)>0.$ Therefore, an admissible strategy $M$ satisfies

$\displaystyle-W_{t-}^{(1)}\leq
M_{t}\leq\frac{W_{t-}^{(0)}}{1+\bar{\epsilon}},\quad t\geq 0.$ (2.4)

For an admissible strategy $M$ and the corresponding solutions $W^{(1)}$ and
$W^{(0)}$ of the SDEs (2.3), let $X_{t}:=W_{t}^{(1)}/W_{t}$ be the fraction of
the total wealth invested in the stock market at time $t$. Then, the
inequalities in (2.4) imply that $0\leq X_{t}\leq 1$ and the application of
Ito’s formula produces the following SDE for $(W,X)$:

$\begin{split}dW_{t}&=\left(r(1-X_{t-})+\mu X_{t-}\right)W_{t-}dt+\sigma
X_{t-}W_{t-}dB_{t}-\left(\bar{\epsilon}M_{t}^{+}+\underline{\epsilon}M_{t}^{-}\right)dP_{t},\\\
dX_{t}&=X_{t-}(1-X_{t-})\left((\mu-r-\sigma^{2}X_{t-})dt+\sigma
dB_{t}\right)+\left(\frac{M_{t}+\left(\bar{\epsilon}M_{t}^{+}+\underline{\epsilon}M_{t}^{-}\right)X_{t-}}{W_{t-}-\bar{\epsilon}M_{t}^{+}-\underline{\epsilon}M_{t}^{-}}\right)dP_{t},\end{split}$
(2.5)

where the initial conditions are $W_{0}=w_{0}:=w_{0}^{(1)}+w_{0}^{(0)}$ and
$X_{0}=x_{0}:=w_{0}^{(1)}/w_{0}$. We assume that the initial total wealth is
strictly positive, $w_{0}>0$. The nonnegativity of $w_{0}^{(1)}$ and
$w_{0}^{(1)}$ imply $0\leq x_{0}\leq 1$.

Let a constant $T>0$ represent the terminal time of trading. We assume that
the investor’s goal is to maximize the expected log-utility of the total
wealth at the terminal time. That is, we analyze the following optimal
investment problem:

$\displaystyle\sup_{(M_{t})_{t\in[0,T]}}{\mathbb{E}}[\ln(W_{T})],$ (2.6)

where the supremum is taken over all admissible trading strategies.

######  Remark 2.1.

If $\lambda=\infty$ and $\bar{\epsilon}=\underline{\epsilon}=0$, then our
model becomes the classical Merton’s portfolio problem. The case of
$\bar{\epsilon}=\underline{\epsilon}=0$ and $\lambda<\infty$ is studied in
[40].

##  3\. Value function

Let $V$ be the value function of the control problem (2.6):

$\displaystyle
V(t,x,w)=\sup_{(M_{s})_{s\in[t,T]}}{\mathbb{E}}[\ln(W_{T})|{\mathcal{F}}_{t}]\Big{|}_{(X_{t},W_{t})=(x,w)}.$
(3.1)

As usual, the scaling property of the wealth process and the property of log-
utility enable us to conjecture the form of the value function as

$V(t,x,w)=\ln(w)+v(t,x)$

for a function $v$ (we verify this in Theorem 3.4). Then, the HJB equation for
(3.1) becomes

$\begin{split}\begin{cases}0=v(T,x),\\\
0=v_{t}+x(1-x)(\mu-r-\sigma^{2}x)v_{x}+\frac{1}{2}\sigma^{2}x^{2}(1-x)^{2}v_{xx}+(\mu-r)x+r-\tfrac{1}{2}\sigma^{2}x^{2}-\lambda
v\\\
\qquad+\lambda\sup_{y\in[0,1]}\left(v(t,y)-\ln\left(\frac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\frac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\right),\end{cases}\end{split}$
(3.2)

where $v_{t},v_{x},v_{xx}$ are partial derivatives. The following lemma
ensures that there exists a unique classical solution of the above PDE
(partial differential equation).

######  Lemma 3.1.

There exists a unique $v\in C([0,T]\times[0,1])\cap C^{1,2}([0,T]\times(0,1))$
that satisfies the followings: (i) $v$ satisfies the HJB equation (3.2) for
$(t,x)\in(0,T)\times(0,1)$.
(ii) For $x\in\\{0,1\\}$, the map $t\mapsto v(t,x)$ is continuously
differentiable on $[0,T]$ and satisfies

$\begin{split}\begin{cases}0=v(T,x),\\\
0=v_{t}(t,x)+(\mu-r)x+r-\tfrac{1}{2}\sigma^{2}x^{2}-\lambda v(t,x)\\\
\qquad+\lambda\sup_{y\in[0,1]}\left(v(t,y)-\ln\left(\frac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\frac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\right).\end{cases}\end{split}$
(3.3)

(iii) $v_{t}(t,x),\,x(1-x)v_{x}(t,x),\,x^{2}(1-x)^{2}v_{xx}(t,x)$ are
uniformly bounded on $(t,x)\in(0,T)\times(0,1)$.

###### Proof.

See Appendix. ∎

######  Remark 3.2.

In Lemma 3.1, we present the differential equations for $x\in(0,1)$ and
$x\in\\{0,1\\}$ as (3.2) and (3.3). The reason we separate these two cases is
that $v_{x}$ and $v_{xx}$ may not be continuously extended to the endpoints
$x\in\\{0,1\\}$.

######  Remark 3.3.

The model with the search frictions can be seen as a generalization of [19].
However, our analysis has some technically easier features than that of [19],
because our solvency region is the first quadrant and we have no subtle issues
regarding the regularity of the value function at $x=1$ or $x=0$.

######  Theorem 3.4.

Let $V$ be the value function in (3.1) and $v$ be as in Lemma 3.1. Then, for
$(t,x)\in[0,T]\times[0,1]$,

$\displaystyle V(t,x,w)=\ln(w)+v(t,x).$ (3.4)

###### Proof.

Without loss of generality, we prove $V(0,x_{0},w_{0})=\ln(w_{0})+v(0,x_{0})$.
Let $M$ be an admissible trading strategy and $(W,X)$ be the corresponding
solution of (2.5). Let $\tau_{n}:=\inf\\{t\geq 0:P_{t}=n\\}$. If
$X_{\tau_{n}}=0$ (resp., $X_{\tau_{n}}=1$), then for $\tau_{n}\leq
t<\tau_{n+1}$,

$\displaystyle X_{t}=0,\,\,dW_{t}=rW_{t}dt\quad\textrm{\big{(}resp.,
}X_{t}=1,\,\,dW_{t}=\mu W_{t}dt+\sigma W_{t}dB_{t}\big{)}.$ (3.5)

We apply Ito’s formula to $\big{(}\ln(W_{t})+v(t,X_{t})\big{)}_{t\in[0,T]}$
and use (2.5) and (3.5) to obtain

$\begin{split}&\ln(W_{T})-\ln(w_{0})-v(0,x_{0})\\\
&=\int_{0}^{T}\bigg{(}\Big{(}v_{t}(t,x)+x(1-x)(\mu-r-\sigma^{2}x)v_{x}(t,x)+\tfrac{1}{2}\sigma^{2}x^{2}(1-x)^{2}v_{xx}(t,x)\\\
&\qquad\qquad+(\mu-r)x+r-\tfrac{1}{2}\sigma^{2}x^{2}\Big{)}\cdot
1_{\\{0<x<1\\}}+\Big{(}v_{t}(t,x)+(\mu-r)x+r-\tfrac{1}{2}\sigma^{2}x^{2}\Big{)}\cdot
1_{\\{x\in\\{0,1\\}\\}}\bigg{)}\bigg{|}_{x=X_{t-}}dt\\\
&\quad+\int_{0}^{T}\sigma\Big{(}x(1-x)v_{x}(t,x)+x\Big{)}\cdot
1_{\\{0<x<1\\}}\bigg{|}_{x=X_{t-}}dB_{t}+\int_{0}^{T}\sigma\cdot
1_{\\{x=1\\}}\bigg{|}_{x=X_{t-}}dB_{t}\\\ &\quad+\sum_{0<t\leq
T}\Big{(}\ln(W_{t})-\ln(W_{t-})+v(t,X_{t})-v(t,X_{t-})\Big{)}.\end{split}$

The stochastic integral term above is a true martingale, because Lemma 3.1
(iii) implies that the integrand is uniformly bounded. The sum of jumps term
above can be written as

$\begin{split}\int_{0}^{T}\Big{(}v(t,y)-v(t,x)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\Big{)}\bigg{|}_{(x,y)=\left(X_{t-},\frac{X_{t-}W_{t-}+M_{t}}{W_{t-}-\bar{\epsilon}M_{t}^{+}-\underline{\epsilon}M_{t}^{-}}\right)}dP_{t}.\end{split}$

Considering that the integrand above is bounded and $(P_{t}-\lambda
t)_{t\in[0,T]}$ is a martingale, we can write the expected value of the above
expression as

$\begin{split}{\mathbb{E}}\left[\int_{0}^{T}\lambda\Big{(}v(t,y)-v(t,x)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\Big{)}\bigg{|}_{(x,y)=\left(X_{t-},\frac{X_{t-}W_{t-}+M_{t}}{W_{t-}-\bar{\epsilon}M_{t}^{+}-\underline{\epsilon}M_{t}^{-}}\right)}dt\right].\end{split}$

Combining these observations, we obtain

$\begin{split}&{\mathbb{E}}[\ln(W_{T})]-\ln(w_{0})-v(0,x_{0})\\\
&={\mathbb{E}}\Bigg{[}\int_{0}^{T}\Bigg{(}\bigg{(}v_{t}(t,x)+x(1-x)(\mu-r-\sigma^{2}x)v_{x}(t,x)+\tfrac{1}{2}\sigma^{2}x^{2}(1-x)^{2}v_{xx}(t,x)+(\mu-r)x\\\
&\qquad+r-\tfrac{1}{2}\sigma^{2}x^{2}+\lambda\Big{(}v(t,y)-v(t,x)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\Big{)}\bigg{)}\cdot
1_{\\{0<x<1\\}}\\\
&\qquad+\bigg{(}v_{t}(t,x)+(\mu-r)x+r-\tfrac{1}{2}\sigma^{2}x^{2}+\lambda\Big{(}v(t,y)-v(t,x)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}\\\
&\qquad-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\Big{)}\bigg{)}\cdot
1_{\\{x\in\\{0,1\\}\\}}\Bigg{)}\Bigg{|}_{(x,y)=\left(X_{t-},\frac{X_{t-}W_{t-}+M_{t}}{W_{t-}-\bar{\epsilon}M_{t}^{+}-\underline{\epsilon}M_{t}^{-}}\right)}dt\Bigg{]},\end{split}$
(3.6)

where the integrability is due to Lemma 3.1 (iii). The equality (LABEL:ito)
and Lemma 3.1 (i) and (ii) imply that for any admissible trading strategy $M$,

${\mathbb{E}}[\ln(W_{T})]\leq\ln(w_{0})+v(0,x_{0}).$ (3.7)

To complete the proof, we find an optimal strategy $\hat{M}$ that satisfies
the equality in (3.7). We observe that the map

$\displaystyle(t,x,y)\mapsto
v(t,y)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}$

is continuous on $[0,T]\times[0,1]^{2}$. Therefore, according to Lemma D.1, we
can choose a measurable function $\hat{y}:[0,T]\times[0,1]\to[0,1]$ such that

$\displaystyle\hat{y}(t,x)\in\operatorname*{argmax}_{y\in[0,1]}\left(v(t,y)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\right).$
(3.8)

Using $\hat{y}$, we define a measurable function
$m:[0,T]\times[0,\infty)\times[0,1]\to{\mathbb{R}}$ as

$\displaystyle
m(t,w,x):=\frac{w(\hat{y}(t,x)-x)}{1+\bar{\epsilon}\,\hat{y}(t,x)}\cdot
1_{\big{\\{}\hat{y}(t,x)>x\big{\\}}}+\frac{w(\hat{y}(t,x)-x)}{1-\underline{\epsilon}\,\hat{y}(t,x)}\cdot
1_{\big{\\{}\hat{y}(t,x)<x\big{\\}}}.$ (3.9)

Let $(\hat{W},\hat{X})$ be the unique solution777Indeed, the SDE in (2.5)
without $dP_{t}$ term has a unique (explicit) solution. The unique solution of
(2.5) can be obtained by patching the unique solutions on time intervals
between the jump times of the Poisson process, with the jump size described by
the coefficient of $dP_{t}$ term. of SDE (2.5) with
$M_{t}=m(t,W_{t-},X_{t-})$. Now we define
$\hat{M}_{t}:=m(t,\hat{W}_{t-},\hat{X}_{t-})$, then it is a predictable
process. From (3.9) and (2.5), we observe that

$\Delta\hat{X}_{t}=\left(\hat{y}(t,\hat{X}_{t-})-\hat{X}_{t-}\right)\Delta
P_{t}.$

Then the structure of (2.5) and $0\leq\hat{y}\leq 1$ imply that
$0\leq\hat{X}_{t}\leq 1$ and $\hat{W}_{t}\geq 0$ for all $t\in[0,T]$.
Therefore, we conclude that $\hat{M}$ is an admissible trading strategy.

Finally, we substitute $(\hat{X},\hat{W},\hat{M})$ for $(X,W,M)$ in
(LABEL:ito). Then, the differential equations (3.2) and (3.3) for $v$, the
optimality of $\hat{y}$ in (3.8), and the observation
$\frac{\hat{X}_{t-}\hat{W}_{t-}+\hat{M}_{t}}{\hat{W}_{t-}-\bar{\epsilon}\hat{M}_{t}^{+}-\underline{\epsilon}\hat{M}_{t}^{-}}=\hat{y}(t,\hat{X}_{t-})$
produce

$\displaystyle{\mathbb{E}}[\ln(\hat{W}_{T})]=\ln(w_{0})+v(0,x_{0}).$ (3.10)

By (3.7) and (3.10), we conclude $V(0,x_{0},w_{0})=\ln(w_{0})+v(0,x_{0})$ and
the optimality of $\hat{M}$. ∎

In the proof of Theorem 3.4, the optimal trading strategy is described by the
function $\hat{y}$ in (3.8). In the next section, we show that the maximizer
in (3.8) is unique, and we characterize the form of the unique optimal
strategy in detail. For this purpose, we use the strict concavity of the value
function. To be more specific, we define
${\tilde{V}}:[0,T]\times\left([0,\infty)^{2}\setminus\\{(0,0)\\}\right)\to{\mathbb{R}}$
as

$\displaystyle{\tilde{V}}(t,a,b):=\sup_{(M_{s})_{s\in[t,T]}}{\mathbb{E}}\left[\ln\left(W_{T}^{a,b,M}\right)\right],$
(3.11)

where $W_{T}^{a,b,M}$ represents the total wealth at time $T$ with
$(W^{(0)}_{t},W^{(1)}_{t})=(a,b)$ and the trading strategy $M$. In other
words, ${\tilde{V}}$ is the value function in terms of the wealth amount
invested in the bond and the stock. Then, (3.1) and (3.11) imply that

$\displaystyle{\tilde{V}}(t,a,b)=V\left(t,\tfrac{b}{a+b},a+b\right).$ (3.12)

######  Proposition 3.5.

For $t\in[0,T)$, the maps $(a,b)\mapsto{\tilde{V}}(t,a,b)$ and $x\mapsto
v(t,x)$ are strictly concave.

###### Proof.

Without loss of generality, we prove the statement for $t=0$ case. Let
$(a_{0},b_{0})\neq(a_{1},b_{1})$ be elements of
$[0,\infty)^{2}\setminus\\{(0,0)\\}$ and $\theta\in(0,1)$. According to the
proof of Theorem 3.4 and the relation (3.12), we can find optimal trading
strategies for the initial positions $(a_{0},b_{0})$ and $(a_{1},b_{1})$, and
we denote them by $\hat{M}^{0}$ and $\hat{M}^{1}$, respectively. Then the
structure of the SDE in (2.3) implies that
$M^{\theta}:=(1-\theta)\hat{M}^{0}+\theta\hat{M}^{1}$ is an admissible trading
strategy with initial position
$(a_{\theta},b_{\theta}):=\left((1-\theta)a_{0}+\theta
a_{1},(1-\theta)b_{0}+\theta b_{1}\right)$ and satisfies888 The inequality
(3.13) becomes strict on the event $\left\\{\omega\in\Omega:\exists
t\in[0,T]\textrm{ such that }\Delta P_{t}(\omega)=1\textrm{ and
}\hat{M}_{t}^{0}(\omega)\hat{M}_{t}^{1}(\omega)<0\right\\}$.

