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 $\mathcal{F}^{(B,W)}=\big\{\mathcal{F}^{(B,W)}_t, t\in[0, T]\big\}$ . On $(\Omega,\mathcal{A},\mathbb{P})$ , we also consider a Hawkes process $N=\{N_t, t\in[0, T]\}$ with stochastic intensity given by

\begin{align*} \lambda_t=\lambda_0+\alpha\int_0^t e^{-\beta(t-s)}dN_s,\end{align*}

or, equivalently,

\begin{align*} d\lambda_t=-\beta(\lambda_t-\lambda_0)dt+\alpha dN_t,\end{align*}

where $\lambda_0>0$ is the initial intensity, $\beta>0$ is the speed of mean reversion, and $\alpha \in (0,\beta)$ is the self-exciting factor. Note that the stability condition

\begin{align*} \alpha\int_0^\infty e^{-\beta s}ds=\frac{\alpha}{\beta}<1\end{align*}

holds. See [Reference Bacry, Mastromatteo and Muzy4, Section 2] and [Reference Jaisson34, Section 3.1.1] for the definition of N. We then consider a sequence of independent and identically distributed (i.i.d.), strictly positive, and integrable random variables $\{J_i\}_{i\geq 1}$ and the compound Hawkes process $L=\{L_t, t\in[0, T]\}$ given by

\begin{align*} L_t=\sum_{i=1}^{N_t}J_i.\end{align*}

We assume that (B, W), N and $\{J_i\}_{i\geq 1}$ are independent of each other. We write $\mathcal{F}^L=\left\{\mathcal{F}^L_t, t\in[0, T]\right\}$ for the minimally augmented filtration generated by L and

\begin{align*} \mathcal{F}=\big\{\mathcal{F}_t=\mathcal{F}_t^{(B,W)}\vee\mathcal{F}_t^L, t\in[0, T]\big\}\end{align*}

for the joint filtration. We assume that $\mathcal{A}=\mathcal{F}_T$ , and we 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

(2.1) \begin{align} \frac{dS_t}{S_t} & =\mu_tdt+\sqrt{v_t}\!\left(\sqrt{1-\rho^2} dB_t+\rho dW_t\right), \notag \\ dv_t & =-\kappa\!\left(v_t-\bar{v}\right)dt+\sigma\sqrt{v_t}dW_t+\eta dL_t,\end{align}

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 $2\kappa\bar{v}\geq\sigma^2$ is satisfied; see [Reference Alfonsi1, Proposition 1.2.15].

Note that our stochastic volatility model is the well-known Heston model but with a compound Hawkes process added in the variance process. The procedure for proving strong existence and pathwise uniqueness of the stochastic differential equation (SDE) in (2.1) is essentially given in [Reference Protter45, Chapter V.10, Theorem 57]. Informally, between jump times, the SDE is just a standard Cox–Ingersoll–Ross (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 uniqueness are proven for the SDE that defines the CIR model (see [Reference Alfonsi1, Theorem 1.2.1]), one can properly interlace the solutions of the continuous-path SDE and the jumps as is done in [Reference Protter45, Chapter V.10, Theorem 57] and prove strong existence and pathwise uniqueness for our SDE in (2.1). In particular, this implies that v is a positive process.

Since the sizes of the jumps of the compound Hawkes process are strictly positive, one expects that our variance is greater than or equal to the Heston variance. We prove this property following the proof of [Reference Karatzas and Shreve38, Chapter 5.2.C, Proposition 2.18]. The procedure is the same as there, but in our case some extra computations will appear because of 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 [Reference Alfonsi1, 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

(2.2) \begin{align} \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.\end{align}

Then

\begin{align*} \mathbb{P}\!\left(\left\{\omega\in\Omega \,:\, \widetilde{v}_t(\omega)\leq v_t(\omega) \ \forall t\in[0, T]\right\}\right)=1,\end{align*}

where v is the pathwise unique strong solution of (2.1).

Proof. See [Reference Alfonsi1, Theorem 1.2.1] for a reference on the solution of the SDE in (2.2). 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

\begin{align*} \int_{a_n}^{a_{n-1}}\frac{du}{u}=n,\end{align*}

for every $n\geq1$ . Precisely, $a_n=\exp\!\left({-}\frac{n(n+1)}{2}\right)$ .

For each $n\geq1$ , there exists a continuous function $\rho_n$ on $\mathbb{R}$ with support in $(a_n,a_{n-1})$ such that

(2.3) \begin{align} 0\leq \rho_n(x)\leq\frac{2}{nx}\end{align}

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

Then the function

(2.4) \begin{align} \psi_n(x)=\int_0^{|x|}\int_0^y\rho_n(u)dudy\end{align}

is even and twice continuously differentiable, with $|\psi^{\prime}_n(x)|\leq1$ and $\lim_{n\rightarrow\infty}\psi_n(x)=|x|$ for $x\in\mathbb{R}$ . Furthermore, $\{\psi_n\}_{n=1}^\infty$ is non-decreasing.

Next, define the non-decreasing function $\varphi_n(x)=\psi_n(x)\mathbb{1}_{(0,\infty)}(x)$ . Note that

\begin{align*} \lim_{n\rightarrow\infty}\varphi_n(x)=\max\{x,0\}\,=\!:\,x^+. \end{align*}

Define $I_t\,:\!=\,\widetilde{v}_t-v_t$ for $t\in[0, T]$ . Applying Itô’s formula, we get

\begin{align*} \varphi_n(I_t) = \ &\varphi_n(0)+\int_0^t\varphi^{\prime}_n(I_{s-})dI_s +\frac{1}{2}\int_0^t\varphi^{\prime\prime}_n(I_{s-})d[I]_s^{\text{c}} \\ & +\sum_{0<s\leq t}\left[\varphi_n(I_s)-\varphi_n(I_{s-})-\varphi^{\prime}_n(I_{s-})\Delta I_s\right].\end{align*}

Note that

\begin{align*} dI_s&=-\kappa(\widetilde{v}_s-v_s)ds+\sigma\!\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)dW_s-\eta dL_s, \\ d[I]_s&=\sigma^2\!\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)^2ds+\eta^2d[L]_t.\end{align*}

