Shadow price in the power utility case
We consider the problem of maximizing expected power utility from consumption over an infinite horizon in the Black-Scholes model with proportional transaction costs, as studied in the paper \cite{SS}. Similarly to \cite{KMK}, we derive a shadow price, that is, a frictionless price process with values in the bid-ask spread which leads to the same optimal policy. In doing so we explore and exploit the strong relationship between the shadow price and the Hamilton-Jacobi-Bellman-equation.
1112.4385.tex
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\begin{document}
\begin{frontmatter}
  
\title{Shadow price in the power utility case}
\runtitle{Shadow price in the power utility case}
\runauthor{A. Herczegh and V. Prokaj}
\begin{aug}
\author{Attila Herczegh \ead[label=e1]{prince@cs.elte.hu}}\and 
\author{Vilmos Prokaj \ead[label=e2]{prokaj@cs.elte.hu}}

\affiliation{E\"otv\"os Lor\'and University}
\address{Department of Probability Theory and Statistics \\
E\"otv\"os Lor\'and University \\
P\'azm\'any P\'eter s\'et\'any 1/C, 1117 Budapest, Hungary \\
\printead{e1}\\
\printead{e2}}
\end{aug}
\thankstext{T1}{The European Union and the European Social Fund have provided
  financial support to the project under the grant agreement no. T\'AMOP
  4.2.1/B-09/KMR-2010-0003.} 

\begin{abstract}
We consider the problem of maximizing expected power utility from consumption
over an infinite horizon in the Black-Scholes model with proportional
transaction costs, as studied in the paper  \cite{SS}. 

Similarly to  \cite{KMK}, we derive a shadow price,
that is, a frictionless price process with values in the bid-ask spread which
leads to the same optimal policy. In doing so we explore and exploit the
strong relationship between the shadow price and the
Hamilton-Jacobi-Bellman-equation. 
\end{abstract}

\begin{keyword}[class=AMS]
\kwd[Primary ]{91B28}
\kwd{91B16}
\kwd[; secondary ]{60H10}
\end{keyword}

\begin{keyword}
\kwd{shadow price process}
\kwd{transaction costs}
\kwd{optimal consumption}
\kwd{power utility}
\end{keyword}
\end{frontmatter}

\maketitle

\section{Introduction}

It is a classical problem of mathematical finance to consider the problem of
maximizing expected utility from consumption. This was initiated by
\cite{M1,M2}, and thus is often referred to as the Merton problem. He found 
that for logarithmic or power utility it is optimal to keep a constant
fraction of wealth in stocks and to consume at a rate proportional to current
wealth. 

This was extended to proportional transaction costs by \cite{MC}. They stated
that it is optimal to restrain from 
trading while the fraction of wealth invested in stocks is inside an interval
$[\theta_1,\theta_2].$ Their heuristic argument was made precise by
\cite{DN}, which was then generalized by   \cite{SS} who
managed to remove a couple of assumptions needed in \cite{DN}. 

These papers use methods from stochastic control. In recent years it seems
there is more and more emphasis on solving portfolio optimization problems
with transaction costs by determining the shadow price of the problem
\cite{GGS,GS,KMK}. This is a process that establishes a link between portfolio
optimization with and without transaction costs as the optimal policy of the
shadow price without frictions must coincide with that of the original problem.

The first article in this context is 
\cite{KMK}. They use this dual approach to come up with a free boundary
problem and solve that to derive the shadow price for logarithmic
utility. They also showed a connection with the original solution of
\cite{DN}. They point out how the optimal consumption derived by Davis and
Norman can be used to determine the shadow value process and from that the
shadow price itself. 

Our paper basically does the same for the power utility case. We start from
the known solution of the problem with transaction costs and consider what
that means for the shadow price. We notice that this optimal policy can only
be derived from one shadow process, so it is quite straightforward to come up
with a candidate. Then we only have to verify that this process satisfies the
required properties of the shadow price. 

The rest of the paper is organized as follows. Section 2 introduces the model
and summarizes the known facts about problems with and without transaction
costs for the power utility case. Section 3 shows how to come up with the
candidate for the shadow price. Section 4 contains the verification, the main
result is in Theorem \ref{thm:sp}. 


\section{Model and known results}

\subsection{The model}

We study the problem of maximizing expected utility from consumption over an
infinite horizon in the presence of proportional transaction costs as in
\cite{DN,SS,KMK}. We consider a market with a bank account or bond whose value
is constant $1$ and a risky asset, a stock, whose price evolution is given by
$$
dS_t=S_t(\mu dt+\sigma dW_t),
$$
with $S_0,\mu,\sigma>0$, where $W$ is a Brownian motion on the filtered
probability space $(\Omega,(\F_t)\tgo,\P)$. Whenever trading occurs, the
investor faces higher ask (buying) and lower bid (selling) prices, namely he
can buy at $\overline{S}_t=(1+\overline{\lambda})S_t$ and sell at
$\underline{S}_t=(1-\underline{\lambda})S_t$ for some
$\overline{\lambda}\in(0,\infty)$ and $\underline{\lambda}\in(0,1).$ As
transactions of infinite variation lead to instantaneous bankruptcy, we limit
ourselves to the following set of trading strategies. 

\begin{defn}
A trading strategy is a predictable $(\varphi^0_t,\varphi^1_t)\in\mathbb{R}^2$
process of finite variation, where $\varphi^0_t$ and $\varphi^1_t$ denote the
number of units in the bond and the stock at time $t$ respectively. A
consumption rate is an adapted $c_t\in\mathbb{R}_{+}$ process, which
statisfies $\int_0^t{c_s}ds<\infty$ a.s. for all $t\geq0.$ A pair
$((\varphi^0_t,\varphi^1_t),c_t)$ of $(\varphi^0_t,\varphi^1_t)$ trading
strategy and $c_t$ consumption rate is called a portfolio/consumption
process. 
\end{defn}

Writing $\varphi^1_t=\varphi^{\uparrow}_t-\varphi^{\downarrow}_t$ as the
difference between the cumulative number of shares bought
$(\varphi^{\uparrow}_t)$ and sold $(\varphi^{\downarrow}_t)$ up to time $t$, a
portfolio/consumption process is called self-financing, if 
$$
d\varphi^0_t=
-\overline{S}_td\varphi^{\uparrow}_t+\underline{S}_td\varphi^{\downarrow}_t-c_tdt 
$$
holds. This way a self-financing strategy can be determined by
$(\varphi_t^1,c_t)$. From now on we only focus on self-financing strategies
and use only the notation $(\varphi_t^1,c_t)$ for them. 

\begin{defn}
A self-financing portfolio/consumption process is admissible if its
liquidation value is non-negative, i.e.,
$$
V_t^{\varphi}=\varphi^0_t+\underline{S}_t\varphi^{+}_t-\overline{S}_t\varphi^{-}_t\geq0, 
\quad\text{a.s. for all $t\geq0.$}
$$
Given an initial endowment $(x_0,y_0),$ referring to the value of bonds and
stocks respectively, an admissible portfolio/consumption process
$(\varphi^1_t,c_t)$ is optimal if it maximizes 
$$
\E{\int_0^{\infty}{e^{-\delta t}u(c_t)}dt}
$$
over all admissible portfolio/consumption processes. Here $\delta>0$ denotes a
fixed given impatience rate, $u$ a utility function, in our case the power
utility, i.e., $u(c)=\frac{c^{\gamma}}{\gamma}$ for some $\gamma\in(0,1)$. 
\end{defn}

The goal of this paper is to determine the shadow price process to this problem.

\begin{defn}
A shadow price is a frictionless price process $\tilde{S}_t$, lying within the
bid-ask spread ($\underline{S}_t\leq\tilde{S}_t\leq\overline{S}_t$ a.s.), such
that the maximal expected utility for price process $S_t$ with transaction costs
and for price process $\tilde{S}_t$ without transaction costs coincide. 
\end{defn}

Obviously, for any process lying in the bid-ask spread, the maximal expected
utility is at least as high as for the original market with price process
$S_t$, since the investor can trade at a smaller ask and a higher bid
price. Indeed, this is what makes the shadow price so special, the optimal
strategy with respect to it must only buy (resp. sell) when the shadow price
coincides with the original ask (resp. bid) price. 