$\displaystyle(1-\theta)W_{T}^{a_{0},b_{0},\hat{M}^{0}}+\theta
W_{T}^{a_{1},b_{1},\hat{M}^{1}}\leq W_{T}^{a_{\theta},b_{\theta},M^{\theta}}.$
(3.13)

We also observe that

$\begin{split}{\mathbb{P}}\left(W_{T}^{a_{0},b_{0},\hat{M}^{0}}\neq
W_{T}^{a_{1},b_{1},\hat{M}^{1}}\right)&\geq{\mathbb{P}}\left(W_{T}^{a_{0},b_{0},\hat{M}^{0}}\neq
W_{T}^{a_{1},b_{1},\hat{M}^{1}}\textrm{ and }P_{T}=0\right)\\\
&={\mathbb{P}}(P_{T}=0)\cdot{\mathbb{P}}\left(W_{T}^{a_{0},b_{0},\hat{M}^{0}}\neq
W_{T}^{a_{1},b_{1},\hat{M}^{1}}\,\big{|}\,P_{T}=0\right)\\\ &=e^{-\lambda
T}\cdot{\mathbb{P}}\left((a_{1}-a_{0})e^{rT}+(b_{1}-b_{0})e^{(\mu-\frac{\sigma^{2}}{2})T+\sigma
B_{T}}\neq 0\right)\\\ &=e^{-\lambda T}>0,\end{split}$ (3.14)

where the second equality is due to the independence of $B$ and $P$, and the
last equality is due to $(a_{0},b_{0})\neq(a_{1},b_{1})$ and the fact that
$B_{T}$ is a continuous random variable. By these observations, we obtain the
strict concavity of ${\tilde{V}}$:

$\begin{split}(1-\theta){\tilde{V}}(0,a_{0},b_{0})+\theta{\tilde{V}}(0,a_{1},b_{1})&={\mathbb{E}}\left[(1-\theta)\ln\left(W_{T}^{a_{0},b_{0},\hat{M}^{0}}\right)+\theta\ln\left(W_{T}^{a_{1},b_{1},\hat{M}^{1}}\right)\right]\\\
&<{\mathbb{E}}\left[\ln\left((1-\theta)W_{T}^{a_{0},b_{0},\hat{M}^{0}}+\theta
W_{T}^{a_{1},b_{1},\hat{M}^{1}}\right)\right]\\\
&\leq{\mathbb{E}}\left[\ln\left(W_{T}^{a_{\theta},b_{\theta},M^{\theta}}\right)\right]\\\
&\leq{\tilde{V}}(0,a_{\theta},b_{\theta}),\end{split}$

where the first inequality is due to the strict concavity of logarithm and
(3.14), and the second inequality is from (3.13). Therefore, the map
$(a,b)\mapsto{\tilde{V}}(0,a,b)$ is strictly concave. This also implies that
the map $x\mapsto{\tilde{V}}(0,1-x,x)$ is strictly concave. Finally, Theorem
3.4 and the relation (3.12) connect ${\tilde{V}}$ and $v$ as

$\displaystyle{\tilde{V}}(0,1-x,x)=V(0,x,1)=v(0,x),$

and we conclude that the map $x\mapsto v(0,x)$ is strict concave. ∎

##  4\. Optimal strategy

In this section, we show that the optimal strategy can be characterized in
terms of the no-trade region. We start with the construction of the candidate
boundary points $\underline{y}$ and $\bar{y}$ of the no-trade region in the
following lemma.

######  Lemma 4.1.

For $t\in[0,T)$, there exist $0\leq\underline{y}(t)\leq\bar{y}(t)\leq 1$ such
that

$\begin{split}\big{\\{}\underline{y}(t)\big{\\}}=\operatorname*{argmax}_{y\in[0,1]}\Big{(}v(t,y)-\ln(1+\bar{\epsilon}y)\Big{)},\\\
\big{\\{}\bar{y}(t)\big{\\}}=\operatorname*{argmax}_{y\in[0,1]}\Big{(}v(t,y)-\ln(1-\underline{\epsilon}y)\Big{)}.\end{split}$

To be more specific, the following statements hold:
(i) The map $y\mapsto v(t,y)-\ln(1+\bar{\epsilon}y)$ strictly increases
(decreses) on $y\in[0,\underline{y}(t)]$ ($y\in[\underline{y}(t),1]$).
If $0<\underline{y}(t)<1$, then $\underline{y}(t)$ is the unique solution of
the equation $v_{x}(t,x)=\frac{\bar{\epsilon}}{1+\bar{\epsilon}x}$.
(ii) The map $y\mapsto v(t,y)-\ln(1-\underline{\epsilon}y)$ strictly increases
(decreses) on $y\in[0,\bar{y}(t)]$ ($y\in[\bar{y}(t),1]$).
If $0<\bar{y}(t)<1$, then $\bar{y}(t)$ is the unique solution of the equation
$v_{x}(t,x)=-\frac{\underline{\epsilon}}{1-\underline{\epsilon}x}$.

###### Proof.

Let $t\in[0,T)$ be fixed. We consider the map
$z\in[0,\tfrac{1}{1+\bar{\epsilon}}]\mapsto{\tilde{V}}(t,1-(1+\bar{\epsilon})z,z)$,
where ${\tilde{V}}$ is defined in (3.11). Theorem 3.4 and (3.12) imply that
this map is differentiable (we denote the partial derivatives as
${\tilde{V}}_{a}$ and ${\tilde{V}}_{b}$), and the map is strictly concave due
to Proposition 3.5. Therefore, its derivative

$\displaystyle
D(t,z):=-(1+\bar{\epsilon}){\tilde{V}}_{a}(t,1-(1+\bar{\epsilon})z,z)+{\tilde{V}}_{b}(t,1-(1+\bar{\epsilon})z,z)$
(4.1)

is strictly decreasing in $z\in(0,\tfrac{1}{1+\bar{\epsilon}})$, and there
exists a unique $\underline{z}(t)\in[0,\tfrac{1}{1+\bar{\epsilon}}]$ such that

$\displaystyle\begin{cases}D(t,z)>0&\textrm{for
$z\in(0,\underline{z}(t))$},\\\ D(t,z)<0&\textrm{for
$z\in(\underline{z}(t),\tfrac{1}{1+\bar{\epsilon}})$}.\\\ \end{cases}$ (4.2)

Obviously, in case $\underline{z}(t)\in(0,\tfrac{1}{1+\bar{\epsilon}})$,
$\underline{z}(t)$ is the unique solution of $D(t,z)=0$.

By Theorem 3.4 and (3.12), we observe that for $y\in[0,1]$,

$\displaystyle
v(t,y)-\ln(1+\bar{\epsilon}y)=V\left(t,y,\tfrac{1}{1+\bar{\epsilon}y}\right)={\tilde{V}}\left(t,\tfrac{1-y}{1+\bar{\epsilon}y},\tfrac{y}{1+\bar{\epsilon}y}\right).$

We take derivative with respect to $y$ above and use (4.1) to obtain

$\displaystyle\tfrac{\partial}{\partial
y}\Big{(}v(t,y)-\ln(1+\bar{\epsilon}y)\Big{)}=\tfrac{1}{(1+\bar{\epsilon}y)^{2}}\,D\left(t,\tfrac{y}{1+\bar{\epsilon}y}\right).$
(4.3)

Now we define
$\underline{y}(t):=\frac{\underline{z}(t)}{1-\bar{\epsilon}\underline{z}(t)}\in[0,1]$.
Since the map $y\mapsto\frac{y}{1+\bar{\epsilon}y}$ is strictly increasing on
$[0,1]$, the definition of $\underline{z}(t)$ in (4.2) implies that

$\displaystyle\begin{cases}D\left(t,\frac{y}{1+\bar{\epsilon}y}\right)>0&\textrm{for
$y\in\left(0,\underline{y}(t)\right)$},\\\
D\left(t,\frac{y}{1+\bar{\epsilon}y}\right)<0&\textrm{for
$y\in\left(\underline{y}(t),1\right)$}.\\\ \end{cases}$ (4.4)

Also, in case $\underline{y}(t)\in(0,1)$, $\underline{y}(t)$ is the unique
solution of $D\left(t,\frac{y}{1+\bar{\epsilon}y}\right)=0$. From (4.3) and
(4.4), we conclude that statement (i) holds and

$\big{\\{}\underline{y}(t)\big{\\}}=\operatorname*{argmax}_{y\in[0,1]}\Big{(}v(t,y)-\ln(1+\bar{\epsilon}y)\Big{)}.$

By the same way, we conclude that
$\max_{y\in[0,1]}\big{(}v(t,y)-\ln(1-\underline{\epsilon}y)\big{)}$ has the
unique maximizer denoted by $\bar{y}(t)$ and statement (ii) holds.

It only remains to check the inequality $\underline{y}(t)\leq\bar{y}(t)$. If
$\bar{y}(t)=1$, then $\underline{y}(t)\leq\bar{y}(t)$ is obvious. If
$\bar{y}(t)<1$, then
$v_{x}(t,\bar{y}(t))\leq\frac{-\underline{\epsilon}}{1-\underline{\epsilon}\bar{y}(t)}\leq\frac{\bar{\epsilon}}{1+\bar{\epsilon}\bar{y}(t)}$
by (ii). From (4.3) and (4.4), we obtain $\underline{y}(t)\leq\bar{y}(t)$. ∎

In the next theorem, we explicitly characterize the optimizer $\hat{y}$ in
(3.8) in terms of $\underline{y}$ and $\bar{y}$ in Lemma 4.1.

######  Theorem 4.2.

For $t\in[0,T)$, the $\operatorname*{argmax}$ in (3.8) is a singleton, and
$\hat{y}$ has the following expression:

$\displaystyle\hat{y}(t,x)=\begin{cases}\underline{y}(t),&\textrm{if
}x\in\left[0,\underline{y}(t)\right)\\\ x,&\textrm{if
}x\in\left[\underline{y}(t),\bar{y}(t)\right]\\\ \bar{y}(t),&\textrm{if
}x\in\left(\bar{y}(t),1\right]\\\ \end{cases}$ (4.5)

where $\underline{y}(t)$ and $\bar{y}(t)$ are uniquely determined in Lemma
4.1.

###### Proof.

We may rewrite the maximization in (3.8) as

$\begin{split}&\max_{y\in[0,1]}\left(v(t,y)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\right)\\\
&=\max\left\\{\max_{y\in[0,x]}\left(v(t,y)-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)\right),\max_{y\in[x,1]}\left(v(t,y)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)\right)\right\\}.\end{split}$
(4.6)

Using Lemma 4.1, we observe that

$\begin{split}\max_{y\in[0,x]}\left(v(t,y)-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)\right)=\begin{cases}v(t,x),&\textrm{if
}x\leq\bar{y}(t)\\\
v(t,\bar{y}(t))-\ln\left(\frac{1-\underline{\epsilon}\bar{y}(t)}{1-\underline{\epsilon}x}\right),&\textrm{if
}x>\bar{y}(t)\end{cases}\\\
\max_{y\in[x,1]}\left(v(t,y)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)\right)=\begin{cases}v(t,\underline{y}(t))-\ln\left(\frac{1+\bar{\epsilon}\underline{y}(t)}{1+\bar{\epsilon}x}\right),&\textrm{if
}x<\underline{y}(t)\\\ v(t,x),&\textrm{if
}x\geq\underline{y}(t)\end{cases}\end{split}$ (4.7)

where the maximizers are unique. Combining (LABEL:max_split1) and (4.7), we
obtain

$\begin{split}&\max_{y\in[0,1]}\left(v(t,y)-\ln\left(\tfrac{1+\bar{\epsilon}y}{1+\bar{\epsilon}x}\right)1_{\\{x<y\\}}-\ln\left(\tfrac{1-\underline{\epsilon}y}{1-\underline{\epsilon}x}\right)1_{\\{x>y\\}}\right)\\\
&=\begin{cases}v(t,\underline{y}(t))-\ln\left(\frac{1+\bar{\epsilon}\underline{y}(t)}{1+\bar{\epsilon}x}\right),&\textrm{if
}x\in\left[0,\underline{y}(t)\right)\\\ v(t,x),&\textrm{if
}x\in\left[\underline{y}(t),\bar{y}(t)\right]\\\
v(t,\bar{y}(t))-\ln\left(\frac{1-\underline{\epsilon}\bar{y}(t)}{1-\underline{\epsilon}x}\right),&\textrm{if
}x\in\left(\bar{y}(t),1\right]\\\ \end{cases}\end{split}$

and conclude that the corresponding unique maximizer is as in (4.5). ∎

$\begin{array}[]{cc}\includegraphics[width=195.12767pt]{oyuy.pdf}&\,\,\,\includegraphics[width=195.12767pt]{haty.pdf}\end{array}$

Figure 1. The left graph shows $\underline{y}(t)$ and $\bar{y}(t)$ as
functions of $t$, and the right graph describes $\hat{y}(t,x)$ as a function
of $x$ for fixed $t=0.5$. In both graphs, the dashed line is the Merton
fraction $y_{\infty}=\frac{\mu-r}{\sigma^{2}}$. The parameters are
$\mu=0.4,\,r=0.1,\sigma=1,\,\lambda=3,\,\underline{\epsilon}=\bar{\epsilon}=0.05$,
and $T=1$.

Theorem 4.2 implies that the optimal trading strategy is characterized by the
no-trade region: if the investor can trade at time $t\in[0,T)$, then the
investor should minimally trade to keep the fraction of wealth invested in the
stock inside of the interval $[\underline{y}(t),\bar{y}(t)]$. To be specific,
if the fraction $X_{t}$ is less (more, resp.) than $\underline{y}(t)$
($\bar{y}(t)$, resp.), then the investor should buy (sell, resp.) the stock
and adjust the fraction to $\underline{y}(t)$ ($\bar{y}(t)$, resp.). If the
fraction $X_{t}$ is inside of the interval $[\underline{y}(t),\bar{y}(t)]$,
then the investor should not trade. Figure 1 illustrates the no-trade region
and the optimal trading strategy.

If $\bar{\epsilon}=\underline{\epsilon}=0$, then Lemma 4.1 implies that
$\underline{y}(t)=\bar{y}(t)$, hence the no-trade region becomes a singleton.
For $\bar{\epsilon}>0$, one may expect that the investor would not want to buy
the stock at times close to the terminal time $T$ due to the transaction
costs. If $\bar{\epsilon}=0$ and the Merton fraction
$\frac{\mu-r}{\sigma^{2}}$ is greater than zero, then one may expect that the
investor would want to hold strictly positive shares of the stock all the
time. Our next task is to examine and prove this type of trading behaviors.

For detailed analysis, we first provide stochastic representations of $v$ and
$v_{x}$. We apply the Feynman-Kac formula (i.e., see Theorem 5.7.6 in [32])
and the expression of the optimizer $\hat{y}$ in Theorem 4.2 to Lemma 3.1, and
obtain the following representation for $v$:

$\displaystyle
v(t,x)=\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[(\mu-r)Y_{s}^{(t,x)}+r-\tfrac{1}{2}\sigma^{2}\left(Y_{s}^{(t,x)}\right)^{2}+\lambda\,L(s,Y_{s}^{(t,x)})\right]ds,$
(4.8)

where for $(s,x)\in[t,T)\times[0,1]$,

$\begin{split}Y_{s}^{(t,x)}&:=\frac{x\cdot\exp{\left(\left(\mu-r-\frac{1}{2}\sigma^{2}\right)(s-t)+\sigma(B_{s}-B_{t})\right)}}{x\cdot\exp{\left(\left(\mu-r-\frac{1}{2}\sigma^{2}\right)(s-t)+\sigma(B_{s}-B_{t})\right)}+(1-x)},\\\
L(s,y)&:=v(s,\hat{y}(s,y))-\ln\left(\tfrac{1+\bar{\epsilon}\,\hat{y}(s,y)}{1+\bar{\epsilon}y}\right)1_{\\{y<\hat{y}(s,y)\\}}-\ln\left(\tfrac{1-\underline{\epsilon}\,\hat{y}(s,y)}{1-\underline{\epsilon}y}\right)1_{\\{y>\hat{y}(s,y)\\}}.\end{split}$
(4.9)

The representation of $v_{x}$ is given in the following lemma.

######  Lemma 4.3.

The function $L$ in (4.9) is continuously differentiable with respect to $y$,

$\displaystyle
L_{y}(t,x)=\begin{cases}\frac{\bar{\epsilon}}{1+\bar{\epsilon}x},&x\in(0,\underline{y}(t)]\\\
v_{x}(t,x),&x\in(\underline{y}(t),\bar{y}(t))\\\
-\frac{\underline{\epsilon}}{1-\underline{\epsilon}x},&x\in[\bar{y}(t),1)\end{cases}\quad\textrm{for
}(t,x)\in[0,T)\times(0,1),$ (4.10)

and $v_{x}(t,x)$ has the following representation: for
$(t,x)\in[0,T)\times(0,1)$,

$\displaystyle
v_{x}(t,x)=\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\left(\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right)\left(\mu-r-\sigma^{2}Y_{s}^{(t,x)}+\lambda\,L_{y}(s,Y_{s}^{(t,x)})\right)\right]ds.$
(4.11)

###### Proof.

Combining (4.5) and (4.9), we rewrite $L$ as

$\displaystyle
L(t,x)=\begin{cases}v(t,\underline{y}(t))-\ln\left(\frac{1+\bar{\epsilon}\,\underline{y}(t)}{1+\bar{\epsilon}x}\right),&x\in(0,\underline{y}(t)]\\\
v(t,x),&x\in(\underline{y}(t),\bar{y}(t))\\\
v(t,\bar{y}(t))-\ln\left(\frac{1-\underline{\epsilon}\,\bar{y}(t)}{1-\underline{\epsilon}x}\right),&x\in[\bar{y}(t),1)\end{cases}.$
(4.12)

We take derivative with respect to $x$ above and obtain the expression (4.10)
for $x\in(0,1)\setminus\\{\underline{y}(t),\bar{y}(t)\\}$. If
$0<\underline{y}(t)<1$ (resp., $0<\bar{y}(t)<1$), then
$v_{x}(t,\underline{y}(t))=\frac{\bar{\epsilon}}{1+\bar{\epsilon}\underline{y}(t)}$
(resp.,
$v_{x}(t,\bar{y}(t))=-\frac{\underline{\epsilon}}{1-\underline{\epsilon}\bar{y}(t)}$).
Therefore, we conclude that $L$ is continuously differentiable with respect to
$y$ and (4.10) is valid.

