# Insider trading with penalties, entropy and quadratic BSDEs

Umut Çetin  Department of Statistics, London School of Economics and Political
Science, 10 Houghton st, London, WC2A 2AE, UK

###### Abstract.

Kyle model in continuous time where the insider may be subject to legal
penalties is considered. In equilibrium the insider internalises this legal
risk by trading less aggressively. The equilibrium is characterised via the
solution of a backward stochastic differential equation (BSDE) whose terminal
condition is determined as the fixed point of a non-linear operator in
equilibrium. The insider’s expected penalties in equilibrium is non-monotone
in the fee structure and is given by the relative entropy of the law of a
particular $h$-transformation of Brownian motion.

##  1\. Introduction

Kyle [14] (together with its formulation in continuous time by Back [1]) is
the canonical model for analysing the dissemination of private information
into prices in financial markets. In a market consisting of noise traders,
competitive market makers and an informed trader (insider), it predicts that
the insider gradually reveals all her private information to the market in
equilibrium. At the end of the finite horizon, the prices become fully
informative in the sense that there is no remaining uncertainty about the
private valuation of the insider.

The Kyle model does not distinguish between legal informed trading, i.e. the
investor paying a fee to obtain private information typically as a result of
fundamental research, and insider trading when the trader receives private
information from an insider via illegal means. As there is no legal penalties
in the Kyle model, both types of informed traders behave identically.

Incorporation of legal risk into the Kyle model is a relatively less studied
extension. The previous works in the literature typically assume a one period
setting (see, e.g., Shin [16] and Carré et al. [7]) with the exception of
Kacperczyk and Pagnotta [12] studying a two-period model111See also DeMarzo et
al. [11] for another study of regulation of insider trading in a one-period
model but outside Kyle’s framework.. Although some insiders trade in blocks,
they typically split their orders over a relatively long trading period.
Indeed, Kacperczyk and Pagnotta [12] find that “the median trader splits
trades over a period equivalent to 71% of the information horizon.” Thus,
understanding the impact of legal risk in continuous time is essential for the
regulation of illegal insider trading.

This paper analyses the impact of legal penalties on insider trading in a
continuous time Kyle model. Another notable distinction from the earlier works
is the fundamental value $V$ of the traded asset is allowed to have an
arbitrary distribution while the aforementioned works assume a Gaussian or
uniform distribution. Exi

Following [16] and [12] the insider pays a quadratic penalty if caught by the
authorities as a result of a successful investigation, in addition to losing
all her previous gains from trading. The presence of legal penalties makes the
resolution of equilibrium in closed form less likely. However, quadratic (or
more general convex) penalties results in the optimal strategy of the insider
being given in the feedback form as a function of the pricing rule to be
determined in equilibrium. Section 3 shows that the value function of the
insider is the sum of two components: while the first component is a solution
of heat equation reminiscent of the value function of the Kyle model with no
penalties, the second component is given by the solution of the following
quadratic BSDE (backward stochastic differential equation)

$dU_{t}=\sigma Z_{t}dB_{t}-\frac{1}{2c}Z_{t}^{2}dt,$ (1.1)

whose terminal condition is to be determined in equilibrium. Moreover, the
optimal strategy of the insider coincides with a constant multiple of the
process $Z$ appearing in the above BSDE, whose explicit solution is available
once its terminal condition is established.

Section 4 describes an operator whose fixed point, if it exists, yields the
terminal condition of the BSDE (1.1). This requires finding two convex
functions $\phi$ and $\Psi$ such that

$\begin{split}\int_{\mathbb{R}}\exp\Big{(}\frac{yv-\phi(y)}{\hat{c}}\Big{)}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy&=\exp\Big{(}\frac{\Psi(v)}{\hat{c}}\Big{)},\\\
\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)&=\exp\Big{(}\frac{\phi(y)}{\hat{c}}\Big{)},\qquad
y\in\mathbb{R},\end{split}$ (1.2)

where $\Pi$ is the distribution of $V$, $v$ is an arbitrary element of the
support of $\Pi$, and $\hat{c}$ and $\sigma$ are constants. The above
describes a non-linear operator whose fixed point will determine $\phi$, and,
consequently $\Psi$. Under the assumption that $V$ is bounded, Theorem 4.1
establishes the existence of a fixed point $\Phi$ that is convex. However,
boundedness of the asset value is not a necessary condition for the existence:
Theorem 5.2 explicitly computes the fixed point when $V$ has a Gaussian
distribution.

Section 5 studies the properties of equilibrium under the assumption that
there exist a fixed point of the operator described by (1.2). A remarkable
property of the equilibrium is that the insider’s strategy depends on the
distribution of the fundamental value of the asset. This is a striking
difference from the previous extensions of the Kyle model with risk neutral
agents and no penalties, where the insider always employs the same bridge
strategy in equilibrium irrespective of the asset distribution222A non-
exhaustive list includes [1], [5], [6], [9], [8], and [2]..

The analysis of the equilibrium in Section 5 also reveals that all equilibrium
parameters such as insider’s wealth and noise traders’ loss have direct
representations in terms of $\Psi$, $\phi$ and its derivatives. In particular,
it shows that equilibrium price is given by the solution of a heat equation
with terminal value given by the first derivative of $\phi$. Moreover,
variance of the fundamental value after all trading is complete is non-zero
and corresponds to an expectation involving the second derivative of $\phi$
acting on Brownian motion.

Another remarkable consequence of the form of the solution $(U,Z)$ of (1.1)
and the description of the optimal strategy of the insider via $Z$ is that it
reveals a connection with the BSDE (1.1) and a particular $h$-transform of
Brownian motion. Proposition 5.2 studies this connection and shows that the
expected penalty of the insider in equilibrium is a constant multiple of the
relative entropy, which is finite.

When the asset value has a Gaussian distribution, the equilibrium can be
computed explicitly. The second part of Section 5 is devoted to the
equilibrium with Gaussian payoffs. As expected, the insider’s wealth is
decreasing as the intensity of penalties decrease. However, this does not
follow from the plausible explanation that expected penalties that are
incurred in equilibrium are very high as the regulators impose higher fines.
Instead, the presence of legal risk forces the insider moderate her trades and
trade very little in case of high legal fines. Although the insider continue
to buy (resp. sell) when the market valuation is below (resp. above) the
fundamentals, the amount that she trades decrease in presence of higher fines.
As a result, the expected penalties in equilibrium remain bounded even if the
regulatory fines increase without bound. That the insider trades less
aggressively aligns with the predictions of Shin [16], Kacperczyk and Pagnotta
[12], as well as Carré et al. [7]), and agrees with the empirical evidence
presented in [12].

The outline of the paper is as follows. Section 2 presents the model
assumptions. Section 3 gives a recipe for constructing the equilibrium and
makes the connection with a quadratic BSDEs. Section 4 studies the fixed point
of a nonlinear operator that yields the terminal condition of the BSDE via
equilibrium constraints. Section 5 analyses the properties of equilibrium and
its connection with an $h$-transform of Brownian motion. Finally, Section 6
concludes and gives directions for future research.

##  2\. The setup

As in [1], the trading will take place over the time interval $[0,1]$ and the
risk free interest rate is set to $0$. Let
$(\Omega,{\mathcal{G}},({\mathcal{G}}_{t})_{t\in[0,1]},\mathbb{Q})$ be a
filtered probability space satisfying the usual conditions. The fundamental
value of this asset equals $V$, which will become public knowledge at time
$1$. Moreover, $V$ has a finite variance with possibility of atoms. This in
particular implies the existence of a non-decreasing $f$ such that
$V=f(\eta)$, where $\eta$ is a standard Normal random variable. The
distribution of $V$ is denoted by $\Pi$. That is, $\mathbb{Q}(V\in
dv)=\Pi(dv)$ for $v$ in the support of $V$, which is denoted by
$f(\mathbb{R})$.

Three types of agents trade in the market. They differ in their information
sets and objectives as follows and the reader is referred to [10] for further
details.

  * •

Noise/liquidity traders are non-strategic and trade for liquidity reasons, and
their total demand at time $t$ is given by $\sigma B$ for a standard
$({\mathcal{G}}_{t})$-Brownian motion $B$ independent of $V$, and constant
$\sigma>0$.

  * •

Market makers only observe the total demand

$Y=\theta+\sigma B,$

where $\theta$ is the demand process of the informed trader. The admissibility
condition imposed later on $\theta$ will entail in particular that $Y$ is a
semimartingale.

The market makers compete in a Bertrand fashion and clear the market. Similar
to [1] the market makers set the price $S$ as a function of the total order
process at time $t$, i.e.

$S_{t}=H(t,Y_{t}),\qquad\forall t\in[0,1).$ (2.3)

  * •

The informed investor observes the price process $S$ given by
$S_{t}=H(t,Y_{t})$ as well as $V$. Different than [1] the informed trader
incurs quadratic transaction costs which in particular implies that her
trading strategy is absolutely continuous, i.e.

$d\theta_{t}=\alpha_{t}dt$

for some $\alpha$ adapted to her own filtration. The accumulated transaction
cost by time $t$ is given by

$C_{t}:=\frac{c}{2}\int_{0}^{t}\alpha^{2}_{s}ds.$

for some $c>0$, that will be called the rate of penalties in the sequel. Since
she is risk-neutral, her objective is to maximize the expected final wealth,
i.e.

$\displaystyle\sup_{\theta\in\mathcal{A}}E^{0,v}\left[W_{1}^{\theta}\right],\mbox{
where}$ (2.4) $\displaystyle
W_{1}^{\theta}=(V-S_{1-})\theta_{1-}+\int_{0}^{1-}\theta_{s}dS_{s}-\frac{c}{2}\int_{0}^{1}\alpha^{2}_{s}ds=\int_{0}^{1}(V-S_{s})\alpha_{s}ds-\frac{c}{2}\int_{0}^{1}\alpha^{2}_{s}ds.$
(2.5)

In above $\mathcal{A}$ is the set of admissible trading strategies for the
given pricing rule, which will be defined in Definition 2.2. Moreover,
$E^{0,v}$ is the expectation with respect to $P^{0,v}$, which is the regular
conditional distribution of $(B_{s},V;s\leq 1)$ given $B_{0}=0$ and $V=v$,
which exists due to Theorem 44.3 in [3].

The expected wealth formulation above can also arise in case the informed
trader is an illegal inside trader who faces the risk of investigation that
will result in penalties. Indeed, suppose that an investigation can
successfully identify illegal inside trading with probability $p$, after which
the insider not only loses her gains from trade but also pay a legal penalty
of $k\int_{0}^{1}\alpha^{2}_{t}dt$. The expected profit of the insider333It is
assumed that the invesitigation takes place after traded are over as in the
case of a whistleblower. under this scenario is

$E^{0,v}\bigg{[}(1-p)\int_{0}^{1}(V-S_{s})\alpha_{s}ds-
pk\int_{0}^{1}\alpha^{2}_{s}ds\bigg{]}=(1-p)E^{0,v}\bigg{[}\int_{0}^{1}(V-S_{s})\alpha_{s}ds-\frac{pk}{1-p}\int_{0}^{1}\alpha^{2}_{s}ds\bigg{]}.$

Thus the coefficient $c$ in (2.5) can be associated with $\frac{pk}{1-p}$,
which gets large as the probability of a successful investigation gets bigger.

Given the above market structure, one can now define the filtrations of the
market makers and of the informed trader. First define
${\mathcal{F}}:=\sigma(B_{t},V;t\leq 1)$. Due to the measurability of regular
conditional distributions one can define the probability measure $\mathbb{P}$
on $(\Omega,{\mathcal{F}})$ by

$\mathbb{P}(E)=\int_{f(\mathbb{R})}P^{0,v}(E)\Pi(dv),$ (2.6)

for any $E\in{\mathcal{F}}$.

The market makers’ filtration, denoted by ${\mathcal{F}}^{M}$, will be the
right-continuous augmentation with the $\mathbb{P}$-null sets of the
filtration generated by $Y$. In particular ${\mathcal{F}}^{M}$ satisfies the
usual conditions.

On the other hand, augmented the insider’s filtration with the $P^{0,v}$-null
sets will create an extra dependence on the value of $V$ purely for technical
reasons. To avoid this, we assume that the informed trader’s filtration,
denoted by ${\mathcal{F}}^{I}$, is the right continuous augmentation444See
Section 3 of [15] for a recipe of the procedure. of the filtration generated
by $S$ and $V$ with the sets of

${\mathcal{N}}^{I}:=\\{E\subset{\mathcal{F}}:P^{0,v}(E)=0,\,\forall
z\in\mathbb{R}\\}.$ (2.7)

Similarly, ${\mathcal{F}}^{B,V}$ will denote the right continuous augmentation
of the filtration generated by $B$ and $V$ with the sets of
${\mathcal{N}}^{I}$. Note that the resulting filtrations are not complete.
Moreover, ${\mathcal{F}}^{I}={\mathcal{F}}^{B,V}$ when $H$ is strictly
increasing in total order, which will be the case when $H$ is an admissible
pricing rule per Definition 2.1.

An equilibrium is a pair consisting of an _admissible_ pricing rule and an
_admissible_ trading strategy such that: a) given the pricing rule the trading
strategy is optimal, b) given the trading strategy, the pricing rule is
rational in the following sense:

$H(t,Y_{t})=S_{t}=\mathbb{E}[V|\mathcal{F}_{t}^{M}],$ (2.8)

where $\mathbb{E}$ corresponds to the expectation operator under $\mathbb{P}$.
To formalize this definition of equilibrium, one needs the notions of
admissible pricing rules and trading strategies.

######  Definition 2.1.

An admissible pricing rule is any function $H$ such that $H\in
C^{1,2}([0,1)\times\mathbb{R})\cap C([0,1]\times\mathbb{R})$, and $x\mapsto
H(t,x)$ is strictly increasing for every $t\in[0,1]$.

######  Definition 2.2.

An ${\mathcal{F}}^{B,Z}$-adapted $\theta$ is said to be an admissible trading
strategy for a given admissible pricing rule $H$ if the following conditions
are stisfied:

  1. (1)

$\theta$ is absolutely continuous and of finite variation; that is

$\theta_{t}=\int_{0}^{t}\alpha_{s}ds\mbox{ and
}\int_{0}^{1}|\alpha_{t}|dt<\infty,\,P^{0,v}\mbox{-a.s.,}$

for some adapted $\alpha$ for each $v\in f(\mathbb{R})$.

