# Thiele’s PIDE for unit-linked policies in the 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

The main purpose of the paper is to derive Thiele’s differential equation for
unit-linked policies in the Heston-Hawkes stochastic volatility model
presented in [13]. This model is an extension of the well-known Heston model
that incorporates the volatility clustering feature by adding a compound
Hawkes process in the volatility. Since the model is arbitrage-free, pricing
unit-linked policies via the equivalence principle under $\mathbb{Q}$ is
possible. Some integrability conditions are checked and a suitable family of
risk neutral probability measures is found to obtain Thiele’s differential
equation. The established and practical method to compute reserves in life
insurance is by solving Thiele’s equation, which is crucial to guarantee the
solvency of the insurance company.

Keywords: Thiele’s equation, reserve, unit-linked policy, life insurance
policy, equivalence principle, stochastic volatility, risk neutral measure,
Hawkes process, volatility with jumps.

AMS classification MSC2020: 60G55, 60H30, 91G05, 91G15.

##  1 Introduction

A life insurance policy is a contract that defines the evolution of the cash
flow between the insurance company and the insured. In traditional life
insurance contracts, the benefit provided to the insured if the pre-agreed
conditions are met is fixed. In contrast, the distinctive property of unit-
linked life insurance policies is that the benefit is linked to the market
value of some fixed portfolio composed of stocks, bonds or other financial
assets. Consequently, two types of risk prevail in unit-linked policies, the
mortality risk that controls the future flow of payments between the two
parties and the financial risk arising from the future returns on the
financial investments.

The computation of the so-called reserve is the basis to guarantee the
solvency of the insurance company against the two aforementioned risks.
Namely, the reserve is the present value of future potential payments from the
insurance company to the insured and is the ordinary procedure to determine
the premiums. The established method to calculate the reserve is by solving
the celebrated Thiele’s differential equation, which dates back to 1875. It
was first published in the obituary on Thiele in [31] and presented in the
scientific paper [38] in 1913.

Literature on pricing unit-linked policies is vast and we mention here a non-
exhaustive list. Brennan and Schwartz [16, 17] and Boyle and Schwartz [15]
conducted, to the best of our knowledge, the first studies of unit-linked
contracts using modern financial techniques. Delbaen [26] and Bacinello and
Ortu [7] examined unit-linked contracts employing the martingale-based theory
developed by Harrison and Kreps [33]. See also [1] where Aase and Persson
revisit the principle of equivalence under a risk neutral probability measure
for the pricing of unit-linked policies after presenting some historical
context on the topic.

It is acknowledged in life insurance that in the case of long-term insurance
contracts interest rate risk plays an important role. To incorporate that risk
into the valuation of life insurance policies, Norberg and Møller obtained
Thiele’s differential equation assuming stochastic interest rates in [41] and,
later on, Persson derived the risk adjusted version of it in [45]. For more
fundamental references about interest rate risk in unit-linked policies see
[44, 8, 10, 40] and some further extensions and generalizations [14, 3, 9,
12].

It is worth pointing out that also policyholder behaviour has a significant
impact in life insurance. For instance, the insured may surrender the
contract, cancel all future cash flows and receive just a single payment.
Similarly, the policyholder may cancel the future premiums and accept a
reduction in the benefits. See [29, 6, 18, 19, 25, 36, 32, 46] for references
where behavioural risk in life insurance policies is studied.

The scope of this paper is to study the financial risk in unit-linked policies
coming from stock returns. Precisely, we derive Thiele’s differential equation
under the assumption that the benefit is linked to a stock that follows the
Heston-Hawkes stochastic volatility model presented in [13]. This model is an
extension of the well-known Heston model that incorporates the volatility
clustering feature by adding a compound Hawkes process component in the
volatility process. A Hawkes process is a self-exciting point process
introduced in 1917 by Hawkes, see [34, 35], with many applications in high-
frequency finance, insurance, seismology and other disciplines. It is worth
mentioning that there is notable literature supporting the inclusion of jumps
in the volatility, for instance, see [42, 27, 28, 22]. By exploiting the
tractability of the model, it is shown in [13] that the Heston-Hawkes model is
arbitrage-free and incomplete. Furthermore, the passage from the historical
probability to the risk neutral one is made explicit via a family of
equivalent martingale measures.

It is a common practice in actuarial science to consider two independent
probability spaces, one to describe the states of the insured and the other
one to describe the evolution of the financial investments. As explained in
[1], the right methodology to price unit-linked policies is through the
principle of equivalence under $\mathbb{Q}$, a risk neutral probability
measure in the financial probability space. On account of that, the arbitrage-
free property of the stochastic volatility model proven in [13] is crucial and
required to make sense of the pricing of unit-linked policies.

The paper is organized as follows. In Section 2 we summarize the Heston-Hawkes
stochastic volatility model presented in [13]. Then, we give some preliminary
results in Section 3 that are needed to obtain Thiele’s differential equation.
Essentially, there are several technical results regarding the compensator of
the Hawkes process under the risk neutral probability measure. The proofs are
postponed in the Appendix for the sake of clarity. Finally, in Section 4 we
obtain the desired Thiele’s differential equation for unit-linked policies
under this stochastic volatility model.

##  2 Stochastic volatility model

In this section, we outline the Heston-Hawkes stochastic volatility model
given in [13]. 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,

$\displaystyle d\lambda_{t}=-\beta(\lambda_{t}-\lambda_{0})dt+\alpha dN_{t},$
(2.1)

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

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

holds. See [11, Section 2] and [37, Section 3.1.1] for the definition of $N$.
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 assume that $(B,W),N$ and $\\{J_{i}\\}_{i\geqslant 1}$ 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

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

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.

We assume that the interest rate is deterministic and constant equal to $r$.
Finally, with all these ingredients, we introduce the Heston-Hawkes model. The
stock price $S=\\{S_{t},t\in[0,T]\\}$ and its variance
$v=\\{v_{t},t\in[0,T]\\}$ are given by

$\displaystyle\frac{dS_{t}}{S_{t}}$
$\displaystyle=\mu_{t}dt+\sqrt{v_{t}}\left(\sqrt{1-\rho^{2}}dB_{t}+\rho
dW_{t}\right),$ (2.2) $\displaystyle dv_{t}$
$\displaystyle=-\kappa\left(v_{t}-\bar{v}\right)dt+\sigma\sqrt{v_{t}}dW_{t}+\eta
dL_{t},$ (2.3)

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 [4, Proposition 1.2.15].
For more details and further results on the model see [13].

##  3 Preliminary results

Some preliminary results are needed to derive Thiele’s PIDE under this
stochastic volatility model. First, we summarize results proven in [13]
regarding the existence and positivity of the variance process and the change
of measure. Then, we study the existence of (positive and negative) moments of
the variance and (positive) moments of the Radon–Nikodym derivative of the
change of measure. To conclude the section, we prove that the compensator of
the Hawkes process is the same under the historic and the risk neutral
probability measures. At first glance, this may seem a straightforward result.
However, independence between the Hawkes process and the Radon-Nikodym
derivative is not obvious since the counting process is part of the latter. As
a consequence, this result is not clear and needs to be proven requiring the
aforementioned integrability results. For the sake of clarity, all the proofs
of this section are postponed to the Appendix.

###  3.1 Arbitrage-free property

Existence and positivity of the variance process are given in Proposition 3.1
and Proposition 3.2, respectively. All the results involving the change of
measure are compacted in Theorem 3.4.

######  Proposition 3.1.

Equation (2.3) has a pathwise unique strong solution.

###### Proof.

See [13, Proposition 2.1]. ∎

######  Proposition 3.2.

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

$\displaystyle\widetilde{v}_{t}=v_{0}-\kappa\int_{0}^{t}\left(\widetilde{v}_{s}-\bar{v}\right)ds+\sigma\int_{0}^{t}\sqrt{\widetilde{v}_{s}}dW_{s}.$
(3.1)

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 (2.3). In particular, $v$
is a strictly positive process.

###### Proof.

See [13, Proposition 2.2]. ∎

######  Assumption 1.

There exists $\epsilon_{J}>0$ such that the moment generating function
$M_{J}(t)=\mathbb{E}[\exp(tJ_{1})]$ of $J_{1}$ is well defined in
$(-\infty,\epsilon_{J})$. Moreover, $(-\infty,\epsilon_{J})$ 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.

######  Proposition 3.3.

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, $c_{l}>0$ and for $c<c_{l}$

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

###### Proof.

See [13, Lemma 3.1] and [13, Proposition 3.5]. ∎

######  Theorem 3.4.

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

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

and $X_{t}^{(a)}:=Y_{t}^{(a)}Z_{t}^{(a)}$. Recall the definition of $c_{l}$ in
Proposition 3.3.

  1. 1.

$X^{(a)}$ is a $(\mathcal{F},\mathbb{P})$-martingale for $|a|<\sqrt{2c_{l}}$.

  2. 2.

The set

$\displaystyle\mathcal{E}:=\left\\{\mathbb{Q}(a)\hskip 5.69046pt\text{given
by}\hskip 5.69046pt\frac{d\mathbb{Q}(a)}{d\mathbb{P}}=X_{T}^{(a)}\hskip
5.69046pt\text{with}\hskip 5.69046pt|a|<\sqrt{2c_{l}}\right\\}$ (3.2)

is a set of equivalent local martingale measures.

  3. 3.

Let $\mathbb{Q}(a)\in\mathcal{E}$, the process
$(B^{\mathbb{Q}(a)},W^{\mathbb{Q}(a)})$ defined by

$\displaystyle dB_{t}^{\mathbb{Q}(a)}$
$\displaystyle:=dB_{t}+\theta_{t}^{(a)}dt,$ $\displaystyle
dW_{t}^{\mathbb{Q}(a)}$ $\displaystyle:=dW_{t}+a\sqrt{v_{t}}dt$ (3.3)

is a a two-dimensional standard $(\mathcal{F},\mathbb{Q}(a))$-Brownian motion.

  4. 4.

Let $\mathbb{Q}(a)\in\mathcal{E}$, the dynamics of $S$ and $v$ are given by

$\displaystyle\frac{dS_{t}}{S_{t}}$
$\displaystyle=rdt+\sqrt{v_{t}}\left(\sqrt{1-\rho^{2}}dB_{t}^{\mathbb{Q}(a)}+\rho
dW_{t}^{\mathbb{Q}(a)}\right),$ (3.4) $\displaystyle dv_{t}$
$\displaystyle=-\kappa^{(a)}(v_{t}-\bar{v}^{(a)})dt+\sigma\sqrt{v_{t}}dW_{t}^{\mathbb{Q}(a)}+\eta
dL_{t}.$ (3.5)

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

  5. 5.

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

$\displaystyle\mathcal{E}_{m}:=\left\\{\mathbb{Q}(a)\in\mathcal{E}:|a|<\min\left\\{\frac{\sqrt{2c_{l}}}{2},\sqrt{c_{l}-\rho^{2}}\right\\}\right\\}$
(3.6)

is a set of equivalent martingale measures.

###### Proof.

See [13, Theorem 3.6], [13, Observation 3.8] and [13, Theorem 3.9]. ∎

###  3.2 Moments of the variance and the Radon-Nikodym derivative

First, we prove the existence and integrability of all positive moments of the
variance process. Essentially, this is a consequence of the fact that all
positive moments of the standard Heston variance and the compound Hawkes
process exist (see Lemma A.1 in the Appendix).

######  Lemma 3.5.

Let $s\geqslant 1$. Then, $\mathbb{E}\left[v_{t}^{s}\right]<\infty$ for all
$t\in[0,T]$ and $\int_{0}^{T}\mathbb{E}\left[v_{t}^{s}\right]dt<\infty$.

###### Proof.

See Lemma A.2 in the Appendix. ∎

Now, we check the existence and integrability of the $s$th negative moment of
the variance process under the condition $2\kappa\bar{v}>s\sigma^{2}$. Note
that the larger the $s$, the larger the fraction
$\frac{\kappa\bar{v}}{\sigma^{2}}$ must be, that is, the product of the mean
reversion speed and the long-term variance divided by the volatility of the
variance. Informally, the larger the fraction
$\frac{\kappa\bar{v}}{\sigma^{2}}$ is, the less likely is that the variance
will approach zero and that contributes to the existence of negative moments.

######  Lemma 3.6.

Let $s\geqslant 1$. If $2\kappa\bar{v}>s\sigma^{2}$,
$\mathbb{E}\left[\frac{1}{v_{t}^{s}}\right]<\infty$ for all $t\in[0,T]$ and
$\int_{0}^{T}\mathbb{E}\left[\frac{1}{v_{t}^{s}}\right]dt<\infty$.

###### Proof.

See Lemma A.4 in the Appendix. ∎

In addition to the positive and negative moments of the variance, we study the
existence of positive moments of the Radon-Nykodim derivative
$\frac{d\mathbb{Q}(a)}{d\mathbb{P}}=X_{T}^{(a)}$, given in Theorem 3.4. The
proof boils down to check that expectations of the type

$\displaystyle\mathbb{E}\left[\exp\left(A\int_{0}^{T}\frac{1}{v_{u}}du\right)\right]\hskip
8.5359pt\text{and}\hskip
8.5359pt\mathbb{E}\left[\exp\left(B(a)\int_{0}^{T}v_{u}du\right)\right]$

are finite, where $A$ is a constant independent of $a$ and $B(a)$ is a
constant depending on $a$. To ensure that the first expectation is finite we
require the two following conditions on the model parameters
$2\kappa\bar{v}>\sigma^{2}$ and
$\frac{1-\rho^{2}}{D(s^{2}-s)}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}>1$
(see Lemma A.5 in the Appendix) where $s>1$ is the moment that we want to
study. To check that the second expectation is finite we use Proposition 3.3
obtaining the two conditions on $a$ in (3.7).

######  Lemma 3.7.

Let $\mathbb{Q}(a)\in\mathcal{E}$, $s>1$,
$D:=\sup_{t\in[0,T]}(\mu_{t}-r)^{2}<\infty$ and $X^{(a)}$ defined in Theorem
3.4. Assume that $2\kappa\bar{v}>\sigma^{2}$ and
$\frac{1-\rho^{2}}{D(s^{2}-s)}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}>1$.
Consider $q_{2}$ such that
$1<q_{2}<\frac{1-\rho^{2}}{D(s^{2}-s)}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}$
and define $q_{1}:=\frac{q_{2}}{q_{2}-1}>1$. If

$\displaystyle|a|<\min\Bigg{\\{}\frac{1}{q_{1}s}\sqrt{\frac{c_{l}}{2}},\sqrt{\frac{(1-\rho^{2})c_{l}}{q_{1}s\left[2q_{1}s(1-\rho^{2})+\rho^{2}s-1\right]}}\Bigg{\\}},$
(3.7)

then

$\displaystyle\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{s}\right]<\infty\hskip
14.22636pt\text{and}\hskip
14.22636pt\mathbb{E}\left[\left(X_{t}^{(a)}\right)^{s}\right]\leqslant\left(\frac{s}{s-1}\right)^{s}\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{s}\right]<\infty,$

for all $t\in[0,T]$.