Since
$-\frac{\underline{\epsilon}}{1-\underline{\epsilon}x}<v_{x}(t,x)<\frac{\bar{\epsilon}}{1+\bar{\epsilon}x}$
for $x\in(\underline{y}(t),\bar{y}(t))$, we observe that for
$(t,x)\in[0,T)\times(0,1)$,

$\displaystyle-\frac{\underline{\epsilon}}{1-\underline{\epsilon}}\leq
L_{y}(t,x)\leq\bar{\epsilon}.$ (4.13)

We also observe that for $(t,x)\in[0,T)\times(0,1)$,

$\begin{split}e^{-\left|r-\mu+\frac{\sigma^{2}}{2}\right|(s-t)-\sigma\left|B_{s}-B_{t}\right|}\leq\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\leq
e^{\left|r-\mu+\frac{\sigma^{2}}{2}\right|(s-t)+\sigma\left|B_{s}-B_{t}\right|}.\end{split}$
(4.14)

Now we take derivative with respect to $x$ in (4.8). The mean value theorem
and the dominated convergence theorem, together with the inequalities (4.13)
and (4.14), allow us to take derivative inside of the expectation. By (4.10)
and the chain rule, we obtain the representation (4.11). ∎

Using these representations, we extract some properties about the boundaries
$\underline{y}(t)$ and $\bar{y}(t)$ of the no-trade region.

######  Proposition 4.4.

(Properties of no-trade region)
Let $y_{\infty}:=\frac{\mu-r}{\sigma^{2}}$ denote the Merton fraction.
(i) If $\bar{\epsilon}>0$ ($\underline{\epsilon}>0$, resp.), then there exists
$t_{0}\in[0,T)$ such that $\underline{y}(t)=0$ ($\bar{y}(t)=1$, resp.) for
$t\in[t_{0},T)$.
(ii) If $0<y_{\infty}$ ($y_{\infty}<1$, resp.), then $\bar{y}(t)>0$
($\underline{y}(t)<1$, resp.) for $t\in[0,T)$.
(iii) If $0<y_{\infty}<1$ and at least one of $\bar{\epsilon}$ and
$\underline{\epsilon}$ is strictly positive, then
$\underline{y}(t)<\bar{y}(t)$ for $t\in[0,T)$.
(iv) If $0<y_{\infty}$ and $\bar{\epsilon}=0$ ($y_{\infty}<1$ and
$\underline{\epsilon}=0$, resp.), then $\underline{y}(t)>0$ ($\bar{y}(t)<1$,
resp.) for $t\in[0,T)$.

###### Proof.

By (4.14), we have

$\displaystyle 0<{\mathbb{E}}\left[\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right]\leq 2e^{(|r-\mu|+\sigma^{2})(s-t)},$ (4.15)

and we apply this inequality and (4.13) to the expression (4.11) to obtain

$\displaystyle\underline{c}\,(T-t)\leq
v_{x}(t,x)\leq\bar{c}\,(T-t)\quad\textrm{for }x\in(0,1),$ (4.16)

where $\underline{c}$ and $\bar{c}$ are constants that only depend on
$\mu,r,\sigma,\lambda,\bar{\epsilon},\underline{\epsilon},T$.

(i) Suppose that $\bar{\epsilon}>0$ (the case of $\underline{\epsilon}>0$ can
be treated similarly). If $\bar{c}\leq 0$, then we choose $t_{0}=0$ and easily
observe that

$\displaystyle\lim_{x\downarrow 0}\tfrac{\partial}{\partial
x}\Big{(}v(t,x)-\ln(1+\bar{\epsilon}x)\Big{)}\leq 0\quad\textrm{for
}t\in[t_{0},T).$ (4.17)

If $\bar{c}>0$, then (4.16) implies that (4.17) holds with
$t_{0}=\left(T-\frac{\bar{\epsilon}}{\bar{c}}\right)^{+}$. Therefore, in any
case, we can choose $t_{0}\in[0,T)$ that satisfies (4.17). By Lemma 4.1 and
(4.17), we conclude $\underline{y}(t)=0$ for $t\in[t_{0},T)$.

(ii) Suppose that $0<y_{\infty}$ (the case of $y_{\infty}<1$ can be treated
similarly). Let $t\in[0,T)$ be fixed. To check $\bar{y}(t)>0$, we observe from
(4.11) and the dominated convergence theorem that

$\begin{split}\lim_{x\downarrow
0}v_{x}(t,x)&=\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\lim_{x\downarrow
0}\left(\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right)\left(\mu-r-\sigma^{2}Y_{s}^{(t,x)}+\lambda
L_{y}(s,Y_{s}^{(t,x)})\right)\right]ds\\\
&=\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[e^{(\mu-r-\tfrac{1}{2}\sigma^{2})(s-t)+\sigma(B_{s}-B_{t})}\right]\\\
&\qquad\cdot\left(\mu-r+\lambda\Big{(}\bar{\epsilon}\cdot
1_{\left\\{\underline{y}(s)>0\right\\}}+\lim_{x\downarrow 0}v_{x}(s,x)\cdot
1_{\left\\{\underline{y}(s)=0<\bar{y}(s)\right\\}}-\underline{\epsilon}\cdot
1_{\left\\{\bar{y}(s)=0\right\\}}\Big{)}\right)ds\\\
&\geq\int_{t}^{T}e^{(\mu-r-\lambda)(s-t)}\left(\mu-r-\lambda\underline{\epsilon}\right)ds,\end{split}$
(4.18)

where the second equality is from (4.10) and (4.14), and the inequality is due
to the fact that
$-\frac{\underline{\epsilon}}{1-\underline{\epsilon}x}<v_{x}(t,x)$ for
$x<\bar{y}(t)$. Obviously, the last integral in (4.18) is nonnegative if
$\mu-r-\lambda\underline{\epsilon}\geq 0$. If
$\mu-r-\lambda\underline{\epsilon}<0$, then $\mu-r-\lambda<0$ due to
$\underline{\epsilon}\in[0,1)$ and we observe that

$\int_{t}^{T}e^{(\mu-r-\lambda)(s-t)}\left(\mu-r-\lambda\underline{\epsilon}\right)ds=\frac{\mu-r-\lambda\underline{\epsilon}}{\mu-r-\lambda}\left(e^{(\mu-r-\lambda)(T-t)}-1\right)>-\underline{\epsilon},$

where we use $y_{\infty}>0$ for the inequality. The above inequality and
(4.18) imply

$\lim_{x\downarrow 0}\tfrac{\partial}{\partial
x}\Big{(}v(t,x)-\ln(1-\underline{\epsilon}x)\Big{)}>0,$

and we conclude $\bar{y}(t)>0$ by Lemma 4.1.

(iii) Suppose that $0<y_{\infty}<1$ and $\bar{\epsilon}>0$ (the case of
$\underline{\epsilon}>0$ can be treated similarly). According to (ii),
$\underline{y}(t)<1$ and $\bar{y}(t)>0$. Therefore, there are only two
possibilities: $\underline{y}(t)=0$ or $0<\underline{y}(t)<1$. In case
$\underline{y}(t)=0$, we immediately obtain $\underline{y}(t)<\bar{y}(t)$
since $\bar{y}(t)>0$. In case $0<\underline{y}(t)<1$, by Lemma 4.1, we have

$v_{x}(t,\underline{y}(t))=\frac{\bar{\epsilon}}{1+\bar{\epsilon}\,\underline{y}(t)}>\frac{-\underline{\epsilon}}{1-\underline{\epsilon}\,\underline{y}(t)},$

and this inequality, together with (4.3) and (4.4), implies
$\underline{y}(t)<\bar{y}(t)$.

(iv) Suppose that $0<y_{\infty}$ and $\bar{\epsilon}=0$ (the case of
$y_{\infty}<1$ and $\underline{\epsilon}=0$ can be treated similarly). Lemma
4.1 implies that

$\displaystyle\underline{y}(t)>0\quad\textrm{if and only
if}\quad\lim_{x\downarrow 0}v_{x}(t,x)>0.$ (4.19)

The second equality in (4.18) can be written as

$\displaystyle\lim_{x\downarrow 0}v_{x}(t,x)$
$\displaystyle=\int_{t}^{T}e^{(\mu-r-\lambda)(s-t)}\left(\mu-r+\lambda\lim_{x\downarrow
0}v_{x}(s,x)\cdot
1_{\left\\{\underline{y}(s)=0<\bar{y}(s)\right\\}}\right)ds,$ (4.20)

where we substitute $\bar{\epsilon}=0$ and use $\bar{y}(s)>0$ by (ii). Define
$t^{*}$ as

$\displaystyle t^{*}:=\inf\left\\{t\in[0,T):\,\,\lim_{x\downarrow
0}v_{x}(s,x)>0\textrm{ for all }s\in[t,T)\right\\}.$ (4.21)

Due to (4.16) and the strict positivity of $\mu-r$, for $t$ close enough to
$T$, the integrand in (4.20) is strictly positive and $\lim_{x\downarrow
0}v_{x}(t,x)>0$. Therefore, the set in (4.21) is non-empty and $t^{*}<T$.
Suppose that $t^{*}>0$. For $0\leq\delta\leq t^{*}$, we apply (4.21) and
(4.19) to expression (4.20) and obtain

$\displaystyle\lim_{x\downarrow 0}v_{x}(t^{*}-\delta,x)$
$\displaystyle>\int_{t^{*}}^{T}e^{(\mu-r-\lambda)(s-t^{*}+\delta)}\left(\mu-r\right)ds$
$\displaystyle\quad+\int_{t^{*}-\delta}^{t^{*}}e^{(\mu-r-\lambda)(s-t^{*}+\delta)}\left(\mu-r+\lambda\lim_{x\downarrow
0}v_{x}(s,x)\cdot
1_{\left\\{\underline{y}(s)=0<\bar{y}(s)\right\\}}\right)ds.$

The first integral above is greater than a strictly positive number
independent of $\delta$, and the second integral can be made arbitrary close
to zero as $\delta\downarrow 0$, due to (4.16). Therefore, there exists
$\delta^{*}\in(0,t^{*}]$ such that $\lim_{x\downarrow
0}v_{x}(t^{*}-\delta,x)>0$ for all $\delta\in[0,\delta^{*}]$. Then,
$\lim_{x\downarrow 0}v_{x}(s,x)>0$ for all $s\in[t^{*}-\delta^{*},T)$, and
this contradicts to the definition of $t^{*}$ in (4.21). Therefore, we
conclude that $t^{*}=0$, and now (4.19) implies that $\underline{y}(t)>0$ for
$t\in(0,T)$. Lastly, $\underline{y}(0)>0$ is obtained by (4.19) and (4.20). ∎

######  Remark 4.5.

Straightforward interpretations of Proposition 4.4 are as follows:
(i) The existence of the transaction costs for selling (buying, resp.) the
stock makes the investor not to sell (buy, resp.) the stock when it is close
to the terminal time. For short period of time, the benefit of rebalancing is
small.
(ii) If $0<y_{\infty}$ ($y_{\infty}<1$, resp), then the investor never
rebalances to the zero-holding of the stock (bond, resp.). However, if the
initial holding of the stock (bond, resp.) is zero, then the investor may not
try to leave the state of zero-holding of the stock (bond, resp.), depending
on the size of the transaction costs.
(iii) If $0<y_{\infty}<1$ and there exist transaction costs, then the no-trade
interval is non-trivial (with strictly positive length) all the time.
(iv) If $0<y_{\infty}$ ($y_{\infty}<1$, resp.) and there is no cost for buying
(selling, resp.) the stock, then just holding the bond (stock, resp.) and
setting zero balance in the stock (bond, resp.) is suboptimal, even with the
search frictions and transaction costs.

######  Remark 4.6.

Obviously, if $y_{\infty}\leq 0$ ($y_{\infty}\geq 1$, resp.), then there is no
reason for buying (selling, resp.) the stock, so $\underline{y}(t)=0$
($\bar{y}(t)=1$, resp.) for $t\in[0,T)$.

##  5\. Asymptotic analysis

In this section, we provide asymptotic analysis for small transaction costs.
To be specific, we focus on the first order approximation of the no-trade
region and the value function with respect to the transaction cost parameter
around zero. To consider non-trivial cases (see Remark 4.6), we assume that
the Merton fraction $y_{\infty}$ is between zero and one, and we set
$\bar{\epsilon}=\underline{\epsilon}$ for convenience.

######  Assumption 5.1.

In this section, we assume that $0<y_{\infty}<1$ and
$\bar{\epsilon}=\underline{\epsilon}=\epsilon$ for $\epsilon\in[0,1)$.

######  Notation 5.2.

(Current section only)
(i) To emphasize their dependence on the transaction cost parameter
$\epsilon$, we denote $v,\underline{y},\bar{y},\hat{y},L,L_{y}$ by
$v^{\epsilon},\underline{y}^{\epsilon},\bar{y}^{\epsilon},\hat{y}^{\epsilon},L^{\epsilon},L_{y}^{\epsilon}$.
In particular, when $\epsilon=0$, they are denoted by
$v^{0},\underline{y}^{0},\bar{y}^{0},\hat{y}^{0},L^{0},L_{y}^{0}$.
(ii) Under Assumption 5.1, Lemma 4.1 and Proposition 4.4 imply that

$\displaystyle
0<\hat{y}^{0}(t,x)=\underline{y}^{0}(t)=\bar{y}^{0}(t)<1\quad\textrm{for}\quad(t,x)\in[0,T)\times[0,1].$
(5.1)

In words, $\hat{y}^{0}(t,x)$ is independent of variable $x$ (just a function
of $t$) and its value equals $\underline{y}^{0}(t)$ and $\bar{y}^{0}(t)$. For
convenience, we abuse notation and write $\hat{y}^{0}(t)$ for
$\hat{y}^{0}(t,x)$ (i.e.,
$\hat{y}^{0}(t)=\underline{y}^{0}(t)=\bar{y}^{0}(t)$).

We start with the technical lemma that is used in the proof of the asymptotic
result.

######  Lemma 5.3.

(i) Let $\epsilon=0$. For $t\in[0,T)$,

$\displaystyle\quad v_{x}^{0}(t,\hat{y}^{0}(t))=0,$ (5.2)
$\displaystyle\sup_{x\in(0,1)}v_{xx}^{0}(t,x)<0.$ (5.3)

(ii) Let $F:[0,T)\times(0,1)\to{\mathbb{R}}$ be defined as

$\displaystyle
F(t,x):=\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\left(\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right)\cdot\operatorname{sgn}\left(\hat{y}^{0}(s)-Y_{s}^{(t,x)}\right)\right]ds.$
(5.4)

Then, for $t\in[0,T)$,

$\displaystyle-1<F(t,\hat{y}^{0}(t))<1.$ (5.5)

(iii) Suppose that $x_{0}\in(0,1)$ and $\,\,\lim_{\epsilon\downarrow
0}x_{\epsilon}=x_{0}$. Then, for $t\in[0,T)$,

$\displaystyle\lim_{\epsilon\downarrow 0}v_{x}^{\epsilon}(t,x_{\epsilon})$
$\displaystyle=v_{x}^{0}(t,x_{0}),$ (5.6)
$\displaystyle\lim_{\epsilon\downarrow 0}v_{xx}^{\epsilon}(t,x_{\epsilon})$
$\displaystyle=v_{xx}^{0}(t,x_{0}).$ (5.7)

###### Proof.

See Appendix. ∎

The following theorem provides the first order approximation of the no-trade
boundaries.

######  Theorem 5.4.

For $t\in[0,T)$,

$\displaystyle\underline{y}^{\epsilon}(t)=\hat{y}^{0}(t)-\frac{F(t,\hat{y}^{0}(t))-1}{v_{xx}^{0}(t,\hat{y}^{0}(t))}\cdot\epsilon+o(\epsilon),$
(5.8)
$\displaystyle\bar{y}^{\epsilon}(t)=\hat{y}^{0}(t)-\frac{F(t,\hat{y}^{0}(t))+1}{v_{xx}^{0}(t,\hat{y}^{0}(t))}\cdot\epsilon+o(\epsilon),$
(5.9)

where $F$ is defined in (5.4). In particular, for small enough $\epsilon>0$,
we have

$\displaystyle
0<\underline{y}^{\epsilon}(t)<\hat{y}^{0}(t)<\bar{y}^{\epsilon}(t)<1.$ (5.10)

###### Proof.

For $(t,x)\in[0,T)\times(0,1)$, the expression (4.11) implies that

$\begin{split}\left|v_{x}^{\epsilon}(t,x)-v_{x}^{0}(t,x)\right|&=\left|\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\left(\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right)L_{y}^{\epsilon}(s,Y_{s}^{(t,x)})\right]ds\right|\\\
&\leq\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\left(\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right)\right]ds\cdot\tfrac{\epsilon}{1-\epsilon}\\\
&\leq\lambda T\cdot
e^{(\mu-r+\sigma^{2})T}\cdot\tfrac{\epsilon}{1-\epsilon},\end{split}$ (5.11)

where the first inequality is due to (4.13) and the positivity of
$\frac{\partial}{\partial x}Y_{s}^{(t,x)}$ in (4.14), and the second
inequality is due to (4.15) and Assumption 5.1.

