# On the discrete-time simulation of the rough Heston model

Alexandre Richard111Université Paris-Saclay, CentraleSupélec, MICS and CNRS
FR-3487, France<EMAIL_ADDRESS>Xiaolu Tan222Department
of Mathematics, The Chinese University of Hong Kong<EMAIL_ADDRESS>research supported by CUHK startup grant and CUHK Faculty of Science Direct
Grant 2020-2021. Fan Yang333Department of Mathematics, The Chinese University
of Hong Kong<EMAIL_ADDRESS>
###### Abstract

We study Euler-type discrete-time schemes for the rough Heston model, which
can be described by a stochastic Volterra equation (with non-Lipschtiz
coefficient functions), or by an equivalent integrated variance formulation.
Using weak convergence techniques, we prove that the limits of the discrete-
time schemes are solution to some modified Volterra equations. Such modified
equations are then proved to share the same unique solution as the initial
equations, which implies the convergence of the discrete-time schemes.
Numerical examples are also provided in order to evaluate different options’
prices under the rough Heston model.

Key words: Rough Heston model, stochastic Volterra equations, Euler scheme,
Monte-Carlo method.

MSC2020 subject classification: 60H20, 45D05, 91G60.

##  1 Introduction

The modelling of rough volatilities is an important subject in mathematical
finance, especially since the paper [22] which brought statistical evidence of
such behaviour in the financial markets (see also the pioneering works [14,
10, 21, 15] in this direction). Rough volatility models have the advantage to
better exhibit the roughness of the volatility time series, to reproduce the
shape and the dynamic of the implied volatility surface, etc. Many of the
rough volatility models consist in replacing the Brownian motions in the
classical models by fractional Brownian motions, which leads to some SDEs or
more general stochastic system driven by fractional Brownian motions, see e.g.
[14, 15, 12, 6, 22, 11] among many others. Another important way to model the
rough volatility process is to use a stochastic Volterra equation, such as the
rough Heston model introduced by El Euch and Rosenbaum [17]:

$S_{t}=S_{0}+\int_{0}^{t}S_{s}\sqrt{V_{s}}~{}dW^{1}_{s},~{}~{}~{}~{}V_{t}=V_{0}+\int_{0}^{t}K(t-s)\Big{(}(\theta-\lambda
V_{s})~{}ds+\nu\,\sqrt{V_{s}}~{}dW^{2}_{s}\Big{)},$ (1)

where $(W^{1},W^{2})$ are two correlated Brownian motions with some
correlation constant $\rho\in(-1,1)$, and $K(t):=Ct^{H-\frac{1}{2}}$ is the
kernel function with some Hurst parameter $H\in(0,\frac{1}{2})$. Above, $S$ is
the risky asset price process under the risk neutral probability, and $V$
represents the volatility process. Further, by considering the integrated
processes:

$X_{t}:=\int_{0}^{t}V_{s}~{}ds,~{}~{}~{}M^{1}_{t}:=\int_{0}^{t}\sqrt{V_{s}}dW^{1}_{s},~{}~{}~{}M^{2}_{t}:=\int_{0}^{t}\sqrt{V_{s}}dW^{2}_{s},$

the rough Heston model (1) is shown to be equivalent (see e.g. Abi Jaber [1])
to the following system

$S_{t}=S_{0}+\int_{0}^{t}S_{s}~{}dM^{1}_{s},~{}~{}~{}~{}X_{t}=V_{0}t+\int_{0}^{t}K(t-s)\big{(}\theta
s-\lambda X_{s}+\nu M^{2}_{s}\big{)}~{}ds,$ (2)

where $(M^{1},M^{2})$ are two continuous martingales with quadratic variation
$\langle M^{1}\rangle=\langle M^{2}\rangle=X$ and quadratic covariation
$\langle M^{1},M^{2}\rangle=\rho X$. Based on the above formulations in (1)
and (2), the super-rough Heston model [16] and the hyper-rough Heston model
[1, 30] have also been developed recently.

In the rough volatility literature, an important topic is to find a good
approximation method (e.g. in order to evaluate option prices), whenever a
closed formula is not available. For different models, some approximation and
asymptotic methods have been introduced and studied, see e.g. [10, 21, 18, 26,
25, 3, 20, 27, 19], etc. In the meantime, for affine models such as (1) and
(2), one can in fact obtain the marginal distributions of $(S_{t},V_{t})$ or
$(S_{t},X_{t})$ at any time $t\geq 0$, by computing their characteristic
function via Riccati-type systems, see e.g. [9, 1, 4, 17]. This allows in
particular to compute efficiently the European call/put options prices under
the affine rough volatility models. Nevertheless, as the path distribution of
the process $(S,V)$ or $(S,X)$ is still unknown, one cannot compute prices of
path-dependent options this way. In this case, a natural and simple solution
would be the Monte-Carlo method based on a discrete-time scheme.

The main objective of the paper is to study the Euler-type discrete-time
scheme for the rough Heston model in both formulations (1) and (2), and to
provide a convergence result. We will stay in a more general rough Heston
setting, i.e. the kernel function $K(t)$ is not necessarily of the form
$Ct^{H-\frac{1}{2}}$. Throughout the paper, we would like to call (1) the
rough Heston model in the stochastic Volterra equation formulation, and (2)
the rough Heston model in the integrated variance formulation (or simply
integrated-rough Heston model).

Notice that Equation (1) satisfied by $V$ is a standard stochastic Volterra
equation (but with non-Lipschitz coefficient). For stochastic Volterra
equations with Lipschitz coefficient equations, the discrete-time schemes such
as Euler scheme and/or Milstein scheme have been studied in [36, 35, 8, 33],
where (sharp) strong convergence rates have been obtained. Nevertheless,
because of the square root term $\sqrt{V_{s}}$, the coefficient function in
(1) is non-Lipschitz. Hence the techniques and results in the aforementioned
papers cannot be applied to obtain a convergence result.

We will apply weak convergence techniques to provide a convergence proof of
the discrete-time scheme for both Equations (1) and (2). The idea is very
classical in the literature on SDEs, see e.g. [31, 28]. First, one shows that
the sequence of discrete time numerical solutions is tight, then that any
limit of the sequence is solution to the continuous-time equation. Next, it is
enough to show that the limit continuous-time equation has a unique weak
solution, so that the numerical solution converges weakly to the unique
solution of the limit equation. In the context of the rough Heston model, such
weak convergence techniques have already been used in [1, 5, 4], in particular
to show the existence of weak solutions of the related equations.

However, for the analysis of the discrete-time numerical solution, it is not
straightforward to apply their techniques and results. First, they usually
consider sequences of continuous-time processes $(V^{n})_{n\geq 1}$ and
$(X^{n})_{n\geq 1}$ which are solutions of equations with smoother
coefficients, where generalized Grönwall lemma applies under conditions on the
kernel $K$. For the discrete-time numerical solution, because of
discretization of the kernel function $K$, it is not trivial to formulate
explicit conditions on $K$ to have the discrete-time generalized Grönwall
lemma. We therefore need to develop different techniques to estimate the
(uniform) moment estimates to obtain the tightness of discrete-time solutions.
Next, their approximating processes are already positive (resp. non-
decreasing), so that the limit process $V$ (resp. $X$) is automatically
positive (resp. non-decreasing). In the discrete-time setting, the numerical
solution $V^{n}$ may not always be positive, we hence need to take its
positive part $(V^{n})_{+}$ before taking the square root, i.e.
$\sqrt{(V^{n})_{+}}$. As for the numerical solution $X^{n}$, we need to
replace $X^{n}_{t}$ by $\overline{X}^{n}_{t}:=\max_{0\leq s\leq t}X^{n}_{s}$
to make it non-decreasing, so that it can be the quadratic variation of some
martingales. Consequently, it turns out that the limit of the numerical
solutions is solution to some modified equations. We will then need to show
that the limit process $V$ is positive and $X$ is non-decreasing, and the
modified equation shares the same unique weak solution as the initial
equations. For this, we will adapt the ideas from [2] to our context. This
allows us to obtain weak convergence results of the discrete-time numerical
solutions. Finally, we also provide some numerical simulation examples to
evaluate (path-dependent) options’ prices in the rough Heston model.

The rest of the paper is organised as follows. In Section 2, we state the two
equivalent formulations of the rough Heston model with more details, and
present the corresponding discrete-time schemes as well as the convergence
results. In Section 3, we provide some numerical examples in order to evaluate
the option prices in the rough Heston model. Proofs of the main (weak)
convergence results in Theorems 2.2 and 2.3 are provided in Section 4.

##  2 Discrete-time simulation of the rough Heston model

We will first restate the two equivalent formulations (1) and (2) of the rough
Heston model with more precise definitions. Based on the two formulations, we
introduce the corresponding Euler-type schemes, which are defined on a
discrete grid. Let us consider a sequence $(\pi_{n})_{n\geq 1}$ of discrete-
time grid on $[0,T]$, with
$\pi_{n}=\\{0=t_{0}^{n}<t_{1}^{n}<t_{2}^{n}<...<t_{n}^{n}=T\\}$ for each
$n\geq 1$. For each $n\geq 1$, we define $\eta_{n}(s):=t^{n}_{k}$ for
$s\in[t^{n}_{k},t^{n}_{k+1})$, $k=0,\cdots,n-1$, and $\eta_{n}(T):=T$.

To provide the convergence results, we will make some assumptions on the
kernel function $K:[0,T]\longrightarrow\mathbb{R}$ used in the model. For a
(measurable) kernel function $K:[0,T]\longrightarrow\mathbb{R}$, let us recall
that the resolvent of the first kind of $K$ is a finite measure $L$ on $[0,T]$
such that

$(K*L)(t):=\int_{[0,t]}K(t-s)L(ds)=1,~{}~{}\mbox{for all}~{}t\in(0,T].$ (3)

Notice that we are in a one-dimensional context, so that the above definition
is much simpler than the general one (see [24]).

