# Change of measure in a Heston-Hawkes stochastic volatility model

David R. Baños, Salvador Ortiz-Latorre, and Oriol Zamora Font Department of
Mathematics, University of Oslo, P.O. Box 1053 Blindern, N-0316 Oslo, Norway,
Email<EMAIL_ADDRESS>of Mathematics, University of Oslo, P.O.
Box 1053 Blindern, N-0316 Oslo, Norway, Email<EMAIL_ADDRESS>of Mathematics, University of Oslo, P.O. Box 1053 Blindern, N-0316 Oslo,
Norway, Email<EMAIL_ADDRESS>
###### Abstract

We consider the stochastic volatility model obtained by adding a compound
Hawkes process to the volatility of the well-known Heston model. A Hawkes
process is a self-exciting counting process with many applications in
mathematical finance, insurance, epidemiology, seismology and other fields. We
prove a general result on the existence of a family of equivalent (local)
martingale measures. We apply this result to a particular example where the
sizes of the jumps are exponentially distributed.

Keywords: stochastic volatility, change of measure, risk neutral measure,
existence of equivalent martingale measure, non-Markovian model, Hawkes
process, volatility with jumps.

AMS classification MSC2020: 60G55, 60J76, 91G15, 60H10.

##  1 Introduction

Valuation of assets and financial derivatives constitutes one of the core
subjects of modern financial mathematics. There have been several approaches
to asset pricing all of which can be classified in two larger groups:
equilibrium pricing and rational pricing. The latter gives rise to the
commonly used methodology of pricing financial instruments by ruling out
arbitrage opportunities. As it is well-known, the absence of arbitrage is
closely related to the existence of a probability measure, so-called risk
neutral measure, hereby denoted by ${\mathbb{Q}}$. Prices of derivatives are
expectations under such measure. This connection is known as the fundamental
theorem of asset pricing [10].

In contrast, markets evolve in time and they do it under the so-called real
world measure, hereby denoted by $\mathbb{P}$. While we can attempt to model
market movements under $\mathbb{P}$, we also need the dynamics under
${\mathbb{Q}}$ for pricing purposes. If we are only interested in pricing,
then modelling under ${\mathbb{Q}}$ and calibrating is possible by matching
market prices to theoretical ones. However, for asset liability management,
investors need often to assess their positions under $\mathbb{P}$. A common
risk management practice is to compute the risk exposure as a way to set
economic and regulatory capital levels. [27] shows that exposures computed
under the risk neutral measure are essentially arbitrary. They depend on the
choice of numéraire and can be manipulated by choosing a different numéraire.
Even when choosing commonly used numéraires, these exposures can differ by a
factor of two or more. Furthermore, a crucial feature when assessing risk
exposures is their distribution. While models under ${\mathbb{Q}}$ may have
well-known tractable distributional properties, this need not be the case when
doing the passage to the $\mathbb{P}$-world and vice versa. For instance, the
Vasicek model for interest rates is invariant under a restricted family of
measure changes. This is a sought property of the Vasicek model, but it does
not need to hold true in other models.

It is a common practice in the literature to assume that such measure change
exists and set up a model under the risk neutral measure. Nonetheless, such
assumption is not innocuous and nonsensical results can occur if it is in fact
not satisfied, see e.g. the discussion in [28] and the references therein. For
instance, [6] and [26] show examples of models where no equivalent local
martingale measure exist. See also [25] for conditions to check existence of
equivalent martingale measures under Markovian models without jumps.

From the modelling perspective, we adopt a stochastic volatility model in a
non-Markovian setting with jumps. The reason for this is twofold. The non-
Markovianity can be justified by the clustering of volatility, see e.g. [9].
This is a well-known stylized fact of volatility that cannot be captured under
Markovianity. Literature on non-Markovian stochastic volatility models is
vast. Here we mention some works on fractional volatility models, which appear
when using the fractional Brownian motion as a driving noise for the
volatility. The main characteristics of such noise are the possibility to
model short and long-range dependence due to the fractional decay of its auto-
correlation function and the ability to generate trajectories which are Hölder
continuous of index different from $1/2$, corresponding to the case of
Brownian motion. Fractional models with long range dependence were first
studied in [8]. Then, in [2], Malliavin calculus is introduced to study the
asymptotics of the implied volatility in general stochastic volatility models,
including fractional models with long and short term dependence. The
popularity of rough models (short term dependence models) started with the
findings of [16]. Their name stemming from the roughness of the underlying
noise in the stochastic evolution of volatility. Several works in this
direction have been conducted, see e.g. [16], [13], [14] to mention a few.
None of the aforementioned works discusses change of measure. In fact, it is
not clear whether the volatility process in the rough Heston model (see e.g.
[14] and [13]) is strictly positive, which is a necessary condition for a
proper change of measure to be possible.

The rough Heston model extends the classical Heston model to a model with
fractional noise. It is moreover proven in [12], [21], [20] that the rough
Heston model can be obtained as a limit of Heston models where the noise is a
pure jump process known as Hawkes process. The latter, justifies again our
choice of model with a self-exciting Hawkes factor.

The inclusion of jumps in the volatility may not be so clear at first glance.
Often, researchers include jumps in the stock process in order to capture
sudden changes of asset returns, see for instance the works of [3], [7], [15].
Classical stochastic volatility models without jumps, such as the Heston
model, have the property that returns conditional on volatility are normally
distributed. This property fails to explain many features of asset price
behaviour. By adding jumps in the returns one can model sudden changes such as
economic crashes as well as having more freedom in the modelling of
distributions of asset returns. Nevertheless, jumps in returns are transient,
in other words, a jump in returns today has no impact on the future
distribution of returns [15]. Hence, Markov models for jumps in the asset
dynamics such as Poisson process are often used. On the other hand, volatility
is highly persistent. If the dynamics is driven by a Brownian motion then
volatility can only increase gradually by a sequence of small normally
distributed increments. Self-exciting jumps in the volatility provide a
rapidly moving persistent factor. This justifies our use of the Hawkes
component in the volatility.

Several works support the presence of jumps in the diffusive volatility. [5],
[11] and [23] provide evidence for the presence of positive jumps in
volatility. For example, the authors in [23, Section 5.4] conduct an empirical
study on the higher moments of the volatility process seeking evidence of
jumps in volatility, as firstly conjectured by [5]. Their findings indicate
the possibility of jumps (with positive mean jump size) in the stochastic-
volatility process or at least fatter tailed innovations in the volatility
process. This justifies our choice of positive self-exciting jumps in our
volatility, which we model by compounding a Hawkes process with independent
jump sizes. The positivity of jumps also allows us to keep volatility from
hitting zero with probability one but, on the other hand, the self-exciting
property of the Hawkes process may cause the process to explode. For this
reason we assume a stability property to prevent explosion, see [19, Section
3.1.1].

In this work we look at a Heston-type stochastic volatility model with
correlated Brownian motions and add a jump part to the volatility process,
namely, a compound Hawkes process. We show that a passage from $\mathbb{P}$ to
${\mathbb{Q}}$ and vice versa is possible for a rich enough family of
probability measures. It is worth noting that our model has three non-tradable
noises and one tradable asset, being thus incomplete. This gives rise to non-
unique choices of measure change. The proof of existence of equivalent
martingale measures for the classical Heston model is conducted in [28].

The paper is organized as follows: in Section 2 we present our stochastic
volatility model with correlated Brownian noises and a compound Hawkes
component in the volatility. We also present in this section some useful
properties on the existence and positivity of the volatility process. In
Section 3 we prove the main results in this paper. First, we study the
integrability of the exponential of the integrated variance and then we prove
the existence of (local) martingale measures in our model. Finally, in
Appendix A, we prove some technical lemmas on the existence of solutions of
ODEs that appear in the proof of the main results.

##  2 Stochastic volatility model

Let $T\in{\mathbb{R}}$ , $T>0$ be a fixed time horizon. On a complete
probability space $(\Omega,\mathcal{A},\mathbb{P})$, we consider a two-
dimensional standard Brownian motion
$(B,W)=\left\\{\left(B_{t},W_{t}\right),t\in[0,T]\right\\}$ and its minimally
augmented filtration
$\mathcal{F}^{(B,W)}=\\{\mathcal{F}^{(B,W)}_{t},t\in[0,T]\\}$. On
$(\Omega,\mathcal{A},\mathbb{P})$, we also consider a Hawkes process
$N=\\{N_{t},t\in[0,T]\\}$ with stochastic intensity given by

$\displaystyle\lambda_{t}=\lambda_{0}+\alpha\int_{0}^{t}e^{-\beta(t-s)}dN_{s},$

or, equivalently,

$\displaystyle d\lambda_{t}=-\beta(\lambda_{t}-\lambda_{0})dt+\alpha dN_{t},$

where $\lambda_{0}>0$ is the initial intensity, $\beta>0$ is the speed of mean
reversion and $0<\alpha<\beta$ is the self-exciting factor. Note that the
stability condition

$\displaystyle\alpha\int_{0}^{\infty}e^{-\beta s}ds=\frac{\alpha}{\beta}<1,$

holds. See [4, Section 2] and [19, Section 3.1.1] for the definition of $N$.
We assume that $(B,W),N$ and $\\{J_{i}\\}_{i\geqslant 1}$ are independent of
each other. Then, we consider a sequence of i.i.d., strictly positive and
integrable random variables $\\{J_{i}\\}_{i\geqslant 1}$ and the compound
Hawkes process $L=\\{L_{t},t\in[0,T]\\}$ given by

$\displaystyle L_{t}=\sum_{i=1}^{N_{t}}J_{i}.$

We write $\mathcal{F}^{L}=\left\\{\mathcal{F}^{L}_{t},t\in[0,T]\right\\}$ for
the minimally augmented filtration generated by $L$ and

$\displaystyle\mathcal{F}=\\{\mathcal{F}_{t}=\mathcal{F}_{t}^{(B,W)}\vee\mathcal{F}_{t}^{L},t\in[0,T]\\},$

for the joint filtration. We assume that $\mathcal{A}=\mathcal{F}_{T}$ and we
will work with $\mathcal{F}$. Since $(B,W)$ and $L$ are independent processes,
$(B,W)$ is also a two-dimensional $(\mathcal{F},\mathbb{P})$-Brownian motion.

Finally, with all these ingredients, we introduce our stochastic volatility
model. We assume that the interest rate is deterministic and constant equal to
$r$, but a non-constant interest rate can easily be fitted into this
framework. The stock price $S=\\{S_{t},t\in[0,T]\\}$ and its variance
$v=\\{v_{t},t\in[0,T]\\}$ are given by

$\displaystyle\frac{dS_{t}}{S_{t}}$
$\displaystyle=\mu_{t}dt+\sqrt{v_{t}}\left(\sqrt{1-\rho^{2}}dB_{t}+\rho
dW_{t}\right),$ (2.1) $\displaystyle dv_{t}$
$\displaystyle=-\kappa\left(v_{t}-\bar{v}\right)dt+\sigma\sqrt{v_{t}}dW_{t}+\eta
dL_{t},$ (2.2)

where $S_{0}>0$ is the initial price of the stock,
$\mu:[0,T]\rightarrow{\mathbb{R}}$ is a measurable and bounded function,
$\rho\in(-1,1)$ is the correlation factor, $v_{0}>0$ is the initial value of
the variance, $\kappa>0$ is the variance’s mean reversion speed, $\bar{v}>0$
is the long-term variance, $\sigma>0$ is the volatility of the variance and
$\eta>0$ is a scaling factor. We assume that the Feller condition
$2\kappa\bar{v}\geqslant\sigma^{2}$ is satisfied, see [1, Proposition 1.2.15].