Proposition 4.4 and Assumption 5.1 imply that $\underline{y}^{\epsilon}(t)<1$
and $\bar{y}^{\epsilon}(t)>0$ for $t\in[0,T)$. By Lemma 4.1, if
$0<\underline{y}^{\epsilon}(t)<1$, then
$v_{x}^{\epsilon}(t,\underline{y}^{\epsilon}(t))=\frac{\epsilon}{1+\epsilon\cdot\underline{y}^{\epsilon}(t)}$,
and if $\underline{y}^{\epsilon}(t)=0$, then $-\frac{\epsilon}{1-\epsilon
x}<v_{x}^{\epsilon}(t,x)<\frac{\epsilon}{1+\epsilon x}$ for
$x\in(0,\bar{y}^{\epsilon}(t))$. In any case, we have

$\displaystyle\left|v_{x}^{\epsilon}(t,\underline{y}^{\epsilon}(t))\right|\leq\tfrac{\epsilon}{1-\epsilon},$
(5.12)

where $v_{x}^{\epsilon}(t,0):=\lim_{x\downarrow 0}v_{x}^{\epsilon}(t,x)$ is
well-defined (see (4.18) for details) for the case of
$\underline{y}^{\epsilon}(t)=0$. We use the mean value theorem and (5.2) to
obtain

$\begin{split}\inf_{x\in(0,1)}\left|v_{xx}^{0}(t,x)\right|\cdot\left|\underline{y}^{\epsilon}(t)-\hat{y}^{0}(t)\right|&\leq\left|v_{x}^{0}(t,\underline{y}^{\epsilon}(t))-v_{x}^{0}(t,\hat{y}^{0}(t))\right|\\\
&\leq\left|v_{x}^{0}(t,\underline{y}^{\epsilon}(t))-v_{x}^{\epsilon}(t,\underline{y}^{\epsilon}(t))\right|+\left|v_{x}^{\epsilon}(t,\underline{y}^{\epsilon}(t))\right|.\end{split}$
(5.13)

The above inequality, together with (5.3), (5.11), and (5.12), we conclude
that

$\displaystyle\underline{y}^{\epsilon}(t)-\hat{y}^{0}(t)=O(\epsilon)\quad\textrm{for}\quad
t\in[0,T).$ (5.14)

By the same way, we also obtain

$\displaystyle\bar{y}^{\epsilon}(t)-\hat{y}^{0}(t)=O(\epsilon)\quad\textrm{for}\quad
t\in[0,T).$ (5.15)

The expression of $L_{y}^{\epsilon}$ in (4.10), together with (5.14) and
(5.15), implies the following limit:

$\displaystyle\lim_{\epsilon\downarrow
0}\frac{L_{y}^{\epsilon}(t,x)}{\epsilon}=\begin{cases}1,&\textrm{if
}x\in(0,\hat{y}^{0}(t))\\\ -1,&\textrm{if }x\in(\hat{y}^{0}(t),1)\\\
\end{cases}\quad\textrm{for}\quad t\in[0,T).$ (5.16)

For $(s,x)\in(t,T)\times(0,1)$, the observation
${\mathbb{P}}\left(Y_{s}^{(t,x)}=\hat{y}^{0}(t)\right)=0$ and (5.16) produce

$\displaystyle\lim_{\epsilon\downarrow
0}\frac{L_{y}^{\epsilon}(s,Y_{s}^{(t,x)})}{\epsilon}=\operatorname{sgn}\left(\hat{y}^{0}(t)-Y_{s}^{(t,x)}\right)\quad\textrm{almost
surely.}$ (5.17)

For $(t,x)\in[0,T)\times(0,1)$, the limit (5.17) and the inequalities (4.13)
and (4.14) enable us to use the dominated convergence theorem to obtain

$\begin{split}\frac{v_{x}^{\epsilon}(t,x)-v_{x}^{0}(t,x)}{\epsilon}&=\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\left(\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right)\tfrac{L_{y}^{\epsilon}(s,Y_{s}^{(t,x)})}{\epsilon}\right]ds\xrightarrow[\epsilon\downarrow
0]{}F(t,x).\end{split}$ (5.18)

The inequality $0<\hat{y}^{0}(t)<1$ and (5.14) imply that
$0<\underline{y}^{\epsilon}(t)<1$ for small enough $\epsilon>0$. Hence, by
Lemma 4.1, we obtain

$\displaystyle
v_{x}^{\epsilon}(t,\underline{y}^{\epsilon}(t))=\frac{\epsilon}{1+\epsilon\,\underline{y}^{\epsilon}(t)}\quad\textrm{for
small enough $\epsilon>0$.}\quad$ (5.19)

By the mean value theorem, there exists $k(\epsilon)$ such that

$\displaystyle
v_{x}^{\epsilon}(t,\underline{y}^{\epsilon}(t))-v_{x}^{\epsilon}(t,\hat{y}^{0}(t))=v_{xx}^{\epsilon}(t,k(\epsilon))\left(\underline{y}^{\epsilon}(t)-\hat{y}^{0}(t)\right)\quad\textrm{and}\quad\lim_{\epsilon\downarrow
0}k(\epsilon)=\hat{y}^{0}(t).$ (5.20)

Now we obtain (5.8) as follows:

$\begin{split}\lim_{\epsilon\downarrow
0}\frac{\underline{y}^{\epsilon}(t)-\hat{y}^{0}(t)}{\epsilon}&=\lim_{\epsilon\downarrow
0}\frac{v_{x}^{\epsilon}(t,\underline{y}^{\epsilon}(t))-v_{x}^{\epsilon}(t,\hat{y}^{0}(t))}{\epsilon\cdot
v_{xx}^{\epsilon}(t,k(\epsilon))}\\\ &=\lim_{\epsilon\downarrow
0}\frac{1}{v_{xx}^{\epsilon}(t,k(\epsilon))}\left(\frac{1}{1+\epsilon\,\underline{y}^{\epsilon}(t)}-\frac{v_{x}^{\epsilon}(t,\hat{y}^{0}(t))-v_{x}^{0}(t,\hat{y}^{0}(t))}{\epsilon}\right)\\\
&=-\frac{F(t,\hat{y}^{0}(t))-1}{v_{xx}^{0}(t,\hat{y}^{0}(t))},\end{split}$
(5.21)

where the first equality is from (5.20), the second equality is due to (5.19)
and (5.2), and the third equality is due to (5.18) and (5.7) with the limit in
(5.20). We also obtain (5.9) by the same way.

Due to (5.3) and (5.5), we observe that
$-\frac{F(t,\hat{y}^{0}(t))-1}{v_{xx}^{0}(t,\hat{y}^{0}(t))}<0$ and
$-\frac{F(t,\hat{y}^{0}(t))+1}{v_{xx}^{0}(t,\hat{y}^{0}(t))}>0$. Therefore,
(5.8) and (5.9) imply that (5.10) holds for small enough $\epsilon>0$. ∎

$\begin{array}[]{cc}\includegraphics[width=195.12767pt]{oyuy_asymp.pdf}&\,\,\,\includegraphics[width=195.12767pt]{value_asymp.pdf}\end{array}$

Figure 2. The left graph shows $\underline{y}^{\epsilon}(t)$ and
$\bar{y}^{\epsilon}(t)$ as functions of $\epsilon$, where the dashed lines are
the linear approximations of them in Theorem 5.4, i.e.,
$\hat{y}^{0}(t)-\frac{F(t,\hat{y}^{0}(t))-1}{v_{xx}^{0}(t,\hat{y}^{0}(t))}\cdot\epsilon$
and
$\hat{y}^{0}(t)-\frac{F(t,\hat{y}^{0}(t))-1}{v_{xx}^{0}(t,\hat{y}^{0}(t))}\cdot\epsilon$.
The right graph describes $v^{\epsilon}(t,\hat{y}^{0}(t))$ as a function of
$\epsilon$, where the dashed line is the linear approximation of it in Theorem
5.5, i.e.,
$v^{0}(t,\hat{y}^{0}(t))-\left(G(t)+\lambda\int_{t}^{T}G(s)ds\right)\cdot\epsilon$.
In both graphs, the parameters are
$\mu=0.4,\,r=0.1,\sigma=1,\,\lambda=3,\,T=1$, and $t=0.75$.

The following theorem provides the first order approximation of the value
function.

######  Theorem 5.5.

For $t\in[0,T)$,

$\displaystyle
v^{\epsilon}(t,\hat{y}^{0}(t))=v^{0}(t,\hat{y}^{0}(t))-\left(G(t)+\lambda\int_{t}^{T}G(s)ds\right)\cdot\epsilon+o(\epsilon),$
(5.22)

where $G:[0,T]\to{\mathbb{R}}$ is defined as

$\displaystyle
G(t):=\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\right]ds$
(5.23)

###### Proof.

For $\epsilon>0$, we define a function
$\psi^{\epsilon}:[0,T)\rightarrow{\mathbb{R}}$ as

$\displaystyle\psi^{\epsilon}(t):=\frac{v^{\epsilon}(t,\hat{y}^{0}(t))-v^{0}(t,\hat{y}^{0}(t))}{\epsilon}.$
(5.24)

Using (4.8) and (4.12), we obtain

$\begin{split}\psi^{\epsilon}(t)&=\lambda\int_{t}^{T}e^{-\lambda(s-t)}\psi^{\epsilon}(s)ds+\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[I_{1}^{(t,s,\epsilon)}+I_{2}^{(t,s,\epsilon)}+I_{3}^{(t,s,\epsilon)}\right]ds,\end{split}$
(5.25)

where the random variables
$I_{1}^{(t,s,\epsilon)},I_{2}^{(t,s,\epsilon)},I_{3}^{(t,s,\epsilon)}$ are
defined as

$\begin{split}I_{1}^{(t,s,\epsilon)}&:=\left(\tfrac{v^{\epsilon}(s,\underline{y}^{\epsilon}(s))-v^{\epsilon}(s,\hat{y}^{0}(s))}{\epsilon}-\frac{1}{\epsilon}\ln\left(\tfrac{1+\epsilon\,\underline{y}^{\epsilon}(s)}{1+\epsilon\,Y_{s}^{(t,\hat{y}^{0}(t))}}\right)\right)\cdot
1_{\left\\{Y_{s}^{(t,\hat{y}^{0}(t))}\leq\underline{y}^{\epsilon}(s)\right\\}},\\\
I_{2}^{(t,s,\epsilon)}&:=\left(\tfrac{v^{\epsilon}(s,Y_{s}^{(t,\hat{y}^{0}(t))})-v^{\epsilon}(s,\hat{y}^{0}(s))}{\epsilon}\right)\cdot
1_{\left\\{\underline{y}^{\epsilon}(s)<Y_{s}^{(t,\hat{y}^{0}(t))}<\bar{y}^{\epsilon}(s)\right\\}},\\\
I_{3}^{(t,s,\epsilon)}&:=\left(\tfrac{v^{\epsilon}(s,\bar{y}^{\epsilon}(s))-v^{\epsilon}(s,\hat{y}^{0}(s))}{\epsilon}-\frac{1}{\epsilon}\ln\left(\tfrac{1-\epsilon\,\bar{y}^{\epsilon}(s)}{1-\epsilon\,Y_{s}^{(t,\hat{y}^{0}(t))}}\right)\right)\cdot
1_{\left\\{\bar{y}^{\epsilon}(s)\leq
Y_{s}^{(t,\hat{y}^{0}(t))}\right\\}}.\end{split}$ (5.26)

The mean value theorem produces

$\displaystyle\left|\tfrac{1}{\epsilon}\ln\left(\tfrac{1+\epsilon\,\underline{y}^{\epsilon}(s)}{1+\epsilon\,x}\right)\right|\leq
1\quad\textrm{and}\quad\left|\tfrac{1}{\epsilon}\ln\left(\tfrac{1-\epsilon\,\bar{y}^{\epsilon}(s)}{1-\epsilon\,x}\right)\right|\leq\tfrac{1}{1-\epsilon}\quad\textrm{for
}x\in(0,1).$ (5.27)

By Assumption 5.1 and Proposition 4.4 (ii), we obtain
$\underline{y}^{\epsilon}(s)<1$ and $\bar{y}^{\epsilon}(s)>0$ for $s\in[0,T)$.
The definition of $Y_{s}^{(t,x)}$ in (4.9) indicates that $0<Y_{s}^{(t,x)}<1$.
Therefore,

$\displaystyle
1_{\left\\{Y_{s}^{(t,\hat{y}^{0}(t))}\leq\underline{y}^{\epsilon}(s)\right\\}}\leq
1_{\big{\\{}0<\underline{y}^{\epsilon}(s)<1\big{\\}}},\quad
1_{\left\\{\bar{y}^{\epsilon}(s)\leq Y_{s}^{(t,\hat{y}^{0}(t))}\right\\}}\leq
1_{\big{\\{}0<\bar{y}^{\epsilon}(s)<1\big{\\}}}.$ (5.28)

The concavity of $v^{\epsilon}$ on $x$ variable (see Proposition 3.5) and
(5.27) and (5.28) imply

$\begin{split}\left|I_{1}^{(t,s,\epsilon)}\right|&\leq\left(\max\left\\{\left|\tfrac{v_{x}^{\epsilon}(s,\underline{y}^{\epsilon}(s))}{\epsilon}\right|,\,\left|\tfrac{v_{x}^{\epsilon}(s,\hat{y}^{0}(s))}{\epsilon}\right|\right\\}\cdot\left|\underline{y}^{\epsilon}(s)-\hat{y}^{0}(s)\right|+1\right)\cdot
1_{\left\\{0<\underline{y}^{\epsilon}(s)<1\right\\}},\\\
\left|I_{2}^{(t,s,\epsilon)}\right|&\leq\max\left\\{\left|\tfrac{v_{x}^{\epsilon}(s,Y_{s}^{(t,\hat{y}^{0}(t))})}{\epsilon}\right|,\,\left|\tfrac{v_{x}^{\epsilon}(s,\hat{y}^{0}(s))}{\epsilon}\right|\right\\}\cdot\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\cdot
1_{\left\\{\underline{y}^{\epsilon}(s)<Y_{s}^{(t,\hat{y}^{0}(t))}<\bar{y}^{\epsilon}(s)\right\\}},\\\
\left|I_{3}^{(t,s,\epsilon)}\right|&\leq\left(\max\left\\{\left|\tfrac{v_{x}^{\epsilon}(s,\bar{y}^{\epsilon}(s))}{\epsilon}\right|,\,\left|\tfrac{v_{x}^{\epsilon}(s,\hat{y}^{0}(s))}{\epsilon}\right|\right\\}\cdot\left|\bar{y}^{\epsilon}(s)-\hat{y}^{0}(s)\right|+\frac{1}{1-\epsilon}\right)\cdot
1_{\left\\{0<\bar{y}^{\epsilon}(s)<1\right\\}}.\end{split}$ (5.29)

Combining (5.11) and (5.2), we obtain

$\displaystyle\left|v_{x}^{\epsilon}(s,\hat{y}^{0}(s))\right|\leq\lambda
T\cdot e^{(\mu-r+\sigma^{2})T}\cdot\tfrac{\epsilon}{1-\epsilon}.$ (5.30)

By Lemma 4.1, we have the following inequalities:

$\begin{split}&\left|v_{x}^{\epsilon}(s,\underline{y}^{\epsilon}(s))\right|\cdot
1_{\left\\{0<\underline{y}^{\epsilon}(s)<1\right\\}}\leq\epsilon,\quad\left|v_{x}^{\epsilon}(s,\bar{y}^{\epsilon}(s))\right|\cdot
1_{\left\\{0<\bar{y}^{\epsilon}(s)<1\right\\}}\leq\tfrac{\epsilon}{1-\epsilon},\\\
&\left|v_{x}^{\epsilon}(s,x)\right|\cdot
1_{\left\\{\underline{y}^{\epsilon}(s)<x<\bar{y}^{\epsilon}(s)\right\\}}\leq\tfrac{\epsilon}{1-\epsilon}\quad\textrm{for}\quad
x\in(0,1).\end{split}$ (5.31)

Now we apply (5.30) and (LABEL:vx_ye_bound) to (5.29) and obtain

$\begin{split}\left|I_{1}^{(t,s,\epsilon)}+I_{2}^{(t,s,\epsilon)}+I_{3}^{(t,s,\epsilon)}\right|\leq
C\frac{1}{1-\epsilon},\end{split}$ (5.32)

where $C$ is a constant independent of $(t,s,\epsilon)$. Also, Theorem 5.4,
the mean value theorem, and (5.6) imply that

$\displaystyle\lim_{\epsilon\downarrow
0}\left(I_{1}^{(t,s,\epsilon)}+I_{2}^{(t,s,\epsilon)}+I_{3}^{(t,s,\epsilon)}\right)=-\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\quad\textrm{almost
surely.}$ (5.33)

The uniform boundedness in (5.32) and the convergence in (5.33) allow us to
apply the dominated convergence theorem and conclude

$\displaystyle\lim_{\epsilon\downarrow
0}J^{\epsilon}(t)=-G(t)\quad\textrm{and}\quad\lim_{\epsilon\downarrow
0}\int_{t}^{T}J^{\epsilon}(s)ds=-\int_{t}^{T}G(s)ds,$ (5.34)

where $G$ is defined in (5.23) and the function
$J^{\epsilon}:[0,T]\to{\mathbb{R}}$ is defined as

$J^{\epsilon}(t):=\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[I_{1}^{(t,s,\epsilon)}+I_{2}^{(t,s,\epsilon)}+I_{3}^{(t,s,\epsilon)}\right]ds.$

We rewrite (5.25) as

$\displaystyle J^{\epsilon}(t)$
$\displaystyle=\psi^{\epsilon}(t)-\lambda\int_{t}^{T}e^{-\lambda(s-t)}\psi^{\epsilon}(s)ds=-\frac{d}{dt}\left(\int_{t}^{T}e^{-\lambda(s-t)}\psi^{\epsilon}(s)ds\right).$

We integrate both sides above to obtain

$\displaystyle\int_{t}^{T}e^{-\lambda(s-t)}\psi^{\epsilon}(s)ds=\int_{t}^{T}J^{\epsilon}(s)ds.$
(5.35)

Using (5.35), the equation (5.25) becomes

$\displaystyle\psi^{\epsilon}(t)$
$\displaystyle=J^{\epsilon}(t)+\lambda\int_{t}^{T}J^{\epsilon}(s)ds.$ (5.36)

Finally, (5.34) and (5.36) imply (5.22). ∎

######  Corollary 5.6.

For $(t,x)\in[0,T)\times(0,1)$,

$\displaystyle
v^{\epsilon}(t,x)=v^{0}(t,x)-\left(G(t)+\lambda\int_{t}^{T}G(s)ds-\int_{\hat{y}^{0}(t)}^{x}F(t,\eta)d\eta\right)\cdot\epsilon+o(\epsilon),$
(5.37)

where $F$ and $G$ are defined in (5.4) and (5.23).