######  Assumption 2.1.

The function $K\in L^{2}([0,T])$ is nonnegative, not identically $0$, non-
increasing and continuous. Its resolvent of the first kind $L$ is nonnegative
and such that $s\longmapsto L([s,s+t])$ is non-increasing for all $t\geq 0$.
Moreover, there exist constants $C>0$ and $H>0$ such that, for all $0\leq
t\leq T$, and $n\geq 1$, $\delta\in(0,T-t]$, one has

$\int_{t}^{t+\delta}\big{|}K(t+\delta-\eta_{n}(s))\big{|}^{2}~{}ds~{}\leq~{}C\delta^{2H},$
(4)

and

$\int_{0}^{t}\big{|}K(t+\delta-\eta_{n}(s))-K(t-\eta_{n}(s))\big{|}^{2}~{}ds~{}\leq~{}C\delta^{2H}.$
(5)

######  Remark 2.1.

$\mathrm{(i)}$ Let $K(t):=Ct^{H-\frac{1}{2}}$ with Hurst constant
$H\in(0,\frac{1}{2})$ and some constant $C>0$. Then (4) and (5) can be checked
by direct computation, while the resolvent of $K$ is
$L(dt)=C_{H}\,t^{-(H+\frac{1}{2})}\,dt$ for some $C_{H}>0$, and thus satisfies
Assumption 2.1.

$\mathrm{(ii)}$ Let $K(t):=K_{1}(t)K_{2}(t)$, where $K_{1}(t)=C\exp(-\beta t)$
and $K_{2}(t)=t^{H-\frac{1}{2}}$ for some constants $C>0$, $\beta>0$ and
$H\in(0,\frac{1}{2})$. One can check by direct computation that $K$ satisfies
(4) and (5). Moreover, the resolvent of $K$ is explicitely given in [4, Table
1] and satisfies Assumption 2.1.

$\mathrm{(iii)}$ More generally, if $K$ is a completely monotone function (as
defined in [24, Section 5.2]) and is not identically $0$, then by [24, Theorem
5.5.4] it admits a resolvent of the first kind $L$ which is nonnegative and is
such that $s\longmapsto L([s,s+t])$ is non-increasing for all $t\geq 0$. So if
$K$ also verifies (4) and (5), it will satisfy Assumption 2.1. This is for
instance the case of $K_{1}$ and $K_{2}$ above, as well as for instance
$K_{3}(t)=\log(1+\frac{1}{t+1})$. In addition, any linear combination and
multiplication of completely monotone function is still a completely monotone
function. Hence Assumption 2.1 covers a wide range of kernels.

###  2.1 The rough Heston model in two equivalent formulations

Let $\rho\in[-1,1]$ be a constant, $S_{0}$, $V_{0}$, $\theta$, $\lambda$ and
$\nu$ be all strictly positive constants. The first formulation of the rough
Heston model is given by

$\begin{split}S_{t}&=S_{0}+\int_{0}^{t}S_{s}\,\sqrt{V_{s}}~{}d\big{(}\rho\,W_{s}+\sqrt{1-\rho^{2}}\,W_{s}^{\bot}\big{)},\\\
V_{t}&=V_{0}+\int_{0}^{t}K(t-s)\Big{(}\big{(}\theta-\lambda
V_{s}\big{)}~{}ds+\nu\sqrt{V_{s}}~{}dW_{s}\Big{)},\end{split}$ (6)

where $(W,W^{\bot})$ are two independent Brownian motions. Namely, $S$
represents the risky asset price under the risk-neutral probability,
$\sqrt{V_{t}}$ is the volatility at time $t\geq 0$. We give immediately a
precise definition of weak solution to (6).

######  Definition 2.2.

We say that Equation (6) has a weak solution if there exists a complete
filtered probability space
$(\Omega,\mathcal{F},(\mathcal{F}_{t})_{t\in[0,T]},\mathbb{P})$ equipped with
two independent Brownian motion $W,~{}W^{\bot}$, and a pair of
$\mathbb{R}_{+}$-valued adapted continuous processes $(V,S)$ such that (6) is
satisfied a.s. for all $t\geq 0$.

######  Remark 2.3.

Under Assumption 2.1, the existence and uniqueness of the weak solution to (6)
can be found in e.g. [4]. Indeed, let $n\longrightarrow\infty$ in (4)-(5), it
follows by Fatou’s lemma that

$\int_{t}^{t+\delta}\big{|}K(t+\delta-s)\big{|}^{2}ds+\int_{0}^{t}\big{|}K(t+\delta-s)-K(t-s)\big{|}^{2}ds\leq
2C\delta^{2H},$

which corresponds to the conditions required in [4].

The process $(\sqrt{V_{t}})_{t\geq 0}$ is called the volatility process. Let
us consider the integrated variance process $X$ given by

$X_{t}=\int_{0}^{t}V_{s}~{}ds,~{}~{}t\geq 0.$

As observed in [1], one can reformulate (6) into an equivalent system on
$(X,S)$. Namely, by applying the stochastic Fubini theorem (see e.g. [34,
p.175]), the processes $(S,X)$ satisfy the following stochastic Volterra
equation

$\displaystyle\begin{split}S_{t}&=S_{0}+\int_{0}^{t}S_{s}~{}d\big{(}\rho
M_{s}+\sqrt{1-\rho^{2}}M_{s}^{\bot}\big{)},\\\
X_{t}&=V_{0}t+\int_{0}^{t}K(t-s)\big{(}\theta s-\lambda X_{s}+\nu
M_{s}\big{)}~{}ds,\end{split}$ (7)

where $M_{s},~{}M^{\bot}_{s}$ are two orthogonal continuous martingales with
quadratic variation $\langle M\rangle=\langle M^{\bot}\rangle=X$, and initial
condition $M_{0}=M^{\bot}_{0}=0$. We will call it the integrated variance
formulation (or simply integrated-rough Heston model). Following [1], let us
introduce the definition of weak solution of Equation (7).

######  Definition 2.4.

We say that Equation (7) has a weak solution if there exists a filtered
probability space $(\Omega,\mathcal{F},(\mathcal{F}_{t})_{t\geq
0},\mathbb{P})$ supporting a pair of orthogonal continuous martingales
$(M,M^{\bot})$, a non-decreasing, non-negative, continuous and adapted process
$X$ and a non-negative continuous and adapted process $S$, such that (7) holds
a.s.

######  Remark 2.5.

Under Assumption 2.1, the existence and uniqueness of the weak solution to
Equation (7) is proved in [1]. In fact, the context in [1] covers the case
with $L^{1}$ kernel functions. For technical reasons, in particular the
equivalence results in Section 4.3, we will stay in the context of $L^{2}$
kernel functions $K$.

######  Remark 2.6.

Let us define $Y_{t}:=\log(S_{t})$, then it is clear that one has

$Y_{t}=Y_{0}-\int_{0}^{t}\frac{1}{2}V_{s}ds+\int_{0}^{t}\sqrt{V_{s}}~{}d\big{(}\rho\,W_{s}+\sqrt{1-\rho^{2}}\,W_{s}^{\bot}\big{)}~{}\mbox{in
the formulation \eqref{eq:RoughHeston}},$

and

$Y_{t}=Y_{0}-\frac{1}{2}X_{t}+\rho
M_{t}+\sqrt{1-\rho^{2}}M^{\bot}_{t}~{}\mbox{in the formulation
\eqref{eq:IntegVarHeston}}.$

###  2.2 The discrete-time schemes and their convergence

Recall that $(\pi_{n})_{n\geq 1}$ is a sequence of discrete-time grids on
$[0,T]$, with $\pi_{n}=\\{0=t_{0}^{n}<t_{1}^{n}<t_{2}^{n}<...<t_{n}^{n}=T\\}$
for each $n\geq 1$. Let us denote $\delta_{n}:=|\pi_{n}|:=\max\limits_{0\leq
k\leq n-1}\Delta t^{n}_{k+1}$, with $\Delta
t^{n}_{k+1}:=t^{n}_{k+1}-t^{n}_{k}$ and $\eta_{n}(s):=t_{k}^{n}$, for
$s\in[t_{k}^{n},t_{k+1}^{n})$, $k=0,\cdots,n-1$. We will simulate $(S,V)$ of
(6) and $(S,X)$ of (7) on the discrete-time grid $\pi_{n}$. More precisely, in
view of Remark 2.6, we would like to simulate the process $(Y,V)$ in place of
$(S,V)$ in (6), and to simulate $(Y,X)$ in place of $(S,X)$ in (7). As
observed in the Black-Scholes model, the simulation of $Y$ permits to avoid
the time discretization of the process $S$ in the dynamics of $S$, and one can
expect a better performance for its simulation.