###### Proof.

See Lemma A.6 in the Appendix. ∎

######  Observation 3.8.

Using that $q_{1},s>1$ and that $\rho^{2}<1$ one can check that the expression
$q_{1}s\left[2q_{1}s(1-\rho^{2})+\rho^{2}s-1\right]$ appearing in the second
expression inside the minimum (3.7) is strictly positive. Indeed,

$\displaystyle 2q_{1}s(1-\rho^{2})+\rho^{2}s-1$
$\displaystyle>2(1-\rho^{2})+\rho^{2}-1=1-\rho^{2}>0.$

The existence of the two following moments is required to prove that the
compensator of $N$ is the same under the historic probability measure and a
suitable family of risk neutral probability measures:

$\displaystyle\mathbb{E}[(X_{T}^{(a)})^{2+\varepsilon_{1}}]<\infty\hskip
8.5359pt\text{and}\hskip
8.5359pt\mathbb{E}\left[\left(\frac{1}{v_{t}}\right)^{1+\varepsilon_{2}}\right]<\infty,$

where $\varepsilon_{1},\varepsilon_{2}>0$ are arbitrarily small. To ensure
that the previous expectations are finite we assume some conditions on the
model parameters and define a suitable family of risk neutral probability
measures.

######  Assumption 2.

We fix $\varepsilon_{1},\varepsilon_{2}>0$ and assume that

  1. 1.

$\rho^{2}<c_{l}$.

  2. 2.

$\frac{1-\rho^{2}}{D\left[(2+\varepsilon_{1})^{2}-(2+\varepsilon_{1})\right]}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}>1$.

  3. 3.

$2\kappa\bar{v}>(1+\varepsilon_{2})\sigma^{2}$.

######  Definition 3.9.

Let $q,s>1$, we define a subset of $\mathcal{E}_{m}$ by

$\displaystyle\mathcal{E}_{m}(q,s):=\left\\{\mathbb{Q}(a)\in\mathcal{E}_{m}:|a|<\min\Bigg{\\{}\frac{1}{qs}\sqrt{\frac{c_{l}}{2}},\sqrt{\frac{(1-\rho^{2})c_{l}}{qs\left[2qs(1-\rho^{2})+\rho^{2}s-1\right]}}\Bigg{\\}}\right\\}.$

From now on, we fix $Q_{2}$ such that
$1<Q_{2}<\frac{1-\rho^{2}}{D\left[(2+\varepsilon_{1})^{2}-(2+\varepsilon_{1})\right]}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}$
and $Q_{1}:=\frac{Q_{2}}{Q_{2}-1}>1$.

######  Observation 3.10.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$. By Lemma 3.6,
Lemma 3.7 and Assumption 2 the following holds

$\displaystyle\mathbb{E}[(X_{T}^{(a)})^{2+\varepsilon_{1}}]<\infty,\hskip
8.5359pt\mathbb{E}\left[\left(\frac{1}{v_{t}}\right)^{1+\varepsilon_{2}}\right]<\infty\hskip
8.5359pt\text{and}\hskip
8.5359pt\int_{0}^{T}\mathbb{E}\left[\left(\frac{1}{v_{t}}\right)^{1+\varepsilon_{2}}\right]dt<\infty.$

###  3.3 Compensator of $N$ under the historic and the risk neutral measures

First, we give the compensators of $N$ and $L$ under $\mathbb{P}$. By
definition, the compensator $\Lambda^{N}$ is a $\mathcal{F}^{N}$-predictable
process such that $N-\Lambda^{N}$ is a $(\mathcal{F}^{N},\mathbb{P})$-local
martingale. Using the independence between $(B,W)$ and $N$ one can prove that
martingale property of $N-\Lambda^{N}$ is conserved when considering the joint
filtration
$\mathcal{F}=\\{\mathcal{F}_{t}=\mathcal{F}_{t}^{(B,W)}\vee\mathcal{F}_{t}^{L},t\in[0,T]\\}$.

######  Lemma 3.11.

The following holds

  1. 1.

Define $\Lambda^{N}_{t}:=\int_{0}^{t}\lambda_{u}du$, then $N-\Lambda^{N}$ is a
square integrable $(\mathcal{F},\mathbb{P})$-martingale.

  2. 2.

Define $\Lambda^{L}_{t}:=\mathbb{E}[J_{1}]\int_{0}^{t}\lambda_{u}du$, then
$L-\Lambda^{L}$ is a square integrable $(\mathcal{F},\mathbb{P})$-martingale.

###### Proof.

See Lemma A.7 in the Appendix. ∎

Finally, we prove that the compensators of $N$ and $L$ under
$\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$ are the same than
under $\mathbb{P}$. As previously mentioned, this would be a straightforward
result if there was independence between the Hawkes procces $N$ and the Radon-
Nykodim derivative $X^{(a)}$. However, this is not clear and we require the
existence of the moments in Observation 3.10.

######  Proposition 3.12.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$, then

  1. 1.

$N-\Lambda^{N}$ is a $(\mathcal{F},\mathbb{Q}(a))$-martingale.

  2. 2.

$L-\Lambda^{L}$ is a $(\mathcal{F},\mathbb{Q}(a))$-martingale.

###### Proof.

See Proposition A.9 in the Appendix. ∎

##  4 Derivation of Thiele’s PIDE for unit-linked policies

The objective of this section is to derive Thiele’s differential equation for
unit-linked policies under the Heston-Hawkes stochastic volatility model.
First, we find the drift of a process of the form $t\to
Z(t,S_{t},v_{t},\lambda_{t})$ under a risk neutral probability measure, where
$Z$ is a regular enough function. To do so, the compensator of $N$ under
$\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$ is needed. In order
to lighten the notation, we first define some space of functions. From now on,
$\mathbb{R}_{+}:=(0,\infty)$.

######  Definition 4.1.

We define $\mathcal{D}:=\mathbb{R}_{+}^{2}\times[\lambda_{0},\infty)$ and
$\mathcal{C}^{1,2}:=\mathcal{C}^{1,2}\left([0,T]\times\mathcal{D}\right)$ the
space of functions $Y\colon[0,T]\times\mathcal{D}\to\mathbb{R}_{+}$ that are
jointly continuous, continuously differentiable on the first variable, twice
continuously differentiable on the last three variables and all derivatives
are jointly continuous.
We define
$\mathcal{C}^{0,1,2}:=\mathcal{C}^{0,1,2}\left([0,T]^{2}\times\mathcal{D}\right)$
the space of functions $Z\colon[0,T]^{2}\times\mathcal{D}\to\mathbb{R}_{+}$
that are jointly continuous, continuous on the first variable, continuously
differentiable on the second variable, twice continuously differentiable on
the last three variables and all derivatives are jointly continuous.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}$ and
$\varphi\colon[0,T]\times\mathbb{R}_{+}\to\mathbb{R}_{+}$ a payoff such that
$\mathbb{E}^{\mathbb{Q}(a)}[|\varphi(s,S_{s})|]<\infty$ for all $s\in[0,T]$.
Recall that the price at time $t\in[0,T]$ of the payoff function
$\varphi(s,S_{s})$ with maturity $s\in[t,T]$ is given by
$e^{-r(s-t)}\mathbb{E}^{\mathbb{Q}(a)}[\varphi(s,S_{s})|\mathcal{F}_{t}]$. One
example of such payoff is $\varphi(s,S_{s})=\max\\{G,S_{s}\\}$, where $G$ is
called the guaranteed amount. This means that at time $s$ the insured is paid
the maximum between the guaranteed amount and the stock.

As a consequence of the Markov property of the process $(N,\lambda)$, see [39,
Remark 1.22], we prove in the next lemma that
$\mathbb{E}^{\mathbb{Q}(a)}[\varphi(s,S_{s})|\mathcal{F}_{t}]$ is a
deterministic function of the joint process $(t,S_{t},v_{t},\lambda_{t})$. Due
to the presence of jumps in $v$ and $\lambda$, such function is the solution
of a partial integro-differential equation, PIDE from now on. By applying
Itô’s formula and using the compensators of $N$ and $L$ under
$\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$ we obtain that
PIDE.

######  Lemma 4.2.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$,
$\varphi\colon[0,T]\times\mathbb{R}_{+}\to\mathbb{R}_{+}$ such that
$\mathbb{E}^{\mathbb{Q}(a)}[|\varphi(s,S_{s})|]<\infty$ for all $s\in[0,T]$.
Then, there exists a function
$Z^{\varphi,a}\colon[0,T]^{2}\times\mathcal{D}\to\mathbb{R}_{+}$ such that

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}[\varphi(s,S_{s})|\mathcal{F}_{t}]=Z_{s}^{\varphi,a}(t,S_{t},v_{t},\lambda_{t}).$

where $s,t\in[0,T]$. Note that $Z^{\varphi,a}_{s}(t,x,y,z)=\varphi(s,x)$ for
$t\in[s,T]$.

Furthermore, fix $s\in[0,T]$, if $Z_{s}^{\varphi,a}\in\mathcal{C}^{1,2}$, it
satisfies the following PIDE

$\displaystyle\partial_{t}Z^{\varphi,a}_{s}(t,x,y,z)+rx\partial_{x}Z^{\varphi,a}_{s}(t,x,y,z)-\kappa^{(a)}(y-\bar{v}^{(a)})\partial_{y}Z^{\varphi,a}_{s}(t,x,y,z)$
$\displaystyle-\beta(z-\lambda_{0})\partial_{z}Z^{\varphi,a}_{s}(t,x,y,z)+\frac{1}{2}x^{2}y\partial_{xx}^{2}Z^{\varphi,a}_{s}(t,x,y,z)+\frac{1}{2}\sigma^{2}y\partial_{yy}^{2}Z^{\varphi,a}_{s}(t,x,y,z)$
$\displaystyle+\sigma\rho
xy\partial_{xy}^{2}Z^{\varphi,a}_{s}(t,x,y,z)+z\int_{(0,\infty)}\left[Z^{\varphi,a}_{s}(t,x,y+\eta
u,z+\alpha)-Z^{\varphi,a}_{s}(t,x,y,z)\right]P_{J_{1}}(du)=0,$ (4.1)

for $(t,x,y,z)\in[0,s]\times\mathcal{D}$ and final condition
$Z^{\varphi,a}_{s}(s,x,y,z)=\varphi(s,x)$.

###### Proof.

By definition $\mathcal{F}_{t}=\mathcal{F}_{t}^{(B,W)}\vee\mathcal{F}_{t}^{L}$
and since $S$ and $v$ are strong solutions, we have
$\mathcal{F}_{t}^{(B,W)}\vee\mathcal{F}_{t}^{L}=\mathcal{F}_{t}^{S}\vee\mathcal{F}_{t}^{v}$.
Moreover, since $\mathcal{F}_{t}^{\lambda}\subset\mathcal{F}_{t}^{v}$ we have
that
$\mathcal{F}_{t}=\mathcal{F}_{t}^{S}\vee\mathcal{F}_{t}^{v}\vee\mathcal{F}_{t}^{\lambda}$.
Since $(N,\lambda)$ is a Markov process, see [39, Remark 1.22],
$(S,v,\lambda)$ is also a Markov process. We conclude that there exists a
function $Z^{\varphi,a}\colon[0,T]^{2}\times\mathcal{D}\to\mathbb{R}_{+}$ such
that

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}[\varphi(s,S_{s})|\mathcal{F}_{t}]=Z^{\varphi,a}_{s}(t,S_{t},v_{t},\lambda_{t}),$

where $s,t\in[0,T]$.

Moreover, fix $s\in[0,T]$, if $Z_{s}^{\varphi,a}\in\mathcal{C}^{1,2}$ we can
apply Itô formula to the process $t\to
Z^{\varphi,a}_{s}(t,S_{t},v_{t},\lambda_{t})$ for $t\in[0,s]$. For
convenience, we define $Y_{t}:=(t,S_{t},v_{t},\lambda_{t})$. Applying Itô
formula to $Z^{\varphi,a}_{s}$ we get

$\displaystyle Z^{\varphi,a}_{s}(Y_{t})=\ $ $\displaystyle
Z^{\varphi,a}_{s}(Y_{0})+\int_{0}^{t}\partial_{t}Z^{\varphi,a}_{s}(Y_{u-})du+\int_{0}^{t}\partial_{x}Z^{\varphi,a}_{s}(Y_{u-})dS_{u}+\int_{0}^{t}\partial_{y}Z^{\varphi,a}_{s}(Y_{u-})dv_{u}$
$\displaystyle+\int_{0}^{t}\partial_{z}Z^{\varphi,a}_{s}(Y_{u-})d\lambda_{u}+\frac{1}{2}\int_{0}^{t}\partial_{xx}^{2}Z^{\varphi,a}_{s}(Y_{u-})d[S]_{u}+\frac{1}{2}\int_{0}^{t}\partial_{yy}^{2}Z^{\varphi,a}_{s}(Y_{u-})d[v]_{u}^{\text{c}}$
$\displaystyle+\frac{1}{2}\int_{0}^{t}\partial_{zz}^{2}Z^{\varphi,a}_{s}(Y_{u-})d[\lambda]_{u}^{\text{c}}+\int_{0}^{t}\partial_{xy}^{2}Z^{\varphi,a}_{s}(Y_{u-})d[S,v]_{u}^{\text{c}}+\int_{0}^{t}\partial_{xz}^{2}Z^{\varphi,a}_{s}(Y_{u-})d[S,\lambda]_{u}^{\text{c}}$
$\displaystyle+\int_{0}^{t}\partial_{yz}^{2}Z^{\varphi,a}_{s}(Y_{u-})d[v,\lambda]_{u}^{\text{c}}$
$\displaystyle+\sum_{0<u\leqslant
t}\left[Z^{\varphi,a}_{s}(Y_{u})-Z^{\varphi,a}_{s}(Y_{u-})-\partial_{y}Z_{s}^{\varphi,a}(Y_{u-})\Delta
v_{u}-\partial_{z}Z_{s}^{\varphi,a}(Y_{u-})\Delta\lambda_{u}\right].$ (4.2)