###### Proof.

The inequality (5.11) and the limit (5.18) allow us to use the dominated
convergence theorem, and we obtain

$\displaystyle\lim_{\epsilon\downarrow
0}\int_{\hat{y}^{0}(t)}^{x}\frac{v_{x}^{\epsilon}(t,\eta)-v_{x}^{0}(t,\eta)}{\epsilon}d\eta=\int_{\hat{y}^{0}(t)}^{x}F(t,\eta)d\eta.$

Using the above limit and Theorem 5.5, we obtain

$\displaystyle\frac{v^{\epsilon}(t,x)-v^{0}(t,x)}{\epsilon}$
$\displaystyle=\frac{v^{\epsilon}(t,\hat{y}^{0}(t))-v^{0}(t,\hat{y}^{0}(t))}{\epsilon}+\int_{\hat{y}^{0}(t)}^{x}\frac{v_{x}^{\epsilon}(t,\eta)-v_{x}^{0}(t,\eta)}{\epsilon}d\eta$
$\displaystyle\xrightarrow[\epsilon\downarrow
0]{}-\left(G(t)+\lambda\int_{t}^{T}G(s)ds\right)+\int_{\hat{y}^{0}(t)}^{x}F(t,\eta)d\eta.$

The above limit is equivalent to (5.37). ∎

Not surprisingly, Theorem 5.4 and Theorem 5.5 indicate that the no-trade
region widens and the value function diminishes as the transaction cost
parameter $\epsilon$ increases. See Figure 2 for numerical illustrations.

Our next task is to investigate some intertwined effects of the search
frictions and transaction costs on the no-trade region and value function. To
be specific, for two different search friction parameters
$\lambda_{1}<\lambda_{2}$, we would like to compare the magnitude of the
widening (deminishing, resp.) effects of the transaction costs on the no-trade
region (the value function, resp.). Figure 3 numerically describes this
comparison result. The following technical lemma turns out to be useful to the
proof of this comparison.

######  Lemma 5.7.

(i) There is a constant $C$ such that

$\displaystyle\left|\hat{y}^{0}(t)-y_{\infty}\right|\leq\tfrac{C}{\lambda}\quad\textrm{for}\quad(t,\lambda)\in[0,T)\times[1,\infty).$
(5.38)

(ii) $\frac{\partial}{\partial\lambda}v_{x}^{0}(t,x)$ and
$\frac{\partial}{\partial\lambda}v_{xx}^{0}(t,x)$ exist for
$(t,x)\in[0,T]\times(0,1)$, and

$\begin{split}&\lim_{\lambda\to\infty}\lambda^{2}\tfrac{\partial}{\partial\lambda}v_{x}^{0}(t,x)\Big{|}_{x=\hat{y}^{0}(t)}=0,\quad\lim_{\lambda\to\infty}\lambda^{2}\tfrac{\partial}{\partial\lambda}v_{xx}^{0}(t,x)\Big{|}_{x=\hat{y}^{0}(t)}=\sigma^{2},\\\
&\lim_{\lambda\to\infty}\lambda
v_{xx}^{0}(t,\hat{y}^{0}(t))=-\sigma^{2},\qquad\,\,\,\,\lim_{\lambda\to\infty}\lambda
v_{xxx}^{0}(t,\hat{y}^{0}(t))=0.\end{split}$ (5.39)

(iii) $\frac{\partial}{\partial\lambda}\hat{y}^{0}(t)$ exists for $t\in[0,T)$,
and there is a constant $C$ such that

$\begin{split}\left|\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(t)\right|\leq\tfrac{C}{\lambda^{2}}\quad\textrm{for}\quad(t,\lambda)\in[0,T)\times[1,\infty).\end{split}$
(5.40)

(iv)
$\frac{\partial}{\partial\lambda}{\mathbb{E}}\left[\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\right]$
exists for $0\leq t\leq s\leq T$, and there is a constant $C$ such that

$\begin{split}\left|\tfrac{\partial}{\partial\lambda}\left({\mathbb{E}}\left[\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\right]\right)\right|\leq\tfrac{C}{\lambda^{2}}\quad\textrm{for}\quad(t,s,\lambda)\in[0,T)\times[t,T]\times[1,\infty).\end{split}$
(5.41)

(v) There is a constant $C$ such that

$\begin{split}\tfrac{{\mathbb{E}}\left[\left|Y_{s}^{(t,x)}-x\right|\right]}{\sqrt{s-t}}\leq
C\quad\textrm{for}\quad(t,s,x)\in[0,T)\times(t,T]\times(0,1),\end{split}$
(5.42)

and for $t\in[0,T)$,

$\begin{split}\lim_{s\downarrow
t}\tfrac{{\mathbb{E}}\left[\left|Y_{s}^{(t,x)}-x\right|\right]}{\sqrt{s-t}}=\sigma\sqrt{\tfrac{2}{\pi}}\,\,x(1-x).\end{split}$
(5.43)

###### Proof.

See Appendix. ∎

$\begin{array}[]{cc}\includegraphics[width=195.12767pt]{NT_compare.pdf}&\,\,\,\includegraphics[width=195.12767pt]{value_compare.pdf}\end{array}$

Figure 3. The left graph compares
$\bar{y}^{\epsilon}(t)-\underline{y}^{\epsilon}(t)$ for $\lambda_{1}=3$
(dashed line) and $\lambda_{2}=12$ (solid line), as functions of $\epsilon$.
The right graph compares
$v^{0}(t,\hat{y}^{0}(t))-v^{\epsilon}(t,\hat{y}^{0}(t))$ for $\lambda_{1}=3$
(dashed line) and $\lambda_{2}=12$ (solid line), as functions of $\epsilon$.
The other parameters are $\mu=0.4,\,r=0.1,\sigma=1,\,T=1$, and $t=0.75$.

######  Proposition 5.8.

For $t\in[0,T)$, there exists $\Lambda(t)>0$ such that
$\Lambda(t)<\lambda_{1}<\lambda_{2}$ implies

$\displaystyle\left\\{\begin{array}[]{c}\left(\bar{y}^{\epsilon}(t)-\underline{y}^{\epsilon}(t)\right)\big{|}_{\lambda=\lambda_{1}}<\,\,\,\left(\bar{y}^{\epsilon}(t)-\underline{y}^{\epsilon}(t)\right)\big{|}_{\lambda=\lambda_{2}}\\\
\left(v^{0}(t,\hat{y}^{0}(t))-v^{\epsilon}(t,\hat{y}^{0}(t))\right)\big{|}_{\lambda=\lambda_{1}}<\,\,\,\left(v^{0}(t,\hat{y}^{0}(t))-v^{\epsilon}(t,\hat{y}^{0}(t))\right)\big{|}_{\lambda=\lambda_{2}}\end{array}\right\\}\,\,\,\,\textrm{for
small enough $\epsilon>0$.}$ (5.46)

###### Proof.

Theorem 5.4 implies that

$\displaystyle\bar{y}^{\epsilon}(t)-\underline{y}^{\epsilon}(t)=-\frac{2}{v_{xx}^{0}(t,\hat{y}^{0}(t))}\cdot\epsilon+o(\epsilon).$

Therefore, to prove the first statement, it is enough to show that
$\frac{\partial}{\partial\lambda}\left(v_{xx}^{0}(t,\hat{y}^{0}(t))\right)>0$
for sufficiently large $\lambda$. Indeed, using (ii) and (iii) in Lemma 5.7,
we obtain

$\begin{split}\lambda^{2}\tfrac{\partial}{\partial\lambda}\left(v_{xx}^{0}(t,\hat{y}^{0}(t))\right)&=\lambda^{2}\tfrac{\partial}{\partial\lambda}v_{xx}^{0}(t,x)\big{|}_{x=\hat{y}^{0}(t)}+\lambda
v_{xxx}^{0}(t,\hat{y}^{0}(t))\cdot\lambda\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(t)\xrightarrow[\lambda\to\infty]{}\sigma^{2}>0.\end{split}$
(5.47)

To prove the second statement, it is enough check
$\frac{\partial}{\partial\lambda}\left(G(t)+\lambda\int_{t}^{T}G(s)ds\right)>0$
for sufficiently large $\lambda$, according to Theorem 5.5. For this purpose,
we focus on proving the following:

$\displaystyle\lim_{\lambda\rightarrow\infty}\sqrt{\lambda}\left(\tfrac{\partial}{\partial\lambda}\left(G(t)+\lambda\int_{t}^{T}G(s)ds\right)\right)>0.$
(5.48)

By (iv) in Lemma 5.7, we can take derivative inside of the expectations and
obtain

$\displaystyle\sqrt{\lambda}\left(\tfrac{\partial}{\partial\lambda}\left(G(t)+\lambda\int_{t}^{T}G(s)ds\right)\right)=\lambda^{\frac{1}{2}}\tfrac{\partial}{\partial\lambda}G(t)+\int_{t}^{T}\left(\lambda^{\frac{1}{2}}G(s)+\lambda^{\frac{3}{2}}\tfrac{\partial}{\partial\lambda}G(s)\right)ds,$
(5.49)

where the partial derivative $\frac{\partial}{\partial\lambda}G(t)$ can be
written as

$\displaystyle\tfrac{\partial}{\partial\lambda}G(t)=\int_{t}^{T}e^{-\lambda(s-t)}\left((1-\lambda(s-t)){\mathbb{E}}\left[\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\right]+\lambda\tfrac{\partial}{\partial\lambda}{\mathbb{E}}\left[\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\right]\right)ds.$

Observe that (i) and (v) in Lemma 5.7 imply

$\displaystyle{\mathbb{E}}\left[\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\right]\leq
C\left(\sqrt{s-t}+\tfrac{1}{\lambda}\right)\quad\textrm{for}\quad(t,s,\lambda)\in[0,T)\times(t,T]\times[1,\infty),$
(5.50)

for a constant $C$. We use (5.38), (5.41), (5.43), and (LABEL:lemma2_result)
to obtain the following limits:

$\displaystyle\lim_{\lambda\to\infty}\lambda^{\frac{1}{2}}G(s)=\tfrac{\sigma}{\sqrt{2}}\,y_{\infty}(1-y_{\infty}),\quad\lim_{\lambda\to\infty}\lambda^{\frac{3}{2}}\tfrac{\partial}{\partial\lambda}G(s)=-\tfrac{\sigma}{2\sqrt{2}}\,y_{\infty}(1-y_{\infty})\quad\textrm{for}\quad
s\in[0,T).$ (5.51)

We apply (5.41), (5.50), and (D.5) to the expressions of $G$ and
$\tfrac{\partial}{\partial\lambda}G$ to obtain the boundedness:

$\displaystyle\left|\lambda^{\frac{1}{2}}G(s)\right|+\left|\lambda^{\frac{3}{2}}\tfrac{\partial}{\partial\lambda}G(s)\right|\leq
C\quad\textrm{for}\quad(s,\lambda)\in[0,T]\times[1,\infty)$ (5.52)

Finally, (5.51) and (5.52) enable us to apply the dominated convergence
theorem to (5.49):

$\displaystyle\lim_{\lambda\to\infty}\sqrt{\lambda}\left(\tfrac{\partial}{\partial\lambda}\left(G(t)+\lambda\int_{t}^{T}G(s)ds\right)\right)=\tfrac{\sigma}{2\sqrt{2}}\,y_{\infty}(1-y_{\infty})(T-t)>0.$
(5.53)

Therefore, we conclude (5.48) and the proof is done. ∎

######  Remark 5.9.

Proposition 5.8 implies that the effects of the transaction costs are more
pronounced (more widening effect of the no-trade region and more diminishing
effect of the value function) in the market with less search frictions. For
better understanding of the result, it would be helpful to observe that the
following two extreme cases ($\lambda=0$ and $\lambda=\infty$) are consistent
with the result:

(i) If we consider an extreme case of $\lambda=0$ (i.e., no trading
opportunity), the transaction costs do not affect the investor anyway. In
words, in the market with very severe search frictions, the effects of the
transaction costs are negligible.

(ii) If we consider an extreme case of $\lambda=\infty$ (i.e., continuous
trading opportunities), it is well known that the width of the no-trade region
is the order of $\epsilon^{\frac{1}{3}}$ and the decrease of the value
function is the order of $\epsilon^{\frac{2}{3}}$. In contrast, when
$\lambda<\infty$, Theorem 5.4 and Theorem 5.5 say that both the width of the
no-trade region and the decrease of the value function are the order of
$\epsilon$. Obviously, for small enough $\epsilon$, we have
$\epsilon<\epsilon^{\frac{1}{3}}$ and $\epsilon<\epsilon^{\frac{2}{3}}$. This
implies that in the market with the search frictions ($\lambda<\infty$), the
effects of the transaction costs are less pronounces, compared to the market
with no search friction ($\lambda=\infty$).

##  6\. Conclusion

This paper presents an utility maximization problem of the terminal wealth, in
a market with two different types of the illiquidity: search frictions and
transaction costs. We show the existence of the solution to the HJB equation
that is regular enough, and provide the verification argument. The optimal
trading strategy is characterized by a no-trade region, where the boundary
points of the no-trade region are uniquely determined by the strict concavity
of the value function. In Proposition 4.4, we provide some conditions under
which the investor wants to achieve zero/positive stock/bond holdings. We
provide the asymptotic expansions of the boundaries of the no-trade region and
the value function for small transaction costs. We further show that reduction
of the search frictions amplifies the effects of the transaction costs (more
widening effect of the no-trade region and more diminishing effect of the
value function).

As a future research, we plan to investigate a joint limiting behavior of the
no-trade region and the value function for small transaction costs and large
arrival rate. For that purpose, we think it could be useful to consider some
stationary versions of the problem with a hope to obtain explicit formulas
independent of the time variable.

## Acknowledgement

This work was supported by the National Research Foundation of Korea (NRF)
grant funded by the Korea government (MSIT) (No. 2020R1C1C1A01014142 and No.
2021R1I1A1A01050679).

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##  Appendix A Proof of Lemma 3.1

Since the parabolic type PDE (3.2) is not uniformly elliptic, we change
variable as $x=h(z):=\frac{e^{z}}{1+e^{z}}$ and consider the PDE for
$v(t,h(z))$.

To handle the nonlinear term in the PDE, we first consider the following map
$\phi$ from $C_{b}([0,T]\times{\mathbb{R}})$ (equipped with the uniform norm)
to itself:

$\displaystyle\phi(f)(t,z):=\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[K_{f}(s,Z_{s}^{(t,z)})\right]ds,$
(A.1)

where for $(s,z)\in[t,T]\times{\mathbb{R}}$,

$\begin{split}Z_{s}^{(t,z)}&:=z+\left(\mu-r-\tfrac{\sigma^{2}}{2}\right)(s-t)+\sigma(B_{s}-B_{t}),\\\
K_{f}(s,z)&:=(\mu-r)h(z)+r-\tfrac{1}{2}\sigma^{2}h(z)^{2}\\\
&\qquad+\lambda\sup_{\zeta\in{\mathbb{R}}}\left(f(s,\zeta)-\ln\left(\tfrac{1+\bar{\epsilon}h(\zeta)}{1+\bar{\epsilon}h(z)}\right)1_{\\{z<\zeta\\}}-\ln\left(\tfrac{1-\underline{\epsilon}h(\zeta)}{1-\underline{\epsilon}h(z)}\right)1_{\\{z>\zeta\\}}\right).\end{split}$
(A.2)

We observe that for $f,g\in C_{b}([0,T]\times{\mathbb{R}})$,

$\begin{split}\|\phi(f)-\phi(g)\|_{\infty}&\leq\lambda\int_{t}^{T}e^{-\lambda(s-t)}\left(\sup_{\zeta\in{\mathbb{R}}}\left|f(s,\zeta)-g(s,\zeta)\right|\right)ds\\\
&\leq\left(1-e^{-\lambda T}\right)\|f-g\|_{\infty},\end{split}$ (A.3)

where the first inequality is due to the triangular inequality for supremum.
Therefore, the map $\phi$ in (A.1) is a contraction map and there exists a
unique $\hat{u}\in C_{b}([0,T]\times{\mathbb{R}})$ such that
$\phi(\hat{u})=\hat{u}$, by the Banach fixed point theorem.