Let us first give the Euler-type scheme for (6). Notice that the process $V$
is $\mathbb{R}_{+}$-valued in the continuous-time setting, but it could become
negative in a discrete-time simulation. For this reason, we use
$(V_{t})_{+}:=\max(V_{t},0)$ in the square root term $\sqrt{(V_{t})_{+}}$ to
define the discrete-time scheme. For the discrete grid $\pi_{n}$, let us write
$t^{n}_{k}$ as $t_{k}$ for simplicity, and denote by
$(S^{n},V^{n})=(S^{n}_{t_{k}},V^{n}_{t_{k}})_{k=0,1,\cdots,n}$ the
corresponding numerical solution, which is given as follows:
$S^{n}_{t_{k}}:=\exp(Y^{n}_{t_{k}})$, $k=0,1,\cdots,n$, and

$\begin{split}Y_{t_{k}}^{n}&=Y_{0}+\sum\limits_{i=0}^{k-1}\Big{(}-\frac{1}{2}(V^{n}_{t_{i}})_{+}\,\Delta
t^{n}_{i+1}+\rho\sqrt{(V^{n}_{t_{i}})_{+}}\big{(}W_{t_{i+1}}-W_{t_{i}}\big{)}+\sqrt{1-\rho^{2}}\sqrt{(V^{n}_{t_{i}})_{+}}\big{(}W^{\bot}_{t_{i+1}}-W^{\bot}_{t_{i}}\big{)}\Big{)},\\\
V^{n}_{t_{k}}&=V_{0}+\sum_{i=0}^{k-1}\Big{(}K(t_{k}-t_{i})\big{(}\theta-\lambda(V^{n}_{t_{i}})_{+}\big{)}\Delta
t^{n}_{i+1}+K(t_{k}-t_{i})\nu\sqrt{(V^{n}_{t_{i}})_{+}}\big{(}W_{t_{i+1}}-W_{t_{i}}\big{)}\Big{)}.\\\
\end{split}$ (8)

For the process $(S^{n},V^{n})=(S^{n}_{t_{k}},V^{n}_{t_{k}})_{k=0,1,\cdots,n}$
defined on the discrete-time grid $\pi_{n}$, one can use linear interpolation
to obtain a continuous time process (with continuous paths), which is still
denoted by $(S^{n},V^{n})=(S^{n}_{t},V^{n}_{t})_{t\in[0,T]}$. We can provide a
weak convergence result of the numerical solution.

######  Theorem 2.2.

$(i)$ For all $p\geq 2$, there exists a constant $C_{p}>0$ such that

$\mathbb{E}\big{[}\big{|}V^{n}_{t}\big{|}^{p}\big{]}\leq
C_{p},~{}~{}~{}\mathbb{E}\big{[}\big{|}V_{t}^{n}-V_{s}^{n}\big{|}^{p}\big{]}~{}\leq~{}C_{p}(t-s)^{p(H\wedge
1)},~{}~{}\mbox{for all}~{}0\leq s\leq t\leq T~{}\mbox{and}~{}n\geq 1.$

$(ii)$ The processes $(S^{n},V^{n})$ defined by (8) converge weakly to $(S,V)$
in $C([0,T],\mathbb{R})\times C([0,T],\mathbb{R})$ as
$n\longrightarrow\infty$, where $(S,V)$ is the unique weak solution to the
first formulation (6) of the rough Heston model.

We now consider the discrete-time simulation problem of Equation (7). Notice
that the process $X$ in the continuous-time setting is a non-decreasing
process, which is the quadratic variation process of the martingales $M$ and
$M^{\bot}$. In discrete-time simulation, $X$ would not be non-decreasing, and
for this reason, we will consider its running maximum
$\overline{X}_{t}:=\max_{0\leq s\leq t}X_{s}$ to define the quadratic
variation process. For each discrete time grid $\pi_{n}$, let us define
$(S^{n},X^{n})=(S^{n}_{t_{k}},X^{n}_{t_{k}})_{k=0,1,\cdots,n}$ as follows:
$S^{n}_{t_{k}}:=\exp(Y^{n}_{t_{k}})$, $k=0,1,\cdots,n$, and

$\begin{split}Y^{n}_{t_{k}}&=Y_{0}-\frac{1}{2}\overline{X}^{n}_{t_{k}}+\rho
M^{n}_{t_{k}}+\sqrt{1-\rho^{2}}M^{n\bot}_{t_{k}},\\\
X_{t_{k}}^{n}&=V_{0}t_{k}+\sum_{i=0}^{k-1}K(t_{k}-t_{i})\Big{(}\theta
t_{i}-\lambda\overline{X}^{n}_{t_{i}}+\nu M^{n}_{t_{i}}\Big{)}\Delta
t^{n}_{i},\\\
M^{n}_{t_{k}}&=\sum_{i=1}^{k}\sqrt{\overline{X}^{n}_{t_{i}}-\overline{X}^{n}_{t_{i-1}}}Z_{i},~{}~{}~{}~{}M^{n\bot}_{t_{k}}=\sum_{i=1}^{k}\sqrt{\overline{X}^{n}_{t_{i}}-\overline{X}^{n}_{t_{i-1}}}Z_{i}^{\bot},\end{split}$
(9)

where $\overline{X}^{n}_{t_{i}}:=\underset{0\leq j\leq i}{\max}X^{n}_{t_{j}}$,
and $(Z_{i},~{}Z_{i}^{\bot})_{i\geq 1}$ is a sequence of i.i.d. random
variables with standard Gaussian distribution $\mathcal{N}(0,1)$.

Similarly, one can interpolate the process
$(S^{n},X^{n})=(S^{n}_{t_{k}},X^{n}_{t_{k}})_{k=0,1,\cdots,n}$ from the
discrete-time grid $\pi_{n}$ to obtain a continuous time process (with
continuous paths) $(S^{n},X^{n})=(S^{n}_{t},X^{n}_{t})_{t\in[0,T]}$.

######  Theorem 2.3.

$(i)$ For each $p\geq 2$, there exists a constant $C_{p}>0$ such that

$\mathbb{E}\big{[}\big{|}X^{n}_{t}\big{|}^{p}\big{]}+\mathbb{E}\big{[}\big{|}\overline{X}^{n}_{t}\big{|}^{p}\big{]}\leq
C_{p},~{}~{}~{}\mathbb{E}\big{[}\big{|}X^{n}_{t}-X^{n}_{s}\big{|}^{p}\big{]}~{}\leq~{}C_{p}(t-s)^{p(H\wedge
1)},~{}~{}\mbox{for all}~{}0\leq s\leq t\leq T~{}\mbox{and}~{}n\geq 1.$

$(ii)$ The processes $(S^{n},X^{n})$ defined by (9) converge weakly to $(S,X)$
in $C([0,T],\mathbb{R})\times C([0,T],\mathbb{R})$ as
$n\longrightarrow\infty$, where $(S,X)$ is the unique weak solution to the
integrated variance formulation (7) of the rough Heston model.

######  Remark 2.7.

$\mathrm{(i)}$ By the weak convergence of $(S^{n},X^{n})$ to $(S,X)$, one has
the convergence $\mathbb{E}[f(S^{n},X^{n})]\longrightarrow\mathbb{E}[f(S,X)]$
for any bounded continuous payoff function $f$ on $(S,X)$. Moreover, since
$\sup_{n\geq 1}\mathbb{E}[|\overline{X}^{n}_{T}|^{p}]<\infty$ for any $p\geq
1$, it follows that $\mathcal{L}(X^{n})$ converges to $\mathcal{L}(X)$ under
the $p$-Wasserstein distance, for any $p\geq 1$. Consequently, one has the
convergence $\mathbb{E}[f(X^{n})]\longrightarrow\mathbb{E}[f(X)]$ for any
continuous function $f:C([0,T],\mathbb{R})\longrightarrow\mathbb{R}$ with
polynomial growth.

Unfortunately, we do not have a uniform moment estimation on $S^{n}$ for all
$n\geq 1$. For the process $S$ in (6), the moment explosion of $S_{t}$ has
been studied in [23]. This is possible since one can find a Volterra type
equation to compute the characteristic function of the marginal distribution
of $\log(S)$. It is nevertheless not clear how to elaborate the same technique
on the discrete time Euler scheme solution $S^{n}$. We would like to leave
this for future research.

$\mathrm{(ii)}$ Our technique does not allow us to obtain a strong convergence
rate. At the same time, as the processes $V$ and $X$ are not semimartingales,
it is not clear how to adapt the error analysis techniques for classical
Heston models, such as in [13, 7], to this rough Heston model context.
Moreover, for Equations (1) and (2), because of the singular kernels and
square-root-type coefficients, the strong existence and uniqueness of the
solution is still an open question. It is therefore not surprising that a
strong convergence rate is left as an open question.

##  3 Numerical examples

In this section, we provide some numerical examples to evaluate option prices
in the rough Heston model with interest rate $r=0$, by using the Monte-Carlo
method based on the schemes (8) and (9). Namely, for an option with payoff
function $f(S,X)$, one aims to estimate its price given by

$\mathbb{E}[f(S,X)].$

We use the uniform discrete-time grid $\pi_{n}=\\{0=t_{0}<\cdots<t_{n}=T\\}$
with $t_{k}:=k\Delta t$, $\Delta t=\frac{T}{n}$ for Schemes (8) and (9). We
use $M$ i.i.d. copies of simulations $(S^{n,m},V^{n,m})_{m=1,\cdots,M}$ or
$(S^{n,m},X^{n,m})_{m=1,\cdots,M}$ to estimate the option price
$\mathbb{E}[f(S,X)]$ by the mean value :

$\overline{U}_{M}~{}:=~{}\frac{1}{M}\sum\limits_{m=1}^{M}f\big{(}S^{n,m}_{.},X^{n,m}_{\cdot}\big{)}.$

Notice that from the simulations $V^{n,m}$, one can use
$X^{n,m}_{t_{k}}:=\sum_{i=0}^{k}V^{n,m}_{t_{i}}$ to compute $X^{n,m}$. We also
compute the empirical standard deviation of the simulations, which (divided by
$\sqrt{M}$) can serve as the statistical error, i.e.