The latter implies that $d[I]_s^{\text{c}}=\sigma^2\!\left(\sqrt{\vphantom{\sum}\,\widetilde{v}_s}-\sqrt{v_s}\right)^2ds$ and $\Delta I_s=-\Delta v_s=-\eta \Delta L_s.$ Therefore,

\begin{align*} \varphi_n(I_t)= \ &-\kappa\int_0^t\varphi^{\prime}_n(I_s)I_sds+\sigma\int_0^t\varphi^{\prime}_n(I_{s-})\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)dW_s \\ &-\eta\int_0^t\varphi^{\prime}_n(I_{s-})dL_s+\frac{\sigma^2}{2}\int_0^t\varphi^{\prime\prime}_n(I_s)\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)^2ds \\ &+\sum_{0<s\leq t}\left[\varphi_n(I_s)-\varphi_n(I_{s-})\right]+\eta\sum_{0<s\leq t}\varphi^{\prime}_n(I_{s-})\Delta L_s.\end{align*}

Note that $\eta\int_0^t\varphi^{\prime}_n(I_{s-})dL_s=\eta\sum_{0<s\leq t}\varphi^{\prime}_n(I_{s-})\Delta L_s.$ Since $I_s-I_{s-}=-\eta\Delta L_s\leq0$ and $\varphi_n$ is a non-decreasing function, we have $\sum_{0<s\leq t}\left[\varphi_n(I_s)-\varphi_n(I_{s-})\right] \leq0.$

By (2.3) and (2.4), and using that $|\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$ for $x,y\geq0$ , we obtain

\begin{align*} \varphi^{\prime\prime}_n(I_s)\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)^2\leq \frac{2}{n}\frac{\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)^2}{\widetilde{v}_s-v_s}\leq\frac{2}{n}.\end{align*}

Since $|\varphi^{\prime}_n(x)|\leq 1$ , we can conclude that

(2.5) \begin{align} \varphi_n(I_t)\leq \ &\kappa\int_0^tI_s^+ds+\sigma\int_0^t\varphi^{\prime}_n(I_{s-})\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)dW_s +\frac{\sigma^2 t}{n}.\end{align}

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

(2.6) \begin{align} \mathbb{E}\!\left[\int_0^t\varphi^{\prime}_n(I_{s-})\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)dW_s\right]=0,\end{align}

we need to check that

(2.7) \begin{align} \mathbb{E}\!\left[\int_0^t\varphi^{\prime}_n(I_s)^2\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)^2ds\right]<\infty.\end{align}

Now,

(2.8) \begin{align} \mathbb{E}\!\left[\int_0^t\varphi^{\prime}_n(I_s)^2\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)^2ds\right]&= \mathbb{E}\!\left[\int_0^t\varphi^{\prime}_n(I_s)^2\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)^2\mathbb{1}_{\{I_s>0\}}ds\right] \notag\\ & \leq \mathbb{E}\!\left[\int_0^t\left(\sqrt{\widetilde{v}_s}-\sqrt{v_s}\right)^2\mathbb{1}_{\{I_s>0\}}ds\right] \notag\\ & \leq \mathbb{E}\!\left[\int_0^t\widetilde{v}_sds\right] \notag\\ & = \int_0^t\mathbb{E}\!\left[\widetilde{v}_s\right]ds.\end{align}

In [Reference Glasserman and Kim27, Section 2, Equation (2.3)], we see that for $t\in (0,T]$ ,

\begin{align*} \widetilde{v}_t \sim \frac{e^{-\kappa t}v_0}{k(t)}\chi_\delta^{\prime 2}\left(k(t)\right) \ \ \text{with} \ \ k(t)\,:\!=\,\frac{4\kappa v_0 e^{-\kappa t}}{\sigma^2\big(1-e^{-\kappa t}\big)} \ \ \text{and} \ \ \delta\,:\!=\,\frac{4\kappa\bar{v}}{\sigma^2},\end{align*}

where $\chi_\delta^{\prime 2}(k(t))$ denotes a non-central chi-squared random variable with $\delta$ degrees of freedom and non-centrality parameter k(t). Then

\begin{align*}\mathbb{E}\!\left[\widetilde{v}_t\right]=\frac{\sigma^2\big(1-e^{-\kappa t}\big)}{4\kappa}\left(\delta+k(t)\right)=v_0e^{-\kappa t}+\bar{v}\big(1-e^{-\kappa t}\big).\end{align*}

Since $t\mapsto\mathbb{E}\!\left[\widetilde{v}_t\right]$ is a continuous function on [0, T], the integral in (2.8) is finite, which implies that the integral in (2.7) is finite and (2.6) holds. Taking expectations in (2.5), we have

\begin{align*} \mathbb{E}\!\left[\varphi_n(I_t)\right]\leq\kappa\int_0^t\mathbb{E}\!\left[I_s^+\right]+\frac{\sigma^2 t}{n}.\end{align*}

Sending n to infinity yields $\mathbb{E}\!\left[I_t^+\right]\leq\kappa\int_0^t\mathbb{E}\!\left[I_s^+\right]ds.$ One can check that

(2.9) \begin{align} \int_0^t\left|\mathbb{E}\!\left[I_s^+\right]\right|ds<\infty.\end{align}

In fact,

$$\int_0^t\left|\mathbb{E}\!\left[I_s^+\right]\right|ds = \int_0^t\mathbb{E}\!\left[I_s^+\right]ds = \int_0^t\mathbb{E}\!\left[I_s\mathbb{1}_{\{I_s>0\}}\right]ds \leq \int_0^t\mathbb{E}\!\left[\widetilde{v}_s\right]ds<\infty,$$

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

Applying a version of Gronwall’s inequality where only the condition (2.9) is required, we get $\mathbb{E}[I_t^+]=0$ for all $t\in[0, T]$ . This means that

\begin{align*} \mathbb{P}\!\left(\{\omega\in\Omega\,:\, \widetilde{v}_t(\omega)\leq v_t(\omega)\}\right)=1 \hspace{1cm} \forall t\in[0, T].\end{align*}

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

\begin{align*} \mathbb{P}\!\left(\left\{\omega\in\Omega \,:\, \widetilde{v}_t(\omega)\leq v_t(\omega) \ \forall t\in[0, T]\right\}\right)=1.\end{align*}

Proof. Recall that we have assumed that the Feller condition $2\kappa \bar{v}\geq\sigma^2$ is satisfied. Therefore the process $\widetilde{v}$ defined by (2.2) is strictly positive; see [Reference Alfonsi1, Proposition 1.2.15]. Finally, by Proposition 2.2 we have that

\begin{align*} \mathbb{P}\!\left(\left\{\omega\in\Omega \,:\, \widetilde{v}_t(\omega)\leq v_t(\omega) \ \forall t\in[0, T]\right\}\right)=1.\end{align*}