Claim: For all $\delta\in(0,1)$, $K_{\hat{u}}\in
C^{\frac{\delta}{2},\delta}([0,T]\times{\mathbb{R}})$.

(Proof of Claim): We first check that $K_{\hat{u}}$ is bounded. Since
$0<h(z)<1$, we observe that

$\displaystyle|K_{\hat{u}}(t,z)|\leq|\mu|+2|r|+\tfrac{\sigma^{2}}{2}+\lambda\left(\|\hat{u}\|_{\infty}+\ln\left(\tfrac{1+\bar{\epsilon}}{1-\underline{\epsilon}}\right)\right),$
(A.4)

i.e., $\|K_{\hat{u}}\|_{\infty}<\infty$. Therefore, to prove the claim, it is
enough to check that $K_{\hat{u}}$ is uniformly Lipschitz with respect to $z$
variable and $\tfrac{1}{2}$-Hölder continuous with respect to $t$ variable.
For $\Delta>0$, we obtain the following inequalities:

$\begin{split}&|K_{\hat{u}}(t,z+\Delta)-K_{\hat{u}}(t,z)|\\\
&\leq(|\mu-r|+\sigma^{2})\Delta+\lambda\sup_{\zeta\in{\mathbb{R}}}\left|1_{\\{z+\Delta<\zeta\\}}\cdot\ln\left(\tfrac{1+\bar{\epsilon}h(z+\Delta)}{1+\bar{\epsilon}h(z)}\right)+1_{\\{z<\zeta\leq
z+\Delta\\}}\cdot\ln\left(\tfrac{1+\bar{\epsilon}h(\zeta)}{1+\bar{\epsilon}h(z)}\right)\right|\\\
&\quad+\lambda\sup_{\zeta\in{\mathbb{R}}}\left|1_{\\{z+\Delta>\zeta\\}}\cdot\ln\left(\tfrac{1-\underline{\epsilon}h(z+\Delta)}{1-\underline{\epsilon}h(z)}\right)-1_{\\{z\leq\zeta<z+\Delta\\}}\cdot\ln\left(\tfrac{1-\underline{\epsilon}h(\zeta)}{1-\underline{\epsilon}h(z)}\right)\right|\\\
&\leq\left(|\mu-r|+\sigma^{2}+\lambda\left(\bar{\epsilon}+\tfrac{\underline{\epsilon}}{1-\underline{\epsilon}}\right)\right)\Delta,\end{split}$
(A.5)

where we used the mean value theorem and the bounds $0<h<1$ and
$0<h^{\prime}<1$. We can treat $\Delta<0$ by the same way, and conclude that
$K_{\hat{u}}$ is uniformly Lipschitz with respect to $z$.

For $\Delta>0$, using (LABEL:K_Lip) and the mean value theorem, we observe
that

$\begin{split}&{\mathbb{E}}\left|e^{\lambda\Delta}K_{\hat{u}}(s,Z_{s}^{(t+\Delta,z)})-K_{\hat{u}}(s,Z_{s}^{(t,z)})\right|\\\
&\leq\lambda e^{\lambda
T}\|K_{\hat{u}}\|_{\infty}\Delta+{\mathbb{E}}\left|K_{\hat{u}}(s,Z_{s}^{(t+\Delta,z)})-K_{\hat{u}}(s,Z_{s}^{(t,z)})\right|\\\
&\leq\lambda e^{\lambda
T}\|K_{\hat{u}}\|_{\infty}\Delta+\left(|\mu-r|+\sigma^{2}+\lambda\left(\bar{\epsilon}+\tfrac{\underline{\epsilon}}{1-\underline{\epsilon}}\right)\right){\mathbb{E}}\left|Z_{s}^{(t+\Delta,z)}-Z_{s}^{(t,z)}\right|\\\
&\leq\lambda e^{\lambda
T}\|K_{\hat{u}}\|_{\infty}\Delta+\left(|\mu-r|+\sigma^{2}+\lambda\left(\bar{\epsilon}+\tfrac{\underline{\epsilon}}{1-\underline{\epsilon}}\right)\right)\left(\left|\mu-r-\tfrac{\sigma^{2}}{2}\right|\Delta+\sigma\sqrt{\tfrac{2}{\pi}}\,\Delta^{\frac{1}{2}}\right),\end{split}$
(A.6)

where the last inequality is due to

${\mathbb{E}}\left|Z_{s}^{(t+\Delta,z)}-Z_{s}^{(t,z)}\right|\leq\left|\mu-r-\tfrac{\sigma^{2}}{2}\right|\Delta+\sigma\,{\mathbb{E}}\left|B_{t+\Delta}-B_{t}\right|=\left|\mu-r-\tfrac{\sigma^{2}}{2}\right|\Delta+\sigma\sqrt{\tfrac{2}{\pi}}\,\Delta^{\frac{1}{2}}.$

Using $\hat{u}=\phi(\hat{u})$ and triangular inequality, we obtain

$\begin{split}&|K_{\hat{u}}(t+\Delta,z)-K_{\hat{u}}(t,z)|\leq\lambda\sup_{\zeta\in
R}\left|\hat{u}(t+\Delta,\zeta)-\hat{u}(t,\zeta)\right|\\\
&=\lambda\sup_{\zeta\in{\mathbb{R}}}\left|\int_{t+\Delta}^{T}e^{-\lambda(s-t-\Delta)}{\mathbb{E}}\left[K_{\hat{u}}(s,Z_{s}^{(t+\Delta,\zeta)})\right]ds-\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[K_{\hat{u}}(s,Z_{s}^{(t,\zeta)})\right]ds\right|\\\
&\leq\lambda\sup_{\zeta\in{\mathbb{R}}}\bigg{(}\int_{t+\Delta}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left|e^{\lambda\Delta}K_{\hat{u}}(s,Z_{s}^{(t+\Delta,\zeta)})-K_{\hat{u}}(s,Z_{s}^{(t,\zeta)})\right|ds\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\int_{t}^{t+\Delta}e^{-\lambda(s-t)}{\mathbb{E}}\left|K_{\hat{u}}(s,Z_{s}^{(t,\zeta)})\right|ds\bigg{)}\\\
&\leq C\Delta^{\frac{1}{2}},\end{split}$

where the generic constant $C$ only depends on the market parameters and
$\|\hat{u}\|_{\infty}$ and the last inequality is due to (A.4) and
(LABEL:K_delta). We can treat $\Delta<0$ by the same way, and conclude that
$K_{\hat{u}}$ is $\tfrac{1}{2}$-Hölder continuous with respect to $t$
variable.
(End of the proof of Claim).

Let $\delta\in(0,1)$ be fixed. Then, the above claim and Theorem 9.2.3 in [35]
ensure that there exists a unique function $u\in
C^{1+\frac{\delta}{2},2+\delta}((0,T)\times{\mathbb{R}})$ such that it
satisfies the following PDE on $(t,z)\in(0,T)\times{\mathbb{R}}$.

$\begin{split}\begin{cases}0=u(T,z),\\\
0=u_{t}+(\mu-r-\frac{\sigma^{2}}{2})u_{z}+\frac{1}{2}\sigma^{2}u_{zz}-\lambda
u+K_{\hat{u}}\end{cases}\end{split}$ (A.7)

Since $u$ admits a unique continuous extension on $[0,T]\times{\mathbb{R}}$
(i.e., see chapter 8.5 in [35]), we let $u\in
C^{1+\frac{\delta}{2},2+\delta}([0,T]\times{\mathbb{R}})$. By the Feynman-Kac
formula (i.e., see Theorem 5.7.6 in [32]), the solution $u$ of the parabolic
PDE (A.7) has the stochastic representation $u=\phi(\hat{u})$, where $\phi$ is
defined in (A.1). Since $\hat{u}$ is chosen as the unique fixed point of the
map $\phi$, we conclude that $u=\hat{u}$.

Our next task is to define $u(t,\pm\infty)$ for $t\in[0,T]$. Using
$\lim_{z\to\infty}h(Z_{s}^{(t,z)})=1$ and
$\lim_{z\to-\infty}h(Z_{s}^{(t,z)})=0$ almost surely, we obtain

$\begin{split}\begin{cases}\lim_{z\to\infty}K_{u}(s,Z_{s}^{(t,z)})=\mu-\tfrac{1}{2}\sigma^{2}+\lambda\sup_{\zeta\in{\mathbb{R}}}\left(u(s,\zeta)-\ln\left(\frac{1-\underline{\epsilon}h(\zeta)}{1-\underline{\epsilon}}\right)\right)\\\
\lim_{z\to-\infty}K_{u}(s,Z_{s}^{(t,z)})=r+\lambda\sup_{\zeta\in{\mathbb{R}}}\left(u(s,\zeta)-\ln\left(1+\bar{\epsilon}h(\zeta)\right)\right)\end{cases}a.s.\end{split}$
(A.8)

The above convergence and $\|K_{u}\|_{\infty}<\infty$ enable us to apply the
dominated convergence theorem:

$\begin{split}\lim_{z\to\pm\infty}u(t,z)&=\lim_{z\to\pm\infty}\phi(u)(t,z)=\lim_{z\to\pm\infty}\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[K_{u}(s,Z_{s}^{(t,z)})\right]ds\\\
&=\begin{cases}\int_{t}^{T}e^{-\lambda(s-t)}\left(\mu-\tfrac{1}{2}\sigma^{2}+\lambda\sup_{\zeta\in{\mathbb{R}}}\left(u(s,\zeta)-\ln\left(\frac{1-\underline{\epsilon}h(\zeta)}{1-\underline{\epsilon}}\right)\right)\right)ds,&\textrm{for
$z\to\infty$}\\\
\int_{t}^{T}e^{-\lambda(s-t)}\left(r+\lambda\sup_{\zeta\in{\mathbb{R}}}\left(u(s,\zeta)-\ln\left(1+\bar{\epsilon}h(\zeta)\right)\right)\right)ds,&\textrm{for
$z\to-\infty$}\end{cases}.\end{split}$

Therefore, we can continuously extend $u$ to $z=\pm\infty$ and $u(t,\infty)$
and $u(t,-\infty)$ are defined by the above limit. We observe that for
$z\in\\{\infty,-\infty\\}$, $u(t,z)$ satisfies

$\begin{split}0&=u_{t}(t,z)+(\mu-r)h(z)+r-\tfrac{1}{2}\sigma^{2}h(z)^{2}-\lambda
u(t,z)\\\
&\qquad+\lambda\sup_{\zeta\in{\mathbb{R}}}\left(u(t,\zeta)-\ln\left(\tfrac{1+\bar{\epsilon}h(\zeta)}{1+\bar{\epsilon}h(z)}\right)1_{\\{z<\zeta\\}}-\ln\left(\tfrac{1-\underline{\epsilon}h(\zeta)}{1-\underline{\epsilon}h(z)}\right)1_{\\{z>\zeta\\}}\right),\end{split}$
(A.9)

where the function $h$ is continuously extended as $h(\infty):=1$ and
$h(-\infty):=0$.

Now we define $v$ as $v(t,x):=u(t,h^{-1}(x))$ for $(t,x)\in[0,T]\times[0,1]$.
Such $v$ is well-defined because
$h:{\mathbb{R}}\cup\\{\infty,-\infty\\}\to[0,1]$ is bijective. We observe that
for $(t,x)\in(0,T)\times(0,1)$ and $z=h^{-1}(x)$,

$\begin{split}v_{t}(t,x)&=u_{t}(t,z),\\\ x(1-x)v_{x}(t,x)&=u_{z}(t,z),\\\
x^{2}(1-x)^{2}v_{xx}(t,x)&=u_{zz}(t,z)-(1-2x)u_{z}(t,z).\end{split}$ (A.10)

The PDE for $u$ in (A.7) with $\hat{u}$ replaced by $u$ and the equalities in
(A.10) produce the PDE for $v$, which is (3.2). Therefore, statement (i) is
valid.

To check statement (ii), we observe that $v(t,0)=u(t,-\infty)$ and
$v(t,1)=u(t,\infty)$. Then, the continuous differentiability of $v(t,0)$ and
$v(t,1)$ with respect to $t$ is followed by that of $u(t,-\infty)$ and
$u(t,\infty)$, and (A.9) produces (3.3).

Finally, statement (iii) is a direct consequence of (A.10) and $u\in
C^{1+\frac{\delta}{2},2+\delta}([0,T]\times{\mathbb{R}})$.

##  Appendix B Proof of Lemma 5.3

(i) When $\epsilon=0$, Lemma 4.1 and (5.1) produce (5.2).

To prove (5.3), we first check that $Y_{s}^{(t,x)}$ in (4.9) safisfies

$\displaystyle
dY_{s}^{(t,x)}=Y_{s}^{(t,x)}(1-Y_{s}^{(t,x)})\left((\mu-r-\sigma^{2}Y_{s}^{(t,x)})ds+\sigma
dB_{s}\right).$ (B.1)

Then application of Ito’s formula produces that for
$(s,x)\in[t,T)\times(0,1)$,

$\displaystyle\left(Y_{s}^{(t,x)}-x\right)^{2}$
$\displaystyle=\int_{t}^{s}Y_{u}^{(t,x)}(1-Y_{u}^{(t,x)})\left(2(Y_{u}^{(t,x)}-x)(\mu-r-\sigma^{2}Y_{u}^{(t,x)})+\sigma^{2}Y_{u}^{(t,x)}(1-Y_{u}^{(t,x)})\right)du$
$\displaystyle\qquad+\int_{t}^{s}2\sigma(Y_{u}^{(t,x)}-x)Y_{u}^{(t,x)}(1-Y_{u}^{(t,x)})dB_{u}.$

Since $0<Y_{s}^{(t,x)}<1$, the local martingale part above is a true
martingale and we obtain

$\displaystyle\tfrac{\partial}{\partial
s}\left({\mathbb{E}}\left[\left(Y_{s}^{(t,x)}-x\right)^{2}\right]\right)$
$\displaystyle={\mathbb{E}}\left[Y_{s}^{(t,x)}(1-Y_{s}^{(t,x)})\left(2(Y_{s}^{(t,x)}-x)(\mu-r-\sigma^{2}Y_{s}^{(t,x)})+\sigma^{2}Y_{s}^{(t,x)}(1-Y_{s}^{(t,x)})\right)\right]$
$\displaystyle=-x^{2}(1-x)^{2}\cdot{\mathbb{E}}\left[\tfrac{\partial}{\partial
x}\left(\tfrac{Y_{s}^{(t,x)}\left(1-Y_{s}^{(t,x)}\right)\left(\mu-r-\sigma^{2}Y_{s}^{(t,x)}\right)}{x(1-x)}\right)\right],$
(B.2)

where the second equality is from elementary computations using the definition
of $Y_{s}^{(t,x)}$ in (4.9).

For $x\in(0,1)$, we observe that

$\displaystyle\tfrac{Y_{s}^{(t,x)}-x}{x(1-x)}=\tfrac{A^{(t,s)}-1}{A^{(t,s)}x+(1-x)}\quad\textrm{with}\quad
A^{(t,s)}:=e^{(\mu-r-\frac{\sigma^{2}}{2})(s-t)+\sigma(B_{s}-B_{t})},$ (B.3)

and the above expression is decreasing in $x$, therefore,

$\displaystyle
1-\tfrac{1}{A^{(t,s)}}<\tfrac{Y_{s}^{(t,x)}-x}{x(1-x)}<A^{(t,s)}-1\quad\textrm{for}\quad
0<x<1.$ (B.4)

When $\epsilon=0$, the representation of $v_{x}$ in (4.11) becomes

$\displaystyle
v_{x}^{0}(t,x)=\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\tfrac{Y_{s}^{(t,x)}\left(1-Y_{s}^{(t,x)}\right)\left(\mu-r-\sigma^{2}Y_{s}^{(t,x)}\right)}{x(1-x)}\right]ds,$
(B.5)

We take derivative with respect to $x$ in the above expression. Then, the mean
value theorem and the dominated convergence theorem, together with the
inequalities (4.14) and (B.4), allow us to take derivative inside of the
expectation and obtain that for $(t,x)\in[0,T)\times(0,1)$,

$\displaystyle v_{xx}^{0}(t,x)$
$\displaystyle=\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\tfrac{\partial}{\partial
x}\left(\tfrac{Y_{s}^{(t,x)}\left(1-Y_{s}^{(t,x)}\right)\left(\mu-r-\sigma^{2}Y_{s}^{(t,x)}\right)}{x(1-x)}\right)\right]ds$
(B.6) $\displaystyle=-\int_{t}^{T}e^{-\lambda(s-t)}\tfrac{\partial}{\partial
s}\left({\mathbb{E}}\left[\left(\tfrac{Y_{s}^{(t,x)}-x}{x(1-x)}\right)^{2}\right]\right)\,ds$
$\displaystyle=-e^{-\lambda(T-t)}\,{\mathbb{E}}\left[\left(\tfrac{Y_{T}^{(t,x)}-x}{x(1-x)}\right)^{2}\right]-\lambda\int_{t}^{T}e^{-\lambda(s-t)}\,{\mathbb{E}}\left[\left(\tfrac{Y_{s}^{(t,x)}-x}{x(1-x)}\right)^{2}\right]\,ds,$
(B.7)

where the second equality is due to (B.2), and the third equality is from
integration by parts. Obviously (B.7) implies that $v_{xx}^{0}(t,x)<0$ for
$(t,x)\in[0,T)\times(0,1)$.