Recall that the dynamics of $S$ is given in (3.4), the dynamics of $v$ in
(3.5) and the dynamics of $\lambda$ in (2.1). Then,

$\displaystyle d[S]_{u}$ $\displaystyle=S_{u}^{2}v_{u}du,$ $\displaystyle
d[v]_{u}$ $\displaystyle=\sigma^{2}v_{u}du+\eta^{2}d[L]_{u}\implies
d[v]_{u}^{\text{c}}=\sigma^{2}v_{u}du,$ $\displaystyle[\lambda]_{u}$
$\displaystyle=\alpha^{2}[N]_{u}=\alpha^{2}N_{u}\implies[\lambda]_{u}^{\text{c}}=0,$
$\displaystyle d[S,v]_{u}$ $\displaystyle=\sigma\rho S_{u}v_{u}du,$
$\displaystyle[S,\lambda]_{u}$ $\displaystyle=0,$
$\displaystyle[v,\lambda]_{u}$ $\displaystyle=\alpha\eta[L,N]_{u}=\alpha\eta
L_{u}\implies[v,\lambda]_{u}^{\text{c}}=0.$

Note that $\Delta v_{u}=\eta\Delta L_{u}$ and $\Delta\lambda_{u}=\alpha\Delta
N_{u}$. Replacing everything in (4) we have

$\displaystyle Z^{\varphi,a}_{s}(Y_{t})=\ $ $\displaystyle
Z^{\varphi,a}_{s}(Y_{0})+\int_{0}^{t}\Big{[}\partial_{t}Z^{\varphi,a}_{s}(Y_{u})+rS_{u}\partial_{x}Z^{\varphi,a}_{s}(Y_{u})-\kappa^{(a)}(v_{u}-\bar{v}^{(a)})\partial_{y}Z^{\varphi,a}_{s}(Y_{u})$
$\displaystyle-\beta(\lambda_{u}-\lambda_{0})\partial_{z}Z^{\varphi,a}_{s}(Y_{u})+\frac{1}{2}S_{u}^{2}v_{u}\partial_{xx}^{2}Z^{\varphi,a}_{s}(Y_{u})+\frac{1}{2}\sigma^{2}v_{u}\partial_{yy}^{2}Z^{\varphi,a}_{s}(Y_{u})$
$\displaystyle+\sigma\rho
S_{u}v_{u}\partial_{xy}^{2}Z_{s}^{\varphi,a}(Y_{u})\Big{]}du+\sqrt{1-\rho^{2}}\int_{0}^{t}S_{u}\sqrt{v_{u}}\partial_{x}Z^{\varphi,a}_{s}(Y_{u-})dB_{u}^{\mathbb{Q}(a)}$
$\displaystyle+\int_{0}^{t}\left[\rho
S_{u}\sqrt{v_{u}}\partial_{x}Z^{\varphi,a}_{s}(Y_{u-})+\sigma\sqrt{v_{u}}\partial_{y}Z^{\varphi,a}_{s}(Y_{u-})\right]dW_{u}^{\mathbb{Q}(a)}$
$\displaystyle+\eta\int_{0}^{t}\partial_{y}Z^{\varphi,a}_{s}(Y_{u-})dL_{u}+\alpha\int_{0}^{t}\partial_{z}Z^{\varphi,a}_{s}(Y_{u-})dN_{u}$
$\displaystyle+\sum_{0<u\leqslant
t}\left[Z^{\varphi,a}_{s}(Y_{u})-Z^{\varphi,a}_{s}(Y_{u-})-\eta\partial_{y}Z_{s}^{\varphi,a}(Y_{u-})\Delta
L_{u}-\alpha\partial_{z}Z_{s}^{\varphi,a}(Y_{u-})\Delta N_{u}\right].$ (4.3)

Note that

$\displaystyle\int_{0}^{t}\partial_{y}Z^{\varphi,a}_{s}(Y_{u-})dL_{u}$
$\displaystyle=\sum_{0<u\leqslant
t}\partial_{y}Z_{s}^{\varphi,a}(Y_{u-})\Delta L_{u}$
$\displaystyle\int_{0}^{t}\partial_{z}Z^{\varphi,a}_{s}(Y_{u-})dN_{u}$
$\displaystyle=\sum_{0<u\leqslant
t}\partial_{z}Z_{s}^{\varphi,a}(Y_{u-})\Delta N_{u}.$

Next, we can write

$\displaystyle\sum_{0<u\leqslant
t}\left[Z^{\varphi,a}_{s}(Y_{u})-Z^{\varphi,a}_{s}(Y_{u-})\right]$
$\displaystyle=\sum_{0<u\leqslant
t}\left[Z^{\varphi,a}_{s}(u,S_{u},v_{u-}+\Delta
v_{s},\lambda_{u-}+\Delta\lambda_{s})-Z^{\varphi,a}_{s}(Y_{u-})\right]$
$\displaystyle=\sum_{0<u\leqslant
t}\left[Z^{\varphi,a}_{s}(u,S_{u},v_{u-}+\eta\Delta
L_{u},\lambda_{u-}+\alpha\Delta N_{u})-Z^{\varphi,a}_{s}(Y_{u-})\right]$
$\displaystyle=\sum_{0<u\leqslant t}g^{\varphi,a}_{s}(u,\Delta L_{u},\Delta
N_{u}),$

where

$\displaystyle
g^{\varphi,a}_{s}(u,b_{1},b_{2}):=Z^{\varphi,a}_{s}(u,S_{u},v_{u-}+\eta
b_{1},\lambda_{u-}+\alpha
b_{2})-Z^{\varphi,a}_{s}(u,S_{u},v_{u-},\lambda_{u-}).$

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

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

We add and subtract the compensator of the counting measure $N^{M}$ to split
the expression into a $(\mathcal{F},\mathbb{Q}(a))$-local martingale plus a
predictable process of finite variation. By Proposition 3.12, the compensators
of $N$ and $L$ under
$\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$ are
$\Lambda_{t}^{N}=\int_{0}^{t}\lambda_{u}du$ and
$\Lambda_{t}^{L}=\mathbb{E}[J_{1}]\int_{0}^{t}\lambda_{u}du$ respectively.
Thus,

$\displaystyle\sum_{0<u\leqslant
t}\left[Z^{\varphi,a}_{s}(Y_{u})-Z^{\varphi,a}_{s}(Y_{u-})\right]=\ $
$\displaystyle\int_{0}^{t}\int_{(0,\infty)^{2}}g^{\varphi,a}_{s}(u,b_{1},b_{2})N^{M}(du,db)$
$\displaystyle=\ $
$\displaystyle\int_{0}^{t}\int_{(0,\infty)^{2}}g^{\varphi,a}_{s}(u,b_{1},b_{2})\left(N^{M}(du,db)-\lambda_{u}P_{J_{1}}(db_{1})\delta_{1}(db_{2})du\right)$
$\displaystyle+\int_{0}^{t}\lambda_{u}\int_{(0,\infty)}g^{\varphi,a}_{s}(u,b_{1},1)P_{J_{1}}(db_{1})du.$

Replacing everything in (4) we finally get

$\displaystyle Z^{\varphi,a}_{s}(Y_{t})=\ $ $\displaystyle
Z^{\varphi,a}_{s}(Y_{0})+\int_{0}^{t}\Big{[}\partial_{t}Z^{\varphi,a}_{s}(Y_{u})+rS_{u}\partial_{x}Z^{\varphi,a}_{s}(Y_{u})-\kappa^{(a)}(v_{u}-\bar{v}^{(a)})\partial_{y}Z^{\varphi,a}_{s}(Y_{u})$
$\displaystyle-\beta(\lambda_{u}-\lambda_{0})\partial_{z}Z^{\varphi,a}_{s}(Y_{u})+\frac{1}{2}S_{u}^{2}v_{u}\partial_{xx}^{2}Z^{\varphi,a}_{s}(Y_{u})+\frac{1}{2}\sigma^{2}v_{u}\partial_{yy}^{2}Z^{\varphi,a}_{s}(Y_{u})$
$\displaystyle+\sigma\rho
S_{u}v_{u}\partial_{xy}^{2}Z_{s}^{\varphi,a}(Y_{u})+\lambda_{u}\int_{(0,\infty)}g_{s}^{\varphi,a}(u,b_{1},1)P_{J_{1}}(db_{1})\Big{]}du$
$\displaystyle+\sqrt{1-\rho^{2}}\int_{0}^{t}S_{u}\sqrt{v_{u}}\partial_{x}Z^{\varphi,a}_{s}(Y_{u-})dB_{u}^{\mathbb{Q}(a)}$
$\displaystyle+\int_{0}^{t}\left[\rho
S_{u}\sqrt{v_{u}}\partial_{x}Z^{\varphi,a}_{s}(Y_{u-})+\sigma\sqrt{v_{u}}\partial_{y}Z^{\varphi,a}_{s}(Y_{u-})\right]dW_{u}^{\mathbb{Q}(a)}$
$\displaystyle+\int_{0}^{t}\int_{(0,\infty)^{2}}g^{\varphi,a}_{s}(u,b_{1},b_{2})\left(N^{M}(du,db)-\lambda_{u}P_{J_{1}}(db_{1})\delta_{1}(db_{2})du\right).$
(4.4)

Recall that $t\to Z^{\varphi,a}_{s}(Y_{t})$ is a
$(\mathcal{F},\mathbb{Q}(a))$-martingale and note that the last three terms in
(4) are $(\mathcal{F},\mathbb{Q}(a))$-local martingales. Moving this three
terms to the left hand side we see that a local martingale is equal to a
continuous process of finite variation. This implies that the integral of the
drift is 0 on every interval $[0,t]\subset[0,T]$ and that the sum of the last
three terms in (4) is a $(\mathcal{F},\mathbb{Q}(a))$-martingale. As a
consequence, the drift is constant equal to $0$ and $Z^{\varphi,a}_{s}$
satisfies the following PIDE

$\displaystyle\partial_{t}Z^{\varphi,a}_{s}(t,x,y,z)+rx\partial_{x}Z^{\varphi,a}_{s}(t,x,y,z)-\kappa^{(a)}(y-\bar{v}^{(a)})\partial_{y}Z^{\varphi,a}_{s}(t,x,y,z)$
$\displaystyle-\beta(z-\lambda_{0})\partial_{z}Z^{\varphi,a}_{s}(t,x,y,z)+\frac{1}{2}x^{2}y\partial_{xx}^{2}Z^{\varphi,a}_{s}(t,x,y,z)+\frac{1}{2}\sigma^{2}y\partial_{yy}^{2}Z^{\varphi,a}_{s}(t,x,y,z)$
$\displaystyle+\sigma\rho
xy\partial_{xy}^{2}Z^{\varphi,a}_{s}(t,x,y,z)+z\int_{(0,\infty)}\left[Z^{\varphi,a}_{s}(t,x,y+\eta
u,z+\alpha)-Z^{\varphi,a}_{s}(t,x,y,z)\right]P_{J_{1}}(du)=0,$

for $(t,x,y,z)\in[0,s]\times\mathcal{D}$ and final condition
$Z^{\varphi,a}_{s}(s,x,y,z)=\varphi(s,x)$. Note that $\mathcal{D}$ is the
support of the process $(S_{t},v_{t},\lambda_{t})$. ∎

######  Definition 4.3.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$,
$f\in\mathcal{C}^{1,2}$ satisfying

$\displaystyle\int_{(0,\infty)}|f(t,x,y+\eta
u,z+\alpha)|P_{J_{1}}(du)<\infty.$ (4.5)

We define the following partial integro-differential operartor
$\mathcal{L}^{a}$ by

$\displaystyle\mathcal{L}^{a}f(t,x,y,z):=\ $ $\displaystyle
rx\partial_{x}f(t,x,y,z)-\kappa^{(a)}(y-\bar{v}^{(a)})\partial_{y}f(t,x,y,z)-\beta(z-\lambda_{0})\partial_{z}f(t,x,y,z)$
$\displaystyle+\frac{1}{2}x^{2}y\partial_{xx}^{2}f(t,x,y,z)+\frac{1}{2}\sigma^{2}y\partial_{yy}^{2}f(t,x,y,z)+\sigma\rho
xy\partial_{xy}^{2}f(t,x,y,z)$
$\displaystyle+z\int_{(0,\infty)}\left[f(t,x,y+\eta
u,z+\alpha)-f(t,x,y,z)\right]P_{J_{1}}(du).$

Since all positive moments of $J_{1}$ exists (see Assumption 1) condition
(4.5) automatically holds if $f$ has polynomial growth in the third variable,
that is, if $f(t,x,y,z)\leqslant\sum_{i=0}^{n}C_{i}(t,x,z)y^{i}$ where the
constants $C_{i}(t,x,z)$ do not depend on $y$. More general, if $|f(t,x,y+\eta
u,z+\alpha)|\leqslant C(t,x,y,z+\alpha)e^{cu}$ with $c<\epsilon_{J}$, then
condition (4.5) holds.

######  Observation 4.4.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$ and
$f\in\mathcal{C}^{1,2}$. Note that we have proved in Lemma 4.2 that the drift
of the Itô differential of $f(t,S_{t},v_{t},\lambda_{t})$ is
$\partial_{t}f(t,S_{t},v_{t},\lambda_{t})+\mathcal{L}^{a}f(t,S_{t},v_{t},\lambda_{t})$.

######  Observation 4.5.

Note that the PIDE in (4.1) is just

$\displaystyle\partial_{t}Z^{\varphi,a}_{s}(t,x,y,z)+\mathcal{L}^{a}Z^{\varphi,a}_{s}(t,x,y,z)=0.$
(4.6)

for $(t,x,y,z)\in[0,s]\times\mathcal{D}$ and final condition
$Z^{\varphi,a}_{s}(s,x,y,z)=\varphi(s,x)$

From the PIDE obtained in Lemma 4.2 we derive the PIDE that satisfies
$e^{-r(s-t)}\mathbb{E}^{\mathbb{Q}(a)}[\varphi(s,S_{s})|\mathcal{F}_{t}]$,
which is the price at time $t$ of the payoff $\varphi(s,S_{s})$ where
$0\leqslant t\leqslant s\leqslant T$.