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, employing Girsanov’s theorem in connection with 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

\begin{align*} \mathbb{E}\!\left[\exp\!\left(c\int_0^Tv_udu\right)\right]<\infty.\end{align*}

By Corollary 2.1, v is strictly positive and the expectation above is finite for $c\leq0$ , but for our applications is essential that c can be strictly positive. Looking at the proof of [Reference Wong and Heyde52, Lemma 3.1], one can see that for $c\leq \frac{\kappa^2}{2\sigma^2}$ the following holds:

(3.1) \begin{align} \mathbb{E}\!\left[\exp\!\left(c\int_0^T\widetilde{v}_udu\right)\right]<\infty,\end{align}

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

\begin{align*} \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.\end{align*}

The procedure for proving that (3.1) holds is to show that

(3.2) \begin{align}\mathbb{E}\!\left[\exp\!\left(c\int_0^T\widetilde{v}_udu\right)\right]\leq\exp\!\left({-}(\kappa \bar{v})\Phi(0)-v_0\psi(0)\right)<\infty,\end{align}

where $\Phi$ and $\psi$ satisfy the following generalized Riccati equations:

(3.3) \begin{align} \psi^{\prime}(t) &=\frac{\sigma^2}{2}\psi^2(t)+\kappa\psi(t)+c, \\ -\Phi^{\prime}(t) & =\psi(t), \notag\\ \psi(T) &=\Phi(T)=0. \notag\end{align}

Because of the jump contribution of the compound Hawkes process, this procedure is more delicate in our model. However, we can obtain a bound similar to that of (3.2), with an additional function that will be the solution of an ODE involving the moment generating function of $J_1$ and the Hawkes parameters $\alpha$ and $\beta$ .

We start by 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 3.1. 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

\begin{align*} \lim_{t\rightarrow\epsilon_J^-}M_J(t)=\infty.\end{align*}

Since $\epsilon_J>0$ , all positive moments of $J_1$ are finite.

Note that this assumption holds for the gamma distribution, the chi-squared distribution, the 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

\begin{align*} M(t)=\exp\!\left(F(t)+G(t)v_t+H(t)\lambda_t+c\int_0^tv_udu\right),\end{align*}

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

\begin{align*} M(T)=\exp\!\left(c\int_0^Tv_udu\right).\end{align*}

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

\begin{align*} \mathbb{E}\!\left[\exp\!\left(c\int_0^Tv_udu\right)\right]=\mathbb{E}[M(T)]\leq M(0)=\exp\!\left(F(0)+G(0)v_0+H(0)\lambda_0\right),\end{align*}

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

The generalized Riccati equations for F, G, and H are obtained by applying Itô’s formula to M and equating the drift terms to 0. This is done formally in Proposition 3.1. In the next lemma we study the existence of solutions of the generalized Riccati equations for our model. Note that the ODE for G is slightly different from the ODE for $\psi$ 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 it is rather technical and based on ODE theory.

Lemma 3.1. For $c\leq\frac{\kappa^2}{2\sigma^2}$ , define $D(c)\,:\!=\,\sqrt{\kappa^2-2\sigma^2c}$ ,

$$\Lambda(c)\,:\!=\,\frac{2\eta c\!\left(e^{D(c)T}-1\right)}{D(c)-\kappa+\left(D(c)+\kappa\right)e^{D(c)T}},$$

and

\begin{align*} c_l\,:\!=\,\sup\left\{c\leq\frac{\kappa^2}{2\sigma^2}\,:\, \Lambda(c)< \epsilon_J \hspace{0.3cm}\text{and}\hspace{0.3cm} M_J\!\left(\Lambda(c)\right)\leq\frac{\beta}{\alpha}\exp\!\left(\frac{\alpha}{\beta}-1\right) \right\}.\end{align*}

Then $0<c_l\leq\frac{\kappa^2}{2\sigma^2}$ , and for $c< c_l$ , the following hold:

1. (i) The ODE

   \begin{align*} G^{\prime}(t) & =-\frac{1}{2}\sigma^2G^2(t)+\kappa G(t)-c, \\ G(T) & = 0 \notag\end{align*}
   has a unique solution in the interval [0, T]. The solution is strictly decreasing and is given by
   \begin{align*} 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)}}.\end{align*}

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

   \begin{align*} 1<U\leq \frac{\beta}{\alpha}\exp\!\left(\frac{\alpha}{\beta}-1\right).\end{align*}

4. (iv) The ODE

   \begin{align*} H^{\prime}(t) & =\beta H(t)-M_J\!\left(\eta G(t)\right)\exp\!\left(\alpha H(t)\right) +1, \\ H(T) & = 0 \notag\end{align*}
   has a unique solution in [0, T].

Observation 3.1. One could also assume that the domain of $M_J$ is big enough so that the function $t\mapsto M_J(\eta G(t))$ is well defined on the interval [0, T] for any value of $c\leq\frac{\kappa^2}{2\sigma^2}$ . Then one can prove the existence of H by applying the Picard–Lindelöf theorem (see [Reference Hartman30, Chapter II, Theorem 1.1]), which yields the following conditions:

\begin{align*} c\leq\frac{\kappa^2}{2\sigma^2} \ \ \text{and} \ \ \beta+f(U)\alpha\leq \frac{1}{T},\end{align*}

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; for the sake of generality, we prefer to avoid this.

Note that, a priori, there is no explicit expression for $c_l$ in the previous lemma, because it is not possible to solve the inequalities

\begin{align*} \Lambda(c)< \epsilon_J \hspace{0.5cm}\text{ and }\hspace{0.5cm} M_J\!\left(\Lambda(c)\right)\leq\frac{\beta}{\alpha}\exp\!\left(\frac{\alpha}{\beta}-1\right).\end{align*}

Moreover, $c_l$ depends on $M_J$ , $\eta$ , $\kappa$ , $\sigma$ , T, $\epsilon_J$ , $\alpha$ , and $\beta$ . However, one can get an explicit value for $c_l$ that is suboptimal but does not depend on T. We obtain an expression for that suboptimal value and give some examples.

Corollary 3.1. Define $c_s$ by

\begin{align*} 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\}.\end{align*}

Then $0<c_s<c_l$ .

Note that we can apply Lemma 3.1 for any $c<c_s$ , because $c_s<c_l$ . We now give some examples of the value $c_s$ .

We have studied the conditions under which the generalized Riccati equations have a well-defined solution on the interval [0, T]. We now proceed to studying the integrability of the exponential of the integrated variance, following the method noted at the beginning of this section.