To conclude (5.3), it only remains to check that $\lim_{x\uparrow
1}v_{xx}^{0}(t,x)<0$ and $\lim_{x\downarrow 0}v_{xx}^{0}(t,x)<0$. Indeed,
(B.3) and (B.4) enable us to apply the dominated convergence theorem to (B.7)
and obtain

$\displaystyle\lim_{x\uparrow 1}v_{xx}^{0}(t,x)$
$\displaystyle=-e^{-\lambda(T-t)}\,{\mathbb{E}}\Big{[}\left(1-\tfrac{1}{A^{(t,T)}}\right)^{2}\Big{]}-\lambda\int_{t}^{T}e^{-\lambda(s-t)}\,{\mathbb{E}}\Big{[}\left(1-\tfrac{1}{A^{(t,s)}}\right)^{2}\Big{]}ds,$
$\displaystyle\lim_{x\downarrow 0}v_{xx}^{0}(t,x)$
$\displaystyle=-e^{-\lambda(T-t)}\,{\mathbb{E}}\Big{[}\left(A^{(t,T)}-1\right)^{2}\Big{]}-\lambda\int_{t}^{T}e^{-\lambda(s-t)}\,{\mathbb{E}}\Big{[}\left(A^{(t,s)}-1\right)^{2}\Big{]}ds,$

and we conclude that $\lim_{x\uparrow 1}v_{xx}^{0}(t,x)<0$ and
$\lim_{x\downarrow 0}v_{xx}^{0}(t,x)<0$.

(ii) The SDE for $Y_{s}^{(t,x)}$ in (B.1) and $0<Y_{s}^{(t,x)}<1$ imply that
for $(s,x)\in[t,T)\times(0,1)$,

$\tfrac{\partial}{\partial
s}\left({\mathbb{E}}\left[Y_{s}^{(t,x)}\right]\right)={\mathbb{E}}\left[Y_{s}^{(t,x)}\left(1-Y_{s}^{(t,x)}\right)\left(\mu-r-\sigma^{2}Y_{s}^{(t,x)}\right)\right]$
(B.8)

We divide both sides of (B.8) by $x(1-x)$ and take derivative with respect to
$x$. Then, we can put the derivative inside of the expectation as in the proof
of part (i), and obtain

$\displaystyle{\mathbb{E}}\left[\tfrac{\partial}{\partial
x}\left(\tfrac{Y_{s}^{(t,x)}\left(1-Y_{s}^{(t,x)}\right)\left(\mu-r-\sigma^{2}Y_{s}^{(t,x)}\right)}{x(1-x)}\right)\right]$
$\displaystyle=\tfrac{\partial}{\partial
s}\left({\mathbb{E}}\left[\tfrac{\partial}{\partial
x}\left(\tfrac{Y_{s}^{(t,x)}}{x(1-x)}\right)\right]\right)$
$\displaystyle=\tfrac{\partial}{\partial
s}\left(\tfrac{{\mathbb{E}}\left[\frac{\partial}{\partial
x}Y_{s}^{(t,x)}\right]}{x(1-x)}-\tfrac{(1-2x){\mathbb{E}}\left[Y_{s}^{(t,x)}\right]}{x^{2}(1-x)^{2}}\right).$

We rearrange the above equation and use (B.8), (B.5), and (B.7) to obtain

$\displaystyle\int_{t}^{T}e^{-\lambda(s-t)}\tfrac{\partial}{\partial
s}\left({\mathbb{E}}\left[\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right]\right)ds=x(1-x)v_{xx}^{0}(t,x)+(1-2x)v_{x}^{0}(t,x).$
(B.9)

Now we conclude that $F(t,\hat{y}^{0}(t))<1$ by the following way:

$\displaystyle F(t,\hat{y}^{0}(t))$
$\displaystyle\leq\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\right]ds\,\Big{|}_{x=\hat{y}^{0}(t)}$
$\displaystyle=\left(1-e^{-\lambda(T-t)}{\mathbb{E}}\left[\tfrac{\partial}{\partial
x}Y_{T}^{(t,x)}\right]+x(1-x)v_{xx}^{0}(t,x)+(1-2x)v_{x}^{0}(t,x)\right)\Big{|}_{x=\hat{y}^{0}(t)}$
$\displaystyle<1,$

where the first inequality is from the definition of $F$ in (5.4) and the
positivity of $\frac{\partial}{\partial x}Y_{s}^{(t,x)}$ (see (4.14)), and the
equality is due to integration by parts and (B.9), and the last inequality is
due to the positivity of $\frac{\partial}{\partial x}Y_{T}^{(t,x)}$,
$v_{x}^{0}(t,\hat{y}^{0}(t))=0$, and $v_{xx}^{0}(t,\hat{y}^{0}(t))<0$.

We can check that $F(t,\hat{y}^{0}(t))>-1$ by the same way as above.

(iii) The expression in (4.11) produces

$\begin{split}v_{x}^{\epsilon}(t,x_{\epsilon})-v_{x}^{0}(t,x_{\epsilon})&=\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\left(\tfrac{\partial}{\partial
x}Y_{s}^{(t,x_{\epsilon})}\right)L_{y}^{\epsilon}(s,Y_{s}^{(t,x_{\epsilon})})\right]ds.\end{split}$
(B.10)

In the above expression, when we take limit as $\epsilon\downarrow 0$, the
inequalities (4.13) and (4.14) enable us to use the dominated convergence
theorem to conclude that

$\lim_{\epsilon\downarrow
0}\left(v_{x}^{\epsilon}(t,x_{\epsilon})-v_{x}^{0}(t,x_{\epsilon})\right)=0.$

The above limit and the continuity of $v_{x}^{0}$ implies (5.6).

To prove (5.7), we first observe that for $(t,x)\in[0,T)\times(0,1)$ and
$(s,z)\in(t,T)\times(0,1)$, the density function for $Y_{s}^{(t,x)}$ is given
by

$\begin{split}\varphi(s,z;t,x)&:=\tfrac{\partial}{\partial
z}{\mathbb{P}}\left(Y_{s}^{(t,x)}\leq
z\right)=\tfrac{\exp\left(-\frac{1}{2\sigma^{2}(s-t)}\left((r-\mu+\frac{\sigma^{2}}{2})(s-t)+\ln\left(\frac{z(1-x)}{(1-z)x}\right)\right)^{2}\right)}{\sigma
z(1-z)\sqrt{2\pi(s-t)}}.\end{split}$ (B.11)

Then, the expression in (4.11) and $\frac{\partial}{\partial
x}Y_{s}^{(t,x)}=\frac{Y_{s}^{(t,x)}(1-Y_{s}^{(t,x)})}{x(1-x)}$ imply that

$\begin{split}v_{x}^{\epsilon}(t,x)-v_{x}^{0}(t,x)&=\lambda\int_{t}^{T}e^{-\lambda(s-t)}{\mathbb{E}}\left[\tfrac{Y_{s}^{(t,x)}\left(1-Y_{s}^{(t,x)}\right)}{x(1-x)}L_{y}^{\epsilon}(s,Y_{s}^{(t,x)})\right]ds\\\
&=\lambda\int_{t}^{T}e^{-\lambda(s-t)}\left(\int_{0}^{1}\tfrac{z(1-z)}{x(1-x)}L_{y}^{\epsilon}(s,z)\varphi(s,z;t,x)dz\right)ds.\end{split}$
(B.12)

For $x\in(0,1)$, direct computations produce

$\begin{split}\tfrac{\partial}{\partial
x}\left(\tfrac{z(1-z)}{x(1-x)}\varphi(s,z;t,x)\right)=\tfrac{(1-z)z\left(r-\mu+(2x-\frac{1}{2})\sigma^{2}+\frac{1}{s-t}\ln\left(\frac{z(1-x)}{(1-z)x}\right)\right)\varphi(s,z;t,x)}{(1-x)^{2}x^{2}\sigma^{2}}.\end{split}$
(B.13)

Assumption 5.1 implies that

$\displaystyle\left|r-\mu+\tfrac{\sigma^{2}}{2}\right|\leq\sigma^{2},\quad\left|r-\mu+(2x-\tfrac{1}{2})\sigma^{2}\right|\leq
2\sigma^{2}\quad\textrm{for}\quad x\in(0,1).$ (B.14)

Then, (B.11) and (B.13) produce the following:

$\begin{split}&\int_{0}^{1}\left|\frac{\partial}{\partial
x}\left(\tfrac{z(1-z)}{x(1-x)}\varphi(s,z;t,x)\right)\right|\,dz\\\
&\leq\int_{0}^{1}\tfrac{\left(2\sigma^{2}+\frac{1}{s-t}\left|\ln\left(\tfrac{z(1-x)}{(1-z)x}\right)\right|\right)}{\sigma^{3}(1-x)^{2}x^{2}\sqrt{2\pi(s-t)}}\exp\left(-\tfrac{\left(\ln\left(\frac{z(1-x)}{(1-z)x}\right)\right)^{2}}{2\sigma^{2}(s-t)}+\left|\ln\left(\tfrac{z(1-x)}{(1-z)x}\right)\right|\right)\,dz\\\
&=\int_{-\infty}^{\infty}\tfrac{\left(2\sigma^{2}+\frac{1}{s-t}\left|\zeta\right|\right)e^{-\frac{\zeta^{2}}{2\sigma^{2}(s-t)}+\left|\zeta\right|}}{\sigma^{3}(1-x)^{2}x^{2}\sqrt{2\pi(s-t)}}\cdot\tfrac{\frac{x}{1-x}e^{\zeta}}{(1+\frac{x}{1-x}e^{\zeta})^{2}}\,d\zeta\\\
&\leq\int_{-\infty}^{\infty}\tfrac{\left(2\sigma^{2}+\frac{1}{s-t}\left|\zeta\right|\right)e^{-\frac{\zeta^{2}}{4\sigma^{2}(s-t)}+4\sigma^{2}(s-t)}}{\sigma^{3}(1-x)^{3}x\sqrt{2\pi(s-t)}}\,d\zeta\\\
&=\tfrac{2\sqrt{2}}{(1-x)^{3}x}\left(1+\tfrac{1}{\sigma\sqrt{\pi(s-t)}}\right)e^{4\sigma^{2}(s-t)},\end{split}$
(B.15)

where the first inequality is due to (B.14), and the first equality is
obtained by change of variables as
$\zeta=\ln\left(\frac{z(1-x)}{(1-z)x}\right)$. The second inequality is due to
the inequality of arithmetic and geometric means, and the second equality is
obtained by direct computations.

By (LABEL:nasty1), we conclude that

$\displaystyle\int_{t}^{T}e^{-\lambda(s-t)}\int_{0}^{1}\left|\tfrac{\partial}{\partial
x}\left(\tfrac{z(1-z)}{x(1-x)}\varphi(s,z;t,x)\right)\right|\,dz\,ds<\infty,$
(B.16)

and the function $H:[0,T)\times(0,1)\to{\mathbb{R}}$ given by

$\displaystyle
H(t,x):=\lambda\int_{t}^{T}e^{-\lambda(s-t)}\int_{0}^{1}\tfrac{\partial}{\partial
x}\left(\tfrac{z(1-z)}{x(1-x)}\varphi(s,z;t,x)\right)L_{y}^{\epsilon}(s,z)\,dz\,ds$
(B.17)

is well-defined due to (B.16) and the boundedness
$|L_{y}^{\epsilon}|\leq\frac{\epsilon}{1-\epsilon}$ in (4.13). Then,
(LABEL:nasty1) implies that

$\displaystyle\left|H(t,x)\right|\leq\tfrac{2\sqrt{2}\lambda
e^{4\sigma^{2}T}}{(1-x)^{3}x}\left(T+\tfrac{2\sqrt{T}}{\sigma\sqrt{\pi}}\right)\cdot\tfrac{\epsilon}{1-\epsilon}\quad\textrm{for}\quad(t,x)\in[0,T)\times(0,1).$
(B.18)

Now, let’s check that

$\displaystyle H(t,x)=v_{xx}^{\epsilon}(t,x)-v_{xx}^{0}(t,x).$ (B.19)

Indeed, for $(t,x)\in[0,T)\times(0,1)$,

$\displaystyle H(t,x)$ $\displaystyle=\lim_{\delta\to
0}\frac{1}{\delta}\int_{x}^{x+\delta}H(t,\eta)d\eta$
$\displaystyle=\lim_{\delta\to
0}\frac{1}{\delta}\,\,\lambda\int_{t}^{T}e^{-\lambda(s-t)}\int_{0}^{1}\left(\tfrac{z(1-z)\varphi(s,z;t,x+\delta)}{(x+\delta)(1-x-\delta)}-\tfrac{z(1-z)\varphi(s,z;t,x)}{x(1-x)}\right)L_{y}^{\epsilon}(s,z)\,dz\,ds$
$\displaystyle=\lim_{\delta\to
0}\frac{1}{\delta}\left(\left(v_{x}^{\epsilon}(t,x+\delta)-v_{x}^{0}(t,x+\delta)\right)-\left(v_{x}^{\epsilon}(t,x)-v_{x}^{0}(t,x)\right)\right)$
$\displaystyle=v_{xx}^{\epsilon}(t,x)-v_{xx}^{0}(t,x),$

where the second equality is due to Fubini’s theorem and the fundamental
theorem of calculus, and the third equality is from (B.12).

Finally, we conclude (5.7) by the following observation:

$\displaystyle\limsup_{\epsilon\downarrow
0}\left|v_{xx}^{\epsilon}(t,x_{\epsilon})-v_{xx}^{0}(t,x_{0})\right|$
$\displaystyle\leq\limsup_{\epsilon\downarrow
0}\left|v_{xx}^{\epsilon}(t,x_{\epsilon})-v_{xx}^{0}(t,x_{\epsilon})\right|+\limsup_{\epsilon\downarrow
0}\left|v_{xx}^{0}(t,x_{\epsilon})-v_{xx}^{0}(t,x_{0})\right|$
$\displaystyle\leq\limsup_{\epsilon\downarrow 0}\tfrac{2\sqrt{2}\lambda
e^{4\sigma^{2}T}}{(1-x_{\epsilon})^{3}x_{\epsilon}}\left(T+\tfrac{2\sqrt{T}}{\sigma\sqrt{\pi}}\right)\cdot\tfrac{\epsilon}{1-\epsilon}=0,$

where the second inequality is due to (B.18), (B.19), and the continuity of
$v_{xx}^{0}$.

##  Appendix C Proof of Lemma 5.7

(i) This result holds due to equation (2.7) in [40] and the inequality
$|\hat{y}^{0}(t)-y_{\infty}|\leq 1$.

(ii) For convenience, we define a function
$\Gamma:[0,T]\times(0,1)\to{\mathbb{R}}$ as

$\displaystyle\Gamma(t,x):={\mathbb{E}}\left[\tfrac{Y_{t}^{(0,x)}\left(1-Y_{t}^{(0,x)}\right)\left(\mu-r-\sigma^{2}Y_{t}^{(0,x)}\right)}{x(1-x)}\right].$
(C.1)

Note that $\Gamma$ does not depend on $\lambda$. The equations (B.5) and (B.6)
can be written as

$\displaystyle v_{x}^{0}(t,x)=\int_{0}^{T-t}e^{-\lambda s}\Gamma(s,x)ds,\quad
v_{xx}^{0}(t,x)=\int_{0}^{T-t}e^{-\lambda s}\Gamma_{x}(s,x)ds$ (C.2)

We can do the similar argument to obtain representations for $v_{xxx}^{0}$ and
partial derivatives of $v_{x}^{0}$ and $v_{xx}^{0}$ with respect to $\lambda$,
with the observation that $\Gamma,\Gamma_{x},\Gamma_{xx}$ are continuous &
bounded maps on $[0,T]\times(0,1)$. The result is summarized as follows:

$\begin{split}&\tfrac{\partial}{\partial\lambda}v_{x}^{0}(t,x)=-\int_{0}^{T-t}e^{-\lambda
s}s\Gamma(s,x)ds,\quad\tfrac{\partial}{\partial\lambda}v_{xx}^{0}(t,x)=-\int_{0}^{T-t}e^{-\lambda
s}s\Gamma_{x}(s,x)ds\\\ &v_{xx}^{0}(t,x)=\int_{0}^{T-t}e^{-\lambda
s}\Gamma_{x}(s,x)ds,\qquad\quad v_{xxx}^{0}(t,x)=\int_{0}^{T-t}e^{-\lambda
s}\Gamma_{xx}(s,x)ds.\end{split}$ (C.3)

Using $Y_{t}^{(t,x)}=x$, direct computations produce

$\begin{split}&\Gamma(0,x)=\mu-r-\sigma^{2}x,\quad\Gamma_{x}(0,x)=-\sigma^{2},\quad\Gamma_{xx}(0,x)=0.\end{split}$
(C.4)

With (LABEL:Gamma_values) and (5.38), we apply Lemma D.2 to
(LABEL:various_rep) and conclude (LABEL:various_limit1).