######  Lemma 4.6.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$,
$\varphi\colon[0,T]\times\mathbb{R}_{+}\to\mathbb{R}_{+}$ such that
$\mathbb{E}^{\mathbb{Q}(a)}[|\varphi(s,S_{s})|]<\infty$ for all $s\in[0,T]$.
Then, there exists a function
$U^{\varphi,a}\colon[0,T]^{2}\times\mathcal{D}\to\mathbb{R}_{+}$ such that

$\displaystyle
e^{-r(s-t)}\mathbb{E}^{\mathbb{Q}(a)}[\varphi(s,S_{s})|\mathcal{F}_{t}]=U_{s}^{\varphi,a}(t,S_{t},v_{t},\lambda_{t})$
(4.7)

where $s,t\in[0,T]$. Note that
$U_{s}^{\varphi,a}(t,x,y,z)=e^{-r(s-t)}\varphi(s,x)$ for $t\in[s,T]$.
Furthermore, fix $s\in[0,T]$, if $U_{s}^{\varphi,a}\in\mathcal{C}^{1,2}$, it
satisfies the following PIDE

$\displaystyle\partial_{t}U_{s}^{\varphi,a}(t,x,y,z)+\mathcal{L}^{a}U_{s}^{\varphi,a}(t,x,y,z)=rU_{s}^{\varphi,a}(t,x,y,z),$
(4.8)

where $\mathcal{L}^{a}$ is defined in Definition 4.3,
$(t,x,y,z)\in[0,s]\times\mathcal{D}$ and final condition
$U_{s}^{\varphi,a}(s,x,y,z)=\varphi(s,x)$.

###### Proof.

See Lemma A.10 in the Appendix. ∎

###  4.1 Thiele’s PIDE for unit-linked policies

We now have all the preliminary results needed to obtain Thiele’s PIDE for
unit-linked policies under the Heston-Hawkes stochastic volatility model. Let
$\mathcal{X}=\\{\mathcal{X}_{t},t\in[0,T]\\}$ be a regular Markov process with
finite state space $\mathcal{J}$ that describes the insured’s state. Let
$\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$ and
$f_{j},g_{j},h_{jk}\colon[0,T]\times\mathbb{R}_{+}\to\mathbb{R}_{+}$ be policy
functions where $j,k\in\mathcal{J}$, $j\neq k$, satisfying
$\mathbb{E}^{\mathbb{Q}(a)}[f_{j}(t,S_{t})]<\infty$,
$\mathbb{E}^{\mathbb{Q}(a)}[g_{j}(t,S_{t})]<\infty$,
$\mathbb{E}^{\mathbb{Q}(a)}[h_{jk}(t,S_{t})]<\infty$ for all $t\in[0,T]$,

$\displaystyle\int_{0}^{T}e^{-rs}p_{ij}(t,s)\mathbb{E}^{\mathbb{Q}(a)}[g_{j}(s,S_{s})|\mathcal{F}_{t}]ds<\infty,$

and

$\displaystyle\int_{0}^{T}e^{-rs}p_{ij}(t,s)\mu_{jk}(s)\mathbb{E}^{\mathbb{Q}(a)}[h_{jk}(s,S_{s})|\mathcal{F}_{t}]ds<\infty,$

a.s. for all $j,k\in\mathcal{J}$, $j\neq k$. The mathematical reserve
$V_{i,\mathcal{F}}^{+,a}(t)$ of a contract with policy functions $f_{j}$,
$g_{j}$ and $h_{jk}$ given that the insured is in state $i\in\mathcal{J}$ at
time $t$ and the information $\mathcal{F}_{t}$, is given by

$\displaystyle V_{i,\mathcal{F}}^{+,a}(t)=\ $ $\displaystyle
e^{rt}\Bigg{[}\sum_{j\in\mathcal{J}}e^{-rT}p_{ij}(t,T)\mathbb{E}^{\mathbb{Q}(a)}[f_{j}(T,S_{T})|\mathcal{F}_{t}]$
$\displaystyle+\sum_{j\in\mathcal{J}}\int_{t}^{T}e^{-rs}p_{ij}(t,s)\mathbb{E}^{\mathbb{Q}(a)}[g_{j}(s,S_{s})|\mathcal{F}_{t}]ds$
$\displaystyle+\sum_{\begin{subarray}{c}j,k\in\mathcal{J}\\\ j\neq
k\end{subarray}}\int_{t}^{T}e^{-rs}p_{ij}(t,s)\mu_{jk}(s)\mathbb{E}^{\mathbb{Q}(a)}[h_{jk}(s,S_{s})|\mathcal{F}_{t}]ds\Bigg{]}.$
(4.9)

In the following result we derive Thiele’s PIDE.

######  Proposition 4.7.

(Thiele’s PIDE) Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$
and $i\in\mathcal{J}$, then, there exists a function
$V_{i}^{a}\colon[0,T]\times\mathcal{D}\to\mathbb{R}_{+}$ such that

$\displaystyle
V_{i,\mathcal{F}}^{+,a}(t)=V_{i}^{a}(t,S_{t},v_{t},\lambda_{t}).$

Furthermore, assume that $U_{T}^{f_{j},a}\in\mathcal{C}^{1,2}$ and
$U^{g_{j},a},U^{h_{jk},a}\in\mathcal{C}^{0,1,2}$ for all $j,k\in\mathcal{J}$,
$j\neq k$. Then, $V_{i}^{a}$ satisfies the following PIDE

$\displaystyle\partial_{t}V_{i}^{a}(t,x,y,z)=\ $ $\displaystyle
rV_{i}^{a}(t,x,y,z)-g_{i}(t,x)-\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(h_{ik}(t,x)+V_{k}^{a}(t,x,y,z)-V_{i}^{a}(t,x,y,z)\right)$
$\displaystyle-\mathcal{L}^{a}V_{i}^{a}(t,x,y,z),$

where $\mathcal{L}^{a}$ is the operator defined in Definition 4.3,
$(t,x,y,z)\in[0,T]\times\mathcal{D}$ and final condition
$V_{i}^{a}(T,x,y,z)=f_{i}(T,x)$.

###### Proof.

Applying Lemma 4.6 there exist functions
$U^{f_{j},a},U^{g_{j},a},U^{h_{jk},a}\colon[0,T]^{2}\times\mathcal{D}\to\mathbb{R}_{+}$
for all $j,k\in\mathcal{J}$, $j\neq k$ such that the mathematical reserve
$V_{i,\mathcal{F}}^{+,a}(t)$ in (4.1) can be rewritten as

$\displaystyle V_{i,\mathcal{F}}^{+,a}(t)=\ $
$\displaystyle\sum_{j\in\mathcal{J}}p_{ij}(t,T)U_{T}^{f_{j},a}(t,S_{t},v_{t},\lambda_{t})+\sum_{j\in\mathcal{J}}\int_{t}^{T}p_{ij}(t,s)U_{s}^{g_{j},a}(t,S_{t},v_{t},\lambda_{t})ds$
$\displaystyle+\sum_{\begin{subarray}{c}j,k\in\mathcal{J}\\\ j\neq
k\end{subarray}}\int_{t}^{T}p_{ij}(t,s)\mu_{jk}(s)U_{s}^{h_{jk},a}(t,S_{t},v_{t},\lambda_{t})ds.$
(4.10)

Defining $V_{i}^{(a)}\colon[0,T]\times\mathcal{D}\to\mathbb{R}_{+}$ by

$\displaystyle V_{i}^{(a)}(t,x,y,z):=\ $
$\displaystyle\sum_{j\in\mathcal{J}}p_{ij}(t,T)U_{T}^{f_{j},a}(t,x,y,z)+\sum_{j\in\mathcal{J}}\int_{t}^{T}p_{ij}(t,s)U_{s}^{g_{j},a}(t,x,y,z)ds$
$\displaystyle+\sum_{\begin{subarray}{c}j,k\in\mathcal{J}\\\ j\neq
k\end{subarray}}\int_{t}^{T}p_{ij}(t,s)\mu_{jk}(s)U_{s}^{h_{jk},a}(t,x,y,z)ds.$

we see that $V_{i,\mathcal{F}}^{+,a}(t)=V_{i}^{a}(t,S_{t},v_{t},\lambda_{t})$
and the first part is proved.

Assume now that $U_{T}^{f_{j},a}\in\mathcal{C}^{1,2}$ and
$U^{g_{j},a},U^{h_{jk},a}\in\mathcal{C}^{0,1,2}$ for all $j,k\in\mathcal{J}$,
$j\neq k$. Since $\mathcal{X}$ is regular, we see that $\partial_{t}V_{i}^{a}$
and $\mathcal{L}^{a}V_{i}^{a}$ are well defined by applying several times
differentiation under the integral sign. For the sake of clarity, we define

$\displaystyle
V_{i,\mathcal{F}}^{+,a}(t)=V_{i}^{a}(t,S_{t},v_{t},\lambda_{t})=G_{i,T}^{a}(t,S_{t},v_{t},\lambda_{t})+\int_{t}^{T}F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})ds,$
(4.11)

where

$\displaystyle G_{i,T}^{a}(t,x,y,z):$
$\displaystyle=\sum_{j\in\mathcal{J}}p_{ij}(t,T)U_{T}^{f_{j},a}(t,x,y,z),$
$\displaystyle F_{i,s}^{a}(t,x,y,z):$
$\displaystyle=\sum_{j\in\mathcal{J}}p_{ij}(t,s)U_{s}^{\theta_{j},a}(t,x,y,z),$
(4.12) $\displaystyle\theta_{j}(s,x):$
$\displaystyle=g_{j}(s,x)+\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\ k\neq
j\end{subarray}}\mu_{jk}(s)h_{jk}(s,x).$

Since $\mathcal{X}$ is regular, note that
$U^{\theta_{j},a},F_{i}^{a}\in\mathcal{C}^{0,1,2}$. Fix $s\in[0,T]$, we can
apply Itô’s formula to the processes $t\to
U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})$ and $t\to
F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})$. Now, we compute the drift of the Itô
differential of $t\to F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})$ in two ways.
First, by direct definition. By Observation 4.4, the drift of the Itô
differential of $t\to F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})$ is

$\displaystyle\partial_{t}F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})+\mathcal{L}^{a}F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t}).$
(4.13)

On the other hand

$\displaystyle
dF_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})=\sum_{j\in\mathcal{J}}\partial_{t}p_{ij}(t,s)U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})dt+\sum_{j\in\mathcal{J}}p_{ij}(t,s)dU_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t}).$

Using Kolmogorov’s backward equation in the first term and then the definition
of $F_{i,s}^{a}(t,x,y,z)$ given in (4.12) we get

$\displaystyle dF_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})=\ $
$\displaystyle\sum_{j\in\mathcal{J}}\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(p_{ij}(t,s)-p_{kj}(t,s)\right)U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})dt$
$\displaystyle+\sum_{j\in\mathcal{J}}p_{ij}(t,s)dU_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})$
$\displaystyle=\ $ $\displaystyle\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})-F_{k,s}^{a}(t,S_{t},v_{t},\lambda_{t})\right)dt$
$\displaystyle+\sum_{j\in\mathcal{J}}p_{ij}(t,s)dU_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t}).$

We know by Observation 4.4 that the drift part of the Itô differential of
$t\to U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})$ is
$\partial_{t}U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})+\mathcal{L}^{a}U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})$.
Moreover, $U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})$ satisfies the PIDE
in (4.8). Thus, the drift part of the Itô differential of $t\to
F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})$ can also be written as

$\displaystyle\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\ k\neq
i\end{subarray}}\mu_{ik}(t)\left(F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})-F_{k,s}^{a}(t,S_{t},v_{t},\lambda_{t})\right)$
$\displaystyle+\sum_{j\in\mathcal{J}}p_{ij}(t,s)\left(\partial_{t}U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})+\mathcal{L}^{a}U_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})\right)$
$\displaystyle=\ $ $\displaystyle\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})-F_{k,s}^{a}(t,S_{t},v_{t},\lambda_{t})\right)+\sum_{j\in\mathcal{J}}p_{ij}(t,s)rU_{s}^{\theta_{j},a}(t,S_{t},v_{t},\lambda_{t})$
$\displaystyle=\ $ $\displaystyle\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})-F_{k,s}^{a}(t,S_{t},v_{t},\lambda_{t})\right)+rF_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t}).$
(4.14)

In the last step we have used again the definition of $F_{i,s}^{a}$ in (4.12).
Equating the two equivalent expressions of the drift of the Itô differential
in (4.13) and (4.1) we get

$\displaystyle\partial_{t}F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})+\mathcal{L}^{a}F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})$
$\displaystyle=\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\ k\neq
i\end{subarray}}\mu_{ik}(t)\left(F_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t})-F_{k,s}^{a}(t,S_{t},v_{t},\lambda_{t})\right)+rF_{i,s}^{a}(t,S_{t},v_{t},\lambda_{t}).$

We deduce that

$\displaystyle\partial_{t}F_{i,s}^{a}(t,x,y,z)+\mathcal{L}^{a}F_{i,s}^{a}(t,x,y,z)$
$\displaystyle=\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\ k\neq
i\end{subarray}}\mu_{ik}(t)\left(F_{i,s}^{a}(t,x,y,z)-F_{k,s}^{a}(t,x,y,z)\right)+rF_{i,s}^{a}(t,x,y,z),$
(4.15)

for $(t,x,y,z)\in[0,s]\times\mathcal{D}$. Since
$F_{i}^{a}\in\mathcal{C}^{0,1,2}$ we can apply differentiation under the
integral sign to get

$\displaystyle\partial_{t}\left(\int_{t}^{T}F_{i,s}^{a}(t,x,y,z)ds\right)=\int_{t}^{T}\partial_{t}F_{i,s}^{a}(t,x,y,z)ds-
F_{i,t}^{a}(t,x,y,z).$ (4.16)

Taking the derivative with respect to $t$ in (4.11) and using (4.16) we get

$\displaystyle\partial_{t}V_{i}^{a}(t,x,y,z)=\partial_{t}G_{i,T}^{a}(t,x,y,z)+\int_{t}^{T}\partial_{t}F_{i,s}^{a}(t,x,y,z)ds-
F_{i,t}^{a}(t,x,y,z).$

Writing the explicit expression of $F_{i,t}^{a}(t,x,y,z)$ we obtain

$\displaystyle\int_{t}^{T}\partial_{t}F_{i,s}^{a}(t,x,y,z)ds=\partial_{t}V_{i}^{a}(t,x,y,z)-\partial_{t}G_{i,T}^{a}(t,x,y,z)+g_{i}(t,x)+\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq i\end{subarray}}\mu_{ik}(t)h_{ik}(t,x).$ (4.17)

We now integrate (4.15) with respect to $s$ on the region $[t,T]$ to obtain

$\displaystyle\int_{t}^{T}\partial_{t}^{T}F_{i,s}^{a}(t,x,y,z)ds+\int_{t}^{T}\mathcal{L}^{a}F_{i,s}^{a}(t,x,y,z)ds$
$\displaystyle=\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\ k\neq
i\end{subarray}}\mu_{ik}(t)\left(\int_{t}^{T}F_{i,s}^{a}(t,x,y,z)ds-\int_{t}^{T}F_{k,s}^{a}(t,x,y,z)\right)+r\int_{t}^{T}F_{i,s}^{a}(t,x,y,z)ds.$