$\Sigma_{M}~{}:=~{}\frac{1}{\sqrt{M}}\bigg{(}\frac{1}{M}\sum\limits_{m=1}^{M}f\big{(}S^{n,m}_{.},X^{n,m}_{\cdot}\big{)}^{2}-\Big{(}\frac{1}{M}\sum\limits_{m=1}^{M}f\big{(}S^{n,m}_{.},X^{n,m}_{\cdot}\big{)}\Big{)}^{2}\bigg{)}.$

We then use the following interval as confidence interval of the estimation:

$\big{[}\overline{U}_{M}-2\Sigma_{M},\overline{U}_{M}-2\Sigma_{M}\big{]}.$

For the rough Heston model, we choose the following parameters: $\lambda=0.3$,
$\nu=0.3$, $V_{0}=0.02$, $\theta=0.02$, $\rho=-0.7$, $S_{0}=1$, and the kernel
function $K(t):=\Gamma(H+\frac{1}{2})^{-1}t^{H-\frac{1}{2}}$ with $H=0.1$. We
will use different discretization parameters $n$ in our simulations. For each
example, we set the number of i.i.d. copies $M=10^{5}$, and display the mean
value, the statistical error as well as the computational time in the tables.

Notice also that in (9), one can compute $\int_{0}^{t_{k}}K(t_{k}-s)\theta
sds$ explicitly instead of approximating it by
$\sum_{i=0}^{k-1}K(t_{k}-t_{i})\theta t_{i}\Delta t^{n}_{i}$. This will be
taken into account in our simulation.

###  3.1 Pricing options on the risky asset

Let us first consider the European Call options, Asian options and Lookback
options with the following payoff:

$(S_{T}-K)_{+},~{}~{}~{}(A_{T}-K)_{+},~{}~{}~{}(M_{T}-K)_{+},$

where $K=1$, $A_{T}:=\int_{0}^{T}S_{t}dt$ and $M_{T}:=\max_{0\leq t\leq
T}S_{t}$. For the Monte-Carlo method, we simply replace $(S_{T},A_{T},M_{T})$
by $(S^{n,m}_{T},A^{n,m}_{T},M^{n,m}_{T})$ in the payoff function to compute
the estimations, where
$A^{n,m}_{T}:=\frac{T}{n}\sum^{n}_{k=1}S^{n,m}_{t_{k}}$,
$M^{n,m}_{T}:=\max\limits_{0\leq k\leq n}S^{n,m}_{t_{k}}$.

For the European call option, one can compute a reference value for
$\mathbb{E}[(S_{T}-K)_{+}]$. Indeed, as described in [17], one can compute the
characteristic function of $\log(S_{T})$ by solving a fractional Riccati
equation, and then use the inverse Fourier transform method to compute
$\mathbb{E}[(S_{T}-K)_{+}]$. With the above parameters of the rough Heston
model, we obtain $0.056832$ as reference value for the European call option.

The numerical results are reported in Figures 1, 2 and 3 (see also
respectively Tables 1, 2 and 3 for the data). We can observe the convergence
of both schemes as $n$ increases. For European call option and Asian option,
it seems that the scheme (9) based on integrated variance formulation (7) has
a better performance. But for the lookback option, the scheme (8) based on the
stochastic Volterra equation (6) seems to perform better.

Figure 1:  Estimation of European call option price Figure 2:  Estimation of
Asian option price Figure 3:  Estimation of lookback option price

###  3.2 Pricing options on the variance process

We next consider the options on the variance process $X$, including the
variance option with payoff $X_{T}$, and the call option on the variance with
payoff $(X_{T}-V_{0})_{+}$. Under the Rough Heston model with interest rate
$r=0$, the price of the options are given respectively by $\mathbb{E}[X_{T}]$
and $\mathbb{E}[(X_{T}-V_{0})_{+}]$. For both options, one can compute the
reference value by deterministic methods.

Indeed, for the variance swap option, one can deduce from the dynamic

$X_{t}~{}=~{}V_{0}T+\int_{0}^{t}K(t-s)\big{(}\theta s-\lambda
X_{s}+W_{X_{s}}\big{)}~{}ds$

that

$\mathbb{E}[X_{t}]~{}=~{}V_{0}t+\int_{0}^{t}K(t-s)\big{(}\theta
s-\lambda\mathbb{E}[X_{s}]\big{)}~{}ds,$

which is a linear Volterra ODE. With the above parameters of the rough Heston
model, one obtains $0.028295$ as the reference value for the variance swap
option price $\mathbb{E}[X_{T}]$.

For the call option on the variance, one can use the results in [1] to compute
the characteristic function of $X_{T}$ by solving the corresponding Riccati
ODE, and then use the inverse Fourier transform method to compute the value of
$\mathbb{E}[(X_{T}-V_{0})_{+}]$. With the above parameters of the rough Heston
model, we obtain $0.013517$ as reference value for
$\mathbb{E}[(X_{T}-V_{0})_{+}]$.

The discrete-time scheme in (9) allows simulating directly the value of
$X^{n}_{T}$. Nevertheless, the discrete-time scheme in (8) provides
simulations of $(V^{n}_{t_{k}})_{k=0,\cdots,n}$, we then set
$X^{n}_{T}:=\sum_{k=0}^{n}V^{n}_{t_{k}}{\cdot\Delta t}$ for the Monte-Carlo
estimation.

The numerical results are reported in Figures 4 and 5 (see respectively Tables
4 and 5 for the data). We can observe the convergence of both scheme as the
number of time steps $n$ increases. At the same time, for the example on
variance swap, the performance of Scheme (9) based on the integrated variance
formulation seems slightly better.

Figure 4: Estimation of variance swap price Figure 5:  Estimation of variance
call price

##  4 Proofs of Theorems 2.2 and 2.3

Recall that, throughout the paper, Assumption 2.1 holds true.

###  4.1 Tightness of solutions to Scheme (8)

Let us first rewrite the system (6) and the corresponding discrete-time scheme
(8) in two-dimensional equations.

Let $(S,V)$ be the weak solution of (6), recall that $Y_{t}:=\log(S_{t})$,
$v_{+}:=\max(v,0)$ for all $v\in\mathbb{R}$. We denote

$\hat{V_{t}}:=\begin{pmatrix}Y_{t}\\\
V_{t}\end{pmatrix},~{}~{}~{}\hat{K}(t-s)\overset{\triangle}{=}\begin{pmatrix}1&0\\\
0&K(t-s)\end{pmatrix},~{}\hat{W_{s}}:=\begin{pmatrix}W_{s}^{\bot}\\\
W_{s}\end{pmatrix},$

and define functions $b:\mathbb{R}^{2}\longrightarrow\mathbb{R}^{2}$,
$\sigma:\mathbb{R}^{2}\longrightarrow{\mathbb{R}^{2\times 2}}$ and
$a:\mathbb{R}^{2}\longrightarrow{\mathbb{R}^{2\times 2}}$ by

$b\begin{pmatrix}y\\\ v\end{pmatrix}:=\begin{pmatrix}-\frac{1}{2}v\\\
\theta-\lambda v_{+}\end{pmatrix},~{}~{}\sigma\begin{pmatrix}y\\\
v\end{pmatrix}:=\sqrt{v_{+}}\begin{pmatrix}\sqrt{1-\rho^{2}}&\rho\\\
0&\nu\end{pmatrix},~{}\mbox{and}~{}a(x)=\sigma\sigma^{\top}(x).$

Notice that $V$ is $\mathbb{R}_{+}$-valued, then it is easy to check that
$\hat{V}=(Y,V)$ satisfies

$\hat{V}_{t}=\hat{V}_{0}+\int_{0}^{t}\hat{K}(t,s)~{}d\hat{Z}_{s},~{}~{}\mbox{with}~{}\hat{Z_{t}}:=\int_{0}^{t}b(\hat{V_{s}})~{}ds+\int_{0}^{t}\sigma(\hat{V_{s}})~{}d\hat{W_{s}}.$
(10)

Next, for $s,t\in[0,T]$, and $\mathbf{x}\in C([0,T],\mathbb{R}^{2})$, we
define the following new coefficients:

$\begin{split}b^{n}(t,\mathbf{x})&:=b(\mathbf{x}(\eta_{n}(t))),~{}~{}~{}\sigma^{n}(t,\mathbf{x}):=\sigma(\mathbf{x}(\eta_{n}(t))),~{}~{}~{}a^{n}(t,\mathbf{x}):=\big{(}\sigma^{n}(\sigma^{n})^{\top}\big{)}(t,\mathbf{x}),\\\
K^{n}(t,s)&:=K(t-\eta_{n}(s)),~{}~{}~{}\hat{K}^{n}(t,s):=\hat{K}(t-\eta_{n}(s)).\\\
\end{split}$ (11)

Let $(S^{n},V^{n})$ be the numerical solution to Scheme (8). Then we can write
$\hat{V}^{n}:=(Y^{n},V^{n})$ as solution to the stochastic integral equation

$\hat{V}^{n}_{t}=\hat{V}_{0}+\int_{0}^{t}\hat{K}^{n}(t,s)~{}d\hat{Z}^{n}_{s},~{}~{}\mbox{with}~{}\hat{Z}_{s}^{n}=\int_{0}^{s}b^{n}(s,\hat{V}^{n})~{}ds+\int_{0}^{s}\sigma^{n}(s,\hat{V}^{n})~{}d\hat{W}_{s}.$
(12)

We next define the infinitesimal generators $\mathcal{L}^{n}$ and
$\mathcal{L}$ as follows: for all $f\in C^{2}_{b}(\mathbb{R}^{2})$ and
$(\mathbf{x},\mathbf{z})\in C([0,T],\mathbb{R}^{2})\times
C([0,T],\mathbb{R}^{2})$ and $t\in[0,T]$, let

$\begin{split}\mathcal{L}^{n}_{t}f(\mathbf{x},\mathbf{z})&~{}=~{}b^{n}(t,\mathbf{x})\cdot\nabla
f(\mathbf{z}_{t})+\frac{1}{2}\text{Tr}\big{(}a^{n}(t,\mathbf{x})\nabla^{2}f(\mathbf{z}_{t})\big{)},\\\
\mathcal{L}_{t}f(\mathbf{x},\mathbf{z})&~{}=~{}b(\mathbf{x}_{t})\cdot\nabla
f(\mathbf{z}_{t})+\frac{1}{2}\text{Tr}\big{(}a(\mathbf{x}_{t})\nabla^{2}f(\mathbf{z}_{t})\big{)}.\end{split}$
(13)

It is easy to check that for some constant $C_{f}>0$ depending on $f$, one has

$\big{|}\mathcal{L}^{n}_{t}f(\mathbf{x},\mathbf{z})\big{|}\leq
C_{f}\big{(}1+|\mathbf{x}_{\eta_{n}(t)})|\big{)},~{}~{}\big{|}\mathcal{L}_{t}f(\mathbf{x},\mathbf{z})\big{|}\leq
C_{f}\big{(}1+|\mathbf{x}_{t}|\big{)},~{}~{}\mbox{for
all}~{}\mathbf{x},\mathbf{z}\in C([0,T],\mathbb{R}^{2}).$ (14)

We next provide a technical lemma.