\subsection{The problem without transaction costs}

In order to determine the shadow price, we need to obtain the optimal
consumption and portfolio for the power utility when the stock price can be
any It\^{o}-process. 
In this section, we follow 
\cite{KS} section 5.8.

Assume that the discounted price process $\tS$ is an \ito/
process of the form 
\begin{displaymath}
  d\tS_t=\tS_t\zjel{\tmu_tdt+\tsigma_tdW_t},
\end{displaymath}
where $W$ is a Brownian motion on the filtered probability space
$(\Omega,(\F_t)\tgo,\P)$. 
We also assume  that there is a local martingale $\tZ$ 
such that $\tZ\tS$ is a local martingale. This amounts to say that
$\tmu$ factorizes as $\tmu_t=\tr_t\tsigma_t$ with the ``Sharpe--ratio'' like
process $\tr$, which is  locally in 
$L^2$ almost surely. Then
\begin{displaymath}
  \tZ_t=\exp{-\int_0^t \tr_sdW_s-\frac12\int_0^t \tr_s^2ds}.
\end{displaymath}
$\tZ$ plays the role of the density process of an equivalent martingale
measure. We use the present form to avoid technical difficulties arising
from the possible nonexistence of an equivalent (local) martingale measure
on $\F_\infty$.

The value process $\tV=\tV^{\varphi^1,c}$ of a self--financing
strategy $(\varphi^1,c)$ is
\begin{displaymath}
  \tV_t=\tV_0+\int_0^t \varphi^1_sd\tS_s-\int_0^t c_sds.
\end{displaymath}
Note that if the strategy is admissible then $\tV_t\geq 0$ holds for all
$t\geq 0.$
Then for the discounted value process $\tZ_t\tV_t$
\begin{multline*}
  \tZ_t\tV_t-\tV_0=\int_0^t \varphi^1_s \tZ_sd\tS_s-\int_0^t \tZ_sc_sds+
  \int_0^t (\varphi^1_s\tS_s+X_s)d\tZ_s+\int_0^t \varphi^1_sd\q{\tS,\tZ}_s=\\
  \int_0^t \varphi^1_s d(\tZ\tS)_s-\int_0^t \tZ_sc_sds+\int_0^t X_sd\tZ_s,
\end{multline*}
where $X_t$ denotes the balance of the bank account at time $t$.
The point here is that $\tZ_t\tV_t+\int_0^t c_s\tZ_sds$ is a local martingale.
By stopping we get that
\begin{displaymath}
  \tV_0-\E{\tZ_{t\wedge\tau}\tV_{t\wedge\tau}|\F_0}=\int_0^t\E{\tZ_sc_s\I{s\leq
      \tau}|\F_0}ds
\end{displaymath}
Then letting $t$ and $\tau$ go to infinity we get that
\begin{equation}
  \label{eq:V0 geq}
  \tV_0\geq \E{\int_0^\infty \tZ_sc_sds|\F_0},
\end{equation}
provided that $c_t,\tV_t\geq 0$ for all $t\geq 0$.
From this we can conclude that if a strategy is self-financing and
admissible, then $\int_0^\infty \tZ_sc_sds$ is finite almost surely and its
generalized conditional expectation with respect to $\F_0$ exists and is
finite almost surely. In the special case when $\E{\tV_0}<\infty$, which is
usually assumed tacitly, we have that 
\begin{displaymath}
  \int_0^\infty \tZ_sc_sds\in L^1. 
\end{displaymath}
From \eqref{eq:V0 geq} an admissible strategy is optimal if
\begin{equation}
  \label{eq:V0=}
  \tV_0=\E{\int_0^\infty \tZ_sc_sds|\F_0}
\end{equation}
and then  
\begin{displaymath}
  \tZ_t\tV_t+\int_0^t c_s\tZ_sds=\E{\int_0^\infty \tZ_sc_sds|\F_t}
\end{displaymath}
is a true martingale.
This also gives us
\begin{equation}\label{eq:V from Q}
  \tZ_t\tV_t=\E{\int_t^\infty \tZ_sc_sds|\F_t},
\end{equation}
that is to say that $\tV_t$ is the present value of
future consumption for the optimal policy,  under the measure given by
density process $\tZ$. 


Observe also that if there exists an optimal strategy $(\varphi^1,c)$ and the
density process is $\tZ$ then
\begin{equation}\label{eq:s from Z}
  \tZ_t\tS_t=\frac{1}{\varphi^1_t}
  \zjel{\E{\int_t^\infty\tZ_sc_sds|\F_t}-\tZ_tX_t}. 
\end{equation}
  
Now 
the optimal consumption is given by
\begin{displaymath}
  c^*_t=I(\eta^*_t), 
\end{displaymath}
where $I=(u')^{-1}$ and 
\begin{displaymath}
  \eta^*_t=\eta^*_0 \tZ_te^{\delta t}.
\end{displaymath}
For details, see \cite{KS}.
The $\F_0$ measurable random variable $\eta^*_0$ is determined by
\eqref{eq:V0=}. For the power utility 
 \begin{align*}
  u(x)&=x^\gamma/\gamma,\quad u'(x)=x^{\gamma-1},\quad I(x)=x^{1/(\gamma-1)}.
\end{align*}
Hence $\eta_0^*$ satisfies
\begin{multline*}
  \tV_0=\E{\int_0^\infty \tZ_t (\eta_0^* e^{\delta t}
    \tZ_t)^{1/(\gamma-1)}dt|\F_0}=\\
  (\eta_0^*)^{1/(\gamma-1)}\int_0^\infty e^{-\frac{1}{1-\gamma}\delta t}
  \E{\tZ_t^{\gamma/(\gamma-1)}|\F_0}dt. 
\end{multline*}
It is assumed that $\int_0^\infty e^{-\frac{1}{1-\gamma}\delta t}
  \E{\tZ_t^{\gamma/(\gamma-1)}}dt$ is finite and therefore $\eta_0^*$
  satisfying this equation exists.
Then 
$$  
c^*_t=c_0^* \tZ_t^{1/(\gamma-1)} e^{\delta t/(\gamma-1)}, 
$$
where $c_0^*=(\eta_0^*)^{1/(\gamma-1)}$.
Later, we need also that $\tZ$ can be expressed using the optimal consumption
rate  
\begin{equation}\label{eq:Z c}
  \tZ_t=e^{-\delta t}\frac{u'(c^*_t)}{u'(c^*_0)}=
  e^{-\delta t} \zfrac{c^*_t}{c^*_0}^{\gamma-1}.
\end{equation}


\subsection{The problem with transaction costs}
\label{subsec:transaction cost}

In this section we
summarize the results for the Merton problem in case of power utility with
geometric Brownian motion price process. For details we refer to the papers
\cite{DN,SS} or the recent monograph \cite{Kabanov}.
  