Note that our stochastic volatility model is the well-known Heston model but
adding a compound Hawkes process in the variance process. The procedure to
prove strong existence and pathwise uniqueness of the SDE in (2.2) is
essentially written in [24, Chapter V.10, Theorem 57]. Informally, between
jump times, the SDE is just a standard CIR model with an initial condition
that depends on the value of the compound Hawkes after the jump has occurred.
Since strong existence and pathwise uniquess is proven for the SDE that
defines the CIR model, see [1, Theorem 1.2.1], one can properly interlace the
solutions of the continuous path SDE and the jumps as they do in [24, Chapter
V.10, Theorem 57] and prove strong existence and pathwise uniqueness for our
SDE in (2.2). In particular, this implies that $v$ is a positive process.

######  Proposition 2.1.

Equation (2.2) has a pathwise unique strong solution.

Since the sizes of the jumps of the compound Hawkes process are strictly
positive one expects that our variance is greater or equal than the Heston
variance. We prove this property following the proof of [22, Chapter 5.2.C,
Proposition 2.18]. The procedure is the same but in our case some extra
computations will appear due to the jump contribution of the compound Hawkes
process. Recall that the variance of the Heston model is strictly positive
because the Feller condition is assumed to hold, see the reference from before
[1, Proposition 1.2.15]. In particular, we will prove that the process $v$ is
also strictly positive.

######  Proposition 2.2.

Let $\widetilde{v}=\\{\widetilde{v}_{t},t\in[0,T]\\}$ be the pathwise unique
strong solution of

$\displaystyle\widetilde{v}_{t}=v_{0}-\kappa\int_{0}^{t}\left(\widetilde{v}_{s}-\bar{v}\right)ds+\sigma\int_{0}^{t}\sqrt{\widetilde{v}_{s}}dW_{s}.$
(2.3)

Then,

$\displaystyle\mathbb{P}\left(\left\\{\omega\in\Omega:\widetilde{v}_{t}(\omega)\leqslant
v_{t}(\omega)\ \forall t\in[0,T]\right\\}\right)=1,$

where $v$ is the pathwise unique strong solution of (2.2).

###### Proof.

See [1, Theorem 1.2.1] for a reference about the solution of the SDE in (2.3).
There exists a strictly decreasing sequence
$\\{a_{n}\\}_{n=0}^{\infty}\subset(0,1]$ with $a_{0}=1$,
$\lim_{n\rightarrow\infty}a_{n}=0$ and

$\displaystyle\int_{a_{n}}^{a_{n-1}}\frac{du}{u}=n,$

for every $n\geqslant 1$. Precisely,
$a_{n}=\exp\left(-\frac{n(n+1)}{2}\right)$.

For each $n\geqslant 1$, there exists a continuous function $\rho_{n}$ on
${\mathbb{R}}$ with support in $(a_{n},a_{n-1})$ so that

$\displaystyle 0\leqslant\rho_{n}(x)\leqslant\frac{2}{nx}$ (2.4)

holds for every $x>0$ and $\int_{a_{n}}^{a_{n-1}}\rho_{n}(x)dx=1$.

Then, the function

$\displaystyle\psi_{n}(x)=\int_{0}^{|x|}\int_{0}^{y}\rho_{n}(u)dudy,$ (2.5)

is even and twice continuously differentiable, with
$|\psi_{n}^{\prime}(x)|\leqslant 1$ and
$\lim_{n\rightarrow\infty}\psi_{n}(x)=|x|$ for $x\in{\mathbb{R}}$.
Furthermore, $\\{\psi_{n}\\}_{n=1}^{\infty}$ is nondecreasing.

Next, define the nondecreasing function
$\varphi_{n}(x)=\psi_{n}(x)\mathds{1}_{(0,\infty)}(x)$. Note that

$\displaystyle\lim_{n\rightarrow\infty}\varphi_{n}(x)=\max\\{x,0\\}=:x^{+}.$

Define $I_{t}:=\widetilde{v}_{t}-v_{t}$ for $t\in[0,T]$. Applying Itô formula
we get

$\displaystyle\varphi_{n}(I_{t})=\ $
$\displaystyle\varphi_{n}(0)+\int_{0}^{t}\varphi_{n}^{\prime}(I_{s-})dI_{s}+\frac{1}{2}\int_{0}^{t}\varphi_{n}^{\prime\prime}(I_{s-})d[I]_{s}^{\text{c}}$
$\displaystyle+\sum_{0<s\leqslant
t}\left[\varphi_{n}(I_{s})-\varphi_{n}(I_{s-})-\varphi_{n}^{\prime}(I_{s-})\Delta
I_{s}\right].$

Note that

$\displaystyle dI_{s}$
$\displaystyle=-\kappa(\widetilde{v}_{s}-v_{s})ds+\sigma\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)dW_{s}-\eta
dL_{s},$ $\displaystyle d[I]_{s}$
$\displaystyle=\sigma^{2}\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}ds+\eta^{2}d[L]_{t}.$

The latter implies that
$d[I]_{s}^{\text{c}}=\sigma^{2}\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}ds$
and $\Delta I_{s}=-\Delta v_{s}=-\eta\Delta L_{s}.$ Therefore,

$\displaystyle\varphi_{n}(I_{t})=\ $
$\displaystyle-\kappa\int_{0}^{t}\varphi_{n}^{\prime}(I_{s})I_{s}ds+\sigma\int_{0}^{t}\varphi_{n}^{\prime}(I_{s-})\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)dW_{s}$
$\displaystyle-\eta\int_{0}^{t}\varphi_{n}^{\prime}(I_{s-})dL_{s}+\frac{\sigma^{2}}{2}\int_{0}^{t}\varphi_{n}^{\prime\prime}(I_{s})\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}ds$
$\displaystyle+\sum_{0<s\leqslant
t}\left[\varphi_{n}(I_{s})-\varphi_{n}(I_{s-})\right]+\eta\sum_{0<s\leqslant
t}\varphi_{n}^{\prime}(I_{s-})\Delta L_{s}.$

Note that
$\eta\int_{0}^{t}\varphi_{n}^{\prime}(I_{s-})dL_{s}=\eta\sum_{0<s\leqslant
t}\varphi_{n}^{\prime}(I_{s-})\Delta L_{s}.$ Since $I_{s}-I_{s-}=-\eta\Delta
L_{s}\leqslant 0$ and $\varphi_{n}$ is a non-decreasing function, we have
$\sum_{0<s\leqslant
t}\left[\varphi_{n}(I_{s})-\varphi_{n}(I_{s-})\right]\leqslant 0.$

By (2.4), (2.5) and using that $|\sqrt{x}-\sqrt{y}|\leqslant\sqrt{|x-y|}$ for
$x,y\geqslant 0$ we obtain

$\displaystyle\varphi_{n}^{\prime\prime}(I_{s})\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}\leqslant\frac{2}{n}\frac{\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}}{\widetilde{v}_{s}-v_{s}}\leqslant\frac{2}{n}.$

Due to the fact that $|\varphi_{n}^{\prime}(x)|\leqslant 1$ we can conclude
that

$\displaystyle\varphi_{n}(I_{t})\leqslant\ $
$\displaystyle\kappa\int_{0}^{t}I_{s}^{+}ds+\sigma\int_{0}^{t}\varphi_{n}^{\prime}(I_{s-})\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)dW_{s}+\frac{\sigma^{2}t}{n}.$
(2.6)

In order to use the zero mean property of the Itô integral

$\displaystyle\mathbb{E}\left[\int_{0}^{t}\varphi_{n}^{\prime}(I_{s-})\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)dW_{s}\right]=0,$
(2.7)

we need to check that

$\displaystyle\mathbb{E}\left[\int_{0}^{t}\varphi_{n}^{\prime}(I_{s})^{2}\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}ds\right]<\infty.$
(2.8)

Now,

$\displaystyle\mathbb{E}\left[\int_{0}^{t}\varphi_{n}^{\prime}(I_{s})^{2}\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}ds\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{t}\varphi_{n}^{\prime}(I_{s})^{2}\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}\mathds{1}_{\\{I_{s}>0\\}}ds\right]$
$\displaystyle\leqslant\mathbb{E}\left[\int_{0}^{t}\left(\sqrt{\widetilde{v}_{s}}-\sqrt{v_{s}}\right)^{2}\mathds{1}_{\\{I_{s}>0\\}}ds\right]$
$\displaystyle\leqslant\mathbb{E}\left[\int_{0}^{t}\widetilde{v}_{s}ds\right]$
$\displaystyle=\int_{0}^{t}\mathbb{E}\left[\widetilde{v}_{s}\right]ds.$ (2.9)

In [17, Section 2, Equation (3)] we see that

$\displaystyle\widetilde{v}_{t}\sim\frac{e^{-\kappa
t}}{k(t)}\chi_{\delta}^{{}^{\prime}2}\left(k(t)\right)\ \ \text{with}\ \
k(t)=\frac{4\kappa e^{-\kappa t}}{\sigma^{2}(1-e^{-\kappa t})}v_{0}\ \
\text{and}\ \ \delta=\frac{4\kappa\bar{v}}{\sigma^{2}},$

where $\chi_{\delta}^{{}^{\prime}2}(k(t))$ denotes a noncentral chi-square
random variable with $\delta$ degrees of freedom and noncentrality parameter
$k(t)$. Then,

$\displaystyle\mathbb{E}\left[\widetilde{v}_{t}\right]=\frac{\sigma^{2}(1-e^{-\kappa
t})}{4\kappa}\left(\delta+k(t)\right)=v_{0}e^{-\kappa t}+\bar{v}(1-e^{-\kappa
t}).$ (2.10)

Since $t\mapsto\mathbb{E}\left[\widetilde{v}_{t}\right]$ is continuous
function on $[0,T]$, the integral in (2) is finite, which implies that the
integral in (2.8) is finite and (2.7) holds. Taking expectations in (2.6) we
have

$\displaystyle\mathbb{E}\left[\varphi_{n}(I_{t})\right]\leqslant\kappa\int_{0}^{t}\mathbb{E}\left[I_{s}^{+}\right]+\frac{\sigma^{2}t}{n}.$

Then, sending $n$ to infinity yields
$\mathbb{E}\left[I_{t}^{+}\right]\leqslant\kappa\int_{0}^{t}\mathbb{E}\left[I_{s}^{+}\right]ds.$
One can check that

$\displaystyle\int_{0}^{t}\left|\mathbb{E}\left[I_{s}^{+}\right]\right|ds<\infty.$
(2.11)

Actually

$\int_{0}^{t}\left|\mathbb{E}\left[I_{s}^{+}\right]\right|ds=\int_{0}^{t}\mathbb{E}\left[I_{s}^{+}\right]ds=\int_{0}^{t}\mathbb{E}\left[I_{s}\mathds{1}_{\\{I_{s}>0\\}}\right]ds\leqslant\int_{0}^{t}\mathbb{E}\left[\widetilde{v}_{s}\right]ds<\infty,$

where the last integral is finite as we have seen before.