Proposition 3.1. Let $c<c_l$ ; then

\begin{align*} \mathbb{E}\!\left[\exp\!\left(c\int_0^Tv_udu\right)\right]<\infty.\end{align*}

Proof. By Corollary 2.1, v is strictly positive and the expectation is finite for $c\leq0$ . We focus on the case when $0<c<c_l$ . We first define the function $f\colon[0, T]\times\mathbb{R}^3\to\mathbb{R}$ by

\begin{align*} f(t,x,y,z)=\exp\!\left(F(t)+G(t)x+H(t)y+cz\right),\end{align*}

where G and H are the solutions of the generalized Riccati equations given in Lemma 3.1, that is,

(3.4) \begin{align} G^{\prime}(t) & =-\frac{1}{2}\sigma^2G^2(t)+\kappa G(t)-c, \\ G(T) & = 0 \notag\end{align}

and

(3.5) \begin{align} H^{\prime}(t) & =\beta H(t)-M_J\!\left(\eta G(t)\right)\exp\!\left(\alpha H(t)\right) +1, \\ H(T) & = 0, \notag\end{align}

and F is given by

(3.6) \begin{align} F^{\prime}(t) & =-\kappa\bar{v}G(t)-\beta\lambda_0H(t), \\ F(T) & =0\notag.\end{align}

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 $V_t\,:\!=\,\int_0^tv_udu$ . We let $Y_t\,:\!=\,(t,v_t,\lambda_t,V_t)$ and define 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’s formula to the process M, we get

\begin{align*} M(t)= \ &M(0)+\int_0^t\partial_tf(Y_{s-})ds+\int_0^t\partial_xf(Y_{s-})dv_s+\int_0^t\partial_y f(Y_{s-})d\lambda_s \\ &+\int_0^t\partial_zf(Y_{s-})dV_s+\frac{1}{2}\int_0^t\partial^2_{xx}f(Y_{s-})d[v]_s^{\text{c}} \\ &+\sum_{0<s\leq t}\left[f(Y_s)-f(Y_{s-})-\partial_xf(Y_{s-})\Delta v_s-\partial_y f(Y_{s-})\Delta\lambda_s\right].\end{align*}

We have used that

\begin{align*} [\lambda]_t & =\alpha^2[N]_t=\alpha^2N_t \implies [\lambda]_t^{\text{c}} =0, \\ [V]_t & = 0, \\ [v,\lambda]_t & =\alpha\eta[L,N]_t=\alpha\eta L_t\implies [v,\lambda]_t^{\text{c}} =0, \\ [v,V]_t & =[\lambda,V]_t =0.\end{align*}

Moreover,

\begin{align*} [v]_t & =\int_0^t\sigma^2v_sds+\eta^2[L]_t \implies [v]_t^{\text{c}}=\int_0^t\sigma^2v_sds, \\ dv_t & =-\kappa(v_t-\bar{v})dt+\sigma\sqrt{v_t}dW_t+\eta dL_t, \\ d\lambda_t & =-\beta(\lambda_t-\lambda_0)dt+\alpha dN_t.\end{align*}

Hence,

\begin{align*} M(t)-M(0) = &\int_0^t\Big[\partial_tf(Y_{s})-\kappa\partial_xf(Y_{s})(v_s-\bar{v})-\beta\partial_y f(Y_{s})(\lambda_s-\lambda_0) \\ &+\partial_zf(Y_{s})v_s+\frac{1}{2}\partial_{xx}^2f(Y_{s})\sigma^2v_s\Big]ds\\ &+\int_0^t\partial_xf(Y_{s-})\sigma\sqrt{v_s}dW_s+\int_0^t\partial_yf(Y_{s-})\eta dL_s+\int_0^t\partial_y f(Y_{s-})\alpha dN_s \\ &+\sum_{0<s\leq t}\left[f(Y_s)-f(Y_{s-})-\partial_xf(Y_{s-})\Delta v_s-\partial_y f(Y_{s-})\Delta \lambda_s\right].\end{align*}

Since $\Delta v_s=\eta\Delta L_s$ and $\Delta \lambda_s=\alpha\Delta N_s$ , we have

\begin{align*} \int_0^t\partial_xf(Y_{s-})\eta dL_s & = \sum_{0<s\leq t}\partial_xf(Y_{s-})\Delta v_s, \\ \int_0^t\partial_y f(Y_{s-})\alpha dN_s & = \sum_{0<s\leq t}\partial_y f(Y_{s-})\Delta \lambda_s,\end{align*}

and we get

\begin{align*} M(t)-M(0) = &\int_0^t\Big[\partial_tf(Y_{s})-\kappa\partial_xf(Y_{s})(v_s-\bar{v})-\beta\partial_y f(Y_{s})(\lambda_s-\lambda_0) \\ &+\partial_zf(Y_{s})v_s+\frac{1}{2}\partial_{xx}^2f(Y_{s})\sigma^2v_s\Big]ds\\ &+\int_0^t\partial_xf(Y_{s-})\sigma\sqrt{v_s}dW_s +\sum_{0<s\leq t}\left[f(Y_s)-f(Y_{s-})\right].\end{align*}

Next, we can write

\begin{align*} \sum_{0<s\leq t}\left[f(Y_s)-f(Y_{s-})\right] & =\sum_{0<s\leq t}\left[f(s,v_{s-}+\Delta v_s, \lambda_{s-}+\Delta \lambda_s,V_s)-f(Y_{s-})\right] \\ & = \sum_{0<s\leq t}\left[f(s,v_{s-}+\eta \Delta L_s, \lambda_{s-}+\alpha\Delta N_s,V_s)-f(Y_{s-})\right] \\ & = \sum_{0<s\leq t} g(s,\Delta L_s,\Delta N_s),\end{align*}

where

\begin{align*} g(s,u_1,u_2)\,:\!=\,f(s,v_{s-}+\eta u_1, \lambda_{s-}+\alpha u_2,V_s)-f(Y_{s-}).\end{align*}

We now define $U_s=(L_s,N_s)$ , and for $t\in[0, T]$ , $A\in\mathcal{B} (\mathbb{R}^2\setminus\{0,0\})$ , we let

\begin{align*} N^U(t,A)=\#\{0<s\leq t, \Delta U_s\in A\}.\end{align*}

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 Hawkes process is given by $\Lambda^N_t=\int_0^t\lambda_udu$ ; see [Reference Bacry, Mastromatteo and Muzy4, Theorem 3]. One can check that the compensator of the compound Hawkes process is given by $\Lambda^L_t=\mathbb{E}[J_1]\int_0^t\lambda_udu$ . Thus,

\begin{align*} \sum_{0<s\leq t}\left[f(Y_s)-f(Y_{s-})\right] & = \int_0^t\int_{(0,\infty)^2}g(s,u_1,u_2)N^U(ds,du) \\ & = \int_0^t\int_{(0,\infty)^2} g(s,u_1,u_2)\left(N^U(ds,du)-\lambda_sP_{J_1}(du_1)\delta_1(du_2)ds\right) \\ &+\int_0^t\int_{(0,\infty)^2} g(s,u_1,u_2)\lambda_sP_{J_1}(du_1)\delta_1(du_2)ds.\end{align*}