######  Lemma 4.1.

Let $(f_{n})_{n\geq 1}$ be a sequence of non-negative functions on $[0,T]$
such that $\sup_{n\geq 1}f_{n}(0)<\infty$. Assume that for some constants
$q\geq 2$ and $C_{0}>0$, one has

$\displaystyle f_{n}(t)~{}\leq~{}$ $\displaystyle
C_{0}\big{(}1+f_{n}(s)\big{)}~{}+~{}C_{0}\Big{(}\int_{s}^{t}\\!\\!K(t-\eta_{n}(r))^{2}\big{(}1+f_{n}(\eta_{n}(r))^{2/q}\big{)}dr\Big{)}^{q/2}$
(15)
$\displaystyle+C_{0}\Big{(}\int_{0}^{s}\\!\\!\big{|}K(t-\eta_{n}(r))-K(s-\eta_{n}(r))\big{|}^{2}\big{(}1+f_{n}(\eta_{n}(r))^{2/q}\big{)}dr\Big{)}^{q/2},~{}\mbox{for
all}~{}n\geq 1,~{}s\leq t.$

Then

$\sup_{n\geq 1}\sup_{0\leq t\leq T}f_{n}(t)~{}<~{}\infty.$

###### Proof.

Let $m\geq 1$ be a fixed integer, and denote $t_{k}:=kT/m$. Let us define the
nondecreasing function $\bar{f}_{n}(t):=\sup_{0\leq s\leq t}f_{n}(s)$. We can
then apply (15) for $s=t_{k}$ and $t\in[t_{k},t_{k+1}]$ to obtain that

$f_{n}(t)~{}\leq~{}(C_{0}+C_{1}+C_{2})+\big{(}C_{0}+C_{1}\big{)}\bar{f}_{n}(t_{k})+C_{2}\bar{f}_{n}(t_{k+1}),~{}\mbox{for
all}~{}t\in[t_{k},t_{k+1}],$

with some positive constants $C_{1}$, $C_{2}$ satisfying that, for each
$k=0,\cdots,m$, and $t\in[t_{k},t_{k+1}]$,

$C_{1}\geq
C_{0}\Big{(}\int_{0}^{t_{k}}\big{|}K(t-\eta_{n}(r))-K(t_{k}-\eta_{n}(r))\big{|}^{2}dr\Big{)}^{q/2},~{}~{}C_{2}\geq
C_{0}\Big{(}\int_{t_{k}}^{t}K(t-\eta_{n}(r))^{2}dr\Big{)}^{q/2}.$

Without loss of generality, we can assume that $C_{0}\geq 1$, so that

$\bar{f}_{n}(t_{k+1})~{}\leq~{}(C_{0}+C_{1}+C_{2})+\big{(}C_{0}+C_{1}\big{)}\bar{f}_{n}(t_{k})+C_{2}\bar{f}_{n}(t_{k+1}).$

Moreover, in view of (4)-(5), one can choose a fixed $m\geq 1$ big enough so
that $C_{2}<1$. Then

$\bar{f}_{n}(t_{k+1})~{}\leq~{}\frac{C_{0}+C_{1}}{1-C_{2}}\bar{f}_{n}(t_{k})+\frac{C_{0}+C_{1}+C_{2}}{1-C_{2}},~{}\mbox{for
all}~{}k=0,\cdots,m-1,$

and, noticing that $C_{0}+C_{1}\neq 1-C_{2}$, it follows that

$\bar{f}_{n}(T)=\bar{f}_{n}(t_{m})~{}\leq~{}\Big{(}\sup_{n\geq
1}f_{n}(0)\Big{)}a^{m}+\frac{a^{m}-1}{a-1}b,~{}\mbox{with
constants}~{}a:=\frac{C_{0}+C_{1}}{1-C_{1}},~{}b:=\frac{C_{0}+C_{1}+C_{2}}{1-C_{1}}.$

Notice that the r.h.s. of the above inequality is independent of $n\geq 1$,
which is enough to conclude the proof. ∎

######  Proposition 4.2.

For each $p\geq 2$, there exist a constant $C_{p}>0$ such that for all $0\leq
s\leq t\leq T$ and $n\geq 1$,

$\displaystyle\mathbb{E}[|\hat{V}_{t}^{n}|^{p}]$ $\displaystyle\leq C_{p},$
(16) $\displaystyle\mathbb{E}[|\hat{V}_{t}^{n}-\hat{V}_{s}^{n}|^{p}]$
$\displaystyle\leq C_{p}(t-s)^{pH},$ (17)
$\displaystyle\mathbb{E}[|\hat{Z}_{t}^{n}-\hat{Z}_{s}^{n}|^{p}]$
$\displaystyle\leq C_{p}(t-s)^{\frac{p}{2}}.$ (18)

###### Proof.

(i) First, one has by Equation (12) that, for all $0\leq s\leq t\leq T$,

$\begin{split}\hat{V}^{n}_{t}-\hat{V}^{n}_{s}=&\int_{0}^{s}\Big{(}\hat{K}(t-\eta_{n}(r))-\hat{K}(s-\eta_{n}(r))\Big{)}b^{n}(r,\hat{V}^{n})~{}dr+\int_{s}^{t}\hat{K}(t-\eta_{n}(r))\,b^{n}(r,\hat{V}^{n})~{}dr\\\
&+\int_{0}^{s}\Big{(}\hat{K}(t-\eta_{n}(r))-\hat{K}(s-\eta_{n}(r))\Big{)}\sigma^{n}(r,\hat{V}^{n})~{}dW_{r}+\int_{s}^{t}\hat{K}(t-\eta_{n}(r))\sigma^{n}(r,\hat{V}^{n})~{}dW_{r}.\end{split}$
(19)

Then using the Burkholder-Davis-Gundy inequality, there is a constant $C>0$
depending only on $p\geq 2$ such that

$\begin{split}\mathbb{E}\big{[}\big{|}\hat{V}^{n}_{t}-\hat{V}_{s}^{n}\big{|}^{p}\big{]}&\leq
C\,\mathbb{E}\Big{[}\Big{|}\int_{0}^{s}\Big{(}\hat{K}(t-\eta_{n}(r))-\hat{K}(s-\eta_{n}(r))\Big{)}b^{n}(r,\hat{V}^{n})~{}dr\Big{|}^{p}\Big{]}\\\
&~{}~{}~{}+C\,\mathbb{E}\Big{[}\Big{|}\int_{0}^{s}\Big{(}\hat{K}(t-\eta_{n}(r))-\hat{K}(s-\eta_{n}(r))\Big{)}^{2}\sigma^{n}(r,\hat{V}^{n})^{2}~{}dr\Big{|}^{\frac{p}{2}}\Big{]}\\\
&~{}~{}~{}+C\,\mathbb{E}\Big{[}\Big{|}\int_{s}^{t}\hat{K}(t-r)\,b^{n}(r,\hat{V}^{n})~{}dr\Big{|}^{p}\Big{]}~{}+C\,\mathbb{E}\Big{[}\Big{|}\int_{s}^{t}\hat{K}(t-\eta_{n}(r))^{2}\,\sigma^{n}(r,\hat{V}^{n})^{2}~{}dr\Big{|}^{\frac{p}{2}}\Big{]}.\end{split}$

Applying Hölder’s inequality and using the linear growth of $b^{n}$ and
$\sigma^{n}$, there is a positive constant $C>0$ (independent of $n\geq 1$),
such that

$\displaystyle\mathbb{E}\big{[}\big{|}\hat{V}_{t}^{n}-\hat{V}_{s}^{n}\big{|}^{p}\big{]}~{}\leq~{}$
$\displaystyle
C\mathbb{E}\Big{[}\Big{|}\int_{0}^{s}\Big{(}\hat{K}(t-\eta_{n}(r))-\hat{K}(s-\eta_{n}(r))\Big{)}^{2}\big{(}1+(\hat{V}^{n}_{\eta_{n}(r)})^{2}\big{)}~{}dr\Big{|}^{\frac{p}{2}}\Big{]}$
$\displaystyle+C\mathbb{E}\Big{[}\Big{|}\int_{s}^{t}\hat{K}(t-\eta_{n}(r))^{2}\,\big{(}1+(\hat{V}^{n}_{\eta_{n}(r)})^{2}\big{)}~{}dr\Big{|}^{\frac{p}{2}}\Big{]}.$