We can apply a version of Gronwall’s inequality where only condition (2.11) is
required and we get $\mathbb{E}[I_{t}^{+}]=0$ for all $t\in[0,T]$. This means
that

$\displaystyle\mathbb{P}\left(\\{\omega\in\Omega:\widetilde{v}_{t}(\omega)\leqslant
v_{t}(\omega)\\}\right)=1\hskip 28.45274pt\forall t\in[0,T].$

Since the sample paths of $\widetilde{v}$ are continuous and the sample paths
of $v$ are càdlàg we get that

$\displaystyle\mathbb{P}\left(\left\\{\omega\in\Omega:\widetilde{v}_{t}(\omega)\leqslant
v_{t}(\omega)\ \forall t\in[0,T]\right\\}\right)=1.$

∎

######  Corollary 2.3.

The variance $v=\\{v_{t},t\in[0,T]\\}$ is a strictly positive process.

###### Proof.

Recall that we have assumed that the Feller condition
$2\kappa\bar{v}\geqslant\sigma^{2}$ is satisfied. Then, the process
$\widetilde{v}$ defined by (2.3) is strictly positive, see [1, Proposition
1.2.15]. Finally, by Proposition 2.2 we have that

$\displaystyle\mathbb{P}\left(\left\\{\omega\in\Omega:\widetilde{v}_{t}(\omega)\leqslant
v_{t}(\omega)\ \forall t\in[0,T]\right\\}\right)=1.$

We conclude that the process $v$ is also strictly positive. ∎

##  3 Risk neutral probability measures

To prove the existence of a family of risk neutral probability measures we
follow the classical approach where we employ Girsanov’s theorem in connection
with the Novikov’s condition. Thus, we need to study the integrability of the
exponential of the integrated variance, that is, what values $c>0$, if any,
satisfy

$\displaystyle\mathbb{E}\left[\exp\left(c\int_{0}^{T}v_{u}du\right)\right]<\infty.$

By Corollary 2.3, $v$ is strictly positive and the previous expectation is
finite for $c\leqslant 0$ but for our applications is essential that $c$ can
be strictly positive. Looking at the proof of [28, Lemma 3.1] one can see that
for $c\leqslant\frac{\kappa^{2}}{2\sigma^{2}}$ the following holds

$\displaystyle\mathbb{E}\left[\exp\left(c\int_{0}^{T}\widetilde{v}_{u}du\right)\right]<\infty,$
(3.1)

where $\widetilde{v}$ is the standard Heston volatility given by

$\displaystyle\widetilde{v}_{t}=v_{0}-\kappa\int_{0}^{t}\left(\widetilde{v}_{s}-\bar{v}\right)ds+\sigma\int_{0}^{t}\sqrt{\widetilde{v}_{s}}dW_{s}.$

The procedure to prove that (3.1) holds is to show that

$\displaystyle\mathbb{E}\left[\exp\left(c\int_{0}^{T}\widetilde{v}_{u}du\right)\right]\leqslant\exp\left(-(\kappa\bar{v})\Phi(0)-v_{0}\psi(0)\right)<\infty,$
(3.2)

where $\Phi$ and $\psi$ satisfy the following Riccati ODE

$\displaystyle\psi^{\prime}(t)$
$\displaystyle=\frac{\sigma^{2}}{2}\psi^{2}(t)+\kappa\psi(t)+c$ (3.3)
$\displaystyle-\Phi^{\prime}(t)$ $\displaystyle=\psi(t)$
$\displaystyle\psi(T)$ $\displaystyle=\Phi(T)=0.$

Due to the jump contribution of the compound Hawkes process, this procedure is
more delicate in our model. However we can obtain a similar bound as in (3.2)
with an additional function that will be the solution of an ODE where the
moment generating function of $J_{1}$ and the Hawkes parameters $\alpha$ and
$\beta$ are involved.

We start making an assumption on the moment generating function of $J_{1}$. We
write $M_{J}$ for the moment generating function of $J_{1}$, that is,
$M_{J}(t)=\mathbb{E}\left[\exp\left(tJ_{1}\right)\right]$. Since $J_{1}$ is
strictly positive, $M_{J}$ is well-defined at least on the interval
$(-\infty,0]$.

######  Assumption 1.

There exists $\epsilon_{J}>0$ such that $M_{J}$ is well defined on
$(-\infty,\epsilon_{J})$ and it is the maximal domain in the sense that

$\displaystyle\lim_{t\rightarrow\epsilon_{J}^{-}}M_{J}(t)=\infty.$

Since $\epsilon_{J}>0$, all positive moments of $J_{1}$ are finite.

Note that this is the case of the gamma distribution, chi-squared
distribution, uniform distribution and others.

We start studying the functions that will appear in a bound like (3.2) for our
variance. To find the ODEs that define those functions, one can consider the
process $M=\\{M(t),t\in[0,T]\\}$ defined by

$\displaystyle
M(t)=\exp\left(F(t)+G(t)v_{t}+H(t)\lambda_{t}+c\int_{0}^{t}v_{u}du\right),$

for some unknown functions $F,G,H\colon[0,T]\to{\mathbb{R}}$ satisfying
$F(T)=G(T)=H(T)=0$. Note that

$\displaystyle M(T)=\exp\left(c\int_{0}^{T}v_{u}du\right).$

Hence, $\mathbb{E}[M(T)]$ is exactly the expectation that we want to study.
Now, if we can find $F,G$ and $H$ such that $M$ is a local martingale, since
it is non-negative, it will be a supermartingale and then

$\displaystyle\mathbb{E}\left[\exp\left(c\int_{0}^{T}v_{u}du\right)\right]=\mathbb{E}[M(T)]\leqslant
M(0)=\exp\left(F(0)+G(0)v_{0}+H(0)\lambda_{0}\right),$

where the last expression will be finite as long as $F,G$ and $H$ exist and
are well defined on $[0,T]$.

To make $M$ a local martingale one can apply Itô’s formula to the process $M$
and equate all the drift terms to $0$. By this procedure one will obtain the
ODEs for $F,G$ and $H$. This procedure is formally done in Proposition 3.5. In
the next lemma we first study the existence of solutions of the ODEs that will
appear later. Note that the ODE for G is slightly different from the ODE of
$\psi$ in (3.3), this depends on whether one considers the expectation in
(3.1) with the parameter $c$ or $-c$ and on how one defines the process $M$.
The proof of the next lemma is deferred to the appendix because is rather
technical and based on ODE theory.

######  Lemma 3.1.

For $c\leqslant\frac{\kappa^{2}}{2\sigma^{2}}$, define
$D(c):=\sqrt{\kappa^{2}-2\sigma^{2}c}$, $\Lambda(c):=\frac{2\eta
c\left(e^{D(c)T}-1\right)}{D(c)-\kappa+\left(D(c)+\kappa\right)e^{D(c)T}}$ and

$\displaystyle
c_{l}:=\sup\left\\{c\leqslant\frac{\kappa^{2}}{2\sigma^{2}}:\Lambda(c)<\epsilon_{J}\hskip
8.5359pt\text{and}\hskip
8.5359ptM_{J}\left(\Lambda(c)\right)\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right\\}.$

Then, $0<c_{l}\leqslant\frac{\kappa^{2}}{2\sigma^{2}}$ and for $c<c_{l}$,

  1. (i)

The ODE

$\displaystyle G^{\prime}(t)$
$\displaystyle=-\frac{1}{2}\sigma^{2}G^{2}(t)+\kappa G(t)-c$ (3.4)
$\displaystyle G(T)$ $\displaystyle=0$

has a unique solution in the interval $[0,T]$. The solution is strictly
decreasing and it is given by

$\displaystyle
G(t)=\frac{2c\left(e^{D(c)(T-t)}-1\right)}{D(c)-\kappa+\left(D(c)+\kappa\right)e^{D(c)(T-t)}}.$

  2. (ii)

The function $t\mapsto M_{J}(\eta G(t))$ is well defined for $t\in[0,T]$.

  3. (iii)

Define $U:=\sup_{t\in[0,T]}M_{J}(\eta G(t))$. Then, $U=M_{J}(\eta G(0))$ and

$\displaystyle
1<U\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right).$

  4. (iv)

The ODE

$\displaystyle H^{\prime}(t)$ $\displaystyle=\beta H(t)-M_{J}\left(\eta
G(t)\right)\exp\left(\alpha H(t)\right)+1$ (3.5) $\displaystyle H(T)$
$\displaystyle=0$

has a unique solution in $[0,T]$.

###### Proof.

See Lemma A.1 in the appendix. ∎

######  Observation 3.2.

One could also assume that the domain of $M_{J}$ is big enough to make the
function $t\mapsto M_{J}(\eta G(t))$ well defined on the interval $[0,T]$ for
any value of $c\leqslant\frac{\kappa^{2}}{2\sigma^{2}}$. Then one can prove
the existence of $H$ by applying the Picard-Lindelöf theorem (see [18, Chapter
II, Theorem 1.1]) which yields the following conditions

$\displaystyle c\leqslant\frac{\kappa^{2}}{2\sigma^{2}}\ \ \text{and}\ \
\beta+f(U)\alpha\leqslant\frac{1}{T},$

where $f(U)$ is some function of $U$. Therefore, there would be more
admissible values of $c$ but $\alpha$ and $\beta$ would have to satisfy an
inequality involving $T$, which is quite restrictive and for the sake of
generality we prefer to avoid.

Note that, a priori, there is not an explicit expression of $c_{l}$ in the
previous lemma because it is not possible to solve the inequalities

$\displaystyle\Lambda(c)<\epsilon_{J}\hskip 14.22636pt\text{ and }\hskip
14.22636ptM_{J}\left(\Lambda(c)\right)\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right).$

Moreover, $c_{l}$ depends on $M_{J}$, $\eta$, $\kappa$, $\sigma$, $T$,
$\epsilon_{J}$, $\alpha$ and $\beta$. However, one can get a suboptimal and
explicit value of $c_{l}$ that does not depend on $T$. We obtain an expression
for that suboptimal value and we give some examples.

######  Corollary 3.3.

Define $c_{s}$ by

$\displaystyle
c_{s}=\min\left\\{\frac{\kappa\epsilon_{J}}{2\eta},\frac{\kappa}{2\eta}M_{J}^{-1}\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

Then, $0<c_{s}<c_{l}$.