Note that

\begin{align*} \int_0^t\int_{(0,\infty)^2} g(s,u_1,u_2)\lambda_sP_{J_1}(du_1)\delta_1(du_2)ds = \int_0^t\int_{(0,\infty)} g(s,u_1,1)\lambda_sP_{J_1}(du_1)ds.\end{align*}

We conclude that

\begin{align*} M(t)-M(0) = &\int_0^t\Big[\partial_tf(Y_{s})-\kappa\partial_xf(Y_{s})(v_s-\bar{v})-\beta\partial_y f(Y_{s})(\lambda_s-\lambda_0) \\ &+\partial_zf(Y_{s})v_s+\frac{1}{2}\partial_{xx}^2f(Y_{s})\sigma^2v_s+\int_{(0,\infty)} g(s,u_1,1)\lambda_sP_{J_1}(du_1)\Big]ds \\ &+\int_0^t\partial_xf(Y_{s-})\sigma\sqrt{v_s}dW_s \\ &+\int_0^t\int_{(0,\infty)} g(s,u_1,1)\left(N^U(ds,du)-\lambda_sP_{J_1}(du_1)ds\right).\end{align*}

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

\begin{align*} \partial_tf(t,x,y,z) & =\left(F^{\prime}(t)+G^{\prime}(t)x+H^{\prime}(t)y\right)f(t,x,y,z), \\ \partial_xf(t,x,y,z) & =G(t)f(t,x,y,z), \\ \partial_y f(t,x,y,z) & =H(t)f(t,x,y,z), \\ \partial_zf(t,x,y,z)& =cf(t,x,y,z).\end{align*}

Furthermore,

\begin{align*} g(s,u_1,1) & =f(s,v_{s-}+\eta u_1,\lambda_{s-}+\alpha,V_s)-f(s,v_{s-},\lambda_{s-},V_s) \\ & = f(s,v_{s-},\lambda_{s-},V_s)\left[\exp\!\left(\eta u_1G(s)\right)\exp\!\left(\alpha H(s)\right)-1\right].\end{align*}

Therefore,

\begin{align*} \int_{(0,\infty)} g(s,u_1,1)\lambda_sP_{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].\end{align*}

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, we have the following:

• Coefficient multiplying v in the drift:

  \begin{align*} f(Y_{t})\left[G^{\prime}(t)-\kappa G(t)+\frac{1}{2}G(t)^2\sigma^2+c\right]=0,\end{align*}
  where we have substituted Equation (3.4).

• Coefficient multiplying $\lambda$ in the drift:

  \begin{align*} 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 ,\end{align*}
  where we have substituted Equation (3.5).

• Free coefficient in the drift:

  \begin{align*} f(Y_{t})\left[F^{\prime}(t)+\kappa\bar{v}G(t)+\beta\lambda_0H(t)\right]=0,\end{align*}
  where we have substituted Equation (3.6).

Therefore, the process M can be written as

\begin{align*} M(t) = & \ M(0) +\int_0^t\partial_vf(Y_{s-})\sigma\sqrt{v_s}dW_s \\ &+\int_0^t\int_{(0,\infty)} g(s,u_1,1)\left(N^U(ds,du)-\lambda_sP_{J_1}(du_1)ds\right).\end{align*}

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

\begin{align*} \mathbb{E}[M(T)] &=\mathbb{E}\!\left[\exp\!\left(F(T)+G(T)v_T+H(T)\lambda_T+cV_T\right)\right] \\ & = \mathbb{E}\!\left[\exp\!\left(c\int_0^Tv_udu\right)\right] \\ & \leq M(0) =\exp\!\left(F(0)+G(0)v_0+H(0)\lambda_0\right) <\infty.\end{align*}

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 an $(\mathcal{F},\mathbb{Q}(a))$ -local martingale.

Theorem 3.1. Let $a\in\mathbb{R}$ and define $\theta_t^{(a)}\,:\!=\,\frac{1}{\sqrt{1-\rho^2}}\left(\frac{\mu_t-r}{\sqrt{v_t}}-a\rho\sqrt{v_t}\right)$ ,

\begin{align*} Y_t^{(a)} &\,:\!=\,\exp\!\left({-}\int_0^t\theta_s^{(a)}dB_s-\frac{1}{2}\int_0^t\big(\theta_s^{(a)}\big)^2ds\right), \\ Z_t^{(a)} &\,:\!=\,\exp\!\left({-}a\int_0^t\sqrt{v_s}dW_s-\frac{1}{2}a^2\int_0^tv_sds\right),\end{align*}

and $X_t^{(a)}\,:\!=\,Y_t^{(a)}Z_t^{(a)}$ . The set

(3.7) \begin{align} \mathcal{E}\,:\!=\,\left\{\mathbb{Q}(a) \hspace{0.2cm} given\ by \hspace{0.2cm} \frac{d\mathbb{Q}(a)}{d\mathbb{P}}=X_T^{(a)} \hspace{0.2cm} with \hspace{0.2cm} |a|<\sqrt{2c_l}\right\}\end{align}

is a set of equivalent local martingale measures.

Proof. Define the process $\big(B^{\mathbb{Q}(a)},W^{\mathbb{Q}(a)}\big)=\big\{\big(B_t^{\mathbb{Q}(a)},W_t^{\mathbb{Q}(a)}\big), t\in[0, T]\big\}$ by

(3.8) \begin{align} dB_t^{\mathbb{Q}(a)}&=dB_t+\theta_t^{(a)}dt, \notag\\ dW_t^{\mathbb{Q}(a)}&=dW_t+a\sqrt{v_t}dt.\end{align}

The dynamics of the stock is now given by

\begin{align*} \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).\end{align*}

Note that

\begin{align*} \mu_t-\sqrt{v_t}\left(\sqrt{1-\rho^2} \theta_t^{(a)}+a\rho \sqrt{v_t}\right)=r,\end{align*}

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 [Reference Wong and Heyde52, Equations (3.4) and (3.7)]. This choice preserves the standard Heston variance dynamics after the change of measure.