The endowment at time $t$ is given by the pair $(X_t,Y_t)$, where $X_t$ is the
balance of the bank account and $Y_t$ is the value of the stocks we hold
evaluated using the mid-price $S_t$. 
Following \cite{SS}, we introduce the so-called Bellman or value function
$v(x,y)$ which denotes the optimal value gained from consumption if we start
with initial endowment $(x,y).$ Next, we denote by $\mathcal{S}$ the solvency
cone, which is the set of those $(x,y)$ for which the liquidation value is
still nonnegative: 
$$
\mathcal{S}=\set{(x,y)\in\real^2}{x+(1+\overline{\lambda})y\geq0\text{ and }
  x+(1-\underline{\lambda})y\geq0}.
$$

Then we have that the function $v$ is $C^2$ in $\mathcal{S}$, except maybe the
positive $y$-axis, and satisfies the following equation 
\begin{multline}
  \label{eq:HJB}
  \min\{\delta v(x,y)-\frac{1}{2}\sigma^2y^2v_{yy}(x,y)-\mu yv_y(x,y)-\tilde{u}(v_x(x,y)),\\
  -(1-\underline{\lambda})v_x(x,y)+v_y(x,y),(1+\overline{\lambda})v_x(x,y)-v_y(x,y)\}=0,
\end{multline}
where $\tilde{u}(p)=\frac{1-\gamma}{\gamma}p^{-\frac{\gamma}{1-\gamma}}$ is
the Legendre transform of $-u$.  

This equation tells us the following. The second and the third parts come
from the fact that for the optimal policy trading at a given situation, or
more precisely trading differently than according to the optimal strategy
cannot improve matters.  
 
The first one expresses that the process
$$
e^{-\delta t}v(X_t,Y_t)+\int_0^t{e^{-\delta s}u(c_s)}ds
$$ 
must be a supermartingale and for the optimal policy a (local)
martingale. This is not surprising if we consider that the first term above is
the discounted value of the value process starting from endowment $(X_t,Y_t)$,
and the second is the discounted value gained from the consumption process up
to time $t$. So the first one is what can be gained from consumption after
time $t$ and the second is what we have gained so far following a strategy up
to time $t$. 

We can partition the solvency cone according to which one of the above
equations is active, this way we get three sections, the no-trade (NT), the
selling-stock (SS), and the buying-stock (BS) (or SMM) regions. 

Our standing assumption is that the no-trade region (NT in the sequel) is
contained in the first quadrant of $\real^2.$ We also suppose that the
initial endowment $(x_0,y_0)$ is in the NT.

In the no-trade region $v$ satisfies the HJB-equation
\begin{equation}
  \label{eq:HJB v}
  \begin{aligned}
    \delta v(x,y)-\frac{1}{2}\sigma^2y^2v_{yy}(x,y)-\mu
    yv_y(x,y)-\tilde{u}(v_x(x,y))&=0,\\
    (1+\overline{\lambda})v_x(x,y)-v_y(x,y)&>0,\\
    -(1-\underline{\lambda})v_x(x,y)+v_y(x,y)&>0.
  \end{aligned}
\end{equation}
The latter two come from the fact that it is not optimal to buy or sell stocks
in the NT, see \cite{SS}. Note that they also mean that the fraction
$\frac{v_y(x,y)}{v_x(x,y)}$ does not leave the interval
$[1-\underline{\lambda},1+\overline{\lambda}].$ This will be important in the
sequel. 

For the SS section we have
$$
(1+\overline{\lambda})v_x(x,y)-v_y(x,y)=0
$$
and for the BS section
$$
-(1-\underline{\lambda})v_x(x,y)+v_y(x,y)=0.
$$
These equations correspond to the fact that when the fraction of wealth invested in stocks is too high (resp. too low), then it is optimal to sell (resp. buy) stocks. Due to continuity, these hold true for the respective boundaries of the NT and are henceforth referred to as the boundary conditions of the HJB-equation.

We can deduce from this by differentiating with respect to $x$ and $y$ that
for the SS section, except maybe for the positive $y$-axis,
$$
(1+\overline{\lambda})v_{xx}(x,y)-v_{xy}(x,y)=0 \quad\text{and}
\quad(1+\overline{\lambda})v_{xy}(x,y)-v_{yy}(x,y)=0
$$
and for the BS section
$$
-(1-\underline{\lambda})v_{xx}(x,y)+v_{xy}(x,y)=0 \quad\text{and}
-(1-\underline{\lambda})v_{xy}(x,y)+v_{yy}(x,y)=0.
$$ 
Thus these equations also hold for the respective boundaries of the NT. 

Using the homotheticity property of the Bellman function, for $y>0$ we can write 
$$v(x,y)=y^{\gamma}h\left(\frac{x}{y}\right)$$
with $h(u)=v(u,1).$
It is easy to see that because of this connection, $v$ and $h$ 
have the same order of smoothness. 

For this $h$ 
function the HJB-equation can be written in the form
\begin{multline}
  \label{eq:HJB h}
  \left(\delta-\gamma\mu+\frac12\gamma(1-\gamma)\sigma^2\right)h(u)+\left(\mu-(1-\gamma)\sigma^2\right)uh'(u)\\
-\frac12\sigma^2u^2h''(u)-\tilde{u}(h'(u))=0.
\end{multline}
For later use, we introduce the notation
\begin{multline}
  \label{eq:defH}
  H(u)=\\
  \zjel{\delta-\gamma\mu+\frac12\gamma(1-\gamma)\sigma^2}h(u)+\left(\mu-(1-\gamma)\sigma^2\right)uh'(u)-\frac12\sigma^2u^2h''(u).
\end{multline}
With this the HJB-equation becomes $H(u)=\tilde{u}(h'(u)).$
By rearranging the above equation, we can express $h''(u)$ 
by the lower order terms, which means that $h$ 
is smooth in 
$\left[\frac{1-\theta_2}{\theta_2},\frac{1-\theta_1}{\theta_1}\right]$, where
$[\theta_1,\theta_2]$ is the interval in which it is optimal to keep the
fraction of wealth invested in stocks. 
Hence we get that $v$ is smooth in the closure of the NT. Thus we can
differentiate the HJB-equation freely with respect to $x$ and $y$. 