###### Proof.

See Lemma A.2 in the appendix. ∎

Note that for any $c<c_{s}$ we can apply Lemma 3.1 because $c_{s}<c_{l}$. We
now give some examples of the value $c_{s}$.

######  Example 3.4.

Some examples:

  1. (i)

If $J_{1}\sim\text{Exponential}(\lambda)$, then

$\displaystyle
c_{s}=\min\left\\{\frac{\kappa\lambda}{2\eta}\left(1-\frac{\alpha}{\beta}\exp\left(1-\frac{\alpha}{\beta}\right)\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

  2. (ii)

If $J_{1}\sim\text{Gamma}(\mu,\lambda)$ with $\mu,\lambda>0$ as the shape and
the rate, respectively. Then

$\displaystyle
c_{s}=\min\left\\{\frac{\kappa\lambda}{2\eta}\left(1-\frac{1}{\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right)^{1/\mu}}\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

  3. (iii)

If $J_{1}=j>0$, then

$\displaystyle c_{s}=\min\left\\{\frac{\kappa}{2\eta
j}\left(\ln\left(\frac{\beta}{\alpha}\right)+\frac{\alpha}{\beta}-1\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

###### Proof.

See Example A.3 in the appendix. ∎

We have studied under which conditions the previous ODEs have a well defined
solution on the interval $[0,T]$. We proceed studying the integrability of the
exponential of the integrated variance following the method commented in the
beginning of this section.

######  Proposition 3.5.

Let $c<c_{l}$, then

$\displaystyle\mathbb{E}\left[\exp\left(c\int_{0}^{T}v_{u}du\right)\right]<\infty.$

###### Proof.

By Corollary 2.3, $v$ is strictly positive and the expectation is finite for
$c\leqslant 0$. We focus on the case when $0<c<c_{l}$. We first define the
function $f\colon[0,T]\times{\mathbb{R}}^{3}\to{\mathbb{R}}$

$\displaystyle f(t,x,y,z)=\exp\left(F(t)+G(t)x+H(t)y+cz\right),$

where $G$ and $H$ are the solution of the differential equations given in
Lemma 3.1, that is,

$\displaystyle G^{\prime}(t)$
$\displaystyle=-\frac{1}{2}\sigma^{2}G^{2}(t)+\kappa G(t)-c$ (3.6)
$\displaystyle G(T)$ $\displaystyle=0$

and

$\displaystyle H^{\prime}(t)$ $\displaystyle=\beta H(t)-M_{J}\left(\eta
G(t)\right)\exp\left(\alpha H(t)\right)+1$ (3.7) $\displaystyle H(T)$
$\displaystyle=0,$

and $F$ is given by

$\displaystyle F^{\prime}(t)$
$\displaystyle=-\kappa\bar{v}G(t)-\beta\lambda_{0}H(t)$ (3.8) $\displaystyle
F(T)$ $\displaystyle=0.$

Note that with the assumption we have made on $c$, the functions $F$, $G$ and
$H$ are well defined on $[0,T]$.

We also define the integrated variance $V_{t}:=\int_{0}^{t}v_{u}du$,
$Y_{t}:=(t,v_{t},\lambda_{t},V_{t})$ and the process $M=\\{M(t),t\in[0,T]\\}$
by $M(t)=f(t,v_{t},\lambda_{t},V_{t})=f(Y_{t})$. Applying Itô formula to the
process $M$ we get

$\displaystyle M(t)=\ $ $\displaystyle
M(0)+\int_{0}^{t}\partial_{t}f(Y_{s-})ds+\int_{0}^{t}\partial_{x}f(Y_{s-})dv_{s}+\int_{0}^{t}\partial_{y}f(Y_{s-})d\lambda_{s}$
$\displaystyle+\int_{0}^{t}\partial_{z}f(Y_{s-})dV_{s}+\frac{1}{2}\int_{0}^{t}\partial^{2}_{xx}f(Y_{s-})d[v]_{s}^{\text{c}}$
$\displaystyle+\sum_{0<s\leqslant
t}\left[f(Y_{s})-f(Y_{s-})-\partial_{x}f(Y_{s-})\Delta
v_{s}-\partial_{y}f(Y_{s-})\Delta\lambda_{s}\right].$

We have used that

$\displaystyle[\lambda]_{t}$
$\displaystyle=\alpha^{2}[N]_{t}=\alpha^{2}N_{t}\implies[\lambda]_{t}^{\text{c}}=0,$
$\displaystyle[V]_{t}$ $\displaystyle=0,$ $\displaystyle[v,\lambda]_{t}$
$\displaystyle=\alpha\eta[L,N]_{t}=\alpha\eta
L_{t}\implies[v,\lambda]_{t}^{\text{c}}=0,$ $\displaystyle[v,V]_{t}$
$\displaystyle=[\lambda,V]_{t}=0.$

Moreover,

$\displaystyle[v]_{t}$
$\displaystyle=\int_{0}^{t}\sigma^{2}v_{s}ds+\eta^{2}[L]_{t}\implies[v]_{t}^{\text{c}}=\int_{0}^{t}\sigma^{2}v_{s}ds,$
$\displaystyle dv_{t}$
$\displaystyle=-\kappa(v_{t}-\bar{v})dt+\sigma\sqrt{v_{t}}dW_{t}+\eta dL_{t},$
$\displaystyle d\lambda_{t}$
$\displaystyle=-\beta(\lambda_{t}-\lambda_{0})dt+\alpha dN_{t}.$

Hence,

$\displaystyle M(t)-M(0)=$
$\displaystyle\int_{0}^{t}\Big{[}\partial_{t}f(Y_{s})-\kappa\partial_{x}f(Y_{s})(v_{s}-\bar{v})-\beta\partial_{y}f(Y_{s})(\lambda_{s}-\lambda_{0})$
$\displaystyle+\partial_{z}f(Y_{s})v_{s}+\frac{1}{2}\partial_{xx}^{2}f(Y_{s})\sigma^{2}v_{s}\Big{]}ds$
$\displaystyle+\int_{0}^{t}\partial_{x}f(Y_{s-})\sigma\sqrt{v_{s}}dW_{s}+\int_{0}^{t}\partial_{y}f(Y_{s-})\eta
dL_{s}+\int_{0}^{t}\partial_{y}f(Y_{s-})\alpha dN_{s}$
$\displaystyle+\sum_{0<s\leqslant
t}\left[f(Y_{s})-f(Y_{s-})-\partial_{x}f(Y_{s-})\Delta
v_{s}-\partial_{y}f(Y_{s-})\Delta\lambda_{s}\right].$

Since $\Delta v_{s}=\eta\Delta L_{s}$ and $\Delta\lambda_{s}=\alpha\Delta
N_{s}$, then

$\displaystyle\int_{0}^{t}\partial_{x}f(Y_{s-})\eta dL_{s}$
$\displaystyle=\sum_{0<s\leqslant t}\partial_{x}f(Y_{s-})\Delta v_{s}$
$\displaystyle\int_{0}^{t}\partial_{y}f(Y_{s-})\alpha dN_{s}$
$\displaystyle=\sum_{0<s\leqslant t}\partial_{y}f(Y_{s-})\Delta\lambda_{s},$

and we get

$\displaystyle M(t)-M(0)=$
$\displaystyle\int_{0}^{t}\Big{[}\partial_{t}f(Y_{s})-\kappa\partial_{x}f(Y_{s})(v_{s}-\bar{v})-\beta\partial_{y}f(Y_{s})(\lambda_{s}-\lambda_{0})$
$\displaystyle+\partial_{z}f(Y_{s})v_{s}+\frac{1}{2}\partial_{xx}^{2}f(Y_{s})\sigma^{2}v_{s}\Big{]}ds$
$\displaystyle+\int_{0}^{t}\partial_{x}f(Y_{s-})\sigma\sqrt{v_{s}}dW_{s}+\sum_{0<s\leqslant
t}\left[f(Y_{s})-f(Y_{s-})\right].$

Next, we can write

$\displaystyle\sum_{0<s\leqslant t}\left[f(Y_{s})-f(Y_{s-})\right]$
$\displaystyle=\sum_{0<s\leqslant t}\left[f(s,v_{s-}+\Delta
v_{s},\lambda_{s-}+\Delta\lambda_{s},V_{s})-f(Y_{s-})\right]$
$\displaystyle=\sum_{0<s\leqslant t}\left[f(s,v_{s-}+\eta\Delta
L_{s},\lambda_{s-}+\alpha\Delta N_{s},V_{s})-f(Y_{s-})\right]$
$\displaystyle=\sum_{0<s\leqslant t}g(s,\Delta L_{s},\Delta N_{s}),$

where

$\displaystyle g(s,u_{1},u_{2}):=f(s,v_{s-}+\eta u_{1},\lambda_{s-}+\alpha
u_{2},V_{s})-f(Y_{s-}).$

We now define $U_{s}=(L_{s},N_{s})$ and for $t\in[0,T]$,
$A\in\mathcal{B}({\mathbb{R}}^{2}\setminus\\{0,0\\})$

$\displaystyle N^{U}(t,A)=\\#\\{0<s\leqslant t,\Delta U_{s}\in A\\}.$

We add and subtract the compensator of the counting measure $N^{U}$ to split
the expression into a local martingale plus a predictable process of finite
variation. Note that the compensator of the Hakwes process is given by
$\Lambda^{N}_{t}=\int_{0}^{t}\lambda_{u}du$, see [4, Theorem 3]. One can check
the the compensator of the compound Hawkes process is given by
$\Lambda^{L}_{t}=\mathbb{E}[J_{1}]\int_{0}^{t}\lambda_{u}du$. Thus,

$\displaystyle\sum_{0<s\leqslant t}\left[f(Y_{s})-f(Y_{s-})\right]$
$\displaystyle=\int_{0}^{t}\int_{(0,\infty)^{2}}g(s,u_{1},u_{2})N^{U}(ds,du)$
$\displaystyle=\int_{0}^{t}\int_{(0,\infty)^{2}}g(s,u_{1},u_{2})\left(N^{U}(ds,du)-\lambda_{s}P_{J_{1}}(du_{1})\delta_{1}(du_{2})ds\right)$
$\displaystyle+\int_{0}^{t}\int_{(0,\infty)^{2}}g(s,u_{1},u_{2})\lambda_{s}P_{J_{1}}(du_{1})\delta_{1}(du_{2})ds.$

Note that

$\displaystyle\int_{0}^{t}\int_{(0,\infty)^{2}}g(s,u_{1},u_{2})\lambda_{s}P_{J_{1}}(du_{1})\delta_{1}(du_{2})ds=\int_{0}^{t}\int_{(0,\infty)}g(s,u_{1},1)\lambda_{s}P_{J_{1}}(du_{1})ds.$