To apply Girsanov’s theorem we need to check that the process $X^{(a)}$ is an $(\mathcal{F},\mathbb{P})$ -martingale. Since $X^{(a)}$ is a positive $(\mathcal{F},\mathbb{P})$ -local martingale with $X^{(a)}_0=1$ , it is an $(\mathcal{F},\mathbb{P})$ -supermartingale and it is an $(\mathcal{F},\mathbb{P})$ -martingale if and only if

\begin{align*} \mathbb{E}\!\left[X^{(a)}_T\right]=1.\end{align*}

Using the fact that $Z^{(a)}_T$ is $\mathcal{F}_T^{W}\vee\mathcal{F}_T^L$ -measurable, we have

(3.9) \begin{align}\mathbb{E}\!\left[X^{(a)}_T\right]&=\mathbb{E}\!\left[Y^{(a)}_TZ^{(a)}_T\right] \notag\\&=\mathbb{E}\!\left[\mathbb{E}\!\left[Y^{(a)}_TZ^{(a)}_T|\mathcal{F}_T^{W}\vee\mathcal{F}_T^L\right]\right] \notag\\&=\mathbb{E}\!\left[Z^{(a)}_T\mathbb{E}\!\left[Y^{(a)}_T|\mathcal{F}_T^{W}\vee\mathcal{F}_T^L\right]\right].\end{align}

By Corollary 2.1, the variance process v is strictly positive. This implies that $\int_0^T\big(\theta_s^{(a)}\big)^2ds<\infty$ , $\mathbb{P}$ -almost surely. Since $\theta^{(a)}$ is $\big\{\mathcal{F}_t^{W}\vee\mathcal{F}_t^L\big\}_{t\in[0, T]}$ -adapted,

\begin{align*} Y^{(a)}_T|\mathcal{F}_T^{W}\vee\mathcal{F}_T^L\sim\text{Lognormal}\!\left({-}\frac{1}{2}\int_0^T\big(\theta_s^{(a)}\big)^2ds,\int_0^T\big(\theta_s^{(a)}\big)^2ds\right),\end{align*}

and we obtain that $\mathbb{E}\big[Y^{(a)}_T|\mathcal{F}_T^{W}\vee\mathcal{F}_T^L\big]=1.$ Therefore, substituting this in the last expression in (3.9), we obtain $\mathbb{E}\big[X^{(a)}_T\big]=\mathbb{E}\big[Z^{(a)}_T\big]$ and we only need to check that $\mathbb{E}\big[Z^{(a)}_T\big]=1$ . Since $|a|<\sqrt{2c_l}$ , $\frac{1}{2}a^2< c_l$ , we can apply Proposition 3.1 and get that

(3.10) \begin{align} \mathbb{E}\!\left[\exp\!\left(\frac{1}{2}a^2\int_0^Tv_udu\right)\right]<\infty.\end{align}

Hence, Novikov’s condition is satisfied, $Z^{(a)}$ is an $(\mathcal{F},\mathbb{P})$ -martingale, and we conclude that $\mathbb{E}\big[X^{(a)}_T\big]=\mathbb{E}\big[Z^{(a)}_T\big]=1.$

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

\begin{align*} \frac{d\mathbb{Q}}{d\mathbb{P}}&=X^{(a)}_T, \\ dX^{(a)}_t&=X^{(a)}_t\!\left[-\theta_t^{(a)}dB_t-a\sqrt{v_t}dW_t\right],\end{align*}

and $\big(B^{\mathbb{Q}(a)},W^{\mathbb{Q}(a)}\big)$ as defined in (3.8) is a two-dimensional standard $(\mathcal{F},\mathbb{Q}(a))$ -Brownian motion. The dynamics of the stock under $\mathbb{Q}(a)$ is given by

\begin{align*} \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).\end{align*}

This implies that the discounted stock $\widehat{S}$ is an $(\mathcal{F},\mathbb{Q}(a))$ -local martingale. Therefore, the set $\mathcal{E}$ defined in (3.7) is a set of equivalent local martingale measures.

Observation 3.2. As pointed out at the beginning of this section, since $\frac{1}{2}a^2$ is multiplying the integrated variance in (3.10), it is important that $c_l$ is strictly positive in Lemma 3.1.

Observation 3.3. The dynamics of the variance under $\mathbb{Q}(a)\in\mathcal{E}$ is given by

(3.11) \begin{align} dv_t & =-\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 \notag\\ & =-\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 \notag\\ & =-\kappa^{(a)}\left(v_t-\bar{v}^{(a)}\right)dt+\sigma\sqrt{v_t}dW_t^{\mathbb{Q}(a)}+\eta dL_t,\end{align}

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 martingale 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, an inequality involving the correlation factor $\rho$ also appears in the proof of the existence of equivalent martingales measures in the standard Heston model (see [Reference Wong and Heyde52, Theorem 3.6]).

Theorem 3.2. If $\rho^2<c_l$ , the set

(3.12) \begin{align}\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\}\end{align}

is a set of equivalent martingale measures.

Proof. Let $\mathbb{Q}(a)\in\mathcal{E}_m\subset\mathcal{E}$ . By Theorem 3.1, $\widehat{S}$ is an $(\mathcal{F},\mathbb{Q}(a))$ -local martingale with the following dynamics:

\begin{align*} \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).\end{align*}

Hence,

\begin{align*} \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^tv_sds\right).\end{align*}

Since $\widehat{S}$ is a positive $\left(\mathcal{F},\mathbb{Q}(a)\right)$ -local martingale with $\widehat{S}_0=S_0$ , it is an $\left(\mathcal{F},\mathbb{Q}(a)\right)$ -supermartingale and it is an $\left(\mathcal{F},\mathbb{Q}(a)\right)$ -martingale if and only if $\mathbb{E}^{\mathbb{Q}(a)}\big[\widehat{S}_T\big]=S_0.$

Similarly as we did in Theorem 3.1, we define the processes

\begin{align*} Y_t^{\mathbb{Q}(a)}&:\!=\,\exp\!\left(\sqrt{1-\rho^2}\int_0^t\sqrt{v_s}dB_s^{\mathbb{Q}(a)}-\frac{1-\rho^2}{2}\int_0^tv_sds\right), \\ Z_t^{\mathbb{Q}(a)}&:\!=\,\exp\!\left(\rho\int_0^t\sqrt{v_s}dW_s^{\mathbb{Q}(a)}-\frac{\rho^2}{2}\int_0^tv_sds\right). \end{align*}