As for the boundary conditions of the HJB-equation, they take the following
form for $h$ 
\begin{equation}
  \label{eq:boundary for h}
  \begin{aligned}
    (-1-u+\underline{\lambda})h'(u)+\gamma h(u)&>  0, 
    & u&\in\left[\tfrac{1-\theta_2}{\theta_2},
      \tfrac{1-\theta_1}{\theta_1}\right),\\
    &=  0, & u&=\tfrac{1-\theta_1}{\theta_1},\\
    (1+u+\overline{\lambda})h'(u)-\gamma h(u)&>  0, 
    & u&\in\left(\tfrac{1-\theta_2}{\theta_2},
      \tfrac{1-\theta_1}{\theta_1}\right],\\
    &=  0, & u&=\tfrac{1-\theta_2}{\theta_2}.
  \end{aligned}
\end{equation}
Shreve and Soner, see \cite{SS}, have also shown that $v$ is in $C^2$ except
maybe the $y$-axis. 
So, these conditions should be supplemented with the so-called smooth pasting
conditions which come from the aformentioned derivatives of the original
boundary conditions
\begin{equation}
  \label{eq:boundary for hprime}
  \begin{aligned}
    (-1-u+\underline{\lambda})h''(u)-(1-\gamma)h'(u)& =  0, &&
    \text{for  $u=\tfrac{1-\theta_1}{\theta_1}$},\\
    (1+u+\overline{\lambda})h''(u)+(1-\gamma)h'(u) &= 0, &&
    \text{for $u=\tfrac{1-\theta_2}{\theta_2}$}.
  \end{aligned}
\end{equation}

Shreve and Soner have also proven that for the optimal policy we have
\begin{equation}
  \label{eq:XY}
  \begin{aligned}
    (X_t,Y_t) &\in \overline{NT},\quad 
    t\geq0,\\
    dX_t&={} -v_x(X_t,Y_t)^{-\frac{1}{1-\gamma}}dt+r_1(X_t,Y_t)dk_t,\\
    dY_t&={} Y_t(\mu dt+\sigma dW_t)+r_2(X_t,Y_t)dk_t,\\
    dk_t&={} \I{(X_t,Y_t)\in\partial{NT}}dk_t,    
  \end{aligned}
\end{equation}
where 
\begin{displaymath}
  r(x,y)=(r_1(x,y),r_2(x,y))=
  \begin{cases}
    (-(1+\overline{\lambda}),1),&
    \text{if $(x,y)\in\partial{NT}\cap\partial{BS}$},\\  
    (1-\underline{\lambda},-1), &
    \text{if $(x,y)\in\partial{NT}\cap\partial{SS}$}.
  \end{cases}
\end{displaymath}
From \eqref{eq:XY} the  consumption rate of the optimal policy is 
\begin{equation}
  \label{eq:c v_x}
  c_t=v_x(X_t,Y_t)^{-\frac{1}{1-\gamma}}.  
\end{equation}
Note also, that $r$ is such that for $(x,y)\in\partial{NT}$
\begin{equation}
  \label{eq:dnt}
  v_{x}(x,y)r_1(x,y)+v_{y}(x,y)r_2(x,y)=0,
\end{equation}
and therefore for all $t\geq 0$
\begin{equation}
  \label{eq:HJB boundary}
  (v_{x}(X_t,Y_t)r_1(X_t,Y_t)+v_{y}(X_t,Y_t)r_2(X_t,Y_t))dk_t=0.
\end{equation}
Again we stress that these hold true if we use $v_x$ or $v_y$ instead of $v$. 
 
\section{Candidate for the shadow price process}

Assume that there is a shadow price $\tS$. Then the optimal policy under the
price  $S$ with transaction cost and under the price $\tS$ 
without transaction cost  have to be the same. Comparing
identities \eqref{eq:Z c} and
\eqref{eq:c v_x} gives that the candidate for the martingale density is 
\begin{displaymath}
  \tZ_t=e^{-\delta t}\frac{v_x(X_t,Y_t)}{v_x(X_0,Y_0)},
\end{displaymath}
where $X,Y$ is the solution of \eqref{eq:XY}. 

Now, given $(\varphi^1,c)$ and $\tZ$ we can use \eqref{eq:s from Z} to get a
candidate for the shadow price:
\begin{displaymath}
  \tZ_t\tS_t=\frac{1}{\varphi^1_t}\zjel{\E{\int_t^\infty c_s\tZ_sds|\F_t}-X_t\tZ_t}
\end{displaymath}
Observe that the special structure of the power utility function makes it
possible to express the conditional expectation as
\begin{displaymath}
  \E{\int_t^\infty c_s\tZ_sds|\F_t}=
  \E{\frac{1}{c_0^{\gamma-1}}\int_t^\infty e^{-\delta s} c_s^{\gamma}ds|\F_t}
  =\frac{e^{-\delta t}\gamma v(X_t,Y_t)}{v_x(X_0,Y_0)}
\end{displaymath}
and 
\begin{displaymath}
  X_t\tZ_t=X_t \frac{e^{-\delta t} v_x(X_t,Y_t)}{v_x(X_0,Y_0)}.
\end{displaymath}
Since 
$v(x,y)=y^\gamma h\zfrac{x}{y}$ 
we have that
\begin{equation}
  \label{eq:gv=xvx+yvy}
  \gamma v(x,y)= xv_x(x,y)+yv_y(x,y).
\end{equation}
Hence, using that $\varphi^1_t=Y_t/S_t$,
\begin{displaymath}
  \tZ_t\tS_t=\frac{S_t}{Y_t}\zfrac{e^{-\delta t} Y_tv_y(X_t,Y_t)}{v_x(X_0,Y_0)}=
  \frac{e^{-\delta t} S_tv_y(X_t,Y_t)}{v_x(X_0,Y_0)},
\end{displaymath}
that is, the shadow price candidate $\tS$ can be written as
\begin{equation}
  \label{eq:tS def}
  \tS_t=\frac{S_tv_y(X_t,Y_t)}{v_x(X_t,Y_t)}.
\end{equation}

\begin{thm}
	\label{thm:sp}
  Assume that the no-trade region is inside the positive quadrant of $\real^2.$ Then
  $\tS$ defined in \eqref{eq:tS def} is the shadow price for the problem with
  transaction costs. 
\end{thm}
\begin{proof}
  First, $\tS$ lies in the bid-ask spread of
  the stock price $S$ thanks to \eqref{eq:HJB}. 
  Next, let us try to identify the local martingale density process. Define
  $$
  Z_t=e^{-\delta t}\frac{v_x(X_t,Y_t)}{v_x(X_0,Y_0)}.
  $$
  Then we have
  \begin{align*}
    \frac{dZ_t}{Z_t}&=  -\delta dt+\frac{v_{xx(X_t,Y_t)}}{v_x(X_t,Y_t)}dX_t+\frac{v_{xy}(X_t,Y_t)}{v_x(X_t,Y_t)}dY_t
    +\frac12\frac{v_{xyy}(X_t,Y_t)}{v_x(X_t,Y_t)}d\q{Y}_t=\\
    &=  -\xi_tdW_t
  \end{align*}
  with $\xi_t=-\sigma Y_t\frac{v_{xy}(X_t,Y_t)}{v_x(X_t,Y_t)}.$ The drift and
  the $dk_t$ terms disapper due to the HJB-equation \eqref{eq:HJB v} and the
  boundary conditions \eqref{eq:HJB boundary} differentiated with respect to $x$.