Applying now Minkowski’s integral inequality yields

$\begin{split}\mathbb{E}\big{[}\big{|}\hat{V}_{t}^{n}-\hat{V}_{s}^{n}\big{|}^{p}\big{]}~{}\leq~{}&C\Big{(}\int_{0}^{s}\Big{(}\hat{K}(t-\eta_{n}(r))-\hat{K}(s-\eta_{n}(r))\Big{)}^{2}\big{(}1+\mathbb{E}\big{[}|\hat{V}^{n}_{\eta_{n}(r)}|^{p}\big{]}^{\frac{2}{p}}\big{)}~{}dr\Big{)}^{\frac{p}{2}}\\\
&+C\Big{(}\int_{s}^{t}\hat{K}(t-\eta_{n}(r))^{2}\,\big{(}1+\mathbb{E}\big{[}|\hat{V}^{n}_{\eta_{n}(r)}|^{p}\big{]}^{\frac{2}{p}}\big{)}~{}dr\Big{)}^{\frac{p}{2}}.\end{split}$
(20)

It follows that there is a positive constant $C_{0}>0$ (independent of $n\geq
1$) such that

$\displaystyle\mathbb{E}\big{[}\big{|}\hat{V}_{t}^{n}\big{|}^{p}\big{]}~{}\leq~{}$
$\displaystyle
C\mathbb{E}\big{[}\big{|}\hat{V}^{n}_{s}\big{|}^{p}\big{]}+C\mathbb{E}\big{[}\big{|}\hat{V}^{n}_{t}-\hat{V}^{n}_{s}\big{|}^{p}\big{]}$
$\displaystyle~{}\leq~{}$ $\displaystyle
C_{0}\mathbb{E}\big{[}\big{|}\hat{V}^{n}_{s}\big{|}^{p}\big{]}+C_{0}\Big{(}\int_{0}^{s}\Big{(}\hat{K}(t-\eta_{n}(r))-\hat{K}(s-\eta_{n}(r))\Big{)}^{2}\big{(}1+\mathbb{E}\big{[}|\hat{V}^{n}_{\eta_{n}(r)}|^{p}\big{]}^{\frac{2}{p}}\big{)}~{}dr\Big{)}^{\frac{p}{2}}$
$\displaystyle+C_{0}\Big{(}\int_{s}^{t}\hat{K}(t-\eta_{n}(r))^{2}\,\big{(}1+\mathbb{E}\big{[}|\hat{V}^{n}_{\eta_{n}(r)}|^{p}\big{]}^{\frac{2}{p}}\big{)}~{}dr\Big{)}^{\frac{p}{2}}.$

Let $f_{n}(t):=\mathbb{E}\big{[}\big{|}\hat{V}^{n}_{t}\big{|}^{p}\big{]}$, one
can then apply Lemma 4.1 to obtain that

$\sup_{n\geq 1}\sup_{0\leq t\leq
T}\mathbb{E}\big{[}\big{|}\hat{V}^{n}_{t}\big{|}^{p}\big{]}~{}=~{}\sup_{n\geq
1}\sup_{0\leq t\leq T}f_{n}(t)~{}<~{}\infty.$

(ii) Plugging the estimate (16) in (20), one deduces from (4)-(5) that, for
some constant $C_{p}>0$ independent of $n$,

$\mathbb{E}\big{[}\big{|}\hat{V}^{n}_{t}-\hat{V}_{s}^{n}\big{|}^{p}\big{]}\leq
C_{p}(t-s)^{pH}.$

(iii) Notice that $Z^{n}$ is a semimartingale, with the uniform estimations in
(16), the estimation in (18) is standard. ∎

As a consequence of Proposition 4.2, we immediately obtain the following
tightness result.

######  Corollary 4.3.

The sequence $\\{(\hat{V}^{n},\hat{Z}^{n})\\}_{n\geq 1}$ is tight in
$C([0,T],\mathbb{R}^{2})\times C([0,T],\mathbb{R}^{2})$.

###### Proof.

By Proposition (4.2), we have by choosing $p$ large enough that

$\mathbb{E}[|\hat{V}_{t}^{n}-\hat{V}_{s}^{n}|^{p}]\leq
C(t-s)^{1+\alpha}~{}~{}\mbox{and}~{}\mathbb{E}[|\hat{Z}_{t}^{n}-\hat{Z}_{s}^{n}|^{p}]\leq
C(t-s)^{1+\alpha^{\prime}},$

for some positive constants $C,\,\alpha,\,\alpha^{\prime}$. Then by Theorem
2.4.11 of [31], one has that $\\{\hat{V}^{n}\\}_{n\geq 1}$ and
$\\{\hat{Z}^{n}\\}_{n\geq 1}$ are tight in $C([0,T],\mathbb{R}^{2})$. ∎

Then we identify the equation satisfied by the limit of
$\\{(\hat{V}^{n},\hat{Z}^{n})\\}_{n\geq 1}$, by considering an appropriate
martingale problem.

######  Lemma 4.4.

Let $A$ be a compact subset of $C([0,T],\mathbb{R}^{2})$ under the uniform
norm, then

$\sup_{\mathbf{x}\in A}\max_{0\leq t\leq
T}\big{(}|a^{n}(t,\mathbf{x})-a(\mathbf{x}_{t})|+|b^{n}(t,\mathbf{x})-b(\mathbf{x}_{t})|\big{)}\longrightarrow
0,~{}\mbox{as}~{}n\longrightarrow\infty.$

###### Proof.

Let $A$ be a compact subset of $C([0,T],\mathbb{R}^{2})$. Recall from the
Arzelà-Ascoli Theorem that the following properties hold:

$M:=\underset{\underset{0\leq t\leq T}{\mathbf{x}\in
A}}{\sup}|\mathbf{x}_{t}|<\infty,~{}~{}~{}\lim_{\delta\rightarrow
0}\sup_{\mathbf{x}\in
A}m(\mathbf{x},\delta)=0,~{}\mbox{with}~{}m(\mathbf{x},\delta):=\max\limits_{|s-t|\leq\delta,\,0\leq
s\leq t\leq T}|\mathbf{x}_{s}-\mathbf{x}_{t}|.$

Since $a$ and $b$ are uniformly continuous on
$\\{y\in\mathbb{R}^{2};~{}|y|\leq M\\}$, one can find, for any
$\varepsilon>0$, an integer $N\geq 1$ such that

$\sup_{\mathbf{x}\in A}\max_{0\leq t\leq
T}\big{(}|a^{n}(t,\mathbf{x})-a(\mathbf{x}_{t})|+|b^{n}(t,\mathbf{x})-b(\mathbf{x}_{t})|\big{)}\leq\varepsilon,~{}\mbox{for
all}~{}n\geq N.$

This concludes the proof. ∎

The next lemma is a convergence result for the operator $\mathcal{L}^{n}$
defined in (13), which is in line with Lemmas 3.5 and 3.6 of [5].

######  Lemma 4.5.

Let $(\mathbf{x}^{n},\mathbf{z}^{n})\longrightarrow(\mathbf{x},\mathbf{z})$ in
$C([0,T],\mathbb{R}^{2})\times C([0,T],\mathbb{R}^{2})$, then for any $f\in
C^{2}_{b}(\mathbb{R}^{2})$, the following uniform convergence holds:

$\lim_{n\to\infty}\sup_{t\in[0,T]}\Big{|}\int_{0}^{t}\mathcal{L}^{n}_{s}f(\mathbf{x}^{n},\mathbf{z}^{n})~{}ds-\int_{0}^{t}\mathcal{L}_{s}f(\mathbf{x},\mathbf{z})~{}ds\Big{|}=0.$

###### Proof.

Let us consider the following decomposition:

$\displaystyle\int_{0}^{t}\big{|}\mathcal{L}^{n}_{s}f(\mathbf{x}^{n},\mathbf{z}^{n})-\mathcal{L}_{s}f(\mathbf{x},\mathbf{z})\big{|}~{}ds$
$\displaystyle\leq\int_{0}^{T}\Big{|}b^{n}(s,\mathbf{x}^{n})\cdot\nabla
f(\mathbf{z}^{n}_{s})-b(\mathbf{x}_{s})\cdot\nabla
f(\mathbf{z}_{s})\Big{|}~{}ds+\int_{0}^{T}\Big{|}\mathrm{Tr}\big{(}a^{n}(s,\mathbf{x}^{n})\nabla^{2}f(\mathbf{z}^{n}_{s})-a(\mathbf{x}_{s})\nabla^{2}f(\mathbf{z}_{s})\big{)}\Big{|}~{}ds$
$\displaystyle=:I_{1}+I_{2}.$

We will detail the convergence of $I_{2}$, as the convergence of $I_{1}$
follows by a simpler argument. We get

$\displaystyle I_{2}$
$\displaystyle\leq\int_{0}^{T}\big{|}\mathrm{Tr}\big{(}\big{(}a^{n}(s,\mathbf{x}^{n})-a(\mathbf{x}^{n}_{s})\big{)}\nabla^{2}f(\mathbf{z}^{n}_{s})\big{)}\big{|}~{}ds$
$\displaystyle~{}+\int_{0}^{T}\big{|}\mathrm{Tr}\big{(}\big{(}a(\mathbf{x}^{n}_{s})-a(\mathbf{x}_{s})\big{)}\nabla^{2}f(\mathbf{z}^{n}_{s})\big{)}\big{|}~{}ds$
$\displaystyle~{}+\int_{0}^{T}\big{|}\mathrm{Tr}\big{(}a(\mathbf{x}_{s})~{}(\nabla^{2}f(\mathbf{z}^{n}_{s})-\nabla^{2}f(\mathbf{z}_{s}))\big{)}\big{|}~{}ds$
$\displaystyle=:I_{21}+I_{22}+I_{23}.$

Since $(\mathbf{x}^{n})$ converges uniformly, it lies in a compact of
$C([0,T],\mathbb{R}^{2})$. Hence by Lemma 4.4, $a^{n}(t,\mathbf{x}^{n})$
converges to $a(\mathbf{x}_{t})$ uniformly in $t\in[0,T]$. Then with the
boundedness of $\nabla^{2}f$, this yields the convergence
$I_{21}\longrightarrow 0$. The convergence of $I_{22}$ to $0$ is a direct
consequence of the Lipschitz continuity of $a$, the boundedness of $\nabla f$
and the uniform convergence of $\mathbf{x}^{n}$ towards $\mathbf{x}$. The
boundedness of $a$, continuity of $\nabla^{2}f$ and uniform convergence of
$\mathbf{z}^{n}$ then give, by an application of the dominated convergence
theorem, the convergence of $I_{23}$ to $0$. ∎

Now we define a martingale problem, which adapts the one from [5, Def. 3.1] to
our framework without jumps. It extends the usual definition of a martingale
problem to take into account the non-Markovian property because of the kernel
$K$. As in the classical setting, there is equivalence between being a weak
solution of an SDE and being a solution to the associated martingale problem.
More discussions are provided in Remark 4.8. We denote by
$C_{c}^{2}(\mathbb{R}^{2})$ the space of twice continuously differentiable
functions which are compactly supported.