  2. (2)

No doubling strategies are allowed, i.e. for all $v\in f(\mathbb{R})$

$E^{0,v}\int_{0}^{1}(H^{2}(s,Y_{s})+\alpha^{2}_{s})ds<\infty.$ (2.9)

The set of admissible trading strategies for the given pricing rule $H$ is
denoted with $\mathcal{A}(H)$.

######  Remark 1.

The square integrability condition on $H$ is standard in the literature on the
Kyle-Back models. The new condition on $\alpha$ is almost redundant since the
expected transaction costs become infinite when it is not satisfied. However,
it could be possible for the insider to manipulate the prices so that the
gains from trading $\int_{0}^{1}(V-S_{t})\alpha_{t}dt$ is bigger than the
loss. The extra condition prevents such doubling strategies.

On the other hand, the total wealth at the end of trading is given by

$\int_{0}^{1}(V-H(t,Y_{t}))\alpha_{t}dt-\frac{c}{2}\int_{0}^{1}\alpha^{2}_{t}dt\leq\int_{0}^{1}(V-H(t,Y_{t}))^{2}dt+\frac{1-c}{2}\int_{0}^{1}\alpha_{t}^{2}dt.$

Thus, if the penalties are high enough, i.e. $c>1$, the insider will never use
strategies that are not square integrable if the prices are. In such case, the
condition on $\alpha$ is indeed redundant.

The market equilibrium can now be defined as follows.

######  Definition 2.3.

A couple $(H^{\ast},\theta^{\ast})$ is said to form an equilibrium if
$H^{\ast}$ is an admissible pricing rule,
$\theta^{\ast}\in{\mathcal{A}}(H^{\ast})$, and the following conditions are
satisfied:

  1. (1)

Market efficiency condition: given $\theta^{\ast}$, $H^{\ast}$ is a rational
pricing rule, i.e. it satisfies (2.8).

  2. (2)

Insider optimality condition: given $H^{\ast}$, $\theta^{\ast}$ solves the
insider optimization problem for all $v\in f(\mathbb{R})$:

$E^{0,v}[W^{\theta^{\ast}}_{1}]=\sup_{\theta\in{\mathcal{A}}(H^{\ast})}E^{0,v}[W^{\theta}_{1}]<\infty.$

##  3\. A recipe for equilibrium

Since the market maker’s pricing rule leads to the martingale property of $S$
and one typically expects the insider’s strategy to be inconspicuous, i.e. $Y$
is a Brownian motion in its own filtration, in view of the past literature, it
is not unreasonable to restrict oneself to $H$ satisfying a heat equation:

$H_{t}+\frac{\sigma^{2}}{2}H_{yy}=0,\qquad t\in[0,1].$ (3.10)

Keeping the above equation in mind a formal HJB equation for the value
function of the insider will be obtained in the first part of this section. To
this end define

$J(t,y)=\sup_{\alpha\in{\mathcal{A}}(H)}E^{0,v}\bigg{[}\int_{t}^{1}(v-H(u,Y_{u}))\alpha_{u}du-\frac{c}{2}\int_{t}^{1}\alpha^{2}_{t}dt\Big{|}Y_{t}=y\bigg{]}.$

Direct calculations lead to

$J_{t}+\frac{\sigma^{2}}{2}J_{yy}+\sup_{\alpha}\Big{\\{}\alpha(J_{y}+v-H)-\frac{c\alpha^{2}}{2}\Big{\\}}=0.$

Note that the presence of the penalty factor $c>0$ yields that the optimal
$\alpha$ can be obtained in the feedback form:

$\alpha^{*}=\frac{J_{y}(t,y)+v-H(t,y)}{c}.$ (3.11)

This leads to the following HJB equation:

$J_{t}+\frac{\sigma^{2}}{2}J_{yy}+\frac{(J_{y}+v-H)^{2}}{2c}=0.$ (3.12)

Recall that in the classical Kyle model $c=0$, i.e. no penalty, the optimal
strategy cannot be obtained in the feedback form from the HJB equation and
instead one has $J_{y}=H-z$, which is one should expect when letting
$c\rightarrow 0$ in (3.11) to arrive at a finite limit (see Section 6.2 in
[9]).

Next suppose there exists a smooth function $J^{0}$ such that

$\begin{split}J^{0}_{t}+\frac{\sigma^{2}}{2}J^{0}_{yy}&=0,\\\
J^{0}_{y}&=H-v,\\\ J^{0}(1,y)&=\int_{h^{-1}(v)}^{y}(h(x)-v)dx+\mbox{
constant},\end{split}$ (3.13)

for some $h$ to be determined and a constant possibly depending on $v$. Such a
function exists and is the value function of the insider in the classical Kyle
model with $c=0$ whenever the pricing rule (not necessarily the equilibrium
one) satisfies (3.10) (see Theorem 5.1 in [10]). In this case, $h=H(1,\cdot)$
and the constant is zero.

Thus, if one defines $u=J-J^{0}$ and conjectures that $J(1,\cdot)\equiv 0$,
one obtains

$u_{t}+\frac{1}{2}\sigma^{2}u_{yy}+\frac{1}{2c}u_{y}^{2}=0,\quad
u(1,y)=-j^{0}(y,v):=-J^{0}(1,y).$ (3.14)

Note that $J^{0}$ would be the value function of the insider had there been no
penalties on her trading strategies, i.e. $c=0$. Thus, $-u$ can be viewed as
the expected additional transaction costs (or legal penalties) to the insider
in equilibrium.

It is now a common folklore that the above has a backward stochastic
differential equation (BSDE) formulation:

$dU_{t}=\sigma Z_{t}dB_{t}-\frac{1}{2c}Z_{t}^{2}dt,\quad U_{1}=u(1,\sigma
B_{1}).$ (3.15)

The above quadratic BSDE is known to have an explicit solution, which in turn
yields

$u(t,x)=c\sigma^{2}\log\rho(t,x,v),\quad\rho(t,x,v):=E^{0,v}\Big{[}\exp\Big{(}-\frac{j^{0}(\sigma
B_{1},v)}{c\sigma^{2}}\Big{)}\Big{|}\sigma B_{t}=x\Big{]}.$ (3.16)

It immediately follows from Theorem 4.3.6 in [13] that for each $v\in
f(\mathbb{R})$, $\rho(\cdot,\cdot,v)\in C^{1,2}([0,1)\times\mathbb{R}])$..

In view of (3.11) and (3.13), the above implies the optimal control of the
insider at time $t\in[0,1]$ is $\alpha^{*}(t,Y_{t},V)$, where

$\alpha^{*}(t,y,v):=\sigma^{2}\frac{\rho_{y}(t,y,v)}{\rho(t,y,v)}.$ (3.17)

Recall that the above assumed that the pricing rules satisfied (3.10).
Combined with the requirement that prices follow martingales in market makers’
filtration in equilibrium this in particular yields that $Y$ must be a
Brownian motion in its own filtration. Observe that if
$\kappa(t)\rho(t,Y_{t},v)\Pi(dv)=\mathbb{P}(V\in dv|{\mathcal{F}}^{Y}_{t})$
for some constant $\kappa(t)$ for each $t$, then $Y$ is a Brownian motion in
its own filtration (see, e.g., Corollary 3.1 in [9]) since

$\mathbb{E}\Big{[}\frac{\rho_{y}(t,Y_{t},V)}{\rho(t,Y_{t},V)}\Big{|}{\mathcal{F}}^{Y}_{t}\Big{]}=\int_{f(\mathbb{R})}\rho_{y}(t,Y_{t},v)\kappa(t)dv=\frac{d}{dy}\int_{f(\mathbb{R})}\rho(t,y,v)\kappa(t)dv\Big{|}_{y=Y_{t}}=0,$

provided one can interchange the order of integration and differentiation.

To this end, suppose that $\rho^{0}(t,y,v):=\kappa(t)\rho(t,y,v)$ is the
conditional density of $V$ with respect to the measure $\Pi$ given
${\mathcal{F}}^{Y}_{t}$ and $Y_{t}=y$. Then, in view of Theorem 3.3 in [9],
one expect $\rho^{0}(\cdot,\cdot,v)$ to satisfy for each $v$

$\rho^{0}_{t}+\frac{1}{2}\sigma^{2}\rho^{0}_{yy}=0.$ (3.18)

On the other hand, $\rho$ satisfies (3.18) by its construction. Therefore,
$\frac{\kappa^{\prime}}{\kappa}\equiv 0$, and consequently, $\kappa$ must be
constant since it must be strictly positive. This leads to the following:

######  Proposition 3.1.

Suppose that there exists a continuous function $j^{0}:\mathbb{R}\times
f(\mathbb{R})\to\mathbb{R}$ such that $\kappa\rho(0,0,\cdot)\equiv 1$ for some
normalising constant $\kappa$, where $\rho$ is defined via (3.16). Assume
further that there exists a unique strong solution on
$(\Omega,{\mathcal{G}},({\mathcal{G}}_{t})_{t\in[0,1]},\mathbb{Q})$ to

$Y_{t}=B_{t}+\int_{0}^{t}\frac{\rho_{y}(s,Y_{s},V)}{\rho(s,Y_{s},V)}ds$

such that

$\mathbb{E}^{\mathbb{Q}}\Big{[}\int_{0}^{t}\Big{(}\frac{\rho_{y}(s,Y_{s},V)}{\rho(s,Y_{s},V)}\Big{)}^{2}\Big{]}<\infty,\quad
t\in[0,1].$ (3.19)

Then,

$\kappa\rho(t,Y_{t},v)\Pi(dv)=\mathbb{P}(V\in dv|{\mathcal{F}}^{Y}_{t}),\qquad
t\in[0,1].$

###### Proof.

This is an immediate consequence of Theorem 3.3 in [9]. Note that the process
$Z$ therein coincides with $V$ at all times under the setting of the present
paper. Thus, the martingale problem $\tilde{A}$, i.e., the martingale problem
for the pair $(Y,V)$ is trivially well-posed. ∎

The seemingly strong integrability condition (3.19) is in fact a condition on
the square integrability of $Z$ in the BSDE representation (3.15). A minimal
integrability condition as in the next result ensures that holds.

######  Proposition 3.2.

Consider $\rho$ defined by (3.16) for some continuous $j^{0}$. Assume that

$\int_{f(\mathbb{R})}E^{0,v}[|j^{0}(\sigma B_{1},v)|]\Pi(dv)<\infty.$ (3.20)

Then, (3.19) holds.

###### Proof.

To ease notation take $\sigma=1$. As observed above, it suffices to show that

$\mathbb{E}^{\mathbb{Q}}\Big{[}\int_{0}^{1}Z_{t}^{2}dt\Big{]}<\infty,$

where $(U,Z)$ is as in (3.15). It follows from the Jensen’s inequality and
(3.16) that

$U_{t}\geq
M_{t}(v):=-E^{0,v}\big{[}j^{0}(B_{1},v)\big{|}B_{t}\big{]},\;P^{0,v}\mbox{-a.s..}.$
(3.21)

Since $M(v)$ is a martingale on the time interval $[0,1]$, we have that $U$ is
bounded from below by a uniformly integrable process. Thus, by Fatou’s lemma
$\int_{0}^{\cdot}Z_{s}dB_{s}$ in (3.15) is a supermartingale. Therefore,

$\frac{1}{2c}E^{0,v}\Big{[}\int_{0}^{1}Z_{t}^{2}dt\Big{]}\leq
E^{0,v}[U_{0}-U_{1}]\leq U_{0}+E^{0,v}[j^{0}(B_{1},v)]<\infty.$

This yields the claim by the independence of $B$ and $V$ under $\mathbb{Q}$,
and the hypothesis (3.20). ∎

The above considerations give the recipe for constructing an equilibrium:

  1. (1)

Find a continuous function $j^{0}:\mathbb{R}\times f(\mathbb{R})\to\mathbb{R}$
such that i) $\rho(1,y,v)\Pi(dv)$ is a probability measure for each $y$, where
$\rho$ is given by (3.16), ii) $\kappa\rho(0,0,\cdot)\equiv 1$ for some
constant $\kappa>0$, and iii) it is differentiable in its first parameter with
$j^{0}_{y}(y,v)=h(y)-v$. Note that the second condition entails

$\kappa\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}\exp\Big{(}-\frac{j^{0}(y,v)}{\sigma^{2}c}\Big{)}dy=1,\qquad\forall
v\in f(\mathbb{R}).$ (3.22)

  2. (2)

Set

$H_{t}+\frac{1}{2}H_{yy}=0,\qquad H(1,y)=h(y).$

  3. (3)

Show that $(H,\theta^{*})$ with
$d\theta^{*}_{t}=\frac{\rho_{y}(t,Y_{t},V)}{\rho(t,Y_{t},V)}dt$ is equilibrium
provided they are admissible by using the candidate value function $J=J^{0}-u$
that satisfies (3.12).

######  Remark 2.

In view of the rationality of the pricing rule, i.e. (2.1), in equilibrium the
above recipe appears to assume that the function $h$ also satisfies

$h(y)=\kappa\int_{f(\mathbb{R})}v\rho(1,y,v)\Pi(dv)=\kappa\int_{f(\mathbb{R})}v\exp\Big{(}-\frac{j^{0}(y,v)}{c\sigma^{2}}\Big{)}\Pi(dv).$

However, this already follows from the properties of $j^{0}$ in Step (1)
alone. Indeed, since $\kappa\rho(1,y,\cdot)$ is a proper density for each $y$,

$0=\frac{d}{dy}1=\frac{d}{dy}\kappa\int_{f(\mathbb{R})}\exp\Big{(}-\frac{j^{0}(y,v)}{\sigma^{2}c}\Big{)}\Pi(dv)=\frac{\kappa}{\sigma^{2}c}\int_{f(\mathbb{R})}(v-h(y))\exp\Big{(}-\frac{j^{0}(y,v)}{\sigma^{2}c}\Big{)}\Pi(dv),$

provided one can differentiate under the integral sign. Thus,

$h(y)=h(y)\kappa\int_{f(\mathbb{R})}\exp\Big{(}-\frac{j^{0}(y,v)}{c\sigma^{2}}\Big{)}\Pi(dv)=\kappa\int_{f(\mathbb{R})}v\exp\Big{(}-\frac{j^{0}(y,v)}{c\sigma^{2}}\Big{)}\Pi(dv)$

as required.