Since $F_{i,s}^{a}\in\mathcal{C}^{1,2}$ and it is positive we can apply
differentiation under the integral sign several times and Tonelli’s theorem to
conclude that
$\int_{t}^{T}\mathcal{L}^{a}F_{i,s}^{a}(t,x,y,z)ds=\mathcal{L}^{a}\int_{t}^{T}F_{i,s}^{a}(t,x,y,z)ds$.
Now, we write the expression of
$\int_{t}^{T}\partial_{t}^{T}F_{i,s}^{a}(t,x,y,z)$ in (4.17), we use that
$\int_{t}^{T}\mathcal{L}^{a}F_{i,s}^{a}(t,x,y,z)ds=\mathcal{L}^{a}\int_{t}^{T}F_{i,s}^{a}(t,x,y,z)ds$
and that
$\int_{t}^{T}F_{i,s}^{a}(t,x,y,z)ds=V_{i}^{a}(t,x,y,z)-G_{i,T}^{a}(t,x,y,z)$
to get

$\displaystyle\partial_{t}\left(V_{i}^{a}(t,x,y,z)-G_{i,T}^{a}(t,x,y,z)\right)+g_{i}(t,x)+\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq i\end{subarray}}\mu_{ik}(t)h_{ik}(t,x)$
$\displaystyle+\mathcal{L}^{a}\left(V_{i}^{a}(t,x,y,z)-G_{i,T}^{a}(t,x,y,z)\right)$
$\displaystyle=\ \sum_{\begin{subarray}{c}k\in\mathcal{J}\\\ k\neq
i\end{subarray}}\mu_{ik}(t)\left(V_{i}^{a}(t,x,y,z)-V_{k}^{a}(t,x,y,z)+G_{k,T}^{a}(t,x,y,z)-G_{i,T}^{a}(t,x,y,z)\right)$
$\displaystyle+r\left(V_{i}^{a}(t,x,y,z)-G_{i,T}^{a}(t,x,y,z)\right).$ (4.18)

Now we prove that the terms involving $G_{T}^{a}$ will cancel each other.
Indeed, observe that using Kolmogorov’s backward equation we have

$\displaystyle\partial_{t}G_{i,T}^{a}(t,x,y,z)=\ $
$\displaystyle\sum_{j\in\mathcal{J}}\partial_{t}p_{ij}(t,T)U_{T}^{f_{j},a}(t,x,y,z)+\sum_{j\in\mathcal{J}}p_{ij}(t,T)\partial_{t}U_{T}^{f_{j},a}(t,x,y,z)$
$\displaystyle=\ $
$\displaystyle\sum_{j\in\mathcal{J}}\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(p_{ij}(t,s)-p_{kj}(t,s)\right)U_{T}^{f_{j},a}(t,x,y,z)$
$\displaystyle+\sum_{j\in\mathcal{J}}p_{ij}(t,T)\partial_{t}U_{T}^{f_{j},a}(t,x,y,z)$
$\displaystyle=\ $ $\displaystyle\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(G_{i,T}^{a}(t,x,y,z)-G_{k,T}^{a}(t,x,y,z)\right)+\sum_{j\in\mathcal{J}}p_{ij}(t,T)\partial_{t}U_{T}^{f_{j},a}(t,x,y,z).$

Moreover, using that
$\mathcal{L}^{a}G_{i,T}^{a}(t,x,y,z)=\sum_{j\in\mathcal{J}}p_{ij}(t,T)\mathcal{L}^{a}U_{T}^{f_{j},a}(t,x,y,z)$
and that $U_{T}^{f_{j},a}$ satisfies the PIDE in (4.8) we have

$\displaystyle\partial_{t}G_{i,T}^{a}(t,x,y,z)+\mathcal{L}^{a}G_{i,T}^{a}(t,x,y,z)=\
$ $\displaystyle\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\ k\neq
i\end{subarray}}\mu_{ik}(t)\left(G_{i,T}^{a}(t,x,y,z)-G_{k,T}^{a}(t,x,y,z)\right)$
$\displaystyle+\sum_{j\in\mathcal{J}}p_{ij}(t,T)\left(\partial_{t}U_{T}^{f_{j},a}(t,x,y,z)+\mathcal{L}^{a}U_{T}^{f_{j},a}(t,x,y,z)\right)$
$\displaystyle=\ $ $\displaystyle\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(G_{i,T}^{a}(t,x,y,z)-G_{k,T}^{a}(t,x,y,z)\right)$
$\displaystyle+r\sum_{j\in\mathcal{J}}p_{ij}(t,T)U_{T}^{f_{j},a}(t,x,y,z)$
$\displaystyle=\ $ $\displaystyle\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(G_{i,T}^{a}(t,x,y,z)-G_{k,T}^{a}(t,x,y,z)\right)$
$\displaystyle+rG_{i,T}^{a}(t,x,y,z).$

Replacing this last equality in (4.18) we obtain

$\displaystyle\partial_{t}V_{i}^{a}(t,x,y,z)=\ $ $\displaystyle
rV_{i}^{a}(t,x,y,z)-g_{i}(t,x)-\sum_{\begin{subarray}{c}k\in\mathcal{J}\\\
k\neq
i\end{subarray}}\mu_{ik}(t)\left(h_{ik}(t,x)+V_{k}^{a}(t,x,y,z)-V_{i}^{a}(t,x,y,z)\right)$
$\displaystyle-\mathcal{L}^{a}V_{i}^{a}(t,x,y,z),$

finishing the proof. ∎

## Acknowledgements

The authors would like to acknowledge financial support by the Research
Council of Norway under the SCROLLER project, project number 299897.

##  Appendix A Appendix: Technical results

We give all the proofs that were postponed.

###  A.1 Moments of the variance and the Radon-Nikodym derivative

######  Lemma A.1.

Let $s\geqslant 1$. Then, $\mathbb{E}\left[L_{t}^{s}\right]<\infty$ for all
$t\in[0,T]$ and $\int_{0}^{T}\mathbb{E}\left[L_{t}^{s}\right]dt<\infty$.

Moreover, $\mathbb{E}\left[[L]_{t}^{s}\right]<\infty$ for all $t\in[0,T]$.

###### Proof.

Applying Hölder’s inequality for sums we get

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

Using that $N$ and $\\{J_{i}\\}_{i\geqslant 1}$ are independent we obtain

$\displaystyle\mathbb{E}[L_{t}^{s}]$
$\displaystyle\leqslant\mathbb{E}\left[N_{t}^{s-1}\sum_{i=1}^{N_{t}}J_{i}^{s}\right]$
$\displaystyle=\mathbb{E}\left[\mathbb{E}\left[N_{t}^{s-1}\sum_{i=1}^{N_{t}}J_{i}^{s}\Big{|}\mathcal{F}_{t}^{N}\right]\right]$
$\displaystyle=\mathbb{E}\left[N_{t}^{s-1}\mathbb{E}\left[\sum_{i=1}^{N_{t}}J_{i}^{s}\Big{|}\mathcal{F}_{t}^{N}\right]\right]$
$\displaystyle=\mathbb{E}\left[N_{t}^{s-1}\sum_{i=1}^{N_{t}}\mathbb{E}[J_{i}^{s}|\mathcal{F}_{t}^{N}]\right]$
$\displaystyle=\mathbb{E}\left[N_{t}^{s-1}\sum_{i=1}^{N_{t}}\mathbb{E}[J_{i}^{s}]\right]$
$\displaystyle=\mathbb{E}\left[N_{s}^{s}J_{1}^{s}\right]$
$\displaystyle=\mathbb{E}\left[N_{s}^{s}\right]\mathbb{E}\left[J_{1}^{s}\right]<\infty,$

where we have used that $\mathbb{E}[N_{t}^{s}]<\infty$ and
$\mathbb{E}[J_{1}^{s}]<\infty$. For a reference of
$\mathbb{E}[N_{t}^{s}]<\infty$ see [23, Theorem 1] or [24, Corollary 3.2]
where their condition $\delta>\mu_{1_{G}}$ is our stability condition
$\beta>\alpha$. For $\mathbb{E}[J_{1}^{s}]<\infty$ see Assumption 1. Note that
$\int_{0}^{T}\mathbb{E}\left[L_{t}^{s}\right]dt<\infty$ is a just a
consequence of $\mathbb{E}[L_{t}^{s}]\leqslant\mathbb{E}[L_{T}^{s}]<\infty$
for all $t\in[0,T]$

Finally, in order to prove that $\mathbb{E}\left[[L]_{t}^{s}\right]<\infty$
one can repeat the same argument with $[L]_{t}=\sum_{i=1}^{N_{t}}J_{i}^{2}$. ∎

######  Lemma A.2.

Let $s\geqslant 1$. Then, $\mathbb{E}\left[v_{t}^{s}\right]<\infty$ for all
$t\in[0,T]$ and $\int_{0}^{T}\mathbb{E}\left[v_{t}^{s}\right]dt<\infty$.

###### Proof.

Recall that

$\displaystyle
v_{t}=v_{0}-\kappa\int_{0}^{t}(v_{u}-\bar{v})du+\sigma\int_{0}^{t}\sqrt{v_{u}}dW_{u}+\eta
L_{u}.$

Applying Hölder’s inequality for sums we get

$\displaystyle v_{t}^{s}\leqslant 4^{s-1}$
$\displaystyle\left[v_{0}^{s}+\kappa^{s}\left|\int_{0}^{t}(v_{u}-\bar{v})du\right|^{s}+\sigma^{s}\left|\int_{0}^{t}\sqrt{v_{u}}dW_{u}\right|^{s}+\eta^{s}L_{t}^{s}\right].$
(A.1)

By applying Jensen’s inequality and again Hölder’s inequality for sums we
obtain

$\displaystyle\left|\int_{0}^{t}(v_{u}-\bar{v})du\right|^{s}$
$\displaystyle\leqslant
t^{s}\left(\frac{1}{t}\int_{0}^{t}|v_{u}-\bar{v}|du\right)^{s}$
$\displaystyle\leqslant t^{s-1}\int_{0}^{t}|v_{u}-\bar{v}|^{s}du$
$\displaystyle\leqslant(2t)^{s-1}\int_{0}^{t}\left(v_{u}^{s}+\bar{v}^{s}\right)du$
$\displaystyle\leqslant(2T)^{s-1}\int_{0}^{T}v_{u}^{s}du+2^{s-1}(T\bar{v})^{s}.$

Replacing the last inequality in (A.1) we get

$\displaystyle v_{t}^{s}\leqslant
4^{s-1}\left[v_{0}^{s}+(2T)^{s-1}\kappa^{s}\int_{0}^{T}v_{u}^{s}du+2^{s-1}(T\bar{v}\kappa)^{s}+\sigma^{s}\left|\int_{0}^{t}\sqrt{v_{u}}dW_{u}\right|^{s}+\eta^{s}L_{T}^{s}\right].$

Define now
$A(t):=4^{s-1}\left[v_{0}^{s}+2^{s-1}(T\bar{v}\kappa)^{s}+\sigma^{s}\left|\int_{0}^{t}\sqrt{v_{u}}dW_{u}\right|^{s}+\eta^{s}L_{T}^{s}\right]$,
$B:=(8T)^{s-1}\kappa^{s}$. Then,

$\displaystyle v_{t}^{s}\leqslant A(t)+B\int_{0}^{t}v_{u}^{s}du,$

for all $t\in[0,T]$. For each $\omega\in\Omega$, the functions $t\to
v^{s}_{t}(\omega)$ and $t\to A(t,\omega)$ are measurable in $t$. Moreover,
$\int_{0}^{T}v_{u}^{s}du<\infty$ a.s. because $v$ is a continuous process
except a finite number of finite jumps. Note that $A(t)>0$ for all
$t\in[0,T]$. Then, all the conditions to apply Grönwall’s inequality hold and
we get

$\displaystyle v_{t}^{s}\leqslant A(t)+B\int_{0}^{t}A(u)e^{B(t-u)}du\leqslant
A(t)+Be^{BT}\int_{0}^{T}A(u)du,$ (A.2)

for all $t\in[0,T]$.

By Proposition 3.3, there exists $c_{l}>0$ such that for $c<c_{l}$,
$\mathbb{E}\left[\exp\left(c\int_{0}^{T}v_{u}du\right)\right]<\infty$. In
particular, $\mathbb{E}\left[\int_{0}^{T}v_{u}du\right]<\infty$ and the
process $t\to\int_{0}^{t}\sqrt{v_{u}}dW_{u}$ is a
$(\mathcal{F},\mathbb{P})$-martingale. By the Burkholder–Davis–Gundy
inequality, there exists a constant $C_{s}$ independent of the martingale and
$t$ such that

$\displaystyle\mathbb{E}\left[\left|\int_{0}^{t}\sqrt{v_{u}}dW_{u}\right|^{s}\right]\leqslant
C_{s}\mathbb{E}\left[\left(\int_{0}^{t}v_{u}du\right)^{\frac{s}{2}}\right]\leqslant
C_{s}\mathbb{E}\left[\left(\int_{0}^{T}v_{u}du\right)^{\frac{s}{2}}\right]<\infty,$

where the last expectation is finite because all positive moments of
$\int_{0}^{T}v_{u}du$ exist. By Lemma A.1 we have that
$\mathbb{E}[L_{T}^{s}]<\infty$ and, therefore

$\displaystyle\mathbb{E}\left[A(t)\right]\leqslant
4^{s-1}\left[v_{0}^{s}+2^{s-1}(T\bar{v}\kappa)^{s}+\sigma^{s}C_{s}\mathbb{E}\left[\left(\int_{0}^{T}v_{u}du\right)^{\frac{s}{2}}\right]+\eta^{s}\mathbb{E}[L_{T}^{s}]\right]=:E<\infty,$

for all $t\in[0,T]$. Finally, by taking expectations in (A.2) we obtain

$\displaystyle\mathbb{E}[v_{t}^{s}]\leqslant E+Be^{BT}TE<\infty,$

for all $t\in[0,T]$ and it follows that
$\int_{0}^{T}\mathbb{E}[v_{t}^{s}]dt<\infty$. ∎

######  Definition A.3.

For $a,b,z\in\mathbb{R}$, we define the confluent hypergeometric function of
the first kind ${}_{1}F_{1}(a,b;z)$ in the following way

$\displaystyle{}_{1}F_{1}(a,b;z):=\sum_{n=0}^{\infty}\frac{a^{(n)}}{b^{(n)}}\frac{z^{n}}{n!}$

where $q^{(0)}=1$ and $q^{(n)}=q\cdot(q+1)\cdot...\cdot(q+n-1)$ for
$n\geqslant 1$ is the rising factorial. See [43, Section 5.3] and [2, Section
13] for more details about the confluent hypergeometric function of the first
kind.