  Similar argument shows that
  \begin{equation*}
    M_t=\frac{1}{v_x(X_0,Y_0)}e^{-\delta t}S_tv_y(X_t,Y_t)
  \end{equation*} 
  is a local martingale. Indeed
  by It\^{o} formula we get 
  dynamics of $M_t$ (we omit the arguments
  of the functions for brevity): 
  \begin{align*}
    \frac{dM_t}{M_t}&= 
    \zjel{-\delta+\mu+\sigma^2Y_t\frac{v_{yy}}{v_y}}dt+\sigma
    dW_t+\frac{v_{xy}}{v_y}dX_t+\frac{v_{yy}}{v_y}dY_t+
    \frac{1}{2}\frac{v_{yyy}}{v_y}d\q{Y}_t\\ 
    &=  \nu_tdW_t
  \end{align*}
  with $\nu_t=\sigma\left(\frac{Y_tv_{yy}(X_t,Y_t)}{v_y(X_t,Y_t)}+1\right).$
  Here the drift and the $dk_t$ term disapper due to the HJB-equation
  \eqref{eq:HJB v} and the 
  boundary conditions \eqref{eq:HJB boundary} differentiated with respect to $y$.
  
  Observe that $\tS_t=\frac{M_t}{Z_t}$, i.e.,  
  $\tS_tZ_t$ is a local martingale, and $Z_t$ is indeed the
  local martingale measure density for the process $\tS_t.$ 
  Hence we get that the optimal consumption with respect to this process is
  $$  
  c^*_t=
  c_0^*Z_t^{1/(\gamma-1)} e^{\delta t/(\gamma-1)}=
  c_0^*\frac{v_x(X_t,Y_t)^{-1/(1-\gamma)}}{v_x(X_0,Y_0)^{-1/(1-\gamma)}}.
  $$
  where $c_0^*$ is determined by the identity 
  \begin{equation}
    \label{eq:c0 star}
    \tV_0=c_0^*\int_0^\infty
    e^{-\frac{1}{1-\gamma}\delta t}\E{Z_t^{\gamma/(\gamma-1)}|\F_0}dt
  \end{equation}
  The left hand side of \eqref{eq:c0 star} is
  \begin{displaymath}
    \tV_0=x_0+\frac{y_0}{S_0}\tilde{S}_0=\frac{xv_x(x_0,y_0)+y_0v_y(x_0,y_0)}{v_x(x_0,y_0)}=\gamma\frac{v(x_0,y_0)}{v_x(x_0,y_0)}.
  \end{displaymath}
  In the last step we used \eqref{eq:gv=xvx+yvy}.
  For the right hand side \eqref{eq:c0 star} we use that $v$ is the value
  function and the  optimal policy is described in terms of $(X,Y)$ and $v$ as
  in Subsection \ref{subsec:transaction cost}:
\begin{multline*}  
   v(X_0,Y_0)=\E{\int_0^\infty e^{-\delta t}u(c_t)dt|\F_0}=\\
  \frac1\gamma 
  \E{\int_0^\infty e^{-\delta t}v_x(X_t,Y_t)^{-\frac{\gamma}{1-\gamma}}dt|\F_0}\\
  = \frac{v^{-\gamma/(1-\gamma)}_x(X_0,Y_0)}{\gamma}
  \E{\int_0^\infty e^{-\frac{\delta}{1-\gamma} t} Z_t^{\gamma/(\gamma-1)} dt|\F_0}.
\end{multline*} 
Then rearranging gives that $c_0^*=v_x(X_0,Y_0)^{-\frac1{1-\gamma}}$.
Hence $c_t^*=v_x(X_t,Y_t)^{-\frac{1}{1-\gamma}}$, 
which is the coefficient of the drift term in the evolution of $X_t$.  

Next we show that the optimal startegy $(\varphi^1,c^*)$ of the problem with
transaction cost is admissible and self-financing for the price process $\tS$
without transaction cost.

The self-financing property follows form the fact that $\varphi^1$ only changes
on the boundary of the no-trade region and due to the boundary conditions for
$v$ the shadow price candidate $\tS$ and $S$ coincide on the support of
$d\varphi^1$. 

The value process of this strategy is
$$
\tilde{V}_t=\varphi_t^0+\varphi_t^1\tilde{S}_t=X_t+Y_t\frac{\tilde{S}_t}{S_t}\geq0
$$
as $(X_t,Y_t)\in\mathcal{S}$ and
$\frac{\tS_t}{S_t}\in[1-\underline{\lambda},1+\overline{\lambda}].$ 
Thus this $\varphi_t$ is an admissible strategy and  $c_t$ is the optimal
consumption, we get that $(\varphi_t,c_t)$ is the optimal
portfolio/consumption pair for the process $\tilde{S}_t$ and as the optimal
strategy only trades on the boundaries, we have shown that $\tilde{S}_t$
really is the shadow price. 
\end{proof}
\begin{remark}
Note, that $Z$ and $M$ are true martingales. Hence $Z$ is martingale density
for $\tS$. 
The first part follows form the fact that 
$\xi$ is uniformly bounded since  
$$
\xi_t=\sigma\left((1-\gamma)+U_t\frac{h''(U_t)}{h'(U_t)}\right)
$$
which is a continuous function of the bounded process 
$U_t=\frac{X_t}{Y_t}$. Here we use the assumption that the no-trade region 
does not meet the $x$--axis. 
This makes $Z_t$ a true martingale. 
 
For $M$ we argue similarly.  
$$
\nu_t=
\sigma\zjel{\frac{-\gamma(1-\gamma)h(U_t)+2(1-\gamma)U_th'(U_t)+U_t^2h''(U_t)}
  {\gamma h(U_t)-U_th'(U_t)}+1}
$$ 
is uniformly bounded, making $M_t$ a true martingale.
\end{remark}

\section{Connection with work of  \cite{KMK}}

For the shadow price, we need the function $v$. If we have such a
function that satisfies the HJB-equation with the appropriate boundary
conditions, then we can define the processes $X$ and $Y$, and with that our
candidate for the shadow price. Due to the homotheticity property, it is
enough to determine the function $h$ for which we have a free boundary
problem, though it is not that simple to solve, see \cite{DN}. In the
following, we show how to transform it in order to come up with a somewhat
simpler problem, which also demonstrates the connection of the HJB-equation
and equation (3.15) in \cite{KMK}. 

To derive the shadow price in the logarithmic utility case, Kallsen and
Muhle-Karbe used the function $g$, which made the connection between $C_t$,
the difference between the logarithm of the shadow and original price process,
and $\beta_t$, the difference between the logarithm of the value held in stock
and bond with respect to the shadow price, i.e. they showed that
$C_t=g(\beta_t)$ and that this $g$ function satisfies a free boundary
problem. 