##  4\. A fixed point algorithm

The recipe from the previous section shows that the equilibrium mainly boils
down to finding a function $j^{0}$ with certain properties. The condition on
its derivative implies

$j^{0}(y,v)=\Psi(v)+\phi(y)-yv,$ (4.23)

where $\psi$ and $\phi$ are continuous functions on $f(\mathbb{R})$ and
$\mathbb{R}$, respectively. This section will describe an operator on the
space of continuous functions, whose fixed point will determine the functions
$\psi$ and $\phi$ that will appear in equilibrium. Although the main theorem
of this section will be proved under a stronger assumption, the condition
below is sufficient for a number of important estimates.

######  Assumption 4.1.

All exponential moments of $V$ exist. That is,

$M_{V}(r):=\int_{f(\mathbb{R})}e^{rv}\Pi(dv)<\infty,\qquad\forall
r\in\mathbb{R}.$ (4.24)

The following notation will also be used to ease the exposition:

Notation:  $\hat{c}:=c\sigma^{2}$.

Since $\exp(-\frac{1}{\hat{c}}j^{0}(y,\cdot))\kappa$ will be a probability
density function itself for all $y$ for some constant $\kappa$ independent of
$y$, one can without loss of generality take $\kappa=1$ by incorporating it
into $\Psi$ or $\phi$. This yields the first relationship:

$\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)=\exp\Big{(}\frac{\phi(y)}{\hat{c}}\Big{)},\qquad
y\in\mathbb{R}.$

The second relationship follows from the initial condition (3.22):

$\int_{\mathbb{R}}\exp\Big{(}\frac{yv-\phi(y)}{\hat{c}}\Big{)}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy=\exp\Big{(}\frac{\Psi(v)}{\hat{c}}\Big{)},\qquad
v\in f(\mathbb{R}).$

Thus one obtains the following two operators on the space of measurable
functions:

$\begin{split}T_{1}(\phi)(v):=\int_{\mathbb{R}}\exp\Big{(}\frac{yv-\phi(y)}{\hat{c}}\Big{)}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy&=\exp\Big{(}\frac{\Psi(v)}{\hat{c}}\Big{)},\qquad
v\in f(\mathbb{R});\\\
T_{2}(\Psi)(y):=\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)&=\exp\Big{(}\frac{\phi(y)}{\hat{c}}\Big{)},\qquad
y\in\mathbb{R}\end{split}$ (4.25)

Consequently, one can define an operator $T_{0}$ on the space of measurable
functions by

$T_{0}\phi=\hat{c}\log T_{2}(T_{1}(\phi)),$ (4.26)

whose domain consists of functions so that $T_{0}(\phi)$ is finite for all
$y\in\mathbb{R}$. Once this is done, one can apply a normalisation so that
$\phi(0)=0$. This leads to the operator $T$ defined on the domain of $T_{0}$
as follows:

$T\phi=\hat{c}\log\frac{T_{2}(T_{1}(\phi))}{T_{2}(T_{1}(\phi))(0)}.$ (4.27)

By similar reasoning the domain ${\mathcal{D}}(T_{i})$ of the operator $T_{i}$
consist of functions $g$ such $T_{i}g$ takes values in $(0,\infty)$.
Consequently,

${\mathcal{D}}(T)={\mathcal{D}}(T_{0})=\\{\phi\in{\mathcal{D}}(T_{1}):T_{1}(\phi)\in{\mathcal{D}}(T_{2})\\}.$

Due to the extra normalisation involved it is not immediately obvious that the
fixed points of $T$ and $T_{0}$ are related. However, as the next proposition
shows, there is a one-to-one correspondence.

######  Proposition 4.1.

There exists a fixed point for $T_{0}$ if and only if there exists a fixed
point for $T$. In particular, any fixed point of $T$ is a fixed point of
$T_{0}$.

###### Proof.

By direct calculations, if $T_{0}\phi=\phi$, $T\phi_{0}=\phi_{0}$ for
$\phi_{0}=\phi-\phi(0)$.

Now, suppose $T\phi=\phi$. Then, $T_{2}\Psi=k\exp(\frac{\phi}{\hat{c}})$ for
some $k$, where $\Psi=\hat{c}\log T_{1}\phi$. That is,

$\begin{split}\int_{\mathbb{R}}\exp\Big{(}\frac{yv-\phi(y)-\Psi(v)}{\hat{c}}\Big{)}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy&=1,\qquad
v\in f(\mathbb{R});\\\
\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv-\Psi(v)-\phi(y)}{\hat{c}}\Big{)}\Pi(dv)&=k,\qquad
y\in\mathbb{R}.\end{split}$

Integrating both sides of the first identity with respect to $\Pi$ yields

$k\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy=1,$

in view of the second identity in the previous display and Fubini’s theorem.
Thus, $k=1$ and $T_{0}\phi=\phi$. ∎

The next result shows that the fixed point of $T$, if it exists, should be a
convex function.

######  Proposition 4.2.

Consider $\phi\in{\mathcal{D}}(T_{1})$ and let $\Psi=\hat{c}\log T_{1}\phi$.
Then, the following statements are valid:

  1. (1)

If $\mathbb{E}[|\phi(\sigma B_{1})|]<\infty$, $\Psi$ is bounded from below:

$\Psi\geq-\int_{\mathbb{R}}\phi(y)\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy.$

  2. (2)

Suppose that $\Psi$ is bounded from below and Assumption 4.1 holds. Then
$\phi\in{\mathcal{D}}(T)$.

Moreover, $T\phi$ is infinitely differentiable and strictly convex. In
particular,

$\displaystyle\frac{d}{dy}T\phi(y)$
$\displaystyle=\frac{\int_{f(\mathbb{R})}v\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)}{\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)},$
(4.28) $\displaystyle\frac{d^{2}}{dy^{2}}T\phi(y)$
$\displaystyle=\frac{1}{\hat{c}}\frac{\int_{f(\mathbb{R})}v^{2}\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)}{\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)}-\frac{1}{\hat{c}}\Bigg{(}\frac{\int_{f(\mathbb{R})}v\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)}{\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)}\Bigg{)}^{2}.$
(4.29)

###### Proof.

  1. (1)

This is a straightforward consequence of Jensen’s inequality applied to the
natural logarithm and definition of $T_{1}$ in (4.25).

  2. (2)

Since $\phi\in{\mathcal{D}}(T_{1})$, $\Psi$ is finite. Thus, $T_{2}\Psi$ never
vanishes. Moreover the lower bound on $\Psi$ yields for some constant $K$ that

$T_{2}\Psi(y)\leq
K\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv}{\hat{c}}\Big{)}\Pi(dv)<\infty.$

Thus, $T_{0}\phi$ is finite everywhere. This yields the claim.

Note that since the moment generating function are finite everywhere, it is
smooth and in particular (see, e.g., (21.24) in Section 21 and the preceding
discussion in [4] )

$M_{V}^{(n)}=\int_{f(\mathbb{R})}v^{n}e^{rv}\Pi(dv)<\infty,\qquad\forall
r\in\mathbb{R}\mbox{ and }n\geq 0.$

Thus, the above integrability, the lower bound on $\Psi$, and the dominated
convergence theorem immediately yield (4.28). By iterating this operation one
obtains that the numerator in (4.28) is infinitely differentiable by the
smoothness of $M_{V}$.

Similarly, the second derivative is given by (4.29), which is positive being a
positive multiple of the variance of a random variable $\tilde{v}$ with the
probability density

$P(\tilde{v}\in
dx)=\frac{\exp\Big{(}\frac{yx-\Psi(x)}{\hat{c}}\Big{)}\Pi(dx)}{\int_{f(\mathbb{R})}\exp\Big{(}\frac{yv-\Psi(v)}{\hat{c}}\Big{)}\Pi(dv)},\qquad
x\in f(\mathbb{R}).$

Since $\tilde{v}$ is not constant, the second derivative is strictly positive,
hence the strict convexity follows.

∎

An analogous result, whose proof is omitted, also holds for $\Psi$. To present
its statement let’s introduce

$\tilde{T}\Psi=\hat{c}\log\frac{T_{1}(T_{2}(\Psi))}{T_{1}(T_{2}(\Psi))(0)}.$
(4.30)

######  Proposition 4.3.

Consider $\Psi\in{\mathcal{D}}(T_{2})$ and let $\tilde{\phi}=\hat{c}\log
T_{2}\Psi$. Then, the following statements are valid:

  1. (1)

If $\mathbb{E}[|\Psi(V)|]<\infty$, $\tilde{\phi}$ is bounded from below:

$\tilde{\phi}(y)\geq y\mathbb{E}[V]-\mathbb{E}[\Psi(V)].$

  2. (2)

Suppose that $\tilde{\phi}$ is bounded from below. Then
$\Psi\in{\mathcal{D}}(\tilde{T})$. Moreover, $\tilde{T}\Psi$ is infinitely
differentiable and strictly convex. In particular,

$\displaystyle\frac{d}{dv}\tilde{T}\Psi(v)$
$\displaystyle=\frac{\int_{\mathbb{R}}y\exp\Big{(}\frac{yv-\tilde{\phi}(y)}{\hat{c}}\Big{)}p(\sigma,y)dy}{\int_{\mathbb{R}}\exp\Big{(}\frac{yv-\tilde{\phi}(y)}{\hat{c}}\Big{)}p(\sigma,y)dy},$
(4.31) $\displaystyle\frac{d^{2}}{dv^{2}}\tilde{T}\Psi(v)$
$\displaystyle=\frac{\int_{\mathbb{R}}y^{2}\exp\Big{(}\frac{yv-\tilde{\phi}(y)}{\hat{c}}\Big{)}p(\sigma,y)dy}{\int_{\mathbb{R}}\exp\Big{(}\frac{yv-\tilde{\phi}(y)}{\hat{c}}\Big{)}p(\sigma,y)dy}-\Bigg{(}\frac{\int_{\mathbb{R}}y\exp\Big{(}\frac{yv-\tilde{\phi}(y)}{\hat{c}}\Big{)}p(\sigma,y)dy}{\int_{\mathbb{R}}\exp\Big{(}\frac{yv-\tilde{\phi}(y)}{\hat{c}}\Big{)}p(\sigma,y)dy}\Bigg{)}^{2},$
(4.32)

where $p(\sigma,\cdot)$ is the density of a normal distribution with variance
$\sigma^{2}$.

The above results indicate that the fixed point of $T$ is likely to be a
smooth convex function. The next lemma improves upon the previous proposition
by providing an upper bound on the function $\Psi$.

######  Lemma 4.1.

Consider a convex and differentiable $\phi\in{\mathcal{D}}(T_{1})$ with
$\phi(0)=0$, and let $\Psi=\hat{c}\log T_{1}\phi$. Then

$\Psi(v)\leq\frac{\big{(}v-\phi^{\prime}(0)\big{)}^{2}}{2c^{2}\sigma^{2}},\qquad
v\in f(\mathbb{R}).$ (4.33)

###### Proof.

Observe that $\phi(y)\geq y\phi^{\prime}(0)$ for all $y\in\mathbb{R}$. Thus,

$T_{1}\phi(v)\leq\int_{\mathbb{R}}\exp\Big{(}\frac{y(v-\phi^{\prime}(0))}{\hat{c}}\Big{)}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy=\exp\Big{(}\frac{\big{(}v-\phi^{\prime}(0)\big{)}^{2}\sigma^{2}}{2\hat{c}^{2}}\Big{)},$

in view of the moment generating function of the normal distribution. This
proves the upper bound.

∎

These estimates leads to the following theorem that establishes the existence
of a fixed point when $V$ is bounded.

######  Theorem 4.1.

Suppose that $\kappa:=\sup\\{|v|:v\in f(\mathbb{R})\\}<\infty$. Then, there
exists a strictly convex $\phi$ such that $T\phi=\phi=T_{0}\phi$.

###### Proof.

Let $\phi_{0}=0$ and define $\phi_{n+1}=T\phi_{n}$ until $\phi_{n}$ no longer
belongs to ${\mathcal{D}}(T)$.

  1. Step 1

$\phi_{n}\in{\mathcal{D}}(T)$, $\phi_{n}(0)=0$, and
$|\phi^{\prime}_{n}|\leq\kappa$ for all $n\geq 0$: Denoting $T_{1}\phi_{0}$ by
$\Psi_{0}$, Proposition 4.2 first shows that
$\Psi_{0}\geq-\mathbb{E}[\phi_{0}(\sigma B_{1})]=0$. Since Assumption 4.1 is
clearly satisfied, this in turn implies $\phi_{0}\in{\mathcal{D}}(T)$.
Clearly, $|\phi^{\prime}_{0}|\leq\kappa$.

Next, suppose $\phi_{n}\in{\mathcal{D}}(T)$, $\phi_{n}(0)=0$ and
$|\phi^{\prime}_{n}|\leq\kappa$ for some $n$. Then, $\phi_{n+1}(0)=0$ by the
construction of $T$ and $|\phi_{n+1}^{\prime}|\leq\kappa$ by (4.28). Thus,
$|\phi_{n+1}(y)|\leq\kappa|y|$, implying $\phi_{n+1}\in{\mathcal{D}}(T_{1})$
and $E[|\phi_{n+1}(\sigma B_{1})|]<\infty$. This yields that
$\phi_{n+1}\in{\mathcal{D}}(T)$ by another application of Proposition 4.2.
Therefore, the claim follows from induction.

  2. Step 2

Existence of a convergent subsequence: Since the collection $(\phi_{n})_{n\geq
1}$ is equicontinuous and bounded uniformly by the function
$y\mapsto\kappa|y|$, one can extract a subsequence that converges (uniformly
in compacts) to some $\phi$.