######  Definition 4.6.

A solution of the (local) martingale problem for $(V_{0},K,\mathcal{L})$,
where $\mathcal{L}$ is the operator given by Equation (13), is a pair $(V,Z)$
of $C([0,T],\mathbb{R}^{2})$-valued processes defined on a filtered
probability space, such that $V$ is predictable, $Z$ is a continuous
semimartingale, the process

$M_{t}^{f}=f(Z_{t})-\int_{0}^{t}\mathcal{L}_{s}f(V,Z)~{}ds,~{}t\in[0,T],$

is a local martingale for every $f\in C_{c}^{2}(\mathbb{R}^{2})$, and one has
the equality

$V_{t}=V_{0}+\int_{0}^{t}K(t-s)\,dZ_{s},\quad t\in[0,T],~{}\mbox{a.s.}$ (21)

######  Proposition 4.7.

Any weak limit of $(\hat{V}^{n},\hat{Z}^{n})$ is a solution of the local
martingale problem for $(\hat{V}_{0},\hat{K},\mathcal{L})$.

###### Proof.

Let
$(\Omega^{n},\mathcal{F}^{n},\mathbb{F}^{n}=(\mathcal{F}_{t}^{n})_{t\in[0,T]},\mathbb{P}^{n})$
be a filtered probability space where $(\hat{V}^{n},\hat{Z}^{n})$ is defined
and $(\hat{V},\hat{Z})$ be a weak limit of $(\hat{V}^{n},\hat{Z}^{n})$. Fix
$f\in C^{2}_{c}(\mathbb{R}^{2})$ and $m\in\mathbb{N}$. For any $\mathbf{x}\in
C([0,T],\mathbb{R}^{2})$, define

$\tau_{m}(\mathbf{x})=\inf\\{t\geq 0:|\mathbf{x}(t)|\geq m\\}\wedge T.$

Then $\tau_{m}(\hat{V}^{n})$ is an $\mathbb{F}^{n}$-stopping time, and we have

$\int_{0}^{t\wedge\tau_{m}(\hat{V}^{n})}\big{|}\mathcal{L}_{s}^{n}f(\hat{V}^{n},\hat{Z}^{n})\big{|}~{}ds~{}\leq~{}C_{f}\int_{0}^{t\wedge\tau_{m}(\hat{V}^{n})}\big{(}1+|\hat{V}_{\eta_{n}(s)}^{n}|\big{)}~{}ds~{}\leq~{}(1+m)\,C_{f},~{}t\in[0,T].$

The Itô formula implies that the following process

$M_{t}^{n}=f\big{(}\hat{Z}^{n}_{t\wedge\tau_{m}(\hat{V}^{n})}\big{)}-\int_{0}^{t\wedge\tau_{m}(\hat{V}^{n})}\mathcal{L}_{s}^{n}f(\hat{V}^{n},\hat{Z}^{n})~{}ds$

is a uniformly bounded martingale. Thus, for any $0\leq
t_{1}<t_{2}...<t_{k}\leq s<t\leq T$ and $g_{i}\in
C_{b}(\mathbb{R}^{2}\times\mathbb{R}^{2}),~{}i=1,2,...,k$, we have

$\displaystyle\mathbb{E}\Big{[}\big{(}M^{n}_{t}-M^{n}_{s}\big{)}\prod_{i=1}^{k}g_{i}\big{(}\hat{V}^{n}_{t_{i}},\hat{Z}^{n}_{t_{i}}\big{)}\Big{]}~{}=~{}0.$
(22)

Then, by Skorokhod’s representation Theorem, we may assume that all
$(\hat{V}^{n},\hat{Z}^{n})$ and $(\hat{V},\hat{Z})$ are defined on a common
probability space $(\Omega,\mathcal{F},\mathbb{P})$, that
$(\hat{V}^{n},\hat{Z}^{n})\longrightarrow(\hat{V},\hat{Z})$ in
$C([0,T],\mathbb{R}^{2})\times C([0,T],\mathbb{R}^{2})$ almost surely, and
that each pair $(\hat{V}^{n},\hat{Z}^{n})$ has the same law under $\mathbb{P}$
as it did under $\mathbb{P}^{n}$.

By Lemma 4.5, one has

$\int_{0}^{t\wedge\tau_{m}(\hat{V}^{n})}\mathcal{L}^{n}_{s}f(\hat{V}^{n},\hat{Z}^{n})~{}ds~{}\longrightarrow~{}\int_{0}^{t\wedge\tau_{m}(\hat{V})}\mathcal{L}_{s}f(\hat{V},\hat{Z})~{}ds,$

almost surely for any $t\in[0,T]$. Define

$M_{t}~{}:=~{}f(Z_{t\wedge\tau_{m}(\hat{V})})-\int_{0}^{t\wedge\tau_{m}(\hat{V})}\mathcal{L}_{s}f(\hat{V},\hat{Z})~{}ds,~{}~{}t\in[0,T].$

We conclude that $M_{t}^{n}\longrightarrow M_{t}$ almost surely for
$t\in[0,T]$. Since $M_{t}^{n}$ is bounded uniformly in $n$, we may use the
dominated convergence theorem to pass to the limit $n\longrightarrow\infty$ in
(22) to get

$\displaystyle\mathbb{E}\Big{[}(M_{t}-M_{s})\prod_{i=1}^{k}g_{i}\big{(}\hat{V}_{t_{i}},\hat{Z}_{t_{i}}\big{)}\Big{]}=0.$
(23)

Thus $M$ is a martingale with respect to the filtration given by

$\mathcal{F}_{t}=\sigma(\hat{V}_{s},\hat{Z}_{s}:s\leq t),~{}t\in[0,T].$

Since $\tau_{m}(\hat{V})$ is a stopping time for this filtration and the
constant $m$ in the definition of $\tau_{m}(\hat{V})$ was arbitrary, the
process $M^{f}$

$M^{f}_{t}=f(\hat{Z}_{t})-\int_{0}^{t}\mathcal{L}_{s}f(\hat{V},\hat{Z})~{}ds,~{}t\in[0,T]$

is a local martingale.

Next, recall that by definition of $\hat{V}^{n}_{t}$, we have

$\begin{split}\hat{V}_{t}^{n}=\hat{V}_{0}&+\int_{0}^{t}\hat{K}(t-\eta_{n}(s))~{}d\hat{Z}^{n}_{s}.\end{split}$

Then it is enough to apply [32, Theorem 2.2] (see also [29]) to conclude that,
for each $t\in[0,T]$,

$\hat{V}_{t}=\hat{V}_{0}+\int_{0}^{t}\hat{K}(t-s)\,d\hat{Z}_{s},~{}\mbox{a.s.}$

Notice that the process $V$ is continuous in $t$, one then proves (21). ∎

######  Remark 4.8.

$\mathrm{(i)}$ Our definition of martingale problem in Definition 4.6 differs
slightly from [5, Def. 3.1]. More precisely, our condition (21) is on the
process $V$, but Condition (3.3) of [5] is on the process
$\int_{0}^{t}V_{s}\,ds$. Nevertheless, as soon as $V$ has continuous paths,
the two conditions are in fact equivalent, so that the two definitions are
equivalent. The formulation based on $\int_{0}^{t}V_{s}\,ds$ has the advantage
to obtain the tightness more easily than that of $V$, but still we are able to
obtain the tightness of $V$ in our context.

As in the classical setting, it is easy to check that $(\hat{V},\hat{Z})$ is a
solution of the local martingale problem (Definition 4.6) for
$(\hat{V}_{0},\hat{K},\mathcal{L})$ if and only if $\hat{V}=(Y,V)$ satisfies

$\begin{split}Y_{t}&=Y_{0}+\int_{0}^{t}-\frac{1}{2}(V_{s})_{+}~{}ds+\int_{0}^{t}\sqrt{(V_{s})_{+}}~{}d(\rho
W_{s}+\sqrt{1-\rho^{2}}W_{s}^{\bot}),\\\
V_{t}&=V_{0}+\int_{0}^{t}K(t-s)(\theta-\lambda(V_{s})_{+})~{}ds+\int_{0}^{t}K(t-s)\nu\sqrt{(V_{s})_{+}}~{}dW_{s}.\end{split}$
(24)

The proof is almost the same as Lemma 3.3 of [5], with some simplifications
due to the absence of jumps here. It is therefore omitted.

Consequently, to prove the convergence result in Theorem 2.2, it is enough to
prove that Equation (24) has the same unique solution as that of (6). This
will be proven in Proposition 4.13.(i).