We conclude that

$\displaystyle M(t)-M(0)=$
$\displaystyle\int_{0}^{t}\Big{[}\partial_{t}f(Y_{s})-\kappa\partial_{x}f(Y_{s})(v_{s}-\bar{v})-\beta\partial_{y}f(Y_{s})(\lambda_{s}-\lambda_{0})$
$\displaystyle+\partial_{z}f(Y_{s})v_{s}+\frac{1}{2}\partial_{xx}^{2}f(Y_{s})\sigma^{2}v_{s}+\int_{(0,\infty)}g(s,u_{1},1)\lambda_{s}P_{J_{1}}(du_{1})\Big{]}ds$
$\displaystyle+\int_{0}^{t}\partial_{x}f(Y_{s-})\sigma\sqrt{v_{s}}dW_{s}$
$\displaystyle+\int_{0}^{t}\int_{(0,\infty)}g(s,u_{1},1)\left(N^{U}(ds,du)-\lambda_{s}P_{J_{1}}(du_{1})ds\right).$

Recall that $f(t,x,y,z)=\exp\left(F(t)+G(t)x+H(t)y+cz\right)$, thus,

$\displaystyle\partial_{t}f(t,x,y,z)$
$\displaystyle=\left(F^{\prime}(t)+G^{\prime}(t)x+H^{\prime}(t)y\right)f(t,x,y,z),$
$\displaystyle\partial_{x}f(t,x,y,z)$ $\displaystyle=G(t)f(t,x,y,z),$
$\displaystyle\partial_{y}f(t,x,y,z)$ $\displaystyle=H(t)f(t,x,y,z),$
$\displaystyle\partial_{z}f(t,x,y,z)$ $\displaystyle=cf(t,x,y,z).$

Furthermore,

$\displaystyle g(s,u_{1},1)$ $\displaystyle=f(s,v_{s-}+\eta
u_{1},\lambda_{s-}+\alpha,V_{s})-f(s,v_{s-},\lambda_{s-},V_{s})$
$\displaystyle=f(s,v_{s-},\lambda_{s-},V_{s})\left[\exp\left(\eta
u_{1}G(s)\right)\exp\left(\alpha H(s)\right)-1\right].$

Therefore,

$\displaystyle\int_{(0,\infty)}g(s,u_{1},1)\lambda_{s}P_{J_{1}}(du_{1})=f(Y_{s-})\lambda_{s}\left[M_{J}\left(\eta
G(s)\right)\exp\left(\alpha H(s)\right)-1\right].$

Using the specific form of the derivatives and the previous result we see that
the drift part of $M(t)-M(0)$ vanishes. That is:

  * •

Coefficient multiplying $v$ in the drift:

$\displaystyle f(Y_{t})\left[G^{\prime}(t)-\kappa
G(t)+\frac{1}{2}G(t)^{2}\sigma^{2}+c\right]=0,$

where we have replaced equation (3.6).

  * •

Coefficient multiplying $\lambda$ in the drift:

$\displaystyle f(Y_{t})\left[H^{\prime}(t)-\beta H(t)+M_{J}\left(\eta
G(t)\right)\exp\left(\alpha H(t)\right)-1\right]=0,$

where we have replaced equation (3.7).

  * •

Free coefficient in the drift:

$\displaystyle
f(Y_{t})\left[F^{\prime}(t)+\kappa\bar{v}G(t)+\beta\lambda_{0}H(t)\right]=0,$

where we have replaced equation (3.8).

Therefore, the process $M$ can be written as

$\displaystyle M(t)=$ $\displaystyle\
M(0)+\int_{0}^{t}\partial_{v}f(Y_{s-})\sigma\sqrt{v_{s}}dW_{s}$
$\displaystyle+\int_{0}^{t}\int_{(0,\infty)}g(s,u_{1},1)\left(N^{U}(ds,du)-\lambda_{s}P_{J_{1}}(du_{1})ds\right).$

We conclude that $M$ is a local martingale. Moreover, since $M$ is non-
negative, it is a supermartingale. Then,

$\displaystyle\mathbb{E}[M(T)]$
$\displaystyle=\mathbb{E}\left[\exp\left(F(T)+G(T)v_{T}+H(T)\lambda_{T}+cV_{T}\right)\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(c\int_{0}^{T}v_{u}du\right)\right]$
$\displaystyle\leqslant
M(0)=\exp\left(F(0)+G(0)v_{0}+H(0)\lambda_{0}\right)<\infty.$

Note that $F(0),G(0),H(0)<\infty$ because the functions $F,G$ and $H$ are well
defined on the entire interval $[0,T]$. ∎

We write $\widehat{S}=\\{\widehat{S}_{t},t\in[0,T]\\}$ for the discounted
stock price, that is, $\widehat{S}_{t}=e^{-rt}S_{t}$. We prove the existence
of a family of equivalent local martingale measures $\mathbb{Q}(a)$
parametrized by a number $a$. Namely, $\mathbb{Q}(a)\sim\mathbb{P}$ is such
that $\widehat{S}$ is a $(\mathcal{F},\mathbb{Q}(a))$-local martingale.

######  Theorem 3.6.

Let $a\in\mathbb{R}$ and define
$\theta_{t}^{(a)}:=\frac{1}{\sqrt{1-\rho^{2}}}\left(\frac{\mu_{t}-r}{\sqrt{v_{t}}}-a\rho\sqrt{v_{t}}\right)$,

$\displaystyle Y_{t}^{(a)}$
$\displaystyle:=\exp\left(-\int_{0}^{t}\theta_{s}^{(a)}dB_{s}-\frac{1}{2}\int_{0}^{t}(\theta_{s}^{(a)})^{2}ds\right),$
$\displaystyle Z_{t}^{(a)}$
$\displaystyle:=\exp\left(-a\int_{0}^{t}\sqrt{v_{s}}dW_{s}-\frac{1}{2}a^{2}\int_{0}^{t}v_{s}ds\right)$

and $X_{t}^{(a)}:=Y_{t}^{(a)}Z_{t}^{(a)}$. The set

$\displaystyle\mathcal{E}:=\left\\{\mathbb{Q}(a)\hskip 5.69046pt\text{given
by}\hskip 5.69046pt\frac{d\mathbb{Q}(a)}{d\mathbb{P}}=X_{T}^{(a)}\hskip
5.69046pt\text{with}\hskip 5.69046pt|a|<\sqrt{2c_{l}}\right\\}$ (3.9)

is a set of equivalent local martingale measures.

###### Proof.

Define the process
$(B^{\mathbb{Q}(a)},W^{\mathbb{Q}(a)})=\\{(B_{t}^{\mathbb{Q}(a)},W_{t}^{\mathbb{Q}(a)}),t\in[0,T]\\}$,
by

$\displaystyle dB_{t}^{\mathbb{Q}(a)}$
$\displaystyle=dB_{t}+\theta_{t}^{(a)}dt,$ $\displaystyle
dW_{t}^{\mathbb{Q}(a)}$ $\displaystyle=dW_{t}+a\sqrt{v_{t}}dt.$ (3.10)

The dynamics of the stock is now given by

$\displaystyle\frac{dS_{t}}{S_{t}}=\left[\mu_{t}-\sqrt{v_{t}}\left(\sqrt{1-\rho^{2}}\theta_{t}^{(a)}+a\rho\sqrt{v_{t}}\right)\right]dt+\sqrt{v_{t}}\left(\sqrt{1-\rho^{2}}dB_{t}^{\mathbb{Q}(a)}+\rho
dW_{t}^{\mathbb{Q}(a)}\right).$

Note that

$\displaystyle\mu_{t}-\sqrt{v_{t}}\left(\sqrt{1-\rho^{2}}\theta_{t}^{(a)}+a\rho\sqrt{v_{t}}\right)=r,$
(3.11)

which is a necessary condition for $\mathbb{Q}(a)$ to be an equivalent local
martingale measure. The choice of the market price of risk processes
$\theta^{(a)}$ and $a\sqrt{v_{t}}$ is the same as in [28, Equation (3.4) and
(3.7)]. One of the reasons to make that choice under the standard Heston model
is to maintain the same variance dynamics after the change of measure.

To apply Girsanov’s theorem we need to check that the process $X^{(a)}$ is a
$(\mathcal{F},\mathbb{P})$-martingale. Since $X^{(a)}$ is a positive
$(\mathcal{F},\mathbb{P})$-local martingale with $X^{(a)}_{0}=1$, it is a
$(\mathcal{F},\mathbb{P})$-supermartingale and it is a
$(\mathcal{F},\mathbb{P})$-martingale if and only if

$\displaystyle\mathbb{E}\left[X^{(a)}_{T}\right]=1.$ (3.12)

Using that $Z^{(a)}_{T}$ is
$\mathcal{F}_{T}^{W}\vee\mathcal{F}_{T}^{L}$-measurable we have

$\displaystyle\mathbb{E}\left[X^{(a)}_{T}\right]=\mathbb{E}\left[Y^{(a)}_{T}Z^{(a)}_{T}\right]=\mathbb{E}\left[\mathbb{E}\left[Y^{(a)}_{T}Z^{(a)}_{T}|\mathcal{F}_{T}^{W}\vee\mathcal{F}_{T}^{L}\right]\right]=\mathbb{E}\left[Z^{(a)}_{T}\mathbb{E}\left[Y^{(a)}_{T}|\mathcal{F}_{T}^{W}\vee\mathcal{F}_{T}^{L}\right]\right].$
(3.13)

By Corollary 2.3 the variance process $v$ is strictly positive. This implies
that $\int_{0}^{T}(\theta_{s}^{(a)})^{2}ds<\infty,\mathbb{P}$-a.s, and since
$\theta^{(a)}$ is
$\\{\mathcal{F}_{t}^{W}\vee\mathcal{F}_{t}^{L}\\}_{t\in[0,T]}$-adapted,

$\displaystyle
Y^{(a)}_{T}|\mathcal{F}_{T}^{W}\vee\mathcal{F}_{T}^{L}\sim\text{Lognormal}\left(-\frac{1}{2}\int_{0}^{T}(\theta_{s}^{(a)})^{2}ds,\int_{0}^{T}(\theta_{s}^{(a)})^{2}ds\right),$

and we obtain that
$\mathbb{E}[Y^{(a)}_{T}|\mathcal{F}_{T}^{W}\vee\mathcal{F}_{T}^{L}]=1.$
Therefore, replacing the last expression in (3.13) we obtain
$\mathbb{E}[X^{(a)}_{T}]=\mathbb{E}[Z^{(a)}_{T}]$ and we only need to check
that $\mathbb{E}[Z^{(a)}_{T}]=1$. Since $|a|<\sqrt{2c_{l}}$,
$\frac{1}{2}a^{2}<c_{l}$, we can use Proposition 3.5 and get that

$\displaystyle\mathbb{E}\left[\exp\left(\frac{1}{2}a^{2}\int_{0}^{T}v_{u}du\right)\right]<\infty.$
(3.14)

Hence, Novikov’s condition is satisfied, $Z^{(a)}$ is a
$(\mathcal{F},\mathbb{P})$-martingale and we conclude that
$\mathbb{E}[X^{(a)}_{T}]=\mathbb{E}[Z^{(a)}_{T}]=1.$

Then, $X^{(a)}$ is a $(\mathcal{F},\mathbb{P})$-martingale,
$\mathbb{Q}(a)\sim\mathbb{P}$ is an equivalent probability measure defined by

$\displaystyle\frac{d\mathbb{Q}}{d\mathbb{P}}$ $\displaystyle=X^{(a)}_{T},$
$\displaystyle dX^{(a)}_{t}$
$\displaystyle=X^{(a)}_{t}\left[-\theta_{t}^{(a)}dB_{t}-a\sqrt{v_{t}}dW_{t}\right]$

and $(B^{\mathbb{Q}(a)},W^{\mathbb{Q}(a)})$ defined in (3) is a two-
dimensional standard $(\mathcal{F},\mathbb{Q}(a))$-Brownian motion. The
dynamics of the stock under $\mathbb{Q}(a)$ is given by

$\displaystyle\frac{dS_{t}}{S_{t}}=rdt+\sqrt{v_{t}}\left(\sqrt{1-\rho^{2}}dB_{t}^{\mathbb{Q}(a)}+\rho
dW_{t}^{\mathbb{Q}(a)}\right).$

This implies that the discounted stock $\widehat{S}$ is a
$(\mathcal{F},\mathbb{Q}(a))$-local martingale. Therefore, the set
$\mathcal{E}$ defined in (3.9) is a set of equivalent local martingale
measures. ∎

######  Observation 3.7.

As we said in the beginning of this section, since $\frac{1}{2}a^{2}$ is
multiplying the integrated variance in (3.14) it is crucial that $c_{l}$ is
strictly positive in Lemma 3.1.