  3. Step 3:

Fixed point property: With an abuse of notation suppose $(\phi_{n})$ converges
and let $\Psi_{n}:=T_{1}\phi_{n}$. Note that the dominated convergence theorem
implies that $\Psi:=\lim_{n\rightarrow\infty}T_{1}\phi_{n}=T_{1}\phi$. Next,
Proposition 4.2 shows that

$\Psi_{n}\geq-\mathbb{E}[\phi_{n}(\sigma
B_{1})]\geq-\kappa\sigma\mathbb{E}[|B_{1}|].$

Moreover, Lemma 4.1 yields that
$\Psi_{n}\leq\frac{\kappa^{2}}{c^{2}\sigma^{2}}$. Thus, $(\Psi_{n})$ is
uniformly bounded. Another application of the dominated convergence theorem
now yields

$\lim_{n\rightarrow\infty}T_{2}\Psi_{n}/T_{2}\Psi_{n}(0)=T_{2}\Psi/T_{2}\Psi(0)=\exp\Big{(}\frac{T\phi}{\hat{c}}\Big{)}$

On the other hand,

$T_{2}\Psi_{n}/T_{2}\Psi_{n}(0)=\exp\Big{(}\frac{\phi_{n+1}}{\hat{c}}\Big{)}.$

Combining the last two displays and taking limits establish that $\phi=T\phi$.
The strict convexity follows from Proposition 4.2. The remaining assertion is
a consequence of Proposition 4.1.

∎

######  Example 4.1.

Consider the case of a Bernoulli $V$, where $\Pi(\\{1\\})=1-\Pi(\\{0\\})=p$.
Existence of a fixed point for $T$ is ensured by Theorem 4.1. As the support
contains only two points, an easier representation of $\phi$ can be found as
follows.

In view of the normalisation that $\Phi(0)=0$, one can conjecture that

$\exp\Big{(}\frac{\phi}{\hat{c}}\Big{)}=\exp\Big{(}\frac{y}{\hat{c}}\Big{)}ap+1-ap,\mbox{
and }\Psi(1)=-\hat{c}\log a,\Psi(0)=-\hat{c}\log\frac{1-ap}{1-p}.$

Thus, it remains to pinpoint the value of $a$, which is fixed by the following
non-linear equation.

$\frac{1}{a}=\int_{\mathbb{R}}\frac{e^{\frac{y}{\hat{c}}}}{ap(e^{\frac{y}{\hat{c}}}-1)+1}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy.$

However, this is equivalent to

$1=\int_{\mathbb{R}}\frac{e^{\frac{y}{\hat{c}}}}{p(e^{\frac{y}{\hat{c}}}-1)+a^{-1}}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\Big{(}-\frac{y^{2}}{2\sigma^{2}}\Big{)}dy.$

Note that $a\in(0,p^{-1})$. Thus, the right hand side is increasing from $0$
to $1/p$ on $(0,p^{-1})$, and there exists a unique solution belonging to
$(0,p^{-1})$.

##  5\. Equilibrium

In the first part of this section, existence of equilibrium for the economy
described in Section 2 will be established under the following assumption.

######  Assumption 5.1.

There exists a pair of continuous functions $(\phi^{*},\Psi^{*})$ such that
$\phi^{*}$ is a fixed point of the operator $T$ in (4.27) and
$\Psi^{*}=c\sigma^{2}\log T_{1}\phi^{*}$, where $T_{1}$ is as in (4.25).
Moreover, $\phi^{*}$ is twice continuously differentiable satisfying (4.28)
and (4.29), and (3.20) is satisfied when $j^{0}$ is replaced by

$j^{*}(y,v):=\Psi^{*}(v)+\phi^{*}(y)-yv.$ (5.34)

Recall from Proposition 4.2 that $\phi^{*}$ is strictly convex, which will be
used frequently in proofs.

In view of the recipe at the end of Section 3 the candidate pricing rule for
equilibrium is the following:

$H^{*}(t,y)=\int_{f(\mathbb{R})}\rho^{*}(t,y,z)z\Pi(dz),$ (5.35)

where

$\rho^{*}(t,y,v):=E^{0,v}\Big{[}\exp\Big{(}-\frac{j^{*}(\sigma
B_{1},v)}{\hat{c}}\Big{)}\Big{|}\sigma B_{t}=y\Big{]}.$ (5.36)

Observe that the expectation above is non-zero and finite due to the
following:

######  Lemma 5.1.

The following bounds apply:

$\displaystyle E^{0,v}\Big{[}\exp\Big{(}-\frac{j^{*}(\sigma
B_{1},v)}{\hat{c}}\Big{)}\Big{|}\sigma B_{t}=x\Big{]}$
$\displaystyle\leq\exp\Big{(}-\frac{\Psi^{*}(v)}{\hat{c}}+\frac{(1-t)(v-h^{*}(0))^{2}}{2c^{2}\sigma^{2}}+(v-h^{*}(0)x\Big{)},$
$\displaystyle E^{0,v}\Big{[}\exp\Big{(}-\frac{j^{*}(\sigma
B_{1},v)}{\hat{c}}\Big{)}\Big{|}\sigma B_{t}=x\Big{]}$
$\displaystyle\geq\exp\Big{(}\frac{vx-\Psi^{*}(v)-E^{0,v}[\phi^{*}(\sigma
B_{1})|\sigma B_{t}=x]}{\hat{c}}\Big{)}.$

###### Proof.

Since $\phi^{*}$ is convex by Proposition 4.2 and vanishes at $0$,
$\phi^{*}(y)\geq yh^{*}(0)$. Thus,

$\displaystyle E^{0,v}\Big{[}\exp\Big{(}-\frac{j^{*}(\sigma
B_{1},v)}{\hat{c}}\Big{)}\Big{|}\sigma B_{t}=x\Big{]}$
$\displaystyle=\exp\Big{(}-\frac{\Psi^{*}(v)}{\hat{c}}\Big{)}E^{0,v}\Big{[}\exp\Big{(}\frac{v\sigma
B_{1}-\phi^{*}(\sigma B_{1})}{\hat{c}}\Big{)}\Big{|}\sigma B_{t}=x\Big{]}$
$\displaystyle\leq\exp\Big{(}-\frac{\Psi^{*}(v)}{\hat{c}}\Big{)}E^{0,v}\Big{[}\exp\Big{(}\frac{\sigma
B_{1}(v-h^{*}(0))}{\hat{c}}\Big{)}\Big{|}\sigma B_{t}=x\Big{]},$

which yields the first claim in view of the moment generation function of
normal distribution. The second claim is a direct consequence of Jensen’s
inequality. ∎

Thus, (4.28) implies in particular that

$H^{*}_{t}+\frac{\sigma^{2}}{2}H^{*}_{yy}=0,\qquad
H^{*}(1,\cdot)=h^{*}=\frac{d\phi^{*}}{dy}.$ (5.37)

Moreover, the discussion in Section 3 predicts that the value function of the
informed trader is given by $J=J^{0}+u$,

$J^{0}_{t}+\frac{\sigma^{2}}{2}J^{0}_{yy}=0,\quad J^{0}(1,y)=j^{*}(y,v),$
(5.38)

and $u$ is given by (3.16) with $j^{0}$ replaced by $j^{*}$. These
considerations yield the following in view of the PDE (3.14) that $u$
satisfies.

######  Proposition 5.1.

Under Assumption 5.1 there exists a classical solution to

$J_{t}+\frac{\sigma^{2}}{2}J_{yy}+\frac{(J_{y}+v-H^{*})^{2}}{2c}=0,\qquad
J(1,\cdot)=0,$ (5.39)

where $H^{*}$ is given by (5.35). In particular,
$J=J^{0}+\hat{c}\log\rho^{*}$, where $J^{0}$ is given by (5.38)555When there
are no penalties, the decomposition of $J$ coincides with the one given in
Theorem 3.2 in [2]..

As expected from the formal calculations of Section 3, the function $J$ above
can be used to show that an equilibrium is given by the pricing rule $H^{*}$
and the strategy $\theta^{*}$ given by

$d\theta^{*}_{t}=\sigma^{2}\frac{\rho^{*}_{y}(t,Y_{t},V)}{\rho^{*}(t,Y_{t},V)}dt,t\in[0,1],\mbox{
and }\theta^{*}_{0}=0.$ (5.40)

The admissibility condition that is required from any equilibrium candidate is
square integrability of the strategy as in (2.9). In case of the above
candidate this is intimately related to the entropy af an $h$-transform.

######  Proposition 5.2.

Suppose that Assumption 5.1 holds. For each $v$, there exists a unique strong
solution to

$Y^{*}_{t}=\sigma
B_{t}+\sigma^{2}\int_{0}^{t}\frac{\rho^{*}_{y}(s,Y^{*}_{s},v)}{\rho^{*}(s,Y_{s},v)}ds,$
(5.41)

where $\rho^{*}$ is given by (5.36).

Let $\mathbb{Q}^{*}$ be the law induced by $Y^{*}$ on the space of continuous
functions on $[0,1]$ vanishing at $0$; that is $C_{0}([0,1])$. Let
$({\mathcal{B}}_{t})_{t\in[0,1]}$ be the right continuous augmentation of the
natural filtration of the coordinate process $X$, and $\mathbb{W}$ be the
Wiener measure. Then $\mathbb{Q}^{*}$ is given by the following $h$-transform
of Brownian motion:

$\mathbb{E}^{\mathbb{Q}^{*}}[F]=\frac{\mathbb{E}^{\mathbb{W}}\Big{[}F\exp\Big{(}\frac{v\sigma
X_{1}-\phi^{*}(\sigma
X_{1})}{\hat{c}}\Big{)}\Big{]}}{\mathbb{E}^{\mathbb{W}}\Big{[}\exp\Big{(}\frac{v\sigma
X_{1}-\phi^{*}(\sigma X_{1})}{\hat{c}}\Big{)}\Big{]}},\quad
F\in{\mathcal{B}}_{1}.$ (5.42)

Moreover,

$\displaystyle\frac{1}{2}E^{0,v}\bigg{[}\sigma^{2}\int_{0}^{1}\Big{(}\frac{\rho^{*}_{y}(s,Y^{*}_{s},v)}{\rho^{*}(s,Y_{s},v)}ds\Big{)}^{2}\bigg{]}$
$\displaystyle=H(\mathbb{Q}^{*}||\mathbb{W}):=\mathbb{E}^{\mathbb{W}}\Big{[}\frac{d\mathbb{Q}^{*}}{d\mathbb{W}}\log\frac{d\mathbb{Q}^{*}}{d\mathbb{W}}\Big{]}$
(5.43)
$\displaystyle=\frac{v(\Psi^{*})^{\prime}(v)-\Psi^{*}(v)-E^{0,v}[\phi^{*}(Y^{*}_{1})]}{\hat{c}}<\infty.$

###### Proof.

Existence of a unique strong solution to (5.41) follows from Theorem 2.2.7 in
[9] since for each $v$, $\rho^{*}(\cdot,\cdot,v)\in C^{1,2}([0,1],\mathbb{R})$
and $\rho^{*}$ never vanishes. Note that the claimed smoothness is valid even
at time $t=1$ since $\phi^{*}$ is smooth by Assumption (5.1). Then, (5.42)
follows from Girsanov’s theorem after noticing that
$\rho(1,y,v)=\exp(-\frac{\Psi(v)}{\hat{c}})\exp(\frac{yv-\phi(y)}{\hat{c}})$
so that $\exp(-\frac{\Psi(v)}{\hat{c}})$ drops out in change of measure
calculations.

Thus, the equality in (5.43) is a consequence of (5.42) and the definition of
relative entropy. To compute its desired expression, note that

$\frac{d\mathbb{Q}^{*}}{d\mathbb{W}}=\exp\Big{(}\frac{v\sigma
X_{1}-\phi^{*}(\sigma X_{1})-\Psi^{*}(v)}{\hat{c}}\Big{)}.$

. Thus,

$\begin{split}H(\mathbb{Q}^{*}||\mathbb{W})&=\mathbb{E}^{\mathbb{W}}\Big{[}\frac{v\sigma
X_{1}-\phi^{*}(\sigma X_{1})-\Psi^{*}(v)}{\hat{c}}\exp\Big{(}\frac{v\sigma
X_{1}-\phi^{*}(\sigma X_{1})-\Psi^{*}(v)}{\hat{c}}\Big{)}\Big{]}\\\
&=\frac{v(\Psi^{*})^{\prime}(v)-\Psi^{*}(v)}{\hat{c}}-\mathbb{E}^{\mathbb{W}}\Big{[}\frac{\phi^{*}(\sigma
X_{1})}{\hat{c}}\exp\Big{(}\frac{v\sigma X_{1}-\phi^{*}(\sigma
X_{1})-\Psi^{*}(v)}{\hat{c}}\Big{)}\Big{]}\\\
&=\frac{v(\Psi^{*})^{\prime}(v)-\Psi^{*}(v)-E^{0,v}[\phi^{*}(Y^{*}_{1})]}{\hat{c}},\end{split}$

where the first equality follows from (4.31). The above is finite by the
hypothesis of this section that $j^{*}$ satisfies (3.20).

∎

######  Theorem 5.1.

Suppose that Assumption 5.1 holds. Then, $(H^{*},\theta^{*})$ is an
equilibrium where $H^{*}$ is given by (5.35) and $\theta^{*}$ follows (5.40).
The total demand in equilibrium follows (5.41). Moreover, the following
properties of equilibrium holds:

  1. (1)

The conditional law of $V$ is given by $\mathbb{P}(V\in
dv|{\mathcal{F}}^{Y^{*}}_{t})=\rho^{*}(t,Y^{*}_{t},v)\Pi(dv)$, where
$\rho^{*}$ is as in (5.36). In particular, $\frac{Y^{*}}{\sigma}$ is a
standard Brownian motion in its own filtration.

  2. (2)

Informed trader’s expected profit is given by

$E^{0,v}[W^{\theta^{*}}_{1}]=J(0,0)=\Psi^{*}(v)+E^{0,v}[\phi^{*}(\sigma
B_{1})].$ (5.44)

  3. (3)

Suppose further that one can differentiate inside the expectation666Recall
from (5.37) that
$H(t,Y^{*}_{t})=\mathbb{E}[\frac{d\phi^{*}}{dy}(Y^{*}_{1})|{\mathcal{F}}^{Y^{*}}_{t}]$.
so that

$H_{y}(t,Y^{*}_{t})=\mathbb{E}\Big{[}\frac{d^{2}\phi^{*}}{dy^{2}}(Y^{*}_{1})\big{|}{\mathcal{F}}^{Y^{*}}_{t}\Big{]}.$

Then, the expected loss of noise traders777That is, assuming they are not
subject to additional transaction costs unlike the informed trader. is given
by

$\sigma^{2}\mathbb{E}\Big{[}\frac{d^{2}\phi^{*}}{dy^{2}}(Y^{*}_{1})\Big{]}.$
(5.45)

  4. (4)

The insider expects to pay the following amount of penalty in equilibrium:

$\frac{c}{2}E^{0,v}\big{[}\int_{0}^{1}(\alpha^{*}_{t})^{2}dt\big{]}=\hat{c}H(\mathbb{Q}^{*}||\mathbb{W})=v(\Psi^{*})^{\prime}(v)-\Psi^{*}(v)-E^{0,v}[\phi^{*}(Y^{*}_{1})],$
(5.46)

where $H(\mathbb{Q}^{*}||\mathbb{W})$ is the relative entropy of
$\mathbb{Q}^{*}$ with respect to the Wiener measure, where $\mathbb{Q}^{*}$ is
the law of the demand process in equilibrium, as given by (5.41), from the
point of view of the insider.