Thus, $\widehat{S}_t=S_0Y_t^{\mathbb{Q}(a)} Z_t^{\mathbb{Q}(a)}$ . Using that $Z_T^{\mathbb{Q}(a)}$ is $\mathcal{F}_T^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_T^L$ -measurable, we have

(3.13) \begin{align} \mathbb{E}^{\mathbb{Q}(a)}\big[\widehat{S}_T\big] & =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] \notag \\ & = 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].\end{align}

Since $\int_0^Tv_udu<\infty$ , ${\mathbb{Q}(a)}$ -almost surely, and v is $\big\{\mathcal{F}_t^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_t^L\big\}_{t\in[0, T]}$ -adapted,

\begin{align*} 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^Tv_sds,(1-\rho^2)\int_0^Tv_sds\right),\end{align*}

and we have that $\mathbb{E}^{\mathbb{Q}(a)}\big[Y_T^{\mathbb{Q}(a)}|\mathcal{F}_T^{W^{\mathbb{Q}(a)}}\vee\mathcal{F}_T^L\big]=1.$ Substituting this in the last expression in (3.13), we obtain that $\mathbb{E}^{\mathbb{Q}(a)}\big[\widehat{S}_T\big]=S_0\mathbb{E}^{\mathbb{Q}(a)}\big[Z_T^{\mathbb{Q}(a)}\big],$ and hence we only need to check $\mathbb{E}^{\mathbb{Q}(a)}\big[Z_T^{\mathbb{Q}(a)}\big]=1$ . We will prove that Novikov’s condition holds, that is,

(3.14) \begin{align} \mathbb{E}^{\mathbb{Q}(a)}\left[\exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)\right]<\infty.\end{align}

Unlike in the proof of [Reference Wong and Heyde52, Theorem 3.6], here we cannot directly apply Proposition 3.1 with the volatility parameters $\kappa^{(a)}$ and $\bar{v}^{(a)}$ given in (3.11) to prove that the expectation above is finite. One reason is that under ${\mathbb{Q}(a)}$ the Brownian motion $W^{\mathbb{Q}(a)}$ and the compound Hawkes process L are no longer independent, because

\begin{align*} dW_t^{\mathbb{Q}(a)}=dW_t+a\sqrt{v_t}dt;\end{align*}

therefore, the dynamics of v is different under ${\mathbb{Q}(a)}$ . In order to check (3.14) we note that

\begin{align*} \mathbb{E}^{\mathbb{Q}(a)}\left[\exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)\right]=\mathbb{E}\!\left[\exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)\frac{d\mathbb{Q}(a)}{d\mathbb{P}}\right],\end{align*}

where

\begin{align*} \frac{d\mathbb{Q}(a)}{d\mathbb{P}}=Y_T^{(a)}Z_T^{(a)},\end{align*}

and $Y_T^{(a)}$ and $Z_T^{(a)}$ are given in the statement of Theorem 3.1.

Repeating the same argument as in Theorem 3.1, we can write

\begin{align*} \mathbb{E}^{\mathbb{Q}(a)}\left[\exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)\right] &=\mathbb{E}\!\left[\exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)Y_T^{(a)}Z_T^{(a)}\right] \\ & = \mathbb{E}\!\left[\mathbb{E}\!\left[ \exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)Y_T^{(a)}Z_T^{(a)}\Big|\mathcal{F}_T^W\vee\mathcal{F}_T^L\right]\right] \\ & = \mathbb{E}\!\left[\exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)Z_T^{(a)}\mathbb{E}\!\left[Y_T^{(a)}|\mathcal{F}_T^W\vee\mathcal{F}_T^L\right]\right] \\ & = \mathbb{E}\!\left[\exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)Z_T^{(a)}\right] \\ & = \mathbb{E}\!\left[\exp\!\left({-}a\int_0^T\sqrt{v_s}dW_s-\frac{1}{2}\big(a^2-\rho^2\big)\int_0^Tv_sds\right)\right],\end{align*}

Then, adding and subtracting $a^2\int_0^Tv_sds$ in the exponential and applying the Cauchy–Schwarz inequality, we obtain

(3.15) \begin{align} &\mathbb{E}\!\left[\exp\!\left({-}a\int_0^T\sqrt{v_s}dW_s-\frac{1}{2}\big(a^2-\rho^2\big)\int_0^Tv_sds\right)\right] = \notag \\ &=\mathbb{E}\!\left[\exp\!\left({-}a\int_0^T\sqrt{v_s}dW_s-a^2\int_0^Tv_sds\right)\exp\!\left(\frac{1}{2}\big(a^2+\rho^2\big)\int_0^tv_sds\right)\right] \notag\\ &\leq \mathbb{E}\!\left[\exp\!\left({-}2a\int_0^T\sqrt{v_s}dW_s-2a^2\int_0^Tv_sds\right)\right]^{\frac{1}{2}}\mathbb{E}\!\left[\exp\!\left(\big(a^2+\rho^2\big)\int_0^tv_sds\right)\right]^{\frac{1}{2}}.\end{align}

Note that the first factor in (3.15) is the expectation of a Doléans–Dade exponential. Since $|a|<\frac{\sqrt{2c_l}}{2}$ (recall the choice of a in (3.12)), $2a^2<c_l$ , and by Proposition 3.1, Novikov’s condition holds:

\begin{align*}\mathbb{E}\!\left[\exp\!\left(2a^2\int_0^Tv_sds\right)\right]<\infty.\end{align*}

Therefore, we have that

\begin{align*} \mathbb{E}\!\left[\exp\!\left({-}2a\int_0^T\sqrt{v_s}dW_s-2a^2\int_0^Tv_sds\right)\right]=1.\end{align*}

For the second term in (3.15), we again apply Proposition 3.1. We need $ a^2+\rho^2<c_l,$ which is true because $|a|<\sqrt{c_l-\rho^2}$ . Thus

\begin{align*}\mathbb{E}\!\left[\exp\!\left(\big(a^2+\rho^2\big)\int_0^tv_sds\right)\right]<\infty\end{align*}

and

\begin{align*} \mathbb{E}^{\mathbb{Q}(a)}\left[\exp\!\left(\frac{\rho^2}{2}\int_0^Tv_udu\right)\right]<\infty.\end{align*}

We conclude that $\mathbb{E}^{\mathbb{Q}(a)}\big[Z_T^{\mathbb{Q}(a)}\big]=1$ , $\mathbb{E}^{\mathbb{Q}(a)}\big[\widehat{S}_T\big]=S_0$ , and $\widehat{S}$ is an $(\mathcal{F},{\mathbb{Q}(a)})$ -martingale. Therefore, $\mathcal{E}_m$ is a set of equivalent martingale measures.