In our case
$$
C_t=\log\zfrac{\tS_t}{S_t}=\log\zfrac{v_y(X_t,Y_t)}{v_x(X_t,Y_t)}
$$
and
\begin{equation}
  \label{eq:betadef}
  \beta_t=\log\zfrac{\varphi_t^1\tS_t}{\varphi_t^0}=
  \log\zfrac{Y_tv_y(X_t,Y_t)}{X_tv_x(X_t,Y_t)},  
\end{equation}
thus
\begin{equation}
  \label{eq:Cbeta}
  C_t=\beta_t+\log\zfrac{X_t}{Y_t}.
\end{equation}

As $v(x,y)=y^{\gamma}h\zfrac{x}{y}$, we have that 
$$
\frac{v_y(x,y)}{v_x(x,y)}=\frac{\gamma h(u)-uh'(u)}{h'(u)},
$$ with $u=\frac{x}{y}$. 
So $g$ can be expressed as $g(z)=z+\log(f^{-1}(z))$,  where
$$
f(u)=\log\zfrac{\gamma h(u)-uh'(u)}{uh'(u)}.
$$
For this $f$ we have
\begin{equation}
  \label{eq:uf'}
  uf'(u)=-1-(1-\gamma)e^{-f(u)}-(1+e^{-f(u)})u\frac{h''(u)}{h'(u)}
\end{equation}
and
\begin{multline}\label{eq:f2 with h}
  u^2f''(u)=  
  -uf'(u)+uf'(u)(1-\gamma)e^{-f(u)}+uf'(u)e^{-f(u)}u\frac{h''(u)}{h'(u)}\\
  -(1+e^{-f(u)})
  \zjel{u\frac{h''(u)}{h'(u)}+u^2\frac{h'''(u)}{h'(u)}-
    u^2\frac{h''^2(u)}{h'^2(u)}}.
\end{multline}
From \eqref{eq:uf'} we have
\begin{equation}\label{eq:h21 with f}
u\frac{h''(u)}{h'(u)}=-\frac{1+(1-\gamma)e^{-f(u)}-uf'(u)}{1+e^{-f(u)}},
\end{equation}
and as for \eqref{eq:f2 with h}, we need to express
$u^2\frac{h'''(u)}{h'(u)}$ in terms of $f(u).$ For this we turn to the
HJB-equation for $h$ and its derivative 
\begin{multline}
  \label{eq:h(3)}
  \zjel{\delta+(1-\gamma)\mu-\frac12(1-\gamma)(2-\gamma)\sigma^2}h'(u)+
  (\mu-(2-\gamma)\sigma^2)uh''(u) \\
  -\frac12\sigma^2u^2h'''(u)+h'(u)^{-\frac{1}{1-\gamma}}h''(u)  =0.
\end{multline}
With the notation \eqref{eq:defH} this becomes
$H'(u)+h'(u)^{-\frac{1}{1-\gamma}}h''(u)=0.$ 

Now from the HJB-equation, we can express
$h'(u)^{-\frac{1}{1-\gamma}}=\frac{\gamma}{1-\gamma}\frac{\tilde{u}(h'(u))}{h'(u)}$
and putting this into \eqref{eq:h(3)}, we get 
\begin{equation}\label{eq:h31 with h21}
  \begin{aligned}
    u^2\frac{h'''(u)}{h'(u)}= &
    \zjel{\frac{2\delta}{\sigma^2}+
      (1-\gamma)\frac{2\mu}{\sigma^2}-(1-\gamma)(2-\gamma)}\\ 
    & + \zjel{\frac{1}{1-\gamma}\frac{2\delta}{\sigma^2}-
      \frac{\gamma}{1-\gamma}\frac{2\mu}{\sigma^2}+\gamma}
    \frac{\gamma h(u)}{uh'(u)}u\frac{h''(u)}{h'(u)}\\
    & + \zjel{\frac{1}{1-\gamma}\frac{2\mu}{\sigma^2}-4}
    u\frac{h''(u)}{h'(u)}- \frac{\gamma}{1-\gamma}u^2\frac{h''^2(u)}{h'^2(u)}.
  \end{aligned}
\end{equation}
Putting this and \eqref{eq:h21 with f} into \eqref{eq:f2 with h} and using
$\frac{\gamma h(u)}{uh'(u)}=1+e^{f(u)}$, we have
\begin{equation}
  \label{eq:f}
  \begin{aligned}
    & u^2f''(u)+uf'(u)=\\
    = & \frac{1}{1-\gamma}\frac{2\delta}{\sigma^2}(1+e^{f(u)})-
    \frac{\gamma}{1-\gamma}\frac{2\mu}{\sigma^2}e^{f(u)}+
    \frac{\gamma^2}{1-\gamma}\frac{1}{1+e^{-f(u)}}+\gamma e^{f(u)}\\
    & + \zjel{\frac{1}{1-\gamma}\frac{2\delta}{\sigma^2}(1+e^{f(u)})-
      \frac{1}{1-\gamma}\frac{2\mu}{\sigma^2}(\gamma(1+e^{f(u)})-1)}uf'(u)\\
    & + \zjel{-1+\frac{\gamma(3-\gamma)}{1-\gamma}\frac{1}{1+e^{-f(u)}}+
      \gamma e^{f(u)}}uf'(u)\\
    & + \zjel{-1+\frac{\gamma}{1-\gamma}\frac{1}{1+e^{-f(u)}}+
      \frac{2}{1+e^{-f(u)}}}u^2f'^2(u).
  \end{aligned}
\end{equation}
The boundary conditions \eqref{eq:boundary for h} and \eqref{eq:boundary for
  hprime} take the form 
\begin{equation}\label{eq:fbound}
  \begin{aligned}
    \log{u}+f(u)&>  \log{(1-\underline{\lambda})}, & 
    u&\in\left[\tfrac{1-\theta_2}{\theta_2},\tfrac{1-\theta_1}{\theta_1}\right),\\
    &=  \log{(1-\underline{\lambda})}, &u&=\tfrac{1-\theta_1}{\theta_1},\\
    \log{u}+f(u)&<  \log{(1+\overline{\lambda})}, &
    u&\in\left(\tfrac{1-\theta_2}{\theta_2},\tfrac{1-\theta_1}{\theta_1}\right],
    \\
    &=  \log{(1+\overline{\lambda})}, & u&=\tfrac{1-\theta_2}{\theta_2},\\
    uf'(u)&=  -1, &
    u&=\tfrac{1-\theta_2}{\theta_2},\tfrac{1-\theta_1}{\theta_1}. 
  \end{aligned} 
\end{equation}

As $g(z)=z+\log{(f^{-1}(z))}$, with $z=f(u)$ we have
\begin{align*}
  uf'(u)&=  \frac{1}{g'(z)-1}\\
  u^2f''(u)+uf'(u)& =  \frac{g''(z)}{(1-g'(z))^3}.
\end{align*}
Thus \eqref{eq:f}, can be recast in terms of $g$ yielding
\begin{equation}\label{eq:g}
  g''(z)=a(z)+b(z)g'(z)+c(z)g'^2(z)+d(z)g'^3(z)
\end{equation}
with coefficients
\begin{align*}
  a(z)= & -\frac{2\mu}{\sigma^2}+2(1-\gamma)\frac{1}{1+e^{-z}},\\
  b(z)= &
  -\frac{1}{1-\gamma}\frac{2\delta}{\sigma^2}(1+e^z)+
  \frac{\gamma}{1-\gamma}\frac{2\mu}{\sigma^2}e^z+\frac{4\mu}{\sigma^2}-1+
  (5\gamma-2)\frac{1}{1+e^{-z}}-\gamma 
  e^z,\\ 
  c(z)= &
  \frac{1}{1-\gamma}\frac{4\delta}{\sigma^2}(1+e^z)-
  \frac{\gamma}{1-\gamma}\frac{4\mu}{\sigma^2}e^z-
  \frac{2\mu}{\sigma^2}+1+
  \frac{\gamma(4\gamma-3)}{1-\gamma}\frac{1}{1+e^{-z}}+2\gamma
  e^z,\\ 
  d(z)= &
  -\frac{1}{1-\gamma}\frac{2\delta}{\sigma^2}(1+e^z)+
  \frac{\gamma}{1-\gamma}\frac{2\mu}{\sigma^2}e^z-
  \frac{\gamma^2}{1-\gamma}\frac{1}{1+e^{-z}}-\gamma e^z. 
\end{align*}
Note that for $\gamma=0$ this is exactly the same as (3.15) in \cite{KMK}. 