###### Proof.

  1. (1)

Market efficiency: Note that given $\theta^{*}$, $Y^{*}$ is the unique strong
solution of (5.41). Then it follows from Propositions 3.1 and 4.2 that
$\mathbb{P}(V\in dv|{\mathcal{F}}^{Y^{*}}_{t})=\rho^{*}(t,Y^{*}_{t},v)\Pi(dv)$
and (3.19) holds with $\rho$ replaced by $\rho^{*}$ since
$\rho^{*}(0,0,\cdot)\equiv 1$ by construction. This in particular implies that
and $\frac{Y^{*}}{\sigma}$ is a Brownian motion in its own filtration by
considering the optional projection of the drift. Moreover,

$H^{*}(t,Y^{*}_{t})=\mathbb{E}[V|{\mathcal{F}}^{Y^{*}}_{t}]$

by (5.35). That is, the market efficiency condition of equilibrium is
satisfied.

  2. (2)

Insider optimality and value function: Suppose $H^{*}$ is the pricing rule of
the market makers, and consider an admissible strategy
$d\theta_{t}=\alpha_{t}dt$. Then, Ito’s formula and the PDE (5.39) yield

$\begin{split}J(u,Y_{u})&=J(0,0)+\int_{0}^{u}J_{y}(t,Y_{t})\\{\sigma
dB_{t}+\alpha_{t}dt\\}-\int_{0}^{u}\frac{J_{y}(t,Y_{t})+v-H^{*}(t,Y_{t})}{2c}dt\\\
&=J(0,0)+\int_{0}^{u}J_{y}(t,Y_{t})\sigma
dB_{t}-\int_{0}^{u}\frac{\big{(}J_{y}(t,Y_{t})+v-H^{*}(t,Y_{t})-\alpha
c\big{)}^{2}}{2c}dt\\\
&+\frac{c}{2}\int_{0}^{u}\alpha_{t}^{2}dt-\int_{0}^{u}(v-H^{*}(t,Y_{t}))\alpha_{t}dt.\end{split}$
(5.47)

First observe that $J\geq 0$ by a simple application of Jensen’s inequality.
Moreover, admissibility condition on $\theta$ implies $\alpha$ and $H^{*}$
satisfies (2.9). Thus, the local martingale in (5.47) is bounded from below by
an integrable random variable. That is, $(\int_{0}^{u}J_{y}(t,Y_{t})dB_{t})$
is a $P^{0,v}$ supermartingale. Using the boundary condition of $J$, one
obtains

$E^{0,v}W^{\theta}_{1}\leq
J(0,0)-E^{0,v}\int_{0}^{u}\frac{\big{(}J_{y}(t,Y_{t}+v)-H^{*}(t,Y_{t})-\alpha
c\big{)}^{2}}{2c}dt\leq J(0,0).$

Note that the first inequality becomes equality if
$(\int_{0}^{u}J_{y}(t,Y_{t})dB_{t})$ is a martingale. Moreover, the second
term will disappear when $\theta=\theta^{*}$. Thus, $\theta^{*}$ is optimal
provided it is admissible and the stochastic integral above is a martingale.
In this case, the expected wealth is given by

$J(0,0)=J^{0}(0,0)+\hat{c}\log\rho^{*}(0,0,v)=E^{0,v}[j^{*}(\sigma B_{1},v)].$

which coincides with (5.44).

  3. (3)

Optimality of $\theta^{*}$: Proposition 5.2 yields that

$E^{0,v}\Big{[}\int_{0}^{1}\Big{(}\frac{\rho_{y}(s,Y_{s},v)}{\rho(s,Y_{s},v)}\Big{)}^{2}ds\Big{]}<\infty,$
(5.48)

for all $v$ in the support of $V$. Thus, $\theta^{*}$ is admissible as soon as
$E^{0,v}\int_{0}^{1}H^{2}(t,Y^{*}_{t})dt<\infty$.

To this end, note that, since
$H(t,Y^{*}_{t})=\mathbb{E}[V|{\mathcal{F}}^{Y^{*}}_{t}]$,
$\mathbb{E}[H^{2}(t,Y^{*}_{t})]\leq\mathbb{E}[V^{2}]<\infty$. Thus,
$E^{0,v}\int_{0}^{1}H^{2}(t,Y^{*}_{t})dt$ is finite for almost all $v$. On the
other hand, it follows from (5.42) that

$E^{0,v}\Big{[}\int_{0}^{1}H^{2}(t,Y^{*}_{t})dt\Big{]}=\frac{\mathbb{E}^{\mathbb{W}}\Big{[}\int_{0}^{1}\exp\Big{(}\frac{v\sigma
X_{1}-\phi^{*}(\sigma X_{1})}{\hat{c}}\Big{)}H^{2}(t,\sigma
X_{t})dt\Big{]}}{\mathbb{E}^{\mathbb{W}}\Big{[}\exp\Big{(}\frac{v\sigma
X_{1}-\phi^{*}(\sigma X_{1})}{\hat{c}}\Big{)}\Big{]}}.$

Note that the denominator is finite for each $v$ by Lemma 5.1, and convex in
$v$. Moreover, it is larger than
$\exp(-\hat{c}^{-1}\mathbb{E}^{\mathbb{W}}[\phi^{*}(\sigma X_{1})])>0$. Thus,
the numerator is finite for almost all $v$. Since it is also convex in $v$,
its finiteness for all $v$ follows. Therefore,
$E^{0,v}\int_{0}^{1}H^{2}(t,Y^{*}_{t})dt<\infty$ for all $v$.

Finally, it remains to show that $(\int_{0}^{u}J_{y}(t,Y_{t})dB_{t})$ is a
$P^{0,v}$-martingale. However,
$J_{y}(t,y,v)=\hat{c}\frac{\rho_{y}(t,y,v)}{\rho(t,y,v}+H^{*}(t,y)-v$. Thus,
$(\int_{0}^{u}J_{y}(t,Y_{t})dB_{t})$ is a $P^{0,v}$-martingale since
$E^{0,v}\int_{0}^{1}H^{2}(t,Y^{*}_{t})dt<\infty$ and (5.48) holds.

  4. (4)

Noise traders loss: It follows from integration by parts that the wealth of
noise traders is given by

$\sigma
B_{1}(V-H^{*}(1,Y_{1}))-\sigma\int_{0}^{1}B_{t}dH^{*}(t,Y^{*}_{t})=\sigma
B_{1}V-\sigma\int_{0}^{1}H^{*}(t,Y^{*}_{t})dB_{t}-\sigma^{2}\int_{0}^{1}H^{*}_{y}(t,Y^{*}_{t})dt.$

The admissibility of $\theta^{*}$ implies in particular that the stochastic
integral with respect to $B$ is a martingale. Moreover,
$H^{*}_{y}(t,Y^{*}_{t})$ is an ${\mathcal{F}}^{Y^{*}}$-martingale. Thus,
taking expectations leads to the following expected wealth:

$-\sigma^{2}\mathbb{E}[H^{*}_{y}(1,Y^{*}_{1})]=-\sigma^{2}\mathbb{E}\Big{[}\frac{d^{2}\phi^{*}}{dy^{2}}(Y^{*}_{1})\Big{]}.$

  5. (5)

Expecgted penalty in equilibrium: This is a direct consequence of Proposition
5.2 since

$\alpha^{*}_{t}=\sigma^{2}\frac{\rho^{*}_{y}(t,Y_{t},V)}{\rho^{*}(t,Y_{t},V)}.$

∎

######  Remark 3.

A simple sufficient condition for differentiability under the expectation in
item (3) above is if the second derivative of $\phi^{*}$ has at most
polynomial growth. This would be the case if $V$ has a bounded support.
Indeed, by Theorem 4.1 a fixed point of $T$ exists and has bounded first and
second derivatives by Proposition 4.2

An intresting connection between the expected loss of the noise traders and
the price efficiency of equilibrium is given in the next corollary. Recall
that in Back [1] the equilibrium price is efficient in the sense that all the
private information of the informed trader is fully disseminated to the market
by the end of trading horizon. That is, the conditional variance of $V$ given
the market information at time $1$ is zero. This is no longer the case in the
present model as the market price does not converge to $V$ as $t\rightarrow
1$. The following defines a measure of price inefficiency of equilibrium and
gives its magnitude.

######  Corollary 5.1.

Consider the equilibrium of Theorem 5.1. Then the price inefficiency of the
equilibrium, denoted by $\delta$, is given by

$\delta:=\mathbb{E}[\mbox{Var}(V|{\mathcal{F}}^{Y^{*}}_{1})]=\hat{c}\mathbb{E}\Big{[}\frac{d^{2}\phi^{*}}{dy^{2}}(Y^{*}_{1})\Big{]}.$
(5.49)

Moreover,

$\mbox{Var}(V|{\mathcal{F}}^{Y^{*}}_{t})=\mathbb{E}\Big{[}\hat{c}\frac{d^{2}\phi^{*}}{dy^{2}}(Y^{*}_{1})+(h^{*}(Y^{*}_{1}))^{2}|{\mathcal{F}}^{Y^{*}}_{t}\Big{]}-(H^{*}(t,Y^{*}_{t}))^{2}.$
(5.50)

###### Proof.

It follows from (4.29) that
$\frac{d^{2}\phi^{*}}{dy^{2}}(y)=\hat{c}^{-1}\mbox{Var}(V|{\mathcal{F}}^{Y^{*}}_{1},Y^{*}_{1}=y)$,
which yields th first claim.

To show the second claim, it suffices to show that
$\mathbb{E}[V^{2}|{\mathcal{F}}^{Y^{*}}_{t}]=\mathbb{E}[\hat{c}\frac{d^{2}\phi^{*}}{dy^{2}}(Y^{*}_{1})+(h^{*}(Y^{*}_{1}))^{2}|{\mathcal{F}}^{Y^{*}}_{t}]$.
However, this follows from the fact that
$\hat{c}\frac{d^{2}\phi^{*}}{dy^{2}}(Y^{*}_{1})=\mbox{Var}(V|{\mathcal{F}}^{Y^{*}}_{1})$
and $\mathbb{E}[V|{\mathcal{F}}^{Y^{*}}_{1}]=h^{*}(Y^{*}_{1})$. ∎

######  Remark 4.

Comparing the above to the loss of noise traders from Theorem 5.1 reveals the
interesting observation that, independent of the distribution of the
fundamental value of the asset, the expected loss of noise traders equals
$c\delta$, where $\delta$ is the price inefficiency.

###  5.1. Gaussian case

Although Theorem 4.1 establishes the existence of a fixed point for $V$ with a
bounded support, there is a fixed point when $V$ has a Gaussian distribution.
This subsection studies the existence of a fixed point and the resulting
equilibrium under the following assumption:

######  Assumption 5.2.

The distribution of $V$ is Gaussian, i.e. $V\sim N(\mu,\gamma^{2})$.

In the Kyle model with Gaussian fundamental value but no transaction costs,
the equilibrium price is an affine function of the demand $Y$. Thus, one
expects $H^{*}(t,y)=\lambda y+\mu$ for some $\lambda>0$. The conjecture that
$H^{*}(t,0)=\mu$ is a simple consequence of the rationality of the pricing
rule that requires $\mu=\mathbb{E}[V]=S_{0}=H^{*}(0,0).$ Combined with the
normalisation that $\phi^{*}(0)=0$, one expects that
$\phi^{*}(y)=\frac{\lambda y^{2}}{2}+\mu y$. This leads to the following:

######  Theorem 5.2.

Suppose that Assumpton 5.2 holds. Then, $T_{0}\phi^{*}=\phi^{*}$ with
$\phi^{*}(y)=\frac{\lambda^{*}y^{2}}{2}+\mu y$ for

$\lambda^{*}=\frac{-c+\sqrt{c^{2}+4\frac{\gamma^{2}}{\sigma^{2}}}}{2},$ (5.51)

which is the positive root of
$\sigma^{2}\lambda^{2}+\hat{c}\lambda-\gamma^{2}=0$. Moreover,

$\Psi^{*}(v)=\hat{c}\log
T_{1}\phi^{*}(v)=\frac{\hat{c}}{2}\log\frac{c}{c+\lambda^{*}}+\frac{(\mu-v)^{2}}{2(c+\lambda^{*})}.$
(5.52)

###### Proof.

By completing the squares, it is easily seen that

$T_{1}\phi^{*}(y)=\frac{\Sigma}{\sigma}\exp\Big{(}\frac{\Sigma^{2}(\mu-v)^{2}}{2\hat{c}^{2}}\Big{)},$

where
$\frac{1}{\Sigma^{2}}=\frac{1}{\sigma^{2}}+\frac{\lambda}{\hat{c}}=\frac{c+\lambda}{\hat{c}}$.
That is,

$T_{1}\phi^{*}(y)=\frac{\Sigma}{\sigma}\exp\Big{(}\frac{(\mu-v)^{2}}{2\hat{c}(c+\lambda)}\Big{)}.$

Thus, to have $T_{0}\phi^{*}=\phi^{*}$, one must have

$\begin{split}\exp\big{(}\frac{\lambda y^{2}+2\mu
y}{2\hat{c}}\big{)}&=\frac{\sigma}{\Sigma\gamma}\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi}}\exp\Big{(}\frac{yv}{\hat{c}}-\frac{(\mu-v)^{2}}{2\hat{c}(c+\lambda)}-\frac{(\mu-v)^{2}}{2\gamma^{2}}\Big{)}dv\\\
&=\frac{\sigma\Sigma_{0}}{\Sigma\gamma}\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi\Sigma_{0}^{2}}}\exp\Big{(}\frac{yv}{\hat{c}}-\frac{(\mu-v)^{2}}{2\Sigma_{0}^{2})}\Big{)}dv\\\
&=\frac{\sigma\Sigma_{0}}{\Sigma\gamma}\exp\Big{(}\frac{y\mu}{\hat{c}}+\frac{y^{2}\Sigma_{0}^{2}}{2\hat{c}^{2}}\Big{)},\end{split}$

where
$\frac{1}{\Sigma_{0}^{2}}=\frac{1}{\hat{c}(c+\lambda)}+\frac{1}{\gamma^{2}}=\frac{\gamma^{2}+\hat{c}(c+\lambda)}{\gamma^{2}\hat{c}(c+\lambda)}$.