######  Observation 3.8.

The dynamics of the variance under $\mathbb{Q}(a)\in\mathcal{E}$ is given by

$\displaystyle dv_{t}$
$\displaystyle=-\kappa\left(v_{t}-\bar{v}\right)dt+\sigma\sqrt{v_{t}}\left(dW_{t}^{\mathbb{Q}(a)}-a\sqrt{v_{t}}dt\right)+\eta
dL_{t}$ $\displaystyle=-\left(\kappa\left(v_{t}-\bar{v}\right)+a\sigma
v_{t}\right)dt+\sigma\sqrt{v_{t}}dW_{t}^{\mathbb{Q}(a)}+\eta dL_{t}$
$\displaystyle=-\kappa^{(a)}\left(v_{t}-\bar{v}^{(a)}\right)dt+\sigma\sqrt{v_{t}}dW_{t}^{\mathbb{Q}(a)}+\eta
dL_{t},$ (3.15)

where $\kappa^{(a)}=\kappa+a\sigma$ and
$\bar{v}^{(a)}=\frac{k\bar{v}}{k+a\sigma}$.

So far, we have proven that there exists a set of equivalent local martingales
measures. However, we need to study when those measures are actually
equivalent martingale measures. We prove that under the condition
$\rho^{2}<c_{l}$, there exists a subset of $\mathcal{E}$ of equivalent
martingale measures. The condition $\rho^{2}<c_{l}$ may look quite
restrictive. However, in [28, Theorem 3.6] an inequality involving the
correlation factor $\rho$ also appears in the proof of the existence of
equivalent martingales measures in the standard Heston model.

######  Theorem 3.9.

If $\rho^{2}<c_{l}$, the set

$\displaystyle\mathcal{E}_{m}:=\left\\{\mathbb{Q}(a)\in\mathcal{E}:|a|<\min\left\\{\frac{\sqrt{2c_{l}}}{2},\sqrt{c_{l}-\rho^{2}}\right\\}\right\\}$
(3.16)

is a set of equivalent martingale measures.

###### Proof.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}\subset\mathcal{E}$, by Theorem 3.6
$\widehat{S}$ is a $(\mathcal{F},\mathbb{Q}(a))$-local martingale with the
following dynamics

$\displaystyle\frac{d\widehat{S}_{t}}{\widehat{S}_{t}}=\sqrt{v_{t}}\left(\sqrt{1-\rho^{2}}dB_{t}^{\mathbb{Q}(a)}+\rho
dW_{t}^{\mathbb{Q}(a)}\right).$

Hence,

$\displaystyle\widehat{S}_{t}=S_{0}\exp\left(\sqrt{1-\rho^{2}}\int_{0}^{t}\sqrt{v_{s}}dB_{s}^{\mathbb{Q}(a)}+\rho\int_{0}^{t}\sqrt{v_{s}}dW_{s}^{\mathbb{Q}(a)}-\frac{1}{2}\int_{0}^{t}v_{s}ds\right).$

Since $\widehat{S}$ is a positive
$\left(\mathcal{F},\mathbb{Q}(a)\right)$-local martingale with
$\widehat{S}_{0}=S_{0}$, it is a
$\left(\mathcal{F},\mathbb{Q}(a)\right)$-supermartingale and it is a
$\left(\mathcal{F},\mathbb{Q}(a)\right)$-martingale if and only if
$\mathbb{E}^{\mathbb{Q}(a)}[\widehat{S}_{T}]=S_{0}.$

Similarly as we did in Theorem 3.6, we define the processes

$\displaystyle Y_{t}^{\mathbb{Q}(a)}:$
$\displaystyle=\exp\left(\sqrt{1-\rho^{2}}\int_{0}^{t}\sqrt{v_{s}}dB_{s}^{\mathbb{Q}(a)}-\frac{1-\rho^{2}}{2}\int_{0}^{t}v_{s}ds\right)$
$\displaystyle Z_{t}^{\mathbb{Q}(a)}:$
$\displaystyle=\exp\left(\rho\int_{0}^{t}\sqrt{v_{s}}dW_{s}^{\mathbb{Q}(a)}-\frac{\rho^{2}}{2}\int_{0}^{t}v_{s}ds\right).$

Thus, $\widehat{S}_{t}=S_{0}Y_{t}^{\mathbb{Q}(a)}Z_{t}^{\mathbb{Q}(a)}$. Using
that $Z_{T}^{\mathbb{Q}(a)}$ is
$\mathcal{F}_{T}^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_{T}^{L}$-measurable we
have

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}[\widehat{S}_{T}]$
$\displaystyle=S_{0}\mathbb{E}^{\mathbb{Q}(a)}\left[Y_{T}^{\mathbb{Q}(a)}Z_{T}^{\mathbb{Q}(a)}\right]=S_{0}\mathbb{E}^{\mathbb{Q}(a)}\left[\mathbb{E}^{\mathbb{Q}(a)}\left[Y_{T}^{\mathbb{Q}(a)}Z_{T}^{\mathbb{Q}(a)}|\mathcal{F}_{T}^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_{T}^{L}\right]\right]$
$\displaystyle=S_{0}\mathbb{E}^{\mathbb{Q}(a)}\left[Z_{T}^{\mathbb{Q}(a)}\mathbb{E}^{\mathbb{Q}(a)}\left[Y_{T}^{\mathbb{Q}(a)}|\mathcal{F}_{T}^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_{T}^{L}\right]\right].$
(3.17)

Since $\int_{0}^{T}v_{u}du<\infty,{\mathbb{Q}(a)}$-a.s. and $v$ is
$\\{\mathcal{F}_{t}^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_{t}^{L}\\}_{t\in[0,T]}$-adapted,

$\displaystyle
Y_{T}^{\mathbb{Q}(a)}|\mathcal{F}_{T}^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_{T}^{L}\sim\text{Lognormal}\left(-\frac{1-\rho^{2}}{2}\int_{0}^{T}v_{s}ds,(1-\rho^{2})\int_{0}^{T}v_{s}ds\right)$

and we have that
$\mathbb{E}^{\mathbb{Q}(a)}[Y_{T}^{\mathbb{Q}(a)}|\mathcal{F}_{T}^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_{T}^{L}]=1.$
Replacing the last expression in (3) we obtain that
$\mathbb{E}^{\mathbb{Q}(a)}[\widehat{S}_{T}]=S_{0}\mathbb{E}^{\mathbb{Q}(a)}[Z_{T}^{\mathbb{Q}(a)}],$
and, hence, we only need to check
$\mathbb{E}^{\mathbb{Q}(a)}[Z_{T}^{\mathbb{Q}(a)}]=1$. We will prove that
Novikov’s condition holds, that is,

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)\right]<\infty.$
(3.18)

Differently from the proof of [28, Theorem 3.6], we can not directly apply
Proposition 3.5 with the volatility parameters $\kappa^{(a)}$ and
$\bar{v}^{(a)}$ given in (3.8) to prove that the previous expectation is
finite. One reason is that under ${\mathbb{Q}(a)}$ the Brownian motion
$W^{\mathbb{Q}(a)}$ and the compound Hawkes process $L$ are not longer
independent because

$\displaystyle dW_{t}^{\mathbb{Q}(a)}=dW_{t}+a\sqrt{v_{t}}dt,$

and, therefore, the dynamics of $v$ is different under ${\mathbb{Q}(a)}$. In
order to check $(\ref{condQ})$ we note that

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)\right]=\mathbb{E}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)\frac{d\mathbb{Q}(a)}{d\mathbb{P}}\right],$

where

$\displaystyle\frac{d\mathbb{Q}(a)}{d\mathbb{P}}=Y_{T}^{(a)}Z_{T}^{(a)},$

and $Y_{T}^{(a)}$ and $Z_{T}^{(a)}$ are given in the statement of Theorem 3.6.

Repeating the same argument as in Theorem 3.6, we can write

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)Y_{T}^{(a)}Z_{T}^{(a)}\right]$
$\displaystyle=\mathbb{E}\left[\mathbb{E}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)Y_{T}^{(a)}Z_{T}^{(a)}\Big{|}\mathcal{F}_{T}^{W}\vee\mathcal{F}_{T}^{L}\right]\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)Z_{T}^{(a)}\mathbb{E}\left[Y_{T}^{(a)}|\mathcal{F}_{T}^{W}\vee\mathcal{F}_{T}^{L}\right]\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)Z_{T}^{(a)}\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(-a\int_{0}^{T}\sqrt{v_{s}}dW_{s}-\frac{1}{2}(a^{2}-\rho^{2})\int_{0}^{T}v_{s}ds\right)\right],$

Then, adding and subtracting $a^{2}\int_{0}^{T}v_{s}ds$ in the exponential and
applying Cauchy-Schwarz inequality, we obtain

$\displaystyle\mathbb{E}\left[\exp\left(-a\int_{0}^{T}\sqrt{v_{s}}dW_{s}-\frac{1}{2}(a^{2}-\rho^{2})\int_{0}^{T}v_{s}ds\right)\right]=$
$\displaystyle=\mathbb{E}\left[\exp\left(-a\int_{0}^{T}\sqrt{v_{s}}dW_{s}-a^{2}\int_{0}^{T}v_{s}ds\right)\exp\left(\frac{1}{2}(a^{2}+\rho^{2})\int_{0}^{t}v_{s}ds\right)\right]$
$\displaystyle\leqslant\mathbb{E}\left[\exp\left(-2a\int_{0}^{T}\sqrt{v_{s}}dW_{s}-2a^{2}\int_{0}^{T}v_{s}ds\right)\right]^{\frac{1}{2}}\mathbb{E}\left[\exp\left((a^{2}+\rho^{2})\int_{0}^{t}v_{s}ds\right)\right]^{\frac{1}{2}}.$
(3.19)