4. Application: efficient computation of risk exposures

A common risk management practice is the computation of exposures, such as potential future exposures, expected exposures, and expected positive exposures, among others. The objective of such computations is to estimate the capital that the firm needs to hold in order to manage its risks and comply with economic regulations. Therefore, computing exposures correctly and efficiently is a fundamental risk management procedure that firms need to deal with.

Stein [Reference Stein51] shows that exposures computed under the risk-neutral measure are essentially arbitrary; they must be calculated under the real-world measure. It is proven in [Reference Stein51] that, under the Black–Scholes model, exposures can differ by a factor of two or more across commonly used numéraires and their corresponding risk-neutral measures.

On the other hand, efficient computation of such exposures is relevant for firms to optimize the time and effort used in those calculations. As explained in [Reference Stein50, Reference Stein51], computing exposures under the real-world measure for a portfolio of derivatives can be computationally expensive. The reason is that to compute exposures on derivative portfolios, one simulates risk factors to the horizon date under the real-world measure, and then the portfolio is repriced under a risk-neutral measure. Essentially, this requires performing a Monte Carlo within a Monte Carlo, which can be extremely time-consuming. Moreover, such exposures are usually computed for a variety of horizon times, making the computations even more costly. This is one of the reasons motivating firms (wrongly) to compute exposures under the risk-neutral measure.

However, if a change of measure exists and the Radon–Nikodym derivative of that change is known, such computations can be done in a far more efficient way. Following the explanation in [Reference Stein51], let V be a portfolio of derivatives and consider a risk exposure of the type

(4.1) \begin{align} R=\mathbb{E}\!\left[Y(V_T)\right]\end{align}

for some function Y. Now, assuming that $\rho^2<c_l$ , let $\mathbb{Q}(a)\in\mathcal{E}_m$ be a risk-neutral measure, and let $\frac{d\mathbb{Q}(a)}{d\mathbb{P}}=X_T^{(a)}$ be as given in Theorem 3.1. Then R can be computed in the following way:

(4.2) \begin{align} R=\mathbb{E}\!\left[Y(V_T)\right]=\mathbb{E}^{\mathbb{Q}(a)}\left[Y(V_T)\frac{d\mathbb{P}}{d\mathbb{Q}(a)}\right]=\mathbb{E}^{\mathbb{Q}(a)}\left[Y(V_T)\frac{1}{X_T^{(a)}}\right].\end{align}

The exposure R can often be calculated much more efficiently using the expression on the right-hand side of (4.2) than using (4.1), because the calculations can be done entirely under $\mathbb{Q}(a)$ . Monte Carlo techniques such as least-squares Monte Carlo [Reference Longstaff and Schwartz42], stochastic mesh [Reference Broadie and Glasserman10], or stochastic grid [Reference Jain and Oosterlee33] can be used to compute $V_T$ , and the same scenarios can used to obtain the expression on the right-hand side of (4.2) [Reference Cesari13].

To summarize, an important practical application of our results to risk management is the correct and efficient computation of exposures on derivative portfolios, which is possible thanks to the existence of risk-neutral measures and an explicit expression for the Radon–Nikodym derivative.

4.1. Numerical example

We give a numerical simulation in which we quantify the error obtained if, under our stochastic volatility model, the exposures are wrongly computed under a risk-neutral measure. The objective is to illustrate the importance of computing exposures under the real-world measure. Assume that $\rho^2<c_l$ . Let $\mathbb{Q}(a)\in\mathcal{E}_m$ , and let V be a portfolio consisting of a single call option with strike K and maturity time T, that is,

\begin{align*} V_t=e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}(a)}\left[\max\{S_T-K,0\}|\mathcal{F}_t\right], \hspace{1cm} t\in[0, T].\end{align*}

Following [Reference Stein51], the expected exposure at time $t\in[0, T]$ is defined by $EE(V,t)\,:\!=\,\mathbb{E}\!\left[\max\{V_t,0\}\right]$ . For the sake of simplicity, we compute the expected exposure at maturity time T. Then $EE(V,T)=\mathbb{E}\!\left[\max\{S_T-K,0\}\right]$ , and we define the (wrong) expected exposure under $\mathbb{Q}(a)$ by $EE^{(a)}(V,T)=\mathbb{E}^{\mathbb{Q}(a)}\left[\max\{S_T-K,0\}\right]$ .

We employ an Euler–Maruyama scheme to approximate the processes $\lambda$ , v, and S; we then run a Monte Carlo simulation to compute EE(V, T) and $EE^{(a)}(V,T)$ for different strikes and values of a. For convenience, we assume that the drift $\mu$ of S is constant and that $J_1\sim\text{Exponential}(k)$ . The parameters used for the simulation are given in Table 1; we have taken as a reference the ones given in [Reference Kokholm and Stisen41, Table IV].

Table 1. Parameters used for the Monte Carlo simulation.

One can check that the stability condition for the Hawkes process, the Feller condition, and the condition $\rho^2<c_l$ are satisfied. In Figure 1 we compare EE(V, T) and $EE^{(0)}(V,T)$ for different strikes, and in Figure 2 we make the same comparison for EE(V, T) and $EE^{(1)}(V,T)$ . In Table 2 we give the average and maximum relative errors between the exposures computed under risk-neutral measures and the correct exposures across the different strikes. Note that the relative errors given in Table 2 are of significant size for risk management purposes. Finally, in Figure 3 we compute $EE^{(a)}(V,T)$ for different values of a and for a fixed strike value $K=1.4$ . We also plot the correct value and the relative error.

Figure 1. Expected exposure computed under $\mathbb{P}$ and under $\mathbb{Q}(0)$ as functions of the strike for a call option.

Table 2. Average and maximum relative errors between exposures computed under risk-neutral measures and exposures computed under the real-world measure.

Figure 2. Expected exposure computed under $\mathbb{P}$ and under $\mathbb{Q}(1)$ as functions of the strike for a call option.

Figure 3. Expected exposure computed under $\mathbb{Q}(a)$ for different values of a and a fixed strike value $K=1.4$ .

It is worth mentioning that it would be of interest to compute EE(V, t) for $t\in[0,T)$ and other types of exposures under our stochastic volatility model for more complex derivative portfolios. Nevertheless, the goal is to show that even with a simple portfolio and exposure one can obtain significantly different exposure values if they are wrongly computed under the risk-neutral measure.