Thus, $\lambda$ must solve

$\lambda=\frac{\Sigma_{0}^{2}}{\hat{c}}=\frac{\gamma^{2}(c+\lambda)}{\gamma^{2}+\hat{c}(c+\lambda)}.$
(5.53)

Note that one must also have

$1=\frac{\sigma^{2}\Sigma_{0}^{2}}{\Sigma^{2}\gamma^{2}}=\frac{\sigma^{2}(c+\lambda)}{\gamma^{2}}\lambda.$
(5.54)

That is, $\lambda$ must seemingly satisfy two constraints at the same time.
However, (5.53) is equivalent to

$\lambda c\sigma^{2}(c+\lambda)=c\gamma^{2}.$

Dividing both sides by $c$ yields (5.54). Thus, $\lambda$ is the unique
positive solution of

$\lambda^{2}\sigma^{2}+\lambda c\sigma^{2}-\gamma^{2}=0,$

which is given by $\lambda^{*}$.

Consequently,

$\Psi^{*}(v)=\frac{\hat{c}}{2}\log\frac{\Sigma^{2}}{\sigma^{2}}+\frac{(\mu-v)^{2}}{2(c+\lambda^{*})}=\frac{\hat{c}}{2}\log\frac{c}{c+\lambda^{*}}+\frac{(\mu-v)^{2}}{2(c+\lambda^{*})}.$

∎

The formula for $\lambda^{*}$ suggests a parametrisation for the cost
coefficient $c$ that will be useful when considering the equilibrium with
Gaussian fundamental value. The easy proof of the following corollary is left
to the reader.

######  Corollary 5.2.

Let $c=\kappa\frac{\gamma}{\sigma}$ for some $\kappa>0$. Then,
$\lambda^{*}=\Lambda(\kappa)\frac{\gamma}{\sigma}$, where
$\Lambda:(0,\infty)\to(0,1)$ is a decreasing function given by

$\Lambda(\kappa)=\frac{\sqrt{\kappa^{2}+4}-\kappa}{2}.$

In particular, $\Lambda(0+)=1=1-\Lambda(\infty)$.

Note that the fixed point of $T_{0}$ given by the above result satisfies all
the conditions of Theorem 5.1. Thus, an equilibrium exists and has the
following dynamics as a straightforward corollary. To understand the
dependency of the equilibrium on the amount of adverse selection
$\frac{\gamma}{\sigma}$, the above parametrisation will be employed.

######  Corollary 5.3.

Suppose that Assumption 5.2 holds and $c=\kappa\frac{\gamma}{\sigma}$ for some
$\kappa>0$. Then $(h^{*},\theta^{*})$ is an equilibrium, where
$h^{*}(y)=\lambda^{*}y+\mu$ with $\lambda^{*}$ from (5.51), and

$d\theta^{*}_{t}=\Lambda(\kappa)\sigma\frac{\frac{V-\mu}{\gamma}-\frac{\Lambda(\kappa)Y^{*}_{t}}{\sigma}}{1-t\Lambda^{2}(\kappa)}dt,$
(5.55)

where $\Lambda$ is the function defined in Corollary 5.2. Moreover,

  1. (1)

The conditional law of $V$ given ${\mathcal{F}}^{Y^{*}_{1}}$ is Gaussian with
mean $\lambda^{*}Y^{*}_{t}+\mu$ and variance
$\gamma^{2}(1-t\Lambda^{2}(\kappa))$. Thus,

$\rho^{*}(t,y,v)=\frac{1}{\sqrt{1-t\Lambda^{2}(\kappa)}}\exp\Big{(}-\frac{1}{2\gamma^{2}}\Big{(}\frac{(v-\lambda^{*}y-\mu)^{2}}{1-t\Lambda^{2}(\kappa)}-v^{2}\Big{)}\Big{)}$

  2. (2)

Equilibrium demand $Y^{*}$ follows

$dY^{*}_{t}=\sigma
dB_{t}+\Lambda(\kappa)\sigma\frac{\frac{V-\mu}{\gamma}-\frac{\Lambda(\kappa)Y^{*}_{t}}{\sigma}}{1-t\Lambda^{2}(\kappa)}dt.$

  3. (3)

Informed trader’s expected profit is given by

$E^{0,v}[W^{\theta^{*}}_{1}]=\frac{\hat{c}}{2}\log\frac{c}{c+\lambda^{*}}+\frac{(\mu-v)^{2}}{2(c+\lambda^{*})}+\frac{\lambda^{*}\sigma^{2}}{2}.$
(5.56)

  4. (4)

The expected loss of noise traders is given by
$\lambda^{*}\sigma^{2}=\Lambda(\kappa)\gamma\sigma$.

  5. (5)

The price inefficiency is given by
$\delta=\hat{c}\lambda^{*}=\kappa\Lambda(\kappa)\gamma^{2}$. Moreover,

$\mbox{Var}(V|{\mathcal{F}}^{Y^{*}}_{t})=\kappa\Lambda(\kappa)\gamma^{2}+\Lambda^{2}(\kappa)\gamma^{2}(1-t)=\gamma^{2}(1-t)+t\kappa\Lambda(\kappa)\gamma^{2}.$

###### Proof.

The only claim that is not explicitly given by Theorems 5.1 and 5.2 is the
conditional distribution of $V$. Note that the conditonal distribution at time
$1$ is given by $\exp(-j^{*}(y,v)/c)\Pi(dv)$, where $j^{*}$ is given by
(5.34). Straightforward computations show that this distribution is Gaussian
with density proportional to

$\exp\Big{(}-(\mu-v)^{2}\frac{\gamma^{2}+\hat{c}(c+\lambda^{*})}{2\gamma^{2}\hat{c}(c+\lambda^{*})}+\frac{vy}{\hat{c}}\Big{)}.$

Since
$\frac{\gamma^{2}(c+\lambda^{*})}{\gamma^{2}+\hat{c}(c+\lambda^{*})}=\lambda^{*}\hat{c}$
by (5.53), it follows that the conditional distribution at time $1$ is
Gaussian with mean $\lambda^{*}Y^{*}_{1}+\mu$ and variance
$\hat{c}\lambda^{*}$. Thus, the conditional density at time $t$ given
$Y^{*}_{t}=y$ equals

$\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi\sigma^{2}(1-t)}}\exp\Big{(}-\frac{(y-z)^{2}}{2\sigma^{2}(1-t)}\Big{)}\frac{1}{\sqrt{2\pi\hat{c}\lambda^{*}}}\exp\Big{(}-\frac{(v-\mu-\lambda^{*}z)^{2}}{2\hat{c}\lambda^{*}}\Big{)}dzdv=\rho^{*}(t,y,v)\Pi(dv).$

This will be a Gaussian density as well. Indeed, by considering its
characteristic function

$\begin{split}\int_{\mathbb{R}}e^{irv}\rho^{*}(t,y,v)\Pi(dv)&=e^{-\frac{r^{2}\hat{c}\lambda^{*}}{2}}\int_{\mathbb{R}}\frac{e^{ir(\mu+\lambda^{*}z)}}{\sqrt{2\pi\sigma^{2}(1-t)}}\exp\Big{(}-\frac{(y-z)^{2}}{2\sigma^{2}(1-t)}\Big{)}dz\\\
&=e^{-\frac{r^{2}}{2}(\hat{c}\lambda^{*}+(\lambda^{*})^{2}\sigma^{2}(1-t))+ir(\mu+\lambda^{*}y)}.\end{split}$

Therefore, the conditional distribution is Gaussian with mean
$\mu+\lambda^{*}Y^{*}_{t}$ and variance
$\hat{c}\lambda^{*}+(\lambda^{*})^{2}\sigma^{2}(1-t)$. Using the quadratic
equation that $\lambda^{*}$ solves from Theorem 5.2, the variance in fact
equals $\gamma^{2}-t(\lambda^{*}\sigma)^{2}$. Using the density of normal,
this readily yields the equilibrium $\theta^{*}$ using Corollary 5.2. ∎

The above shows that the insider buys when the market valuation is lower than
the fundamental value and sells otherwise as in the Kyle model without
penalties. Thus, from the point of view of the insider, prices still mean
revert around insider’s valuation. However, the speed of mean reversion as
measured by

$r(t):=\frac{\Lambda(\kappa)}{1-t\Lambda^{2}(\kappa)}$ (5.57)

is considerably smaller than what it would be in the model without penalties,
where the corresponding speed equals $\frac{1}{1-t}$. Thus, the insider trades
less aggressively when there is a legal risk. This in particular results in
prices not fully revealing the insider’s private information at the end of the
trading horizon.

Recall that the Kyle’s lambda equals $\frac{\gamma}{\sigma}$ when there are no
penalties. Thus,

$L(\kappa):=\frac{1}{\Lambda(\kappa)}-1=\frac{1-\Lambda(\kappa)}{\Lambda(\kappa)}$
(5.58)

can be used to represent the gain in liquidity when the regulators impose a
quadratic penalty with rate given by $\kappa\frac{\gamma}{\sigma}$. This gain
is clearly decreasing in $\Lambda$, which is a decreasing function of
$\kappa$. The market thus becomes perfectly liquid in the limit but it will
also be completely inefficient.

Another interesting consequence of the reparametrisation manifests itself in
the representation of insider’s ex-ante profit. Note that $\Lambda(\kappa)$ is
the positive solution of

$\Lambda^{2}(\kappa)+\Lambda(\kappa)\kappa=1.$ (5.59)

######  Corollary 5.4.

Consider the equilibrium in Corollary 5.3. Then, the following statements are
valid:

  1. (1)

The ex-ante profit of the insider equals

${\mathcal{W}}(\kappa):=\mathbb{E}[W^{\theta^{*}}_{1}]=\frac{\gamma\sigma}{2}\big{(}\kappa\log(\kappa\Lambda(\kappa))+2\Lambda(\kappa)\big{)}.$

Moreover, it is convex and decreasing in $\kappa$ and vanishes in the limit as
$\kappa\rightarrow\infty$.

  2. (2)

Derivative of $\Lambda(\kappa)$ is given by

$\Lambda^{\prime}(\kappa)=-\frac{\Lambda^{2}(\kappa)}{1+\Lambda^{2}(\kappa)}.$

  3. (3)

The total welfare that measures the total expected profit of all traders is
given by

$\frac{\gamma\sigma}{2}\kappa\log(\kappa\Lambda(\kappa))=\frac{\gamma\sigma}{2}\kappa\log(1-\Lambda^{2}(\kappa)).$
(5.60)

###### Proof.

The first and the last claims follow from Corollary 5.3 upon noticing that
$(c+\lambda^{*})\frac{\sigma}{\gamma}=\kappa+\Lambda(\kappa)=\frac{1}{\Lambda(\kappa)}$
as well as $\kappa\Lambda(\kappa)=1-\Lambda^{2}(\kappa)$ in view of (5.59).

Differentiating ${\mathcal{W}}$ with respect to $\kappa$ and noting
$1-\Lambda^{2}(\kappa)=\Lambda(\kappa)\kappa$ yield

${\mathcal{W}}^{\prime}(\kappa)=\frac{\gamma\sigma}{2}\log(1-\Lambda^{2}(\kappa)),\quad{\mathcal{W}}^{\prime\prime}(\kappa)=-\gamma\sigma\Lambda^{\prime}(\kappa)\kappa^{-1}.$
(5.61)

Moreover, (5.59) yields

$\Lambda^{\prime}(\kappa)=-\frac{\Lambda(\kappa)}{2\Lambda+\kappa}=-\frac{\Lambda(\kappa)}{\Lambda(\kappa)+\frac{1}{\Lambda(\kappa)}}=-\frac{\Lambda^{2}(\kappa)}{1+\Lambda^{2}(\kappa)}<0.$

This completes the proof the claims on derivatives. To show that the limiting
wealth is $0$, first recall from Corollary 5.2 that $\Lambda(\infty)=0$. On
the other hand,

$\kappa\log(\kappa\Lambda(\kappa))=\frac{1-\Lambda^{2}(\kappa)}{\Lambda(\kappa)}\log(1-\Lambda^{2}(\kappa)).$

Thus, the proof will be complete if $\lim_{x\rightarrow
0}\frac{\log(1-x^{2})}{x}=0$, which follows by a quick application of the
L’Hospital rule. ∎

As $\Lambda(K)\in(0,1)$, the above shows that the welfare is negative. This is
no surprise since the market makers are competitive and, therefore, the total
welfare equals in magnitude to the expected penalty to the insider.

Similarly, one also expect that the insider’s total wealth vanishes as rate of
penalties increases without bound. The following result shows that the
expected penalty in equilibrium is bounded in $\kappa$. Thus, the vanishing
wealth in the limit is not a consequence of large expected costs but insider
trading very little in the presence of legal risk. This is consistent with the
representation of the speed of trading via (5.57). Indeed, $r$ is a decreasing
function of $\kappa$ and vanishes as $\kappa\rightarrow\infty$.

The expected penalty is also not monotone as can be deduced from the following
result or seen in Figure 1.

######  Corollary 5.5.

Consider the equilibrium in Corollary 5.3. The insider’s expected penalties in
equilibrium equals

$P(\kappa)=-\frac{\gamma\sigma}{2}\kappa\log(\kappa\Lambda(\kappa)).$

There exists a $\kappa_{0}$ such that $P$ is concave on $(0,\kappa_{0})$ and
convex otherwise. In particular, it is bounded by $\frac{\gamma\sigma}{2}$.
Moreover, $P(0+)=P(\infty)=0$.