Note that the first factor in (3) is the expectation of a Doléans-Dade
exponential. Since $|a|<\frac{\sqrt{2c_{l}}}{2}$ (recall the choice of $a$ in
(3.16)), $2a^{2}<c_{l}$ and by Proposition 3.5, Novikov’s condition holds

$\displaystyle\mathbb{E}\left[\exp\left(2a^{2}\int_{0}^{T}v_{s}ds\right)\right]<\infty.$

Therefore, we have that

$\displaystyle\mathbb{E}\left[\exp\left(-2a\int_{0}^{T}\sqrt{v_{s}}dW_{s}-2a^{2}\int_{0}^{T}v_{s}ds\right)\right]=1.$

For the second term in (3) we apply again Proposition 3.5. We need
$a^{2}+\rho^{2}<c_{l},$ which is true because $|a|<\sqrt{c_{l}-\rho^{2}}$.
Thus

$\displaystyle\mathbb{E}\left[\exp\left((a^{2}+\rho^{2})\int_{0}^{t}v_{s}ds\right)\right]<\infty,$

and

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}\left[\exp\left(\frac{\rho^{2}}{2}\int_{0}^{T}v_{u}du\right)\right]<\infty.$

We conclude that $\mathbb{E}^{\mathbb{Q}(a)}[Z_{T}^{\mathbb{Q}(a)}]=1$,
$\mathbb{E}^{\mathbb{Q}(a)}[\widehat{S}_{T}]=S_{0}$ and $\widehat{S}$ is a
$(\mathcal{F},{\mathbb{Q}(a)})$-martingale. Therefore, $\mathcal{E}_{m}$ is a
set of equivalent martingale measures. ∎

##  Appendix A Appendix: Technical lemmas

We now give the proofs that were postponed in Section 3.

######  Lemma A.1.

For $c\leqslant\frac{\kappa^{2}}{2\sigma^{2}}$, define
$D(c):=\sqrt{\kappa^{2}-2\sigma^{2}c}$, $\Lambda(c):=\frac{2\eta
c\left(e^{D(c)T}-1\right)}{D(c)-\kappa+\left(D(c)+\kappa\right)e^{D(c)T}}$ and

$\displaystyle
c_{l}:=\sup\left\\{c\leqslant\frac{\kappa^{2}}{2\sigma^{2}}:\Lambda(c)<\epsilon_{J}\hskip
8.5359pt\text{and}\hskip
8.5359ptM_{J}\left(\Lambda(c)\right)\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right\\}.$

Then, $0<c_{l}\leqslant\frac{\kappa^{2}}{2\sigma^{2}}$ and for $c<c_{l}$,

  1. (i)

The ODE

$\displaystyle G^{\prime}(t)$
$\displaystyle=-\frac{1}{2}\sigma^{2}G^{2}(t)+\kappa G(t)-c$ (A.1)
$\displaystyle G(T)$ $\displaystyle=0$

has a unique solution in the interval $[0,T]$. The solution is strictly
decreasing and it is given by

$\displaystyle
G(t)=\frac{2c\left(e^{D(c)(T-t)}-1\right)}{D(c)-\kappa+\left(D(c)+\kappa\right)e^{D(c)(T-t)}}.$

  2. (ii)

The function $t\mapsto M_{J}(\eta G(t))$ is well defined for $t\in[0,T]$.

  3. (iii)

Define $U:=\sup_{t\in[0,T]}M_{J}(\eta G(t))$. Then, $U=M_{J}(\eta G(0))$ and

$\displaystyle
1<U\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right).$

  4. (iv)

The ODE

$\displaystyle H^{\prime}(t)$ $\displaystyle=\beta H(t)-M_{J}\left(\eta
G(t)\right)\exp\left(\alpha H(t)\right)+1$ (A.2) $\displaystyle H(T)$
$\displaystyle=0$

has a unique solution in $[0,T]$.

###### Proof.

We first check that $c_{l}>0$. Since

$\displaystyle\lim_{c\rightarrow 0^{+}}\Lambda(c)=0,$ (A.3)

there exist positive values of $c$ satisfying the inequality
$\Lambda(c)<\epsilon_{J}$. Using that $\alpha<\beta$, one can check that
$\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)>1$. Since the
limit in (A.3) holds, $M_{J}(0)=1$ and $M_{J}$ is a continuous function there
exist positive values of $c$ satisfying the inequality
$M_{J}\left(\Lambda(c)\right)\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right).$
Therefore, $c_{l}>0$. From now on, let $c<c_{l}$.

(i) To find the solution we can transform equation (A.1) to a second order
linear equation with constant coefficients and then apply the standard method
to solve it using that $c<\frac{\kappa^{2}}{2\sigma^{2}}$. To see that $G$ is
strictly decreasing one can check that

$\displaystyle
G^{\prime}(t)=\frac{-4cD(c)^{2}e^{D(c)(T-t)}}{\left(D(c)-\kappa+\left(D(c)+\kappa\right)e^{D(c)(T-t)}\right)^{2}}<0.$

(ii) Since $G$ is strictly decreasing and $\eta>0$

$\displaystyle\sup_{t\in[0,T]}\eta G(t)=\eta G(0)=\frac{2\eta
c\left(e^{D(c)T}-1\right)}{D(c)-\kappa+(D(c)+\kappa)e^{D(c)T}}=\Lambda(c).$

By definition of $c_{l}$ we have $\Lambda(c)<\epsilon_{J}$. Then, $\eta
G(t)<\epsilon_{J}$ for $t\in[0,T]$ and $M_{J}(\eta G(t))$ is well defined for
$t\in[0,T]$.

(iii) Since $G$ is strictly decreasing and $M_{J}$ is strictly increasing we
have $M_{J}(\eta G(0))\geqslant M_{J}(\eta G(t))$ for all $t\in[0,T]$.
Therefore, $U=M_{J}(\eta G(0)).$ Since $\eta G(0)>\eta G(T)=0$, we have that

$\displaystyle U=M_{J}(\eta G(0))>M_{J}(\eta G(T))=M_{J}(0)=1.$

Moreover, by definition of $c_{l}$ we have

$\displaystyle U=M_{J}(\eta G(0))=M_{J}\left(\frac{2\eta
c(e^{D(c)T}-1)}{D(c)-\kappa+(D(c)+\kappa)e^{D(c)T}}\right)=M_{J}(\Lambda(c))\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right).$

(iv) Let us make the change of variables $h(t):=H(T-t)$. Then, the ODE in
(A.2) is transformed to

$\displaystyle h^{\prime}(t)$ $\displaystyle=f(t,h(t))=M_{J}\left(\eta
G(T-t)\right)\exp\left(\alpha h(t)\right)-\beta h(t)-1$ (A.4) $\displaystyle
h(0)$ $\displaystyle=0$

where $f(t,x):=M_{J}(\eta G(T-t))\exp\left(\alpha x\right)-\beta x-1$. Note
that for $t\in[0,T]$ and $x\in{\mathbb{R}}$

$\displaystyle f_{m}(x):=-\beta x-1\leqslant f(t,x)\leqslant U\exp\left(\alpha
x\right)-\beta x-1=:f_{M}(x).$ (A.5)

First, we focus on the ODE

$\displaystyle h_{M}^{\prime}(t)$
$\displaystyle=f_{M}(h_{M}(t))=U\exp\left(\alpha h_{M}(t)\right)-\beta
h_{M}(t)-1$ (A.6) $\displaystyle h_{M}(0)$ $\displaystyle=0$

Since $f_{M}$ is continuously differentiable in $\mathbb{R}$, it is Lipschitz
continuous on bounded intervals and there exists a unique local solution for
every initial condition, see [18, Chapter II, Theorem 1.1]. We want
$f_{M}(0)>0$ and the existence of $x_{p}>0$ such that $f_{M}(x_{p})\leqslant
0$, that would imply the existence of a stable equilibrium point in the
interval $(0,x_{p})$. Therefore, the solution of (A.6) would be well defined
on $[0,\infty)$ and $h_{M}(t)<x_{p}$ for all $t\in[0,\infty)$.

Note that $f_{M}(0)=U-1>0$ as it was seen in the previous part.

The minimum of $f_{M}$ is achieved at
$x_{\text{min}}=\frac{1}{\alpha}\ln\left(\frac{\beta}{\alpha U}\right)$. Note
that $x_{\text{min}}>0$ if and only if $U<\frac{\beta}{\alpha}$. Then,

$\displaystyle
f_{M}(x_{\text{min}})=\frac{\beta}{\alpha}\left(1-\ln\left(\frac{\beta}{\alpha
U}\right)\right)-1\leqslant 0$

if, and only if
$U\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right),$ which
is fulfilled by (iii). Moreover, note that
$\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)<\frac{\beta}{\alpha}$
and then $x_{\text{min}}>0$.

We conclude that the point we were searching is $x_{p}=x_{\text{min}}$. This
guarantees that $h_{M}$ is well defined on $[0,\infty)$ and $h_{M}(t)<x_{p}$
for all $t\in[0,T]$.

Now we focus on the ODE

$\displaystyle h_{m}^{\prime}(t)$ $\displaystyle=f_{m}(h_{m}(t))=-\beta
h_{m}(t)-1$ (A.7) $\displaystyle h_{m}(0)$ $\displaystyle=0$

The solution is given by $h_{m}(t)=\frac{e^{-\beta t}-1}{\beta}.$ The function
$h_{m}$ is well defined on $[0,\infty)$ and
$h_{m}(t)\geqslant\frac{-1}{\beta}$ for all $t\in[0,\infty)$.

Consider again to the ODE in (A.4) and recall that

$\displaystyle f(t,x)=M_{J}\left(\eta G(T-t)\right)\exp\left(\alpha
x)\right)-\beta x-1.$

Define the open interval $V:=(-\frac{1}{\beta}-1,x_{p}+1)$. Note that
$f\colon[0,T]\times V\to{\mathbb{R}}$ is a continuous function, Lipschitz in
$x$ because the exponential function is Lipschitz on bounded intervals. For
$(t,x_{1}),(t,x_{2})\in[0,T]\times V$ we have

$\displaystyle|f(t,x_{1})-f(t,x_{2})|$ $\displaystyle\leqslant|M_{J}(\eta
G(T-t))||\exp\left(\alpha x_{1}\right)-\exp\left(\alpha
x_{2}\right)|+\beta|x_{1}-x_{2}|$ $\displaystyle\leqslant U|\exp\left(\alpha
x_{1}\right)-\exp\left(\alpha x_{2}\right)|+\beta|x_{1}-x_{2}|$
$\displaystyle\leqslant K_{1}|x_{1}-x_{2}|,$

for some constant $K_{1}>0$. Then, by the Picard-Lindelöf theorem (see [18,
Chapter II, Theorem 1.1]) there is a unique solution $h\colon I\to V$ for some
interval $I\subset[0,T]$. Moreover, by [18, Chapter II, Theorem 3.1] only two
cases are possible:

1) $I=[0,T]$. In that case, there is nothing more to prove.