As for the boundary conditions we have the following:
\begin{equation}\label{eq:gbound}
  g(\underline{\beta})=\log{(1+\overline{\lambda})}, 
  \quad g(\overline{\beta})=\log{(1-\underline{\lambda})}, 
  \quad g'(\underline{\beta})=0, \quad g'(\overline{\beta})=0.
\end{equation}

In order to construct the value function, we start with the solution of
\eqref{eq:g}.  

\begin{prop}\label{prop:g}
Suppose that 
\begin{list}{\hbox to 0pt{\hss(\theenumi)}\unskip\ignorespaces}{%
      \usecounter{enumi}}
    \def\theenumi{\roman{enumi}}
\item\label{prop:g:it:1} $(1-\gamma)\sigma^2>\mu>0$ and
\item\label{prop:g:it:2}  $\delta-\gamma\mu+\frac12\gamma(1-\gamma)\sigma^2>0.$
\end{list}
Then the free boundary problem \eqref{eq:g} with boundary conditions \eqref{eq:gbound} has a strictly decreasing solution.
\end{prop}

\begin{proof}
The proof is practically the same as that of Proposition 4.2 in \cite{KMK}. We
repeat the beginning of the proof as there is a misleading typo in it, the
rest goes through with the obvious modifications, so we only sketch the main
steps.  

By
assumption (\ref{prop:g:it:1}) 
there is a unique $z_0$  such that $a(z_0)=0$. 
For any
$\underline{\beta}_\Delta=z_0-\Delta$ with $\Delta>0$, there is a local
solution $g_\Delta$ to \eqref{eq:g} with initial conditions
$g_\Delta(\underline{\beta}_\Delta)=\log{(1+\overline{\lambda})}$ and
$g'_\Delta(\underline{\beta}_\Delta)=0.$  
Note that 
\begin{align*}
  b(z)&=  d(z)+\frac{4\mu}{\sigma^2}-1+
  \frac{-4\gamma^2+7\gamma-2}{1-\gamma}\frac{1}{1+e^{-z}}\\
  c(z)&=  -2d(z)-\frac{2\mu}{\sigma^2}+1+
  \frac{2\gamma^2-3\gamma}{1-\gamma}\frac{1}{1+e^{-z}},
\end{align*}
and $\sup_{z\in\real} d(z)<0$ by assumption (\ref{prop:g:it:2}). 
Then for sufficiently large $M'$ the sign of $g''$ is determined by $d(z)(
g'(z))^3$ whenever $\abs{g'(z)}>M'$. 
That is, we have $g''_{\Delta}(z)<0$ for $g'_{\Delta}(z)>M'$ and
$g''_{\Delta}(z)>0$ for $g'_{\Delta}(z)<-M'.$ Therefore $g'_{\Delta}$ can
never leave the interval $[-M',M']$ for $z\geq\underline{\beta}_{\Delta}.$ 

\smallskip
Following \cite{KMK} we let
$\overline{\beta}_{\Delta}=\inf\set{z>\underline{\beta}_{\Delta}}{g'(z)=0}$. 
Since $a(\underline{\beta}_{\Delta})<0$,
the function  
$g_{\Delta}$ is strictly decreasing on
$[\underline{\beta}_\Delta,\overline{\beta}_\Delta].$ 

Then it remains to show that for any
$L<\log{(1+\overline{\lambda})}$  there is a $\Delta$ such that
$g_{\Delta}\zjel{\overline{\beta} _\Delta}=L$.
The continuity argument of \cite{KMK} applies to our case.
First, it can be shown that
$g_{\Delta}\left(\overline{\beta}_{\Delta}\right)\to\log{(1+\overline{\lambda})}$
as $\Delta\to0$.
Next, it can be established that
$\overline{\beta}_{\Delta}>z_0$. After that, it can be seen that
$g_{\Delta}\left(\overline{\beta}_{\Delta}\right)\to-\infty$ as
$\Delta\to\infty$. Moreover, $(g_{\Delta},g'_{\Delta})$
converges toward $(g_{\Delta_0},g'_{\Delta_0})$ uniformly on compacts as
$\Delta\to\Delta_0.$ Finally, from the precedings it follows that
$g_{\Delta}\left(\overline{\beta}_{\Delta}\right)$ depends continuously on
$\Delta,$ which completes the proof.  
\end{proof}

Even though the proof works with
$\delta-\gamma\mu+\frac12\gamma(1-\gamma)\sigma^2>0,$ we are going to need a
stricter version of it, which is a common standing assumption in related works 
\begin{equation}\label{eq:assum}
  \delta>\frac12\frac{\gamma}{1-\gamma}\frac{\mu^2}{\sigma^2}. 
\end{equation}
This assumption is a necessary and sufficient condition for the value function
in the problem without transaction costs to be finite and thus also a
sufficient one when transaction costs are present. Observe that
\eqref{eq:assum} implies $\delta-\gamma\mu+\frac12\gamma(1-\gamma)\sigma^2>0.$ 

Now that we have $g$, we can define 
\begin{equation}\label{eq:def f}
  f(u)= \left(e^{g-id}\right)^{-1}(u)
\end{equation}
Note that  the domain of $f$ is $[u_1,u_2]$, where 
\begin{align*}
u_1&=  \frac{1-\underline{\lambda}}{e^{\overline{\beta}}},& 
u_2&=  \frac{1+\overline{\lambda}}{e^{\underline{\beta}}}.
\end{align*}
As $g$ is strictly decreasing and
$g\colon[\underline{\beta},\overline{\beta}]\to[\log{(1-\underline{\lambda})},\log{(1+\overline{\lambda})}]$,
the function $f\colon[u_1,u_2]\to[\underline{\beta},\overline{\beta}]$ is also
strictly decreasing and satisfies the free boundary problem \eqref{eq:f} with
boundary conditions \eqref{eq:fbound} by taking $\theta_1=\frac{1}{1+u_2}$ and
$\theta_2=\frac{1}{1+u_1}.$ 

Next we define $h$ as
\begin{equation}\label{eq:def h}
  h(u)=Ke^{\gamma\int_{u_1}^u{\frac{1}{v\left(1+e^{f(v)}\right)}}dv}
\end{equation}
where $K$ is chosen such that $H(u_1)=\tilde{u}(h'(u_1))$ is satisfied. For
this $H(u_1)>0$ is needed and as we are going to need it later on, we prove
now that $H(u)$ is always positive.  