###### Proof.

Note that for $x\in(0,1)$,

$\log(1-x)\geq-\frac{x}{1-x}$

in view of the mean value theorem. Thus,

$\frac{1-\Lambda^{2}(\kappa)}{\Lambda(\kappa)}\log(1-\Lambda^{2}(\kappa))\geq-\Lambda(\kappa)\geq-1.$

This shows the bound on $P$.

Moreover, (5.61) implies

$P^{\prime\prime}(\kappa)=\gamma\sigma(\Lambda^{\prime}(\kappa)\kappa^{-1}+2\Lambda^{\prime\prime}(\kappa))=\gamma\sigma\Big{(}-\frac{\kappa^{-1}\Lambda^{2}(\kappa)}{1+\Lambda^{2}(\kappa)}+4\frac{\Lambda^{3}(K)}{(1+\Lambda^{2}(\kappa))^{3}}\Big{)},$

which is positive if and only if

$\frac{(1+\Lambda^{2}(\kappa))^{2}}{\kappa}\leq 4.$

Note that the left side is decreasing from $\infty$ to $0$ as $\kappa$
increases to $\infty$. Thus, there exists a $\kappa_{0}$ such that $P$ is
concave on $(0,\kappa_{0})$ and convex otherwise.

The limiting values at $0$ and $\infty$ have already been observed in
computations leading to the respective limits for ${\mathcal{W}}$ in the proof
of the previous corollary. ∎

######  Remark 5.

In view of the relationship between relative entropy and expected penalties
via (5.46), the ‘average’ entropy, i.e. $H(\mathbb{Q}^{*}||\mathbb{W})$
integrated with respect to the law of $V$, equals
$-\frac{1}{2}\log(\kappa\Lambda(\kappa))=-\frac{1}{2}\log(1-\Lambda^{2}(\kappa))$.
This is increasing as $\kappa\rightarrow 0$, i.e. smaller quadratic penalties
are imposed by the regulators. The limiting value is infinite, which coincides
with the relative entropy of the Kyle equilibrium without penalties. Note that
when there are no penalties, $Y^{*}/\sigma$ is a Brownian bridge:

$dY^{*}_{t}=\sigma
dB_{t}+\sigma\frac{\frac{V-\mu}{\gamma}-\frac{Y^{*}_{t}}{\sigma}}{1-t}dt.$

The law of $Y^{*}$ is then singular with respect to the Wiener measure.

Figure 1. Expected penalty paid by the insider in equilibrium as a function of
$\kappa$. The values on the $y$-axis are normalised by $\gamma\sigma$.

Although the expected penalties (hence the total welfare) are non-monotone,
the loss of the noise traders and price efficiency decrease as the rate of
penalties, i.e. $\kappa$, increase. These can be verified by direct
differentiation of the relevant expressions and are illustrated in Figure 2.

$\begin{array}[]{cc}\scalebox{0.26}{\includegraphics{lambda.pdf}}&\scalebox{0.26}{\includegraphics{inefficiency.pdf}}\end{array}$

Figure 2. The left pane shows the expected loss of noise traders normalised by
$\gamma\sigma$ whereas the right one is the price inefficiency of equilibrium
normalised by $\gamma^{2}$.

##  6\. Conclusion

When there are quadratic penalties, the equilibrium in the continuous time
version of the Kyle model changes drastically. The insider no longer finds
optimal to bring the market prices to her own valuation at the end ot trading.
The noise traders lose less on average as the potential penalties on the
insider increase; however, the total welfare, i.e the sum of insider and noise
profit, is not monotone in the rate of penalties when the fundamental value is
normally distributed. Interestingly, the maximum expected penalty that is
accrued by the insider occurs when the rate of penalties is sufficiently
small.

The solution of the equilibrium reveals an interesting connection between a
class of quadratic BSDEs and h-transforms. The interesting extension to
equilibrium with general convex costs are left for future research. Such an
extension will be valuable to regulators searching for the optimal penalty
structure to control illegal insider trading as discussed by [7] in a one-
period setting.

## References

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11institutetext: Institute of Technology Guwahati, Guwahati, India,
11email<EMAIL_ADDRESS>11email<EMAIL_ADDRESS>
# On the Representation Number of Bipartite Graphs

Khyodeno Mozhui  K. V. Krishna

###### Abstract

A word-representable graph is a simple graph $G$ which can be represented by a
word $w$ over the vertices of $G$ such that any two vertices are adjacent in
$G$ if and only if they alternate in $w$. It is known that the class of
comparability graphs – the graphs which admit a transitive orientation – is
precisely the class of graphs that can be represented by a concatenation of
permutations of vertices. The class of bipartite graphs is a subclass of
comparability graphs. While it is an open problem to determine the
representation number of comparability graphs, it was conjectured that the
representation number of bipartite graphs on $n$ vertices is at most $n/4$. In
this paper, we propose a polynomial time relabeling algorithm to produce a
word representing a given bipartite graph which is a concatenation of
permutations of the graph’s vertices. Thus we obtain an upper bound for the
representation number of bipartite graphs, which in turn gives us an upper
bound for the dimension of the posets corresponding to bipartite graphs.

###### Keywords:

Word-representable graphs, representation number, comparability graphs,
bipartite graphs

##  1 Introduction

A word-representable graph is a simple graph $G$ which can be represented by a
word $w$ over the vertices of $G$ such that any two vertices are adjacent in
$G$ if and only if they alternate in $w$. While the origin of word-
representable graphs lies in the study of Perkins semigroups [5], this notion
covers many important classes of graphs including comparability graphs, circle
graphs and 3-colorable graphs. The literature has several important
contributions to the theory of word-representable graphs and their connections
to various concepts. A comprehensive introduction to word-representable graphs
and their connections and contributions can be found in the monograph [3].

The representation number of a word-representable graph is the minimum number
$k$ such that the graph is represented by a word containing exactly $k$ copies
of each letter. The class of complete graphs is the class of graphs with
representation number 1 and it was established in [2] that the class of circle
graphs characterizes the graphs with representation number 2. Although there
are examples of graphs with higher representation number, no characterization
of graphs is available for other numbers.

The class of comparability graphs – the graphs which admit a transitive
orientation – play a vital role in the theory of word-representable graphs. In
fact, the neighborhood of each vertex in a word-representable graph is a
comparability graph [4]. Furthermore, the class of comparability graphs is
precisely the class of permutationally representable graphs, that is, they can
be represented by a concatenation of permutations of their vertices [5]. We
call the minimum number of such permutations of vertices of a comparability
graph its permutation representation number. Every comparability graph
corresponds to a partially ordered set (poset) based on one of its transitive
orientations. Due to their correspondence with posets, comparability graphs
received importance in the literature. It is known that finding the
permutation representation number of a comparability graph is equivalent to
finding the dimension of the corresponding poset [2]. However, the problem of
finding the dimension of a poset is NP-hard [7].

The class of bipartite graphs is a subclass of the class of comparability
graphs. While it is an open problem to determine the representation number of
a comparability graph, it was conjectured in [1] that the representation
number of a bipartite graph on $n$ vertices is at most $n/4$. In this paper,
we propose a polynomial time relabeling algorithm to produce a word that is a
concatenation of permutations of vertices representing a given bipartite
graph. Thus we obtain an upper bound for the permutation representation number
of bipartite graphs and consequently an upper bound for their representation
number. This indeed gives us an upper bound for the dimension of the posets
corresponding to bipartite graphs. In fact, the upper bound for the
permutation representation number is tight for the class of bipartite graphs.

##  2 Preliminaries

In this section, we provide necessary background material on words, graphs,
and word-representable graphs. For the concepts that are not presented here,
we may refer to [3, 6].

Let $A$ be a finite set. A word over $A$ is a finite sequence of elements of
$A$. The words are written by juxtaposing the symbols of the sequence. The
empty sequence, called the empty word, is denoted by $\varepsilon$. A subword
$u$ of a word $w$ is a subsequence of the sequence $w$ and it is denoted by
$u\leq w$. For instance, $abaa\leq aabbaba$. Clearly, $\leq$ is a partial
order relation on the set of all words over $A$. We say the letters $a$ and
$b$ alternate in the word $w$ if the subword consisting of all occurrences of
$a$ and $b$, that is the subword obtained by deleting all other letters of
$w$, is of the form $abab\cdots$ or $baba\cdots$ (of even or odd length). A
word $w$ is said to be $k$-uniform if the number of occurrences of each letter
in $w$ is $k$.

Let $A=\\{a_{i_{1}},a_{i_{2}},\ldots,a_{i_{k}}\\}$ be a set and
$i_{1}<i_{2}<\cdots<i_{k}$. We write $\text{Dec}(A)$ to represent the word
$a_{i_{k}}a_{i_{k-1}}\cdots a_{i_{1}}$ in which the symbols of $A$ appear
exactly once and are arranged such that the subscripts are in decreasing
order.

In this paper, the term graph represents only a simple and undirected graph,
i.e., a graph without loops or parallel edges. To distinguish between two-
letter words and edges of graphs, we write $\overline{ab}$ to denote an
undirected edge between vertices $a$ and $b$. A directed edge from $a$ to $b$
is denoted by $\overrightarrow{ab}$. The neighborhood of a vertex $a$ in a
graph, denoted by $N(a)$, is the set of all vertices adjacent to $a$ in the
graph. An orientation of a graph is an assignment of direction to each edge so
that the resulting graph is a directed graph. An orientation of a graph is
said to be a transitive orientation, if the adjacency relation on the vertices
in the resulting directed graph is transitive. A graph $G$ is said to be a
bipartite graph if its vertex set can be partitioned into $\\{V_{1},V_{2}\\}$,
called bipartition, such that every edge in $G$ connects a vertex in $V_{1}$
to a vertex in $V_{2}$. A bipartite graph with bipartition
$\\{V_{1},V_{2}\\}$, where $|V_{1}|=m$ and $|V_{2}|=n$, is said to be
complete, denoted by $K_{m,n}$, if every vertex in $V_{1}$ is adjacent to all
vertices of $V_{2}$.

A graph $G=(V,E)$ is said to be word-representable if there is a word $w$ over
the set $V$ such that two vertices $a$ and $b$ are adjacent in $G$ if and only
if $a$ and $b$ alternate in $w$. A graph is said to be $k$-word-representable
if there is $k$-uniform word representing the graph. Every word-representable
graph is $k$-word-representable for some $k$ [4]. The representation number of
a word-representable graph $G$, denoted by $\mathcal{R}(G)$, is the smallest
number $k$ such that $G$ is $k$-word-representable. A graph $G$ is said to be
permutationally representable if it can be represented by a word of the form
$w_{1}w_{2}\cdots w_{k}$, where each $w_{i}$ is a permutation of vertices of
$G$. Furthermore, if a graph is permutationally representable involving $k$
permutations, then the graph is called permutationally $k$-representable.

###### Definition 1

Let $G$ be a permutationally representable graph. The permutation
representation number of $G$, denoted by $\mathcal{R}^{p}(G)$, is the smallest
$k$ such that $G$ is permutationally $k$-representable.

###### Remark 1

If $G$ is a permutationally representable graph, then
$\mathcal{R}(G)\leq\mathcal{R}^{p}(G)$.

The class of complete graphs has permutation representation number 1, i.e.,
$\mathcal{R}^{p}(K_{n})=\mathcal{R}(K_{n})=1$ for all $n$, where $K_{n}$ is
the complete graph on $n$ vertices. It was shown in [2, see the proof of
Theorem 4] that $\mathcal{R}^{p}(H_{n,n})=n$ (for $n\geq 2$) and in [1] that
$\mathcal{R}(H_{n,n})=\lceil n/2\rceil$ (for $n\geq 5$), where $H_{n,n}$ is
the graph, called a crown graph, obtained by removing a perfect matching in
$K_{n,n}$. Further, it was conjectured in [1, see Conjecture 1 in page 93]
that every bipartite graph on $n$ vertices has representation number at most
$n/4$.

A graph is said to be a comparability graph if it admits a transitive
orientation. Every comparability graph is word-representable. In fact, the
class of comparability graphs is precisely the class of permutationally
representable graphs [5]. Every bipartite graph is a comparability graph.
Based on its transitive relation, every comparability graph corresponds to a
partially ordered set (poset). In fact, the class of bipartite graphs is
precisely the class of comparability graphs that are isomorphic to the Hasse
diagrams of the corresponding posets.

###### Remark 2

Let $G$ be a comparability graph and $H$ be the Hasse diagram of the
corresponding poset. If the graphs $G$ and $H$ are isomorphic, then $G$ is a
bipartite graph, and vice versa.

The dimension of a poset is the minimum number of linear orders of its
elements such that their intersection is the partial order of the poset. It
was shown in [2] that a comparability graph $G$ is permutationally
$k$-representable if and only if the poset corresponding to $G$ has dimension
at most $k$. However, since the problem of finding the dimension of a poset is
NP-hard [7], finding the permutation representation number of a comparability
graph is NP-hard. Hence, in the direction of finding the permutation
representation number of some subclasses of comparability graphs, in [2], the
authors obtained the permutation representation number of crown graph
$H_{k,k}$. Here, the authors constructed linear orders whose intersection
happens to be the corresponding poset of $H_{k,k}$. In the following section,
we extend the idea to arbitrary bipartite graphs and construct linear orders
of the elements of corresponding poset.

##  3 Main Result

In this section, we devise an algorithmic procedure to construct a word
representing permutationally a given bipartite graph. Subsequently, we obtain
an upper bound for the permutation representation number of bipartite graphs.
Consequently, in view of the correspondence between comparability graphs and
posets, we also present an upper bound for the dimension of a special class of
posets which correspond to bipartite graphs.