######  Lemma A.4.

Let $s\geqslant 1$. If $2\kappa\bar{v}>s\sigma^{2}$,
$\mathbb{E}\left[\frac{1}{v_{t}^{s}}\right]<\infty$ for all $t\in[0,T]$ and
$\int_{0}^{T}\mathbb{E}\left[\frac{1}{v_{t}^{s}}\right]dt<\infty$.

###### Proof.

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

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

In [30, Section 2, Equation (2.3)], we see that for $t\in(0,T]$

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

where $\chi_{\delta}^{{}^{\prime}2}(k(t))$ denotes a noncentral chi-square
random variable with $\delta$ degrees of freedom and noncentrality parameter
$k(t)$. Since $2\kappa\bar{v}>s\sigma^{2}$,
$-s>-\frac{2\kappa\bar{v}}{\sigma^{2}}=-\frac{\delta}{2}$. Then, we can use
[43, Section 10.1, Equation (10.9)] to get that for $t>0$

$\displaystyle\mathbb{E}\left[\left(\frac{1}{\widetilde{v}_{t}}\right)^{s}\right]=\left(\frac{k(t)}{e^{-\kappa
t}v_{0}}\right)^{s}\frac{1}{2^{s}e^{k(t)/2}}\frac{\Gamma\left(\frac{\delta}{2}-s\right)}{\Gamma\left(\frac{\delta}{2}\right)}{}_{1}F_{1}\left(\frac{\delta}{2}-s,\frac{\delta}{2};\frac{k(t)}{2}\right)<\infty,$

where ${}_{1}F_{1}$ is the confluent hypergeometric function of the first kind
given in Definition A.3. By Proposition 3.2, this proves that we have
$\mathbb{E}\left[\frac{1}{v_{t}^{s}}\right]<\infty$ for all $t\in[0,T]$.

To prove that the $s$th negative moment is integrable we check that
$t\to\mathbb{E}\left[\left(\frac{1}{\widetilde{v}_{t}}\right)^{s}\right]$ is a
continuous function for $t\in[0,T]$. Since $\frac{\delta}{2}$ is strictly
positive, and $k\colon(0,T]\to\mathbb{R}$ is continuous, the function
$t\to{}_{1}F_{1}\left(\frac{\delta}{2}-s,\frac{\delta}{2};\frac{k(t)}{2}\right)$
is continuous for $t\in(0,T]$. Therefore,
$t\to\mathbb{E}\left[\left(\frac{1}{\widetilde{v}_{t}}\right)^{s}\right]$ is
continuous at least for $t\in(0,T]$. To check continuity at $t=0$ observe that
$\lim_{t\rightarrow 0^{+}}k(t)=\infty$ and by [2, Section 13, Equation 13.1.4]
it is known that

$\displaystyle\lim_{t\rightarrow
0^{+}}\frac{{}_{1}F_{1}\left(\frac{\delta}{2}-s,\frac{\delta}{2};\frac{k(t)}{2}\right)}{\frac{\Gamma\left(\frac{\delta}{2}\right)}{\Gamma\left(\frac{\delta}{2}-s\right)}e^{k(t)/2}\left(\frac{k(t)}{2}\right)^{-s}}=1.$

Then,

$\displaystyle\lim_{t\rightarrow
0^{+}}\mathbb{E}\left[\left(\frac{1}{\widetilde{v}_{t}}\right)^{s}\right]$
$\displaystyle=\lim_{t\rightarrow 0^{+}}\left(\frac{k(t)}{e^{-\kappa
t}v_{0}}\right)^{s}\frac{1}{2^{s}e^{k(t)/2}}\frac{\Gamma\left(\frac{\delta}{2}-s\right)}{\Gamma\left(\frac{\delta}{2}\right)}{}_{1}F_{1}\left(\frac{\delta}{2}-s,\frac{\delta}{2};\frac{k(t)}{2}\right)$
$\displaystyle=\lim_{t\rightarrow 0^{+}}\left(\frac{k(t)}{e^{-\kappa
t}v_{0}}\right)^{s}\frac{1}{2^{s}e^{k(t)/2}}\frac{\Gamma\left(\frac{\delta}{2}-s\right)}{\Gamma\left(\frac{\delta}{2}\right)}\frac{\Gamma\left(\frac{\delta}{2}\right)}{\Gamma\left(\frac{\delta}{2}-s\right)}e^{k(t)/2}\left(\frac{k(t)}{2}\right)^{-s}$
$\displaystyle=\frac{1}{v_{0}^{s}}=\mathbb{E}\left[\left(\frac{1}{\widetilde{v}_{0}}\right)^{s}\right].$

This proves that
$t\to\mathbb{E}\left[\left(\frac{1}{\widetilde{v}_{t}}\right)^{s}\right]$ is a
continuous function for $t\in[0,T]$. Thus,
$\int_{0}^{T}\mathbb{E}\left[\left(\frac{1}{\widetilde{v}_{t}}\right)^{s}\right]dt<\infty$,
which by Proposition 3.2 implies that
$\int_{0}^{T}\mathbb{E}\left[\frac{1}{v_{t}^{s}}\right]dt<\infty$. ∎

######  Lemma A.5.

If $2\kappa\bar{v}>\sigma^{2}$ and
$c\leqslant\frac{1}{2}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}$,
then

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

###### Proof.

We define the process $Z=\\{Z_{t}=\frac{1}{\widetilde{v}_{t}},t\in[0,T]\\}$
where $\widetilde{v}=\\{\widetilde{v}_{t},t\in[0,T]\\}$ is the pathwise unique
strong solution of

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

Note that the Feller condition $2\kappa\bar{v}\geqslant\sigma^{2}$ ensures
that the process $\widetilde{v}$ is strictly positive. Therefore, the process
$Z$ is well defined. Applying Itô formula we get

$\displaystyle dZ_{t}=\left[\kappa
Z_{t}+\left(\sigma^{2}-\kappa\bar{v}\right)Z_{t}^{2}\right]dt-\sigma
Z_{t}^{3/2}dW_{t}.$ (A.3)

Following the notation in [20, Theorem 3], $Z$ is a quadratic drift $3/2$
process with parameters $p(t)=\kappa$, $q=\sigma^{2}-\kappa\bar{v}$ and
$\epsilon=-\sigma$. The condition $q<\frac{\epsilon^{2}}{2}$ is satisfied
because $2\kappa\bar{v}>\sigma^{2}$. Applying [20, Theorem 3] with $u=0$ and
$s=-c$ we get that for
$c\leqslant\frac{1}{2}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}$

$\displaystyle\mathbb{E}\left[\exp\left(c\int_{0}^{T}\frac{1}{\widetilde{v}_{u}}du\right)\right]=\frac{\Gamma(\gamma-\widetilde{\alpha})}{\Gamma(\widetilde{\alpha})}\left(\frac{2}{\sigma^{2}y(0,1/v_{0})}\right)^{\widetilde{\alpha}}{}_{1}F_{1}\left(\widetilde{\alpha},\gamma;-\frac{2}{\sigma^{2}y(0,1/v_{0})}\right)<\infty,$

where

$\displaystyle\widetilde{\alpha}$
$\displaystyle=-\left(\frac{\kappa\bar{v}}{\sigma^{2}}-\frac{1}{2}\right)+\sqrt{\left(\frac{\kappa\bar{v}}{\sigma^{2}}-\frac{1}{2}\right)^{2}-2\frac{c}{\sigma^{2}}},$
$\displaystyle\gamma$
$\displaystyle=2\left(\widetilde{\alpha}+\frac{k\bar{v}}{\sigma^{2}}\right),$
$\displaystyle y(0,1/v_{0})$ $\displaystyle=\frac{e^{\kappa
T}-1}{v_{0}\kappa}.$

and ${}_{1}F_{1}$ is the confluent hypergeometric function defined in
Definition A.3. Note that $\widetilde{\alpha}\in\mathbb{R}$ because
$c\leqslant\frac{1}{2}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}$
by hypothesis. Finally, by Proposition 3.2, we conclude that

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

See also [21, Footnote 10 on page 136] for another reference about the inverse
CIR. Note that [21, Equation (24) on page 137] is the same condition we have
on $c$. ∎

######  Lemma A.6.

Let $\mathbb{Q}(a)\in\mathcal{E}$, $s>1$,
$D:=\sup_{t\in[0,T]}(\mu_{t}-r)^{2}<\infty$ and $X^{(a)}$ defined in Theorem
3.4. Assume that $2\kappa\bar{v}>\sigma^{2}$ and
$\frac{1-\rho^{2}}{D(s^{2}-s)}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}>1$.
Consider $q_{2}$ such that
$1<q_{2}<\frac{1-\rho^{2}}{D(s^{2}-s)}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}$
and define $q_{1}:=\frac{q_{2}}{q_{2}-1}>1$. If

$\displaystyle|a|<\min\Bigg{\\{}\frac{1}{q_{1}s}\sqrt{\frac{c_{l}}{2}},\sqrt{\frac{(1-\rho^{2})c_{l}}{q_{1}s\left[2q_{1}s(1-\rho^{2})+\rho^{2}s-1\right]}}\Bigg{\\}},$
(A.4)

then

$\displaystyle\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{s}\right]<\infty\hskip
14.22636pt\text{and}\hskip
14.22636pt\mathbb{E}\left[\left(X_{t}^{(a)}\right)^{s}\right]\leqslant\left(\frac{s}{s-1}\right)^{s}\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{s}\right]<\infty,$

for all $t\in[0,T]$.

###### Proof.

Recall that $\theta^{(a)},Y^{(a)}$ and $Z^{(a)}$ are defined in Theorem 3.4
and $X_{t}^{(a)}=Y_{t}^{(a)}Z_{t}^{(a)}$. By Proposition 3.2, the variance
process $v$ is strictly positive. This implies that
$\int_{0}^{T}(\theta_{u}^{(a)})^{2}ds<\infty$, $\mathbb{P}$-a.s, and since
$\theta^{(a)}$ is
$\\{\mathcal{F}_{t}^{W}\vee\mathcal{F}_{t}^{L}\\}_{t\in[0,T]}$-adapted,

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

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

$\displaystyle\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{s}\right]$
$\displaystyle=\mathbb{E}\left[\left(Z_{T}^{(a)}\right)^{s}\mathbb{E}\left[\left(Y_{T}^{(a)}\right)^{s}|\mathcal{F}_{T}^{W}\vee\mathcal{F}_{T}^{L}\right]\right]$
$\displaystyle=\mathbb{E}\left[\left(Z_{T}^{(a)}\right)^{s}\exp\left(\left(\frac{s^{2}-s}{2}\right)\int_{0}^{T}(\theta_{u}^{(a)})^{2}du\right)\right].$

Using that
$(\theta_{u}^{(a)})^{2}=\frac{1}{1-\rho^{2}}\left(\frac{(\mu_{u}-r)^{2}}{v_{u}}+a^{2}\rho^{2}v_{u}-2a\rho(\mu_{u}-r)\right)$
we obtain

$\displaystyle\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{s}\right]=e^{O^{(a)}\int_{0}^{T}(\mu_{u}-r)du}\mathbb{E}\left[\exp\left(P^{(a)}\int_{0}^{T}\sqrt{v_{u}}dW_{u}+Q^{(a)}\int_{0}^{T}v_{u}du+R\int_{0}^{T}\frac{(\mu_{u}-r)^{2}}{v_{u}}du\right)\right],$

where

$\displaystyle O^{(a)}$ $\displaystyle:=-\frac{(s^{2}-s)a\rho}{1-\rho^{2}}$
$\displaystyle P^{(a)}$ $\displaystyle:=-as$ $\displaystyle Q^{(a)}$
$\displaystyle:=\frac{(s^{2}-s)a^{2}\rho^{2}}{2(1-\rho^{2})}-\frac{1}{2}sa^{2}=\frac{sa^{2}(\rho^{2}s-1)}{2(1-\rho^{2})}$
(A.5) $\displaystyle R$ $\displaystyle:=\frac{s^{2}-s}{2(1-\rho^{2})},$ (A.6)

We apply Hölder’s inequality with $q_{1}=\frac{q_{2}}{q_{2}-1}>1$ and
$q_{2}>1$ to get

$\displaystyle\mathbb{E}\left[\exp\left(P^{(a)}\int_{0}^{T}\sqrt{v_{u}}dW_{u}+Q^{(a)}\int_{0}^{T}v_{u}du+R\int_{0}^{T}\frac{(\mu_{u}-r)^{2}}{v_{u}}du\right)\right]$
$\displaystyle\leqslant\mathbb{E}\left[\exp\left(q_{1}P^{(a)}\int_{0}^{T}\sqrt{v_{u}}dW_{u}+q_{1}Q^{(a)}\int_{0}^{T}v_{u}du\right)\right]^{\frac{1}{q_{1}}}\mathbb{E}\left[\exp\left(q_{2}R\int_{0}^{T}\frac{(\mu_{u}-r)^{2}}{v_{u}}du\right)\right]^{\frac{1}{q_{2}}}.$
(A.7)

Now, we focus on the first term in (A.1). We add and substract the constant
$q_{1}^{2}(P^{(a)})^{2}$ and we apply Cauchy-Schwarz inequality

$\displaystyle\mathbb{E}\left[\exp\left(q_{1}P^{(a)}\int_{0}^{T}\sqrt{v_{u}}dW_{u}+q_{1}Q^{(a)}\int_{0}^{T}v_{u}du\right)\right]$
$\displaystyle=\mathbb{E}\left[\exp\left(q_{1}P^{(a)}\int_{0}^{T}\sqrt{v_{u}}dW_{u}-q_{1}^{2}(P^{(a)})^{2}\int_{0}^{T}v_{u}du\right)\exp\left((q_{1}Q^{(a)}+q_{1}^{2}(P^{(a)})^{2})\int_{0}^{T}v_{u}du\right)\right]$
$\displaystyle\leqslant\mathbb{E}\left[\exp\left(2q_{1}P^{(a)}\int_{0}^{T}\sqrt{v_{u}}dW_{u}-2q_{1}^{2}(P^{(a)})^{2}\int_{0}^{T}v_{u}du\right)\right]^{\frac{1}{2}}\mathbb{E}\left[\exp\left(2(q_{1}Q^{(a)}+q_{1}^{2}(P^{(a)})^{2})\int_{0}^{T}v_{u}du\right)\right]^{\frac{1}{2}}.$
(A.8)