From the definition of $h$, we have
\begin{align*}
  \frac{h'(u)}{h(u)}&=  \frac{\gamma}{u\zjel{1+e^{f(u)}}}\\
  \frac{h''(u)}{h(u)}&=  
  \frac{\gamma}{u^2\zjel{1+e^{f(u)}}^2}\zjel{\gamma-1-(1+uf'(u))e^{f(u)}}.
\end{align*}
Using that $uf'(u)\geq-1$, we get
\begin{align*}
  \frac{H(u)}{h(u)}&=  \zjel{\delta-\gamma\mu+\frac12\gamma(1-\gamma)\sigma^2}
  +(\mu-(1-\gamma)\sigma^2)\frac{\gamma}{1+e^{f(u)}}\\
  &\hphantom{{} ={}} - \frac12\sigma^2\frac{\gamma}{\zjel{1+e^{f(u)}}^2}
  \zjel{\gamma-1-(1+uf'(u))e^{f(u)}}\\
  &\geq  \zjel{\delta-\gamma\mu+\frac12\gamma(1-\gamma)\sigma^2}+
  \frac{\gamma(\mu-(1-\gamma)\sigma^2)}{1+e^{f(u)}}+
  \frac12\sigma^2\frac{\gamma(1-\gamma)}{\zjel{1+e^{f(u)}}^2}\\
  &=  \delta-\gamma\mu\frac{1}{1+e^{-f(u)}}+
  \frac12\gamma(1-\gamma)\sigma^2\frac{1}{\zjel{1+e^{-f(u)}}^2}
\end{align*}
with equality at $u=u_1,u_2.$
This is a second-order polynomial of $\frac{1}{1+e^{-f(u)}}$.
As the minimum is attained at
$\frac{\mu}{(1-\gamma)\sigma^2}\in(\frac{1}{1+e^{-\underline{\beta}}},\frac{1}{1+e^{-\overline{\beta}}})$,
we need that this polynomial is always positive, which holds if and only if 
$$\mu^2\gamma^2<2\delta\gamma(1-\gamma)\sigma^2\Leftrightarrow\frac12
\frac{\gamma}{1-\gamma}\frac{\mu^2}{\sigma^2}<\delta.$$ This is why we need
this assumption.  

The fact that $f$ satisfies \eqref{eq:f}, yields that the function $h$
satisfies \eqref{eq:h31 with h21}. From this it can be shown that $h$
satisfies the HJB-equation. Let us start with the boundary conditions. The
definition of $h$ implies that  
\begin{equation}\label{eq:f with h}
f(u)=\log{\left(\frac{\gamma h(u)}{uh'(u)}-1\right)}
\end{equation}
meaning that 
$$\frac{\gamma h(u)}{uh'(u)}=1+e^{f(u)}$$
which is again strictly decreasing, giving that 
\begin{align*}
  \frac{\gamma h(u_1)}{u_1h'(u_1)}&=  1+e^{\underline{\beta}},\\
  1+e^{\underline{\beta}}< \frac{\gamma h(u)}{uh'(u)}&<  1+e^{\overline{\beta}},\quad u\in(u_1,u_2),\\
  \frac{\gamma h(u_2)}{u_2h'(u_2)}&=  1+e^{\overline{\beta}}.
\end{align*}
Using the definitions of $u_1$ and $u_2$, it is clear that \eqref{eq:boundary
  for h} holds as $uh'(u)$ is always positive. To get the smooth pasting
conditions, we take a look at the boundary conditions of $f$, namely
$uf'(u)=-1$ for $u=u_1,u_2.$ By differentiating \eqref{eq:f with h}, we get  
$$
u\frac{h''(u)}{h'(u)}=-(1-\gamma)\frac{1}{1+e^{f(u)}}-\frac{1+uf'(u)}{1+e^{-f(u)}}
$$ 
meaning that
\begin{align*}
  u_1\frac{h''(u_1)}{h'(u_1)}&= 
  -(1-\gamma)\frac{u_1}{1+u_1-\underline{\lambda}}\\
  u_2\frac{h''(u_2)}{h'(u_2)}&=  -(1-\gamma)\frac{u_2}{1+u_2+\overline{\lambda}}.
\end{align*}
Thus \eqref{eq:boundary for hprime} holds as well.

The only thing left to be shown is that $h$ satisfies the equation \eqref{eq:HJB h}. This follows from \eqref{eq:h31 with h21}. If we multiply it by $\frac12\sigma^2h'(u)$ and rearrange it, we have
$$
-H'(u)=\frac{\gamma h''(u)}{(1-\gamma)h'(u)}H(u),
$$
which we can divide by $H$ to get
$$\frac{H'(u)}{H(u)}=-\frac{\gamma}{1-\gamma}\frac{h''(u)}{h'(u)}.$$
This yields $K_1H(u)=\tilde{u}(h'(u))$ with some contant $K_1$. 
Since $H(u_1)=\tilde{u}(h'(u_1))$ by the definition of $h$ we have that $K_1=1$.
Hence the HJB-equation \eqref{eq:HJB h} holds for $h$, which in turn means
that $v (x,y) = y^{\gamma}h\zfrac{x}{y}$ satisfies the HJB-equation
\eqref{eq:HJB v} on $\set{(x,y)\in\real^2}{y>0,\,u_1<x/y<u_2}$.  

To summarize, we obtained the following result
\begin{prop}\label{prop:4.2}
  Assume that \eqref{eq:assum} and $(1-\gamma)\sigma^2>\mu>0$ hold.
  Let $g$ be the solution to the free boundary problem \eqref{eq:g},
  \eqref{eq:gbound}.  
  Define $f$, $h$  as in \eqref{eq:def f}, 
  \eqref{eq:def h}, respectively and let 
  \begin{displaymath}
    v(x,y)=y^{\gamma}h\zfrac{x}{y},\quad\text{for $(x,y)\in
      NT$},   
  \end{displaymath}
  where $NT=\set{(x,y)\in \real^2}{y>0,\, u_1< \frac xy<  u_2}$. The function
  $v$ and its derivatives extend continuously to $\overline{NT}$.

  Then $v$ is a solution to \eqref{eq:HJB v} on $NT$, satisfies 
  \begin{equation}
    \label{eq:vx,vy}
    (1-\underline\lambda)v_x(x,y)\leq v_y(x,y)\leq
    (1+\overline\lambda)v_x(x,y),\quad\text{for $(x,y)\in \overline{NT}$}
  \end{equation}
  and the
  boundary conditions 
  \begin{align*}
    (1+\overline{\lambda})v_x(u_1y,y)&=v_y(u_1y,y), &
    (1-\underline{\lambda})v_x(u_2y,y)&=v_y(u_2y,y), \\
    (1+\overline{\lambda})v_{xx}(u_1y,y)&=v_{yx}(u_1y,y),
    &(1-\underline{\lambda})v_{xx}(u_2y,y)&=v_{yx}(u_2y,y),\\
    (1+\overline{\lambda})v_{xy}(u_1y,y)&=v_{yy}(u_1y,y),
    &(1-\underline{\lambda})v_{xy}(u_2y,y)&=v_{yy}(y_2y,y),
  \end{align*}
  for $y>0$. 
\end{prop}
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1112.4385.extracted_bib
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\end{thebibliography}