###### Remark 3

The permutation representation number of a graph does not change by the
inclusion or deletion of isolated vertices. If a graph $G=(V,E)$ (with at
least 3 vertices) has isolated vertices, say $a_{1},a_{2},\ldots,a_{t}$, then
consider the induced subgraph $H$ of $G$ on the vertex set
$V\setminus\\{a_{1},a_{2},\ldots,a_{t}\\}$. That is, $H$ is obtained by
deleting the vertices $a_{1},a_{2},\ldots,a_{t}$ from $G$. Note that $G$ is
permutationally representable if and only if $H$ is permutationally
representable. In fact, if $w_{1}$, $w_{2}$, …, $w_{k}$ are permutations of
vertices of $H$ such that the word $w_{1}w_{2}\cdots w_{k}$ represents the
graph $H$ permutationally, then we can see that $G$ is also permutationally
$k$-representable. For instance, the following word $w$ with $k$ permutations
on the vertices of $G$ represents $G$ permutationally:

$w=\begin{cases*}w_{1}ur(u)w_{2}w_{3}ur(u)w_{4}\cdots w_{k}u,\;\text{ if $k$
is odd};\\\ w_{1}ur(u)w_{2}w_{3}ur(u)w_{4}\cdots r(u)w_{k},\;\text{ if $k$ is
even,}\end{cases*}$

where $u=a_{1}a_{2}\cdots a_{t}$ and $r(u)=a_{t}a_{t-1}\cdots a_{1}$, the
reversal of $u$.

In what follows, $G$ always denotes a bipartite graph with bipartition
$\\{V_{1},V_{2}\\}$ and $m=|V_{1}|\leq|V_{2}|$. In view of Remark 3, we assume
that $G$ has no isolated vertices.

###  3.1 Relabeling algorithm

We now present an algorithm to determine permutations of vertices of a given
bipartite graph $G$ such that their concatenation represents $G$. It is
achieved by relabeling the vertices of $G$ in a particular order based on
their nonadjacent vertices and construct certain permutations of the vertices
of $G$. The algorithm is presented in the following.

  1. 1.

Given a bipartite graph $G$, suppose $V_{1}=\\{a_{1},\ldots,a_{m}\\}$ and
$V_{2}=\\{b_{1},\ldots,b_{n}\\}$.

  2. 2.

If $N(a)=V_{2}$ for all $a\in V_{1}$, then consider the following word and
exit:

$w=a_{1}a_{2}\cdots\ a_{m}b_{1}b_{2}\cdots b_{n}a_{m}a_{m-1}\cdots
a_{1}b_{n}b_{n-1}\cdots\ b_{1}$.

  3. 3.

Else, choose $a$ in $V_{1}$ such that $V_{2}\setminus N(a)\neq\varnothing$ and
label it as $c_{1}$.

  4. 4.

Relabel the remaining vertices of $V_{1}$ as $c_{2},\ldots,c_{m}$,
arbitrarily.

  5. 5.

Relabel the vertices in $V_{2}$ as follows:

    1. (a)

Set $m^{\prime}=m$; $A=V_{2}$.

    2. (b)

For $i=1$ to $m$,

      1. i.

Let $A_{i}=A\setminus N(c_{i})$, say $\\{b_{j_{1}},...,b_{j_{k_{i}}}\\}$.

      2. ii.

Relabel the vertices of $A_{i}$ as
$c_{m^{\prime}+1},\ldots,c_{m^{\prime}+k_{i}}$, arbitrarily.

      3. iii.

Set $m^{\prime}=m^{\prime}+k_{i}$; $A=A\setminus A_{i}$.

    3. (c)

If $A\neq\varnothing$, then relabel the remaining vertices of $A$ as
$c_{m^{\prime}+1},\ldots,c_{m+n}$, arbitrarily.

  6. 6.

Create a list of permutations of the vertices $c_{1},\ldots,c_{m+n}$ as per
the following:

    1. (a)

Set $w_{1}=c_{m}c_{m-1}\cdots c_{2}c_{m+1}c_{m+2}\cdots
c_{m+k_{1}}c_{1}c_{m+k_{1}+1}c_{m+k_{1}+2}\cdots c_{m+n}$.

    2. (b)

For $i=2$ to $m$,

      1. If $V_{2}\setminus N(c_{i})\neq\varnothing$, then set

$w_{i}=c_{1}\cdots c_{i-1}c_{i+1}\cdots c_{m}\text{Dec}(V_{2}\setminus
N(c_{i}))c_{i}\text{Dec}(N(c_{i}))$;

else, set $w_{i}=\varepsilon$.

    3. (c)

If $N(c)=V_{2}$ for some $c\in V_{1}$ then set

$w_{m+1}=c_{1}c_{2}\cdots c_{m}c_{m+n}\cdots c_{m+1}$;

else, set $w_{m+1}=\varepsilon$.

  7. 7.

Report the word $w$ obtained by replacing the original labels of the vertices
of $G$ in the word $w_{1}\cdots w_{m}w_{m+1}$.

###### Remark 4

The relabeling algorithm runs in a polynomial time. In fact, its complexity is
$O(mn)$. Since the relabeling algorithm works by identifying the set of
nonadjacent vertices (adjacent vertices) for each vertex of $V_{1}$, so the
only possible nonadjacent vertices would be the vertices from $V_{2}$.

###### Example 1

We demonstrate the relabeling algorithm on the bipartite graph $G$ given in
Figure 2. Although the selection of vertices can be arbitrary, here we
implemented the algorithm based on their degrees. Accordingly, the vertices of
$V_{1}$, viz., $a_{1},a_{2},a_{3},a_{4}$ are relabeled, respectively, as
$c_{4},c_{2},c_{3},c_{1}$. Further, the vertices of $V_{2}$, viz.,
$b_{1},b_{2},b_{3},b_{4},b_{5}$ are relabeled as
$c_{7},c_{9},c_{8},c_{6},c_{5}$, respectively. The relabeled graph
$G^{\prime}$ is presented in Figure 2. As per the algorithm, the permutations
$w_{1},w_{2},w_{3},w_{4}$ and $w_{5}$ of vertices of $G^{\prime}$ are produced
below. Note that, as $N(c_{3})=N(c_{4})=V_{2}$, $w_{3}=w_{4}=\varepsilon$ and
further $w_{5}$ will be nonempty.

  * $w_{1}=c_{4}c_{3}c_{2}c_{5}c_{6}c_{7}c_{1}c_{8}c_{9}$

  * $w_{2}=c_{1}c_{3}c_{4}c_{6}c_{5}c_{2}c_{9}c_{8}c_{7}$

  * $w_{5}=c_{1}c_{2}c_{3}c_{4}c_{9}c_{8}c_{7}c_{6}c_{5}$

  * $w_{3}=w_{4}=\varepsilon$

Concatenating all these permutations we get

$w^{\prime}=w_{1}w_{2}w_{3}w_{4}w_{5}=c_{4}c_{3}c_{2}c_{5}c_{6}c_{7}c_{1}c_{8}c_{9}c_{1}c_{3}c_{4}c_{6}c_{5}c_{2}c_{9}c_{8}c_{7}c_{1}c_{2}c_{3}c_{4}c_{9}c_{8}c_{7}c_{6}c_{5}$

which represents $G^{\prime}$. Relabeling the vertices in $w^{\prime}$ with
their original labels we get

$w=a_{1}a_{3}a_{2}b_{5}b_{4}b_{1}a_{4}b_{3}b_{2}a_{4}a_{3}a_{1}b_{4}b_{5}a_{2}b_{2}b_{3}b_{1}a_{4}a_{2}a_{3}a_{1}b_{2}b_{3}b_{1}b_{4}b_{5}$

which represents $G$.

$a_{1}$$a_{2}$$a_{3}$$a_{4}$$b_{1}$$b_{2}$$b_{3}$$b_{4}$$b_{5}$ Figure 1: A
bipartite graph $G$

$c_{4}$$c_{2}$$c_{3}$$c_{1}$$c_{7}$$c_{9}$$c_{8}$$c_{6}$$c_{5}$ Figure 2:
Relabeled bipartite graph $G^{\prime}$

###  3.2 Correctness of the algorithm

In the following theorem we establish the correctness of the relabeling
algorithm.

###### Theorem 3.1

The word $w$ generated by the relabeling algorithm permutationally represents
the bipartite graph $G$.

###### Proof

If $G$ is a complete bipartite graph, then the algorithm produces the word $w$
given in Step 3. It is a routine verification to observe that the word $w$
represents $G$. If $G$ is not a complete bipartite graph, let $G^{\prime}$ be
the graph obtained after relabeling in Step 6. We show that the word
$w^{\prime}=w_{1}\cdots w_{m}w_{m+1}$ represents $G^{\prime}$. Accordingly, it
is straightforward that $w$ represents $G$.

We now prove that $\overline{ab}$ is an edge in $G^{\prime}$ if and only if
$a$ and $b$ alternate in the word $w^{\prime}$.

Suppose $\overline{ab}$ is an edge in $G^{\prime}$. Without loss of generality
assume $a\in V_{1}$ and $b\in V_{2}$. If $a=c_{1}$ then clearly $ab\leq w_{i}$
(when $w_{i}\neq\varepsilon$) for all $i\geq 2$. Further, since $b\in N(a)$ we
have $ab\leq w_{1}$ too. Hence, $a$ and $b$ alternate in $w^{\prime}$. If
$a=c_{p}$ for $p\geq 2$, then $ab\leq w_{1}$ and also $ab\leq w_{m+1}$ (when
$w_{m+1}\neq\varepsilon$). Further $ab\leq a\text{Dec}(N(a))\leq w_{p}$ (when
$w_{p}\neq\varepsilon$). Hence, $ab\leq w_{i}$ for all nonempty $w_{i}$ so $a$
and $b$ alternate in $w^{\prime}$.

Conversely, suppose $\overline{ab}$ is not an edge in $G^{\prime}$. We deal
with this part in three cases, as the case $b\in V_{1}$ and $a\in V_{2}$ is
symmetric to Case 2. In each case we identify two permutations produced by the
algorithm: that one with the subword $ab$ and the other with the subword $ba$.

  * •

Case 1: $a,b\in V_{1}$. Let $a=c_{p}$ and $b=c_{q}$ with $p>q$. Note $p\neq
1$. Clearly $ab=c_{p}c_{q}\leq w_{1}$. If $N(c_{p})\neq V_{2}$, then
$w_{p}\neq\varepsilon$ and $ba=c_{q}c_{p}\leq w_{p}$. Otherwise,
$N(c_{p})=V_{2}$ so that $w_{m+1}\neq\varepsilon$ and $ba=c_{q}c_{p}\leq
w_{m+1}$.

  * •

Case 2: $a\in V_{1}$ and $b\in V_{2}$.

    * -

Subcase 2.1: $a=c_{1}$. Since $b\notin N(a)$, $ba\leq w_{1}$. Since there are
no isolated vertices, $N(b)\neq\varnothing$ and hence there exists $c_{q}\in
V_{1}$ such that $b\in N(c_{q})$. Depending on $N(c_{q})=V_{2}$ or not,
$w_{m+1}\neq\varepsilon$ or $w_{q}\neq\varepsilon$, respectively. Accordingly,
$ab\leq w_{m+1}$ or $ab\leq w_{q}$.

    * -

Subcase 2.2: $a=c_{p}$ for some $p\geq 2$. Clearly, $ab\leq w_{1}$. Since
$b\notin N(a)$, we have $N(c_{p})\neq V_{2}$ so that $w_{p}\neq\varepsilon$.
Then $ba=bc_{p}\leq\text{Dec}(V_{2}\setminus N(c_{p}))c_{p}\leq w_{p}$.

  * •

Case 3: $a,b\in V_{2}$. Let $a=c_{p}$ and $b=c_{q}$ with $p>q$. Clearly,
$ba\leq w_{1}$. As there are no isolated vertices, $N(b)\neq\varnothing$.
Hence, there exists $c\in V_{1}$ such that $b\in N(c)$.

    * -

If $N(c)=V_{2}$, then again $w_{m+1}\neq\varepsilon$ and $ab\leq w_{m+1}$.

    * -

If $N(c)\neq V_{2}$, let $c=c_{r}$ and note that $w_{r}\neq\varepsilon$. If
$a\in N(c)$, then $ab=c_{p}c_{q}\leq\text{Dec}(N(c_{r}))\leq w_{r}$.
Otherwise, $a\in V_{2}\setminus N(c)$ so that $ab\leq\text{Dec}(V_{2}\setminus
N(c_{r}))c_{r}\text{Dec}(N(c_{r}))\leq w_{r}$.

Hence, $a$ and $b$ do not alternate in $w^{\prime}$. This completes the proof.

###  3.3 Upper bounds

We now obtain an upper bound for the permutation representation number of
arbitrary bipartite graphs and also an improved upper bound in case of
bipartite graphs with a special property.

First note that, for each vertex in $V_{1}$, the relabeling algorithm produces
at most one permutation of the vertices of $G$. Hence, the algorithm produces
a word consisting of at most $m$ permutations of vertices of $G$. Accordingly,
we obtain an upper bound for the representation number of bipartite graphs in
the following corollary of Theorem 3.1.

###### Corollary 1

For a bipartite graph $G$, the permutation representation number
$\mathcal{R}^{p}(G)\leq m$. Consequently, $\mathcal{R}(G)\leq m$.

Since $\mathcal{R}^{p}(H_{n,n})=n$ (for $n\geq 2$), the upper bound obtained
in Corollary 1 is tight for the permutation representation number of the class
of bipartite graphs.

###### Corollary 2

The dimension of a poset corresponding to a bipartite graph is bounded above
by $m$.

Considering the possibility of occurrence of the empty word $\varepsilon$ in
Step 6(b) of the relabeling algorithm, the upper bound given in Corollary 1
can be improved for a special type of bipartite graphs as per the following
corollary.

###### Corollary 3

Let $\\{a_{i_{1}},\ldots,a_{i_{k}}\\}\subseteq V_{1}$ be the set of vertices
each of which is adjacent to all vertices of $V_{2}$ in a bipartite graph $G$,
then $\mathcal{R}^{p}(G)\leq m-k+1$.

##  4 Conclusion

In this paper, we obtained a word $w$ that is a concatenation of permutations
of the vertices of a given bipartite graph $G$ through a polynomial time
algorithm such that $w$ represent $G$. Accordingly, we obtain a tight upper
bound for the permutation representation number of bipartite graphs and hence
an upper bound for the representation number of bipartite graphs. We believe
that the relabeling algorithm introduced in this paper has a scope to extend
to a larger class of comparability graphs.

## Acknowledgment

We are thankful to the referees for their comments which improved the
presentation of the paper. The first author is thankful to the Council of
Scientific and Industrial Research (CSIR), Government of India, for awarding
the research fellowship for pursuing Ph.D. at IIT Guwahati.

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