2) $I=[0,\epsilon)$ with $\epsilon\leqslant T$ and

$\displaystyle\lim_{t\rightarrow\epsilon^{-}}h(t)\in\left\\{-\frac{1}{\beta}-1,x_{p}+1\right\\}.$
(A.8)

That is, $h$ approaches the boundary of $V$ when $t$ approaches $\epsilon$.
However, as a consequence of (A.5), we will prove that $h_{m}(t)\leqslant
h(t)\leqslant h_{M}(t),$ for all $t\in[0,\epsilon)$.

First, we prove that $h_{m}(t)\leqslant h(t)$ for all $t\in[0,\epsilon)$.
Define the function $g(t):=h_{m}(t)-h(t)$ for $t\in[0,\epsilon)$. Note that
$g(0)=0$ and we want to prove that $g(t)\leqslant 0$ for all
$t\in[0,\epsilon)$. Assume there exists $s\in(0,\epsilon)$ such that $g(s)>0$.
Since $g$ is continuous and $g(0)=0$, there exists $r\in[0,s)$ with $g(r)=0$
and $g(t)>0$ for $t\in(r,s]$. Now, for $t\in[r,s]$ we have

$\displaystyle g^{\prime}(t)$ $\displaystyle=h_{m}^{\prime}(t)-h^{\prime}(t)$
$\displaystyle=f_{m}(h_{m}(t))-f(t,h(t))$ $\displaystyle\leqslant
f(t,h_{m}(t))-f(t,h(t))$ $\displaystyle=M_{J}(\eta
G(T-t))\left(\exp\left(\alpha h_{m}(t)\right)-\exp\left(\alpha
h(t)\right)\right)-\beta\left(h_{m}(t)-h(t)\right)$ $\displaystyle\leqslant
U\left(\exp\left(\alpha h_{m}(t)\right)-\exp\left(\alpha
h(t)\right)\right)-\beta\left(h_{m}(t)-h(t)\right)$ $\displaystyle\leqslant
K_{2}|h_{m}(t)-h(t)|=K_{2}|g(t)|=K_{2}g(t),$

for some constant $K_{2}>0$, where in the last inequality we have used that
since $h_{m}$ and $h$ are continuous, they are bounded in $[r,s]$ and we can
use the Lipschitz property because the exponential function is Lipschitz on
bounded intervals. Applying Gronwall’s inequality we have that $g(s)\leqslant
g(r)e^{K_{2}(s-r)}=0,$ which is a contradiction. We conclude that
$h_{m}(t)\leqslant h(t)$ for all $t\in[0,\epsilon)$. A similar same argument
can be employed to prove that $h(t)\leqslant h_{M}(t)$ for all
$t\in[0,\epsilon)$.

For $t\in[0,\epsilon)$ we have

$\displaystyle h_{m}(t)\leqslant h(t)\leqslant
h_{M}(t)\implies\frac{-1}{\beta}\leqslant h(t)\leqslant
x_{p}\implies\frac{-1}{\beta}\leqslant\lim_{t\rightarrow\epsilon^{-}}h(t)\leqslant
x_{p}.$

This contradicts (A.8) and we conclude that the only possible situation is
that $h$ is well defined on $[0,T]$. ∎

######  Corollary A.2.

Define $c_{s}$ by

$\displaystyle
c_{s}:=\min\left\\{\frac{\kappa\epsilon_{J}}{2\eta},\frac{\kappa}{2\eta}M_{J}^{-1}\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$
(A.9)

Then, $0<c_{s}<c_{l}$.

###### Proof.

We first check that $c_{s}>0$. The function
$M_{J}\colon(-\infty,\epsilon_{J})\to(0,\infty)$ is well defined and strictly
increasing. Therefore, $M_{J}^{-1}\colon(0,\infty)\to(-\infty,\epsilon_{J})$
is also a well defined function and it is strictly increasing.

Since $\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)>1$ and
$M_{J}(0)=1$ we have
$M_{J}^{-1}\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right)>0,$
and we can conclude that $c_{s}>0$.

Recall that $0<c_{l}\leqslant\frac{\kappa^{2}}{2\sigma^{2}}$ and its
definition

$\displaystyle
c_{l}=\sup\left\\{c\leqslant\frac{\kappa^{2}}{2\sigma^{2}}:\Lambda(c)<\epsilon_{J}\hskip
8.5359pt\text{and}\hskip
8.5359ptM_{J}\left(\Lambda(c)\right)\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right\\},$
(A.10)

where $D(c)=\sqrt{\kappa^{2}-2\sigma^{2}c}$ and $\Lambda(c)=\frac{2\eta
c\left(e^{D(c)T}-1\right)}{D(c)-\kappa+\left(D(c)+\kappa\right)e^{D(c)T}}$.

To prove that $c_{s}<c_{l}$ we check that $\Lambda(c_{s})<\epsilon_{J}$ and
$M_{J}\left(\Lambda(c_{s})\right)\leqslant\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)$.
The following inequality holds

$\displaystyle\Lambda(c)=\frac{2\eta
c\left(e^{\sqrt{\kappa^{2}-2\sigma^{2}c}T}-1\right)}{\sqrt{\kappa^{2}-2\sigma^{2}c}-\kappa+\left(\sqrt{\kappa^{2}-2\sigma^{2}c}+\kappa\right)e^{\sqrt{\kappa^{2}-2\sigma^{2}c}T}}<\frac{2\eta
c}{\kappa}.$ (A.11)

Actually

$\displaystyle\Lambda(c)=\frac{2\eta
c\left(e^{\sqrt{\kappa^{2}-2\sigma^{2}c}T}-1\right)}{\sqrt{\kappa^{2}-2\sigma^{2}c}-\kappa+\left(\sqrt{\kappa^{2}-2\sigma^{2}c}+\kappa\right)e^{\sqrt{\kappa^{2}-2\sigma^{2}c}T}}<\frac{2\eta
c}{\kappa}$
$\displaystyle\iff\kappa\left(e^{\sqrt{\kappa^{2}-2\sigma^{2}c}T}-1\right)<\sqrt{\kappa^{2}-2\sigma^{2}c}-\kappa+\left(\sqrt{\kappa^{2}-2\sigma^{2}c}+\kappa\right)e^{\sqrt{\kappa^{2}-2\sigma^{2}c}T}$
$\displaystyle\iff
0<\sqrt{\kappa^{2}-2\sigma^{2}c}\left(1+e^{\sqrt{\kappa^{2}-2\sigma^{2}c}T}\right).$

Now, by definition of $c_{s}$ we have $\Lambda(c_{s})<\frac{2\eta
c_{s}}{\kappa}\leqslant\epsilon_{J},$ and

$\displaystyle M_{J}\left(\Lambda(c_{s})\right)<M_{J}\left(\frac{2\eta
c_{s}}{\kappa}\right)\leqslant
M_{J}\left(M_{J}^{-1}\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right)\right)=\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right).$

We conclude that $c_{s}<c_{l}$. Note that the inequality in $c_{s}<c_{l}$ is
strict because the inequality in (A.11) is strict. ∎

######  Example A.3.

Some examples:

  1. (i)

If $J_{1}\sim\text{Exponential}(\lambda)$, then

$\displaystyle
c_{s}=\min\left\\{\frac{\kappa\lambda}{2\eta}\left(1-\frac{\alpha}{\beta}\exp\left(1-\frac{\alpha}{\beta}\right)\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

  2. (ii)

If $J_{1}\sim\text{Gamma}(\mu,\lambda)$ with $\mu,\lambda>0$ as the shape and
the rate, respectively. Then

$\displaystyle
c_{s}=\min\left\\{\frac{\kappa\lambda}{2\eta}\left(1-\frac{1}{\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right)^{1/\mu}}\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

  3. (iii)

If $J_{1}=j>0$, then

$\displaystyle c_{s}=\min\left\\{\frac{\kappa}{2\eta
j}\left(\ln\left(\frac{\beta}{\alpha}\right)+\frac{\alpha}{\beta}-1\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

###### Proof.

(i) The moment generating function is given by
$M_{J}(t)=\frac{\lambda}{\lambda-t},t<\lambda.$ Hence, with the notation of
Assumption 1, $\epsilon_{J}=\lambda$ and the inverse of the $M_{J}$ is given
by $M_{J}^{-1}(t)=\lambda\left(1-\frac{1}{t}\right),t>0$. Then, applying
Corollary A.2 we have the following expression of $c_{s}$

$\displaystyle
c_{s}=\min\left\\{\frac{\kappa\lambda}{2\eta},\frac{\kappa\lambda}{2\eta}\left(1-\frac{1}{\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)}\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

Note that since
$\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)>1$, we have

$\displaystyle
0<1-\frac{1}{\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)}<1.$

We conclude that

$\displaystyle c_{s}$
$\displaystyle=\min\left\\{\frac{\kappa\lambda}{2\eta},\frac{\kappa\lambda}{2\eta}\left(1-\frac{1}{\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)}\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}$
$\displaystyle=\min\left\\{\frac{\kappa\lambda}{2\eta}\left(1-\frac{\alpha}{\beta}\exp\left(1-\frac{\alpha}{\beta}\right)\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

(ii) The moment generating function is given by
$M_{J}(t)=\left(1-\frac{t}{\lambda}\right)^{-\mu},t<\lambda.$ Thus,
$\epsilon_{J}=\lambda$ and the inverse of $M_{J}$ is given by
$M_{J}^{-1}(t)=\lambda\left(1-t^{-1/\mu}\right),t>0.$ Then, applying Corollary
A.2 we have the following expression of $c_{s}$

$\displaystyle
c_{s}=\min\left\\{\frac{\kappa\lambda}{2\eta},\frac{\kappa\lambda}{2\eta}\left(1-\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right)^{-1/\mu}\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

Note that since
$\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)>1$, we have

$\displaystyle
0<1-\frac{1}{\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right)^{1/\mu}}<1.$

We conclude that

$\displaystyle
c_{s}=\min\left\\{\frac{\kappa\lambda}{2\eta}\left(1-\left(\frac{\beta}{\alpha}\exp\left(\frac{\alpha}{\beta}-1\right)\right)^{-1/\mu}\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

(iii) The moment generating function is given by $M_{J}(t)=e^{tj}$,
$t\in\mathbb{R}$. Thus, $\epsilon_{J}=\infty$ and the inverse of $M_{J}$ is
given by $M_{J}^{-1}(t)=\ln(t)/j,t>0$. Then, applying Corollary A.2 we
conclude that

$\displaystyle c_{s}=\min\left\\{\frac{\kappa}{2\eta
j}\left(\ln\left(\frac{\beta}{\alpha}\right)+\frac{\alpha}{\beta}-1\right),\frac{\kappa^{2}}{2\sigma^{2}}\right\\}.$

∎

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