Note that the first term in (A.1) is the expectation of a Doléans-Dade
exponential. By (A.4), $|a|<\frac{1}{q_{1}s}\sqrt{\frac{c_{l}}{2}}$ and we
have that $2q_{1}^{2}(P^{(a)})^{2}=2q_{1}^{2}a^{2}s^{2}<c_{l}$. Then, by
Proposition 3.3, Novikov’s condition is satisfied, that is

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

Therefore,

$\displaystyle\mathbb{E}\left[\exp\left(2q_{1}P^{(a)}\int_{0}^{T}\sqrt{v_{u}}dW_{u}-2q_{1}^{2}(P^{(a)})^{2}\int_{0}^{T}v_{u}du\right)\right]<\infty.$

For the second term in (A.1) we need to check again that Proposition 3.3 is
satisfied. One can check that

$\displaystyle
2(q_{1}Q^{(a)}+q_{1}^{2}(P^{(a)})^{2})=\frac{q_{1}sa^{2}}{1-\rho^{2}}\left[2q_{1}s(1-\rho^{2})+\rho^{2}s-1\right].$

By Observation 3.8, $2q_{1}s(1-\rho^{2})+\rho^{2}s-1>0$ and by (A.4),
$|a|<\sqrt{\frac{(1-\rho^{2})c_{l}}{q_{1}s\left[2q_{1}s(1-\rho^{2})+\rho^{2}s-1\right]}}$.
Thus, $2(q_{1}Q^{(a)}+q_{1}^{2}(P^{(a)}))^{2}<c_{l}$ and applying again
Proposition 3.3 we obtain

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

We conclude that the two terms in (A.1) are finite and, therefore, the first
expectation in (A.1) is finite as well, that is,

$\displaystyle\mathbb{E}\left[\exp\left(q_{1}P^{(a)}\int_{0}^{T}\sqrt{v_{u}}dW_{u}+q_{1}Q^{(a)}\int_{0}^{T}v_{u}du\right)\right]<\infty.$

We check the second term in (A.1). Recall that we have defined
$D=\sup_{u\in[0,T]}(\mu_{u}-r)^{2}<\infty$. Then

$\displaystyle\mathbb{E}\left[\exp\left(q_{2}R\int_{0}^{T}\frac{(\mu_{u}-r)^{2}}{v_{u}}du\right)\right]\leqslant\mathbb{E}\left[\exp\left(q_{2}RD\int_{0}^{T}\frac{1}{v_{u}}du\right)\right].$

Since $2\kappa\bar{v}>\sigma^{2}$ and $q_{2}$ is such that
$1<q_{2}<\frac{1-\rho^{2}}{D(s^{2}-s)}\left(\frac{2\kappa\bar{v}-\sigma^{2}}{2\sigma}\right)^{2}$
and $q_{2}RD=\frac{q_{2}(s^{2}-s)D}{2(1-\rho^{2})}$, we can apply Lemma A.5 to
obtain

$\displaystyle\mathbb{E}\left[\exp\left(q_{2}R\int_{0}^{T}\frac{(\mu_{u}-r)^{2}}{v_{u}}du\right)\right]<\infty.$

This proves that the second term in (A.1) is also finite and we can conclude
that $\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{s}\right]<\infty$.

By Theorem 3.4, $X^{(a)}$ is a positive $(\mathcal{F},\mathbb{P})$-martingale.
We can apply Doob’s martingale inequality, see [5, Section 2.1.2, Theorem
2.1.5], to get

$\displaystyle\mathbb{E}\left[\left(X_{t}^{(a)}\right)^{s}\right]\leqslant\mathbb{E}\left[\sup_{0\leqslant
u\leqslant
T}\left(X_{u}^{(a)}\right)^{s}\right]\leqslant\left(\frac{s}{s-1}\right)^{s}\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{s}\right]<\infty$

for any $t\in[0,T]$. This finishes the proof of this lemma. ∎

###  A.2 Compensator of $N$ under the historic and the risk neutral measures

######  Lemma A.7.

The following holds

  1. 1.

Define $\Lambda^{N}_{t}:=\int_{0}^{t}\lambda_{u}du$, then $N-\Lambda^{N}$ is a
square integrable $(\mathcal{F},\mathbb{P})$-martingale.

  2. 2.

Define $\Lambda^{L}_{t}:=\mathbb{E}[J_{1}]\int_{0}^{t}\lambda_{u}du$, then
$L-\Lambda^{L}$ is a square integrable $(\mathcal{F},\mathbb{P})$-martingale.

###### Proof.

(1) By [11, Theorem 3] the process is a $N-\Lambda^{N}$ is a
$(\mathcal{F}^{N},\mathbb{P})$-martingale.

Let $0\leqslant s\leqslant t\leqslant T$, since
$\sigma(N_{t}-\Lambda^{N}_{t})\vee\mathcal{F}_{s}^{L}$ is independent of
$\mathcal{F}^{(B,W)}_{s}$,
$\mathcal{F}_{s}^{L}\subset\mathcal{F}_{s}^{N}\vee\sigma(\\{J_{i}\\}_{i\geqslant
1})$, $\sigma(N_{t}-\Lambda^{N}_{t})\vee\mathcal{F}_{s}^{N}$ is independent of
$\sigma(\\{J_{i}\\}_{i\geqslant 1})$ and
$\mathcal{F}_{s}^{N}\subset\mathcal{F}_{s}^{L}$ we have

$\displaystyle\mathbb{E}[N_{t}-\Lambda^{N}_{t}|\mathcal{F}_{s}]$
$\displaystyle=\mathbb{E}[N_{t}-\Lambda^{N}_{t}|\mathcal{F}_{s}^{L}]$
$\displaystyle=\mathbb{E}[\mathbb{E}[N_{t}-\Lambda^{N}_{t}|\mathcal{F}_{s}^{N}\vee\sigma(\\{J_{i}\\}_{i\geqslant
1})]|\mathcal{F}_{s}^{L}]=$
$\displaystyle=\mathbb{E}[\mathbb{E}[N_{t}-\Lambda^{N}_{t}|\mathcal{F}_{s}^{N}]|\mathcal{F}_{s}^{L}]$
$\displaystyle=\mathbb{E}[N_{t}-\Lambda^{N}_{t}|\mathcal{F}_{s}^{N}]$
$\displaystyle=N_{s}-\Lambda_{s}^{N}.$

This proves that $N-\Lambda^{N}$ is a $(\mathcal{F},\mathbb{P})$-martingale.
Moreover, by [23, Theorem 1] and [23, Section 3.1] we have that
$\mathbb{E}[N_{t}^{2}]<\infty$, $\mathbb{E}[\lambda_{t}^{2}]<\infty$ and
$t\to\mathbb{E}[\lambda_{t}^{2}]$ is a continuous function for $t\in[0,T]$.
Then, by applying Jensen’s inequality we see

$\displaystyle\mathbb{E}\left[\left(N_{t}-\Lambda_{t}^{N}\right)^{2}\right]$
$\displaystyle\leqslant
2\mathbb{E}[N_{T}^{2}]+2\mathbb{E}\left[\left(\int_{0}^{T}\lambda_{u}du\right)^{2}\right]$
$\displaystyle=2\mathbb{E}[N_{T}^{2}]+2\mathbb{E}\left[T^{2}\left(\frac{1}{T}\int_{0}^{T}\lambda_{u}du\right)^{2}\right]$
$\displaystyle\leqslant
2\mathbb{E}[N_{T}^{2}]+2T\mathbb{E}\left[\int_{0}^{T}\lambda_{u}^{2}du\right]$
$\displaystyle=2\mathbb{E}[N_{T}^{2}]+2T\int_{0}^{T}\mathbb{E}[\lambda_{u}^{2}]du<\infty,$

for all $t\in[0,T]$. This proves that $N-\Lambda^{N}$ is a square integrable
$(\mathcal{F},\mathbb{P})$-martingale. Actually, one can prove that all
moments exist but for our purposes it is enough with the second moment.

(2) Let $0\leqslant s\leqslant t\leqslant T$. Since
$J_{N_{s}+1},...,J_{N_{t}}$ are independent of
$\mathcal{F}_{s}\vee\mathcal{F}_{t}^{N}$ we have

$\displaystyle\mathbb{E}\left[L_{t}|\mathcal{F}_{s}\right]$
$\displaystyle=\mathbb{E}\left[L_{s}+(L_{t}-L_{s})|\mathcal{F}_{s}\right]$
$\displaystyle=L_{s}+\mathbb{E}\left[\sum_{i=N_{s}+1}^{N_{t}}J_{i}\Big{|}\mathcal{F}_{s}\right]$
$\displaystyle=L_{s}+\mathbb{E}\left[\mathbb{E}\left[\sum_{i=N_{s}+1}^{N_{t}}J_{i}\Big{|}\mathcal{F}_{s}\vee\mathcal{F}_{t}^{N}\right]\Big{|}\mathcal{F}_{s}\right]$
$\displaystyle=L_{s}+\mathbb{E}\left[\sum_{i=N_{s}+1}^{N_{t}}\mathbb{E}\left[J_{i}|\mathcal{F}_{s}\vee\mathcal{F}_{t}^{N}\right]\Big{|}\mathcal{F}_{s}\right]$
$\displaystyle=L_{s}+\mathbb{E}\left[\sum_{i=N_{s}+1}^{N_{t}}\mathbb{E}\left[J_{i}\right]\Big{|}\mathcal{F}_{s}\right]=L_{s}+\mathbb{E}\left[J_{1}\right]\mathbb{E}\left[N_{t}-N_{s}|\mathcal{F}_{s}\right].$

Thus,

$\displaystyle\mathbb{E}[L_{t}-\Lambda^{L}_{t}|\mathcal{F}_{s}]$
$\displaystyle=L_{s}+\mathbb{E}[J_{1}]\mathbb{E}[N_{t}-\Lambda_{t}^{N}-N_{s}|\mathcal{F}_{s}]$
$\displaystyle=L_{s}-\mathbb{E}[J_{1}]\Lambda_{s}^{N}$
$\displaystyle=L_{s}-\Lambda^{L}_{s}.$

This proves that $L-\Lambda^{L}$ is a $(\mathcal{F},\mathbb{P})$-martingale.
Moreover, by Lemma A.1 and the same argument in the previous part one can
prove that $L-\Lambda^{L}$ is a square integrable
$(\mathcal{F},\mathbb{P})$-martingale. Actually, one can prove that all
moments exist but for our purposes it is enough with the second moment. ∎

######  Lemma A.8.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$, the following
holds

  1. 1.

The process $t\to\int_{0}^{t}X_{u}^{(a)}d(L-\Lambda^{L})_{u}$ is a square
integrable $(\mathcal{F},\mathbb{P})$-martingale.

  2. 2.

The process $t\to\int_{0}^{t}L_{u-}dX_{u}^{(a)}$ is a square integrable
$(\mathcal{F},\mathbb{P})$-martingale.

  3. 3.

Let $0\leqslant s\leqslant t\leqslant T$, then
$\mathbb{E}[L_{t}X_{t}^{(a)}|\mathcal{F}_{s}]=L_{s}X_{s}^{(a)}+\mathbb{E}[J_{1}]\int_{s}^{t}\mathbb{E}[\lambda_{u}X_{u}^{(a)}|\mathcal{F}_{s}]du$.

###### Proof.

(1) Recall that $X^{(a)}$ is defined in Theorem 3.4. By Lemma A.7,
$L-\Lambda^{L}$ is a square integrable $(\mathcal{F},\mathbb{P})$-martingale.
To prove that $t\to\int_{0}^{t}X_{u}^{(a)}d(L-\Lambda^{L})_{u}$ is a square
integrable $(\mathcal{F},\mathbb{P})$-martingale we check that

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\left(X_{u}^{(a)}\right)^{2}d[L-\Lambda^{L}]_{u}\right]<\infty.$

The quadratic variation of the compound Hawkes process is given by
$[L]_{t}=\sum_{i=1}^{N_{t}}J_{i}^{2}$ and $[L-\Lambda^{L}]_{t}=[L]_{t}$. Using
Hölder’s inequality with $p=1+\frac{\varepsilon_{1}}{2}>1$ and
$q=\frac{p}{p-1}>1$ (recall Assumption 2 and Observation 3.10) and Doob’s
martingale inequality, see [5, Section 2.1.2, Theorem 2.1.5], in the last step
we have

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\left(X_{u}^{(a)}\right)^{2}d[L-\Lambda^{L}]_{u}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\left(X_{u}^{(a)}\right)^{2}d[L]_{u}\right]$
$\displaystyle\leqslant\mathbb{E}\left[\sup_{t\in[0,T]}\left(X_{t}^{(a)}\right)^{2}[L]_{T}\right]$
$\displaystyle\leqslant\mathbb{E}\left[\sup_{t\in[0,T]}\left(X_{t}^{(a)}\right)^{2p}\right]^{1/p}\mathbb{E}\left[[L]_{T}^{q}\right]^{1/q}$
$\displaystyle\leqslant\mathbb{E}\left[\sup_{t\in[0,T]}\left(X_{t}^{(a)}\right)^{2+\varepsilon_{1}}\right]^{1/p}\mathbb{E}\left[[L]_{T}^{q}\right]^{1/q}$
$\displaystyle\leqslant
q\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{2+\varepsilon_{1}}\right]^{1/p}\mathbb{E}\left[[L]_{T}^{q}\right]^{1/q}<\infty,$

where
$\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{2+\varepsilon_{1}}\right]<\infty$
by Observation 3.10 and $\mathbb{E}\left[[L]_{T}^{q}\right]<\infty$ by Lemma
A.1. We conclude that the process
$t\to\int_{0}^{t}X_{u}^{(a)}d(L-\Lambda^{L})_{u}$ is a square integrable
$(\mathcal{F},\mathbb{P})$-martingale.