(iii) The mean value theorem and $\eqref{vx0_bound}$ produce

$\begin{split}\hat{y}^{0,\lambda+\delta}(t)-\hat{y}^{0,\lambda}(t)=-\tfrac{v_{x}^{0,\lambda+\delta}(t,\hat{y}^{0,\lambda}(t))-v_{x}^{0,\lambda}(t,\hat{y}^{0,\lambda}(t))}{v_{xx}^{0,\lambda+\delta}(t,z(\lambda,\delta))}\quad\textrm{for
$z(\lambda,\delta)$ between $\hat{y}^{0,\lambda}(t)$ and
$\hat{y}^{0,\lambda+\delta}(t)$},\end{split}$

where we specify the dependence on $\lambda$ for clarity. Since
$\frac{\partial}{\partial\lambda}v_{x}^{0}$ exists (see (LABEL:various_rep)),
the above equality and (5.3) ensure that $\hat{y}^{0,\lambda}(t)$ is
differentiable with respect to $\lambda$ and

$\begin{split}\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(t)=-\frac{\frac{\partial}{\partial\lambda}v_{x}^{0}(t,x)\big{|}_{x=\hat{y}^{0}(t)}}{v_{xx}^{0}(t,\hat{y}^{0}(t))}.\end{split}$
(C.5)

Observe that the bounds for $\frac{\partial}{\partial x}Y_{t}^{(0,x)}$ and
$\frac{Y_{t}^{(0,x)}-x}{x(1-x)}$ in (4.14) and (B.4) do not depend on the
variable $x$. Therefore, the following convergence is uniform on $x\in(0,1)$:

$\displaystyle\Gamma_{x}(t,x)=-{\mathbb{E}}\left[\sigma^{2}\left(\tfrac{\partial}{\partial
x}Y_{t}^{(0,x)}\right)^{2}+2\left(\mu-r-\sigma^{2}Y_{t}^{(0,x)}\right)\left(\tfrac{Y_{t}^{(0,x)}-x}{x(1-x)}\right)\tfrac{\partial}{\partial
x}Y_{t}^{(0,x)}\right]\xrightarrow[t\downarrow 0]{}-\sigma^{2}.$

Hence, there is a constant $\tilde{T}\in(0,T)$ such that
$\Gamma_{x}(t,x)\leq-\frac{\sigma^{2}}{2}$ for
$(t,x)\in[0,\tilde{T}]\times(0,1)$. This observation and the expression of
$v_{xx}$ in (LABEL:various_rep) imply

$\displaystyle\lambda
v_{xx}(t,\hat{y}^{0}(t))\leq-\tfrac{\sigma^{2}}{2}\left(1-e^{-\lambda(T-t)}\right)\quad\textrm{for}\quad(t,x)\in[T-\tilde{T},T)\times(0,1).$
(C.6)

We obtain $\|\Gamma_{xt}\|_{\infty}<\infty$ by using Ito’s formula with (B.1)
and the bounds (4.14) and (B.4). Then, (LABEL:various_rep) and
(LABEL:Gamma_values) imply

$\displaystyle\left|\lambda v_{xx}(t,\hat{y}^{0}(t))+\sigma^{2}\right|$
$\displaystyle=\left|\lambda\int_{0}^{T-t}e^{-\lambda
s}s\,\left(\tfrac{\Gamma_{x}(s,\hat{y}^{0}(t))-\Gamma_{x}(0,\hat{y}^{0}(t))}{s}\right)ds+\sigma^{2}e^{-\lambda(T-t)}\right|$
$\displaystyle\leq\|\Gamma_{xt}\|_{\infty}\left(\tfrac{1-e^{-\lambda(T-t)}}{\lambda}-(T-t)e^{-\lambda(T-t)}\right)+\sigma^{2}e^{-\lambda(T-t)}.$

This implies that there exists a constant $\tilde{\Lambda}$ (may depend on
$\tilde{T}$) such that

$\displaystyle\lambda
v_{xx}(t,\hat{y}^{0}(t))\leq-\tfrac{\sigma^{2}}{2}\quad\textrm{for}\quad(t,\lambda)\in[0,T-\tilde{T}]\times[\tilde{\Lambda},\infty).$
(C.7)

Using (LABEL:various_rep) and (LABEL:Gamma_values), we obtain

$\displaystyle\left|\lambda^{3}\tfrac{\partial}{\partial\lambda}v_{x}^{0}(t,x)\big{|}_{x=\hat{y}^{0}(t)}\right|$
$\displaystyle\leq\left|\lambda^{3}\int_{0}^{T-t}e^{-\lambda
s}s^{2}\left(\tfrac{\Gamma(s,\hat{y}^{0}(t))-\Gamma(0,\hat{y}^{0}(t))}{s}\right)ds\right|+\left|\sigma^{2}(y_{\infty}-\hat{y}^{0}(t))\lambda^{3}\int_{0}^{T-t}e^{-\lambda
s}s\,ds\right|$ $\displaystyle\leq
C\left(1-e^{-\lambda(T-t)}+\lambda(T-t)\big{(}1+\lambda(T-t)\big{)}e^{-\lambda(T-t)}\right)\quad\textrm{for}\quad(t,\lambda)\in[0,T]\times[1,\infty),$
(C.8)

where the second inequality is due to $\|\Gamma_{t}\|_{\infty}<\infty$ and
(5.38).

From (C.5), we obtain the boundedness of
$\left|\lambda^{2}\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(t)\right|$:

$\displaystyle\sup_{(t,\lambda)\in[0,T-\tilde{T}]\times[\tilde{\Lambda},\infty)}\left|\lambda^{2}\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(t)\right|<\infty\quad\textrm{due
to \eqref{lambda_vxx_bound}, \eqref{lambda3_vx_bound},
$\sup_{x>0}x(1+x)e^{-x}<\infty$,}$
$\displaystyle\sup_{(t,\lambda)\in[0,T-\tilde{T}]\times[1,\tilde{\Lambda}]}\left|\lambda^{2}\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(t)\right|<\infty\quad\textrm{due
to the continuity on compact set},$
$\displaystyle\sup_{(t,\lambda)\in[T-\tilde{T},T)\times[1,\infty)}\left|\lambda^{2}\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(t)\right|<\infty\quad\textrm{due
to \eqref{lambda_vxx_bound2}, \eqref{lambda3_vx_bound},
$\sup_{x>0}\tfrac{x(1+x)e^{-x}}{1-e^{-x}}<\infty$.}$

Therefore, we conclude (5.40).

(iv) The bounds (4.14) and (5.40) enable us to use Leibniz integral rule to
obtain

$\displaystyle\tfrac{\partial}{\partial\lambda}{\mathbb{E}}\left[\left|Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right|\right]={\mathbb{E}}\left[\left(\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(t)\cdot\tfrac{\partial}{\partial
x}Y_{s}^{(t,x)}\Big{|}_{x=\hat{y}^{0}(t)}-\tfrac{\partial}{\partial\lambda}\hat{y}^{0}(s)\right)\cdot\operatorname{sgn}\left(Y_{s}^{(t,\hat{y}^{0}(t))}-\hat{y}^{0}(s)\right)\right].$

The above expression, together with the bounds (4.14) and (5.40), implies
(5.41).

(v) Explicit computations using the expression of $Y_{s}^{(t,x)}$ in (4.9)
produce

$\displaystyle\tfrac{{\mathbb{E}}\left[\left|Y_{s}^{(t,x)}-x\right|\right]}{\sqrt{s-t}}$
$\displaystyle=\int_{\mathbb{R}}\tfrac{x(1-x)}{x+(1-x)e^{(r-\mu+\frac{\sigma^{2}}{2})u^{2}-\sigma
uz}}\cdot\tfrac{\big{|}1-e^{(r-\mu+\frac{\sigma^{2}}{2})u^{2}-\sigma
uz}\big{|}}{u}\cdot\tfrac{e^{-\frac{z^{2}}{2}}}{\sqrt{2\pi}}dz\bigg{|}_{u=\sqrt{s-t}}$
(C.9)
$\displaystyle\leq\int_{\mathbb{R}}\frac{e^{|r-\mu+\frac{\sigma^{2}}{2}|u^{2}+\sigma
u|z|}-e^{-|r-\mu+\frac{\sigma^{2}}{2}|u^{2}-\sigma
u|z|}}{u}\cdot\tfrac{e^{-\frac{z^{2}}{2}}}{\sqrt{2\pi}}dz\bigg{|}_{u=\sqrt{s-t}}$
$\displaystyle=\frac{e^{(|r-\mu+\frac{\sigma^{2}}{2}|+\frac{\sigma^{2}}{2})u^{2}}\left(1+\int_{0}^{\frac{\sigma
u}{\sqrt{2}}}\frac{2e^{-z^{2}}}{\sqrt{\pi}}dz\right)-e^{(-|r-\mu+\frac{\sigma^{2}}{2}|+\frac{\sigma^{2}}{2})u^{2}}\left(1-\int_{0}^{\frac{\sigma
u}{\sqrt{2}}}\frac{2e^{-z^{2}}}{\sqrt{\pi}}dz\right)}{u}\Bigg{|}_{u=\sqrt{s-t}},$

where we use $|1-e^{a}|\leq e^{|a|}-e^{-|a|}$ for $a\in{\mathbb{R}}$ to obtain
the inequality. The last expression converges to
$\tfrac{2\sqrt{2}\sigma}{\sqrt{\pi}}$ as $u\downarrow 0$, therefore, it is
bounded on $u\in(0,\sqrt{T}]$. Then, we conclude (5.42).

Observe that the integrand in (C.9) is bounded by
$e^{C|z|}\cdot\tfrac{e^{-\frac{z^{2}}{2}}}{\sqrt{2\pi}}$ for a constant $C$
independent of $(u,x,z)\in(0,\sqrt{T}]\times(0,1)\times{\mathbb{R}}$. Since
this bound is integrable with respect to $z$, we apply the dominated
convergence theorem to (C.9) and obtain

$\displaystyle\lim_{s\downarrow
t}\tfrac{{\mathbb{E}}\left[\left|Y_{s}^{(t,x)}-x\right|\right]}{\sqrt{s-t}}$
$\displaystyle=\int_{\mathbb{R}}x(1-x)\sigma|z|\cdot\tfrac{e^{-\frac{z^{2}}{2}}}{\sqrt{2\pi}}dz=\sigma\sqrt{\tfrac{2}{\pi}}\,x(1-x).$

Therefore, we conclude (5.43).

##  Appendix D Supplementary Lemmas

######  Lemma D.1.

Let $F:[0,T]\times[0,1]^{2}\to{\mathbb{R}}$ be a continuous function. We
define $f:[0,T]\times[0,1]\to[0,1]$ as

$\displaystyle
f(t,x):=\max\left\\{z:z\in\operatorname*{argmax}_{y\in[0,1]}F(t,x,y)\right\\},$
(D.1)

then $f$ is upper semicontinuous (which is obviously Borel-measurable).

###### Proof.

This type of result is well-known (e.g., see p. 153 in [7]), but we give a
short proof here for the sake of self-containedness.

Since $F$ is continuous, $\operatorname*{argmax}_{y\in[0,1]}F(t,x,y)$ is a
nonempty closed subset of $[0,1]$, so the maximum element of
$\operatorname*{argmax}_{y\in[0,1]}F(t,x,y)$ exists and $f$ is well-defined.
Let $\\{(t_{n},x_{n})\\}_{n\in{\mathbb{N}}}\subset[0,T]\times[0,1]$ be a
sequence converging to $(t_{\infty},x_{\infty})$ such that
$\lim_{n\to\infty}f(t_{n},x_{n})$ exists. Then, by definition of $f$,

$\displaystyle F(t_{n},x_{n},f(t_{\infty},x_{\infty}))\leq
F(t_{n},x_{n},f(t_{n},x_{n})).$ (D.2)

We let $n\to\infty$ above and using the continuity of $F$ to obtain

$\displaystyle F(t_{\infty},x_{\infty},f(t_{\infty},x_{\infty}))\leq
F(t_{\infty},x_{\infty},\lim_{n\to\infty}f(t_{n},x_{n})).$ (D.3)

This implies that
$\lim_{n\to\infty}f(t_{n},x_{n})\in\operatorname*{argmax}_{y\in[0,1]}F(t_{\infty},x_{\infty},y)$,
and the definition of $f$ ensures

$f(t_{\infty},x_{\infty})\geq\lim_{n\to\infty}f(t_{n},x_{n}).$

Therefore, $f$ is upper semicontinuous. ∎

######  Lemma D.2.

Let $f(s,x):[0,t]\times(0,1)\rightarrow{\mathbb{R}}$ be a continuous function,
and $g(\lambda):[1,\infty)\rightarrow(0,1)$ be a function satisfying
$\lim_{\lambda\rightarrow\infty}g(\lambda)=x_{\infty}\in(0,1)$. Then, for
$\alpha\in\\{0,1,2,3,4\\}$ and $t>0$,

$\begin{split}&\lim_{\lambda\to\infty}\lambda^{\frac{\alpha}{2}+1}\int_{0}^{t}e^{-\lambda
s}s^{\frac{\alpha}{2}}f(s,g(\lambda))ds=c_{\alpha}\cdot
f(0,x_{\infty}),\end{split}$ (D.4)

where $c_{0}=1$, $c_{1}=\frac{\sqrt{\pi}}{2}$, $c_{2}=1$,
$c_{3}=\frac{3\sqrt{\pi}}{4}$, $c_{4}=2$. Also, there exists a constant $C$
such that

$\lambda^{\frac{\alpha}{2}+1}\int_{0}^{t}e^{-\lambda
s}s^{\frac{\alpha}{2}}ds\leq
C,\quad\textrm{for}\quad(t,\lambda,\alpha)\in[0,\infty)\times[1,\infty)\times\\{0,1,2,3,4\\}.$
(D.5)

###### Proof.

Let $\eta>0$ be a given constant. The uniform continuity of $f$ on a compact
set containing the point $(0,x_{\infty})$, together with
$\lim_{\lambda\rightarrow\infty}g(\lambda)=x_{\infty}\in(0,1)$, implies that
there exists $\delta>0$ such that

$\displaystyle\left|f(s,g(\lambda))-f(0,x_{\infty})\right|\leq\eta\quad\textrm{for
any}\quad(s,\lambda)\in[0,\delta]\times\left[\tfrac{1}{\delta},\infty\right).$
(D.6)

Simple computations and (D.6) produce

$\begin{split}\limsup_{\lambda\to\infty}\Big{|}\lambda\int_{0}^{t}e^{-\lambda
s}f(s,g(\lambda))ds-&f(0,x_{\infty})\Big{|}=\limsup_{\lambda\to\infty}\bigg{|}\int_{0}^{\delta}\lambda
e^{-\lambda s}(f(s,g(\lambda))ds-f(0,x_{\infty}))ds\\\
&+\int_{\delta}^{t}\lambda e^{-\lambda
s}(f(s,g(\lambda))ds-f(0,x_{\infty}))ds-e^{-\lambda
t}f(0,x_{\infty})\bigg{|}\leq\eta.\end{split}$

Since $\eta>0$ can be arbitrary small, we conclude (LABEL:lemma2_result) for
the case of $\alpha=0$. The other cases in (LABEL:lemma2_result) can be
obtained by the same way as above, using the following expressions:

$\begin{split}&\lambda^{\frac{3}{2}}\int_{0}^{t}e^{-\lambda
s}s^{\frac{1}{2}}ds=-\sqrt{\lambda t}e^{-\lambda t}+\int_{0}^{\sqrt{\lambda
t}}e^{-s^{2}}ds\xrightarrow[\lambda\to\infty]{}\frac{\sqrt{\pi}}{2},\\\
&\lambda^{2}\int_{0}^{t}e^{-\lambda s}s\,ds=1-(1+\lambda t)e^{-\lambda
t}\xrightarrow[\lambda\to\infty]{}1,\\\
&\lambda^{\frac{5}{2}}\int_{0}^{t}e^{-\lambda
s}s^{\frac{3}{2}}ds=-\tfrac{(3+2\lambda t)\sqrt{\lambda t}e^{-\lambda
t}}{2}+\frac{3}{2}\int_{0}^{\sqrt{\lambda
t}}e^{-s^{2}}ds\xrightarrow[\lambda\to\infty]{}\frac{3\sqrt{\pi}}{4},\\\
&\lambda^{3}\int_{0}^{t}e^{-\lambda s}s^{2}ds=2-(2+\lambda t(2+\lambda
t))e^{-\lambda t}\xrightarrow[\lambda\to\infty]{}2.\\\ \end{split}$ (D.7)

One can easily observe that $\sup_{x>0}x^{\frac{\alpha}{2}}e^{-x}<\infty$ for
$\alpha\in\\{0,1,2,3,4\\}$. This observation and the explicit expressions in
(LABEL:lambda_integrals) produce the bound (D.5). ∎