(2) By Theorem 3.4 and Observation 3.10, $X^{(a)}$ is a square integrable
martingale. To prove that $t\to\int_{0}^{t}L_{u-}dX_{u}^{(a)}$ is a square
integrable $(\mathcal{F},\mathbb{P})$-martingale we check that

$\displaystyle\mathbb{E}\left[\int_{0}^{T}L_{u-}^{2}d[X^{(a)}]_{u}\right]<\infty.$

The quadratic variation of $X^{(a)}$ is given by
$d[X^{(a)}]_{t}=\left[\left(\theta_{t}^{(a)}\right)^{2}+a^{2}v_{t}\right]\left(X_{t}^{(a)}\right)^{2}dt$,
where $\theta^{(a)}$ is defined in Theorem 3.4. Then,

$\displaystyle\mathbb{E}\left[\int_{0}^{T}L_{u-}^{2}d[X^{(a)}]_{u}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}L_{u-}^{2}\left[\left(\theta_{u}^{(a)}\right)^{2}+a^{2}v_{u}\right]\left(X_{u}^{(a)}\right)^{2}du\right]$
$\displaystyle=\int_{0}^{T}\mathbb{E}\left[L_{u-}^{2}\left(\theta_{u}^{(a)}\right)^{2}\left(X_{u}^{(a)}\right)^{2}\right]du+a^{2}\int_{0}^{T}\mathbb{E}\left[L_{u-}^{2}v_{u}\left(X_{u}^{(a)}\right)^{2}\right]du.$
(A.9)

We focus on the first term in (A.2), applying Hölder’s inequality with
$p_{1}=\frac{p_{2}p_{3}}{p_{2}p_{3}-p_{2}-p_{3}}>1$,
$p_{2}=1+\varepsilon_{2}>1$, $p_{3}=1+\frac{\varepsilon_{1}}{2}>1$, and then
Doob’s martingale inequality, see [5, Section 2.1.2, Theorem 2.1.5], to the
last expectation we get

$\displaystyle\mathbb{E}\left[L_{u-}^{2}\left(\theta_{u}^{(a)}\right)^{2}\left(X_{u}^{(a)}\right)^{2}\right]$
$\displaystyle\leqslant\mathbb{E}[L_{u}^{2p_{1}}]^{\frac{1}{p_{1}}}\mathbb{E}\left[\left(\theta_{u}^{(a)}\right)^{2p_{2}}\right]^{\frac{1}{p_{2}}}\mathbb{E}\left[\left(X_{u}^{(a)}\right)^{2p_{3}}\right]^{\frac{1}{p_{3}}}$
$\displaystyle=\mathbb{E}[L_{u}^{2p_{1}}]^{\frac{1}{p_{1}}}\mathbb{E}\left[\left(\theta_{u}^{(a)}\right)^{2+2\varepsilon_{2}}\right]^{\frac{1}{p_{2}}}\mathbb{E}\left[\left(X_{u}^{(a)}\right)^{2+\varepsilon_{1}}\right]^{\frac{1}{p_{3}}}$
$\displaystyle\leqslant\left(\frac{2+\varepsilon_{1}}{1+\varepsilon_{1}}\right)^{2}\mathbb{E}[L_{T}^{2p_{1}}]^{\frac{1}{p_{1}}}\mathbb{E}\left[\left(\theta_{u}^{(a)}\right)^{2+2\varepsilon_{2}}\right]^{\frac{1}{p_{2}}}\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{2+\varepsilon_{1}}\right]^{\frac{1}{p_{3}}}.$
(A.10)

Note that $\mathbb{E}[L_{T}^{2p_{1}}]<\infty$ by Lemma A.1 and
$\mathbb{E}\left[\left(X_{T}^{(a)}\right)^{2+\varepsilon_{1}}\right]<\infty$
by Observation 3.10. Applying Hölder’s inequality for sums we get

$\displaystyle\mathbb{E}\left[\left(\theta_{u}^{(a)}\right)^{2+2\varepsilon_{2}}\right]\leqslant\frac{2^{1+2\varepsilon_{2}}}{(1-\rho^{2})^{1+\varepsilon_{2}}}\left[D^{1+\varepsilon_{2}}\mathbb{E}\left[\left(\frac{1}{v_{u}}\right)^{1+\varepsilon_{2}}\right]+(a\rho)^{2+2\varepsilon_{2}}\mathbb{E}\left[v_{u}^{1+\varepsilon_{2}}\right]\right]$

where $D=\sup_{t\in[0,T]}(\mu_{t}-r)^{2}$ is defined in Lemma 3.7. By
Observation 3.10 and Lemma 3.5 we have that

$\displaystyle\int_{0}^{T}\mathbb{E}\left[\left(\theta_{u}^{(a)}\right)^{2+2\varepsilon_{2}}\right]<\infty.$

Applying Hölder’s inequality with $p_{2}>1$ and
$q_{2}=\frac{p_{2}}{p_{2}-1}>1$ we obtain

$\displaystyle\int_{0}^{T}\mathbb{E}\left[\left(\theta_{u}^{(a)}\right)^{2+2\varepsilon_{2}}\right]^{\frac{1}{p_{2}}}du\leqslant
T^{\frac{p_{2}-1}{p_{2}}}\left(\int_{0}^{T}\mathbb{E}\left[\left(\theta_{u}^{(a)}\right)^{2+2\varepsilon_{2}}\right]du\right)^{\frac{1}{p_{2}}}<\infty.$

By (A.2) this implies that

$\displaystyle\int_{0}^{T}\mathbb{E}\left[L_{u-}^{2}\left(\theta_{u}^{(a)}\right)^{2}\left(X_{u}^{(a)}\right)^{2}\right]du<\infty.$

Similarly, using Lemma A.1, Lemma A.2 and Observation 3.10 we can show that
the second term in (A.2) is finite. We conclude that the process
$t\to\int_{0}^{t}L_{u-}dX_{u}^{(a)}$ is a square integrable
$(\mathcal{F},\mathbb{P})$-martingale.

(3) Applying Itô formula, using that $L_{0}=0$, that $L$ is of finite
variation and $X^{(a)}$ is continuous we have

$\displaystyle
L_{t}X_{t}^{(a)}=\int_{0}^{t}L_{u-}dX_{u}^{(a)}+\int_{0}^{t}X_{u}^{(a)}dL_{u}.$

Using that the processes $t\to\int_{0}^{t}L_{u-}dX_{u}^{(a)}$ and
$t\to\int_{0}^{t}X_{u}^{(a)}d(L-\Lambda^{L})_{u}$ are square integrable
$(\mathcal{F},\mathbb{P})$-martingales and the expression of $\Lambda^{L}$
given in Lemma A.7 we obtain

$\displaystyle\mathbb{E}[L_{t}X_{t}^{(a)}|\mathcal{F}_{s}]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{t}L_{u-}dX_{u}^{(a)}\Big{|}\mathcal{F}_{s}\right]+\mathbb{E}\left[\int_{0}^{t}X_{u}^{(a)}dL_{u}\Big{|}\mathcal{F}_{s}\right]$
$\displaystyle=\int_{0}^{s}L_{u-}dX_{u}^{(a)}+\int_{0}^{s}X_{u}^{(a)}dL_{u}+\mathbb{E}\left[\int_{s}^{t}X_{u}^{(a)}dL_{u}\Big{|}\mathcal{F}_{s}\right]$
$\displaystyle=L_{s}X_{s}^{(a)}+\mathbb{E}\left[\int_{s}^{t}X_{u}^{(a)}d(L-\Lambda^{L})_{u}\Big{|}\mathcal{F}_{s}\right]+\mathbb{E}\left[\int_{s}^{t}X_{u}^{(a)}d\Lambda^{L}_{u}\Big{|}\mathcal{F}_{s}\right]$
$\displaystyle=L_{s}X_{s}^{(a)}+\mathbb{E}[J_{1}]\int_{s}^{t}\mathbb{E}[\lambda_{u}X_{u}^{(a)}|\mathcal{F}_{s}]du.$

∎

######  Proposition A.9.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$, then

  1. 1.

$N-\Lambda^{N}$ is a $(\mathcal{F},\mathbb{Q}(a))$-martingale.

  2. 2.

$L-\Lambda^{L}$ is a $(\mathcal{F},\mathbb{Q}(a))$-martingale.

###### Proof.

First of all, note that for $0\leqslant s\leqslant t\leqslant T$,

$\displaystyle\mathbb{E}\left[\int_{0}^{t}\lambda_{u}X_{t}^{(a)}du\Big{|}\mathcal{F}_{s}\right]$
$\displaystyle=\int_{0}^{s}\mathbb{E}[\lambda_{u}X_{t}^{(a)}|\mathcal{F}_{s}]du+\int_{s}^{t}\mathbb{E}[\lambda_{u}X_{t}^{(a)}|\mathcal{F}_{s}]du$
$\displaystyle=X_{s}^{(a)}\int_{0}^{s}\lambda_{u}du+\int_{s}^{t}\mathbb{E}[\mathbb{E}[\lambda_{u}X_{t}^{(a)}|\mathcal{F}_{u}]\mathcal{F}_{s}]du$
$\displaystyle=X_{s}^{(a)}\int_{0}^{s}\lambda_{u}du+\int_{s}^{t}\mathbb{E}[\lambda_{u}X_{u}^{(a)}|\mathcal{F}_{s}]du.$

By Lemma A.8 we know that
$\mathbb{E}[L_{t}X_{t}^{(a)}|\mathcal{F}_{s}]=L_{s}X_{s}^{(a)}+\mathbb{E}[J_{1}]\int_{s}^{t}\mathbb{E}[\lambda_{u}X_{u}^{(a)}|\mathcal{F}_{s}]du$.
Using the previous equalities we obtain,

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}\left[L_{t}-\mathbb{E}[J_{1}]\int_{0}^{t}\lambda_{u}du\Big{|}\mathcal{F}_{s}\right]=$
$\displaystyle\
\frac{1}{X_{s}^{(a)}}\mathbb{E}\left[L_{t}X_{t}^{(a)}-\mathbb{E}[J_{1}]\int_{0}^{t}\lambda_{u}X_{t}^{(a)}du\Big{|}\mathcal{F}_{s}\right]$
$\displaystyle=\ $
$\displaystyle\frac{1}{X_{s}^{(a)}}\Bigg{[}L_{s}X_{s}^{(a)}+\mathbb{E}[J_{1}]\int_{s}^{t}\mathbb{E}[\lambda_{u}X_{u}^{(a)}|\mathcal{F}_{s}]du$
$\displaystyle-\mathbb{E}[J_{1}]X_{s}^{(a)}\int_{0}^{s}\lambda_{u}du-\mathbb{E}[J_{1}]\int_{s}^{t}\mathbb{E}[\lambda_{u}X_{u}^{(a)}|\mathcal{F}_{s}]du\Bigg{]}$
$\displaystyle=\ $ $\displaystyle
L_{s}-\mathbb{E}[J_{1}]\int_{0}^{s}\lambda_{u}du.$

This finishes the proof. ∎

###  A.3 Derivation of Thiele’s PIDE for unit-linked policies

######  Lemma A.10.

Let $\mathbb{Q}(a)\in\mathcal{E}_{m}(Q_{1},2+\varepsilon_{1})$,
$\varphi\colon[0,T]\times\mathbb{R}_{+}\to\mathbb{R}_{+}$ such that
$\mathbb{E}^{\mathbb{Q}(a)}[|\varphi(s,S_{s})|]<\infty$ for all $s\in[0,T]$.
Then, there exists a function
$U^{\varphi,a}\colon[0,T]^{2}\times\mathcal{D}\to\mathbb{R}_{+}$ such that

$\displaystyle
e^{-r(s-t)}\mathbb{E}^{\mathbb{Q}(a)}[\varphi(s,S_{s})|\mathcal{F}_{t}]=U_{s}^{\varphi,a}(t,S_{t},v_{t},\lambda_{t})$
(A.11)

where $s,t\in[0,T]$. Note that
$U_{s}^{\varphi,a}(t,x,y,z)=e^{-r(s-t)}\varphi(s,x)$ for $t\in[s,T]$.
Furthermore, fix $s\in[0,T]$, if $U_{s}^{\varphi,a}\in\mathcal{C}^{1,2}$, it
satisfies the following PIDE

$\displaystyle\partial_{t}U_{s}^{\varphi,a}(t,x,y,z)+\mathcal{L}^{a}U_{s}^{\varphi,a}(t,x,y,z)=rU_{s}^{\varphi,a}(t,x,y,z),$
(A.12)

where $\mathcal{L}^{a}$ is defined in Definition 4.3,
$(t,x,y,z)\in[0,s]\times\mathcal{D}$ and final condition
$U_{s}^{\varphi,a}(s,x,y,z)=\varphi(s,x)$.

###### Proof.

By Lemma 4.2 there exists a function
$Z^{\varphi,a}\colon[0,T]^{2}\times\mathcal{D}\to\mathbb{R}_{+}$ such that

$\displaystyle\mathbb{E}^{\mathbb{Q}(a)}[\varphi(s,S_{s})|\mathcal{F}_{t}]=Z_{s}^{\varphi,a}(t,S_{t},v_{t},\lambda_{t}),$

where $s,t\in[0,T]$. Define the function
$U^{\varphi,a}\colon[0,T]^{2}\times\mathcal{D}\to\mathbb{R}_{+}$ by
$U_{s}^{\varphi,a}(t,x,y,z):=e^{-r(s-t)}Z_{s}^{\varphi,a}(t,x,y,z)$. Then,
(A.11) is satisfied.
Moreover, fix $s\in[0,T]$, if $U_{s}^{\varphi,a}\in\mathcal{C}^{1,2}$,
$Z_{s}^{\varphi,a}\in\mathcal{C}^{1,2}$ and it satisfies the following PIDE

$\displaystyle\partial_{t}Z^{\varphi,a}_{s}(t,x,y,z)+\mathcal{L}^{a}Z^{\varphi,a}_{s}(t,x,y,z)=0.$
(A.13)

for $(t,x,y,z)\in[0,T]\times\mathcal{D}$ and final condition
$Z^{\varphi,a}_{s}(s,x,y,z)=\varphi(s,x)$. Note that

$\displaystyle\partial_{t}Z^{\varphi,a}_{s}(t,x,y,z)=e^{r(s-t)}\left(-rU^{\varphi,a}_{s}(t,x,y,z)+\partial_{t}U^{\varphi,a}_{s}(t,x,y,z)\right).$

and

$\displaystyle\mathcal{L}^{a}Z^{\varphi,a}_{s}(t,x,y,z)=e^{r(s-t)}\mathcal{L}^{a}U^{\varphi,a}_{s}(t,x,y,z).$

Replacing that in (A.13) we get that $U_{s}^{\varphi,a}$ satisfies the
following PIDE

$\displaystyle\partial_{t}U_{s}^{\varphi,a}(t,x,y,z)+\mathcal{L}^{a}U_{s}^{\varphi,a}(t,x,y,z)=rU_{s}^{\varphi,a}(t,x,y,z).$

for $(t,x,y,z)\in[0,s]\times\mathcal{D}$ and final condition
$U_{s}^{\varphi,a}(s,x,y,z)=\varphi(s,x)$. ∎

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