###  4.2 Tightness of solutions to Scheme (9)

Let us first rewrite the discrete scheme (9) as an integrated-rough Volterra
equation in a continuous-time form: $S^{n}=\exp(Y^{n})$, and

$\begin{cases}Y_{t}^{n}&\\!\\!=~{}Y_{0}+\rho
M^{n}_{t}+\sqrt{1-\rho^{2}}M^{n,\bot}_{t}-\frac{1}{2}\overline{X}^{n}_{t},\\\
X_{t}^{n}&\\!\\!=~{}V_{0}t+\int_{0}^{t}K(t-\eta_{n}(s))\Big{(}\theta\eta_{n}(s)-\lambda\overline{X}^{n}_{\eta_{n}(s)}+\nu
M^{n}_{\eta_{n}(s)}\Big{)}~{}ds,\\\
\overline{X}^{n}_{t_{k}}&\\!\\!=~{}\max\limits_{0\leq j\leq
k}X^{n}_{t_{j}},\\\
\overline{X}^{n}_{t}&\\!\\!=~{}\overline{X}^{n}_{t_{k-1}}+(\overline{X}^{n}_{t_{k}}-\overline{X}^{n}_{t_{k-1}})\frac{t-t_{k-1}}{t_{k}-t_{k-1}}~{},~{}\mbox{for}~{}t\in[t_{k-1},t_{k}),\\\
M^{n}_{t}&\\!\\!=~{}M^{n}_{t_{k-1}}+\sqrt{\overline{X}^{n}_{t_{k}}-\overline{X}^{n}_{t_{k-1}}}\frac{B_{t}-B_{t_{k-1}}}{\sqrt{t_{k}-t_{k-1}}}~{},~{}\mbox{for}~{}t\in[t_{k-1},t_{k}),\\\
M^{n,\bot}_{t}&\\!\\!=~{}M^{n,\bot}_{t_{k-1}}+\sqrt{\overline{X}^{n}_{t_{k}}-\overline{X}^{n}_{t_{k-1}}}\frac{B^{\bot}_{t}-B^{\bot}_{t_{k-1}}}{\sqrt{t_{k}-t_{k-1}}}~{},~{}\mbox{for}~{}t\in[t_{k-1},t_{k}),\end{cases}$
(25)

where $M^{n}_{t_{0}}=M^{n,\bot}_{t_{0}}=0$, $B$ and $B^{\bot}$ are independent
standard Brownian motions. In view of the above definition, $M^{n}$ and
$M^{n,\bot}$ are continuous martingales with quadratic variation

$\langle M^{n}\rangle_{t}=\langle
M^{n,\bot}\rangle_{t}=\overline{X}^{n}_{t_{k-1}}+(\overline{X}^{n}_{t_{k}}-\overline{X}^{n}_{t_{k-1}})\frac{t-t_{k-1}}{t_{k}-t_{k-1}}~{},~{}\mbox{for}~{}t\in[t_{k-1},t_{k}),$

which means that for any $t\in[0,T]$,

$\langle M^{n}\rangle_{t}=\langle M^{n,\bot}\rangle_{t}=\overline{X}^{n}_{t}.$
(26)

Recall the following BDG inequality: for some constant $C>0$, one has

$\mathbb{E}\big{[}\big{|}M_{t}^{n}-M_{s}^{n}\big{|}^{p}\big{]}+\mathbb{E}\big{[}\big{|}M^{n,\bot}_{t}-M^{n,\bot}_{s}\big{|}^{p}\big{]}\leq
C\,\mathbb{E}\big{[}\big{|}\overline{X}^{n}_{t}-\overline{X}^{n}_{s}\big{|}^{\frac{p}{2}}\big{]},~{}~{}\mbox{for
all}~{}0\leq s\leq t\leq T.$ (27)

######  Lemma 4.9.

Assume $(\mathbf{z}^{n})_{n\geq 1}\subseteq C([0,T],\mathbb{R}^{2})$ and
$\mathbf{z}^{n}\longrightarrow z$ in $C([0,T],\mathbb{R}^{2})$. Then the
following uniform convergence holds:

$\lim_{n\to\infty}\sup_{t\in[0,T]}\Big{|}\int_{0}^{t}\hat{K}(t-\eta_{n}(s))\mathbf{z}^{n}_{s}~{}ds-\int_{0}^{t}\hat{K}(t-s)\mathbf{z}_{s}~{}ds\Big{|}=0.$

###### Proof.

One has

$\begin{split}&\Big{|}\int_{0}^{t}\hat{K}(t-s)\mathbf{z}_{s}~{}ds-\int_{0}^{t}\hat{K}(t-\eta_{n}(s))\mathbf{z}^{n}_{s}~{}ds\Big{|}\\\
\leq&\Big{|}\int_{0}^{t}\hat{K}(t-s)(\mathbf{z}_{s}-\mathbf{z}^{n}_{s})~{}ds\Big{|}+\Big{|}\int_{0}^{t}\Big{(}\hat{K}(t-s)-\hat{K}(t-\eta_{n}(s))\Big{)}\mathbf{z}^{n}_{s}~{}ds\Big{|}.\\\
\end{split}$

The term $|\int_{0}^{t}\hat{K}(t-s)(\mathbf{z}_{s}-\mathbf{z}^{n}_{s})~{}ds|$
converges to 0 uniformly since $\mathbf{z}^{n}\longrightarrow\mathbf{z}$
uniformly and ${|\int_{0}^{t}\hat{K}(t-s)~{}ds|<\infty}$ by the fact that
$K\in L^{2}([0,T])$.

As for the second term, it follows from the boundedness of $(\mathbf{z}^{n})$
that

$\Big{|}\int_{0}^{t}\Big{(}\hat{K}(t-s)-\hat{K}(t-\eta_{n}(s))\Big{)}\mathbf{z}^{n}_{s}~{}ds\Big{|}\leq\sup_{n\in\mathbb{N}}\|\mathbf{z}^{n}\|_{\infty}\,\int_{0}^{t}\Big{|}\hat{K}(t-s)-\hat{K}(t-\eta_{n}(s))\Big{|}~{}ds~{}\longrightarrow~{}0.$

∎

Since the processes $\overline{X}^{n}$ and
$\\{\sup_{s\in[0,t]}X^{n}_{s},\,t\in[0,T]\\}$ may differ, we now introduce the
auxiliary process

$\widetilde{X}^{n}_{t}=\sup_{s\in[0,t]}X^{n}_{s}.$

We will eventually show that $\overline{X}^{n}$ and $\widetilde{X}^{n}$
converge to the same limit, which is $X$.

######  Proposition 4.10.

For all $p\geq 2$, there exists a constant $C_{p}>0$ such that for all $0\leq
s\leq t\leq T$ and $n\geq 1$,

$\begin{split}(i)~{}&\mathbb{E}\big{[}\big{|}X^{n}_{T}\big{|}^{p}\big{]}+\mathbb{E}\big{[}\big{|}\overline{X}_{T}^{n}\big{|}^{p}\big{]}~{}\leq~{}C_{p},\\\
(ii)~{}&\mathbb{E}\big{[}\big{|}X^{n}_{t}-X^{n}_{s}\big{|}^{p}\big{]}+\mathbb{E}\big{[}\big{|}\overline{X}^{n}_{t}-\overline{X}^{n}_{s}\big{|}^{p}\big{]}+\mathbb{E}\big{[}\big{|}\widetilde{X}^{n}_{t}-\widetilde{X}^{n}_{s}\big{|}^{p}\big{]}~{}\leq~{}C_{p}\,(t-s)^{pH},\\\
(iii)~{}&\mathbb{E}\big{[}\big{|}M^{n}_{t}-M^{n}_{s}\big{|}^{2p}\big{]}+\mathbb{E}\big{[}\big{|}M^{n,\bot}_{t}-M^{n,\bot}_{s}\big{|}^{2p}\big{]}~{}\leq~{}C_{p}\,(t-s)^{pH}.\\\
\end{split}$

###### Proof.

$(i)$ Recall that $K$ is non-increasing, so that

$\int_{0}^{t}|K(t-\eta_{n}(s))|^{2}\,ds~{}\leq~{}\int_{0}^{T}|K(s)|^{2}ds~{}<~{}\infty,~{}~{}\mbox{for
all}~{}n\geq 1~{}\mbox{and}~{}t\in[0,T].$

Then by the equation on $X^{n}$ in (25), together with Cauchy-Schwarz and
Hölder’s inequality, one has for some positive constants $C_{0}$ and $C_{1}$
(independent of $n\geq 1$) that

$\displaystyle\big{|}{X}^{n}_{t}\big{|}^{p}$ $\displaystyle\leq
C_{0}\,|V_{0}|^{p}+C_{0}\,\Big{|}\int_{0}^{t}K(t-\eta_{n}(s))\Big{(}\theta\eta_{n}(s)-\lambda\overline{X}^{n}_{\eta_{n}(s)}+\nu
M^{n}_{\eta_{n}(s)}\Big{)}~{}ds\Big{|}^{p}$ $\displaystyle\leq
C_{0}\,|V_{0}|^{p}+C_{1}\int_{0}^{t}\big{(}\theta\eta_{n}(s)+\lambda|\overline{X}^{n}_{\eta_{n}(s)}|+\nu|M^{n}_{\eta_{n}(s)}|\big{)}^{p}~{}ds.$

Using the inequality $\sqrt{x}\leq 1+x$, one obtains that for some constant
$C_{2}>0$ (independent of $n$),

$\mathbb{E}\Big{[}\sup_{s\in[0,t]}\big{|}X_{s}^{n}\big{|}^{p}\Big{]}~{}\leq~{}C_{2}+C_{2}\int_{0}^{t}\mathbb{E}\big{[}\big{|}\overline{X}^{n}_{\eta_{n}(s)}\big{|}^{p}\big{]}~{}ds.$

Since $\sup_{t\in[0,T]}X^{n}_{t}=\overline{X}^{n}_{T}$ and
$\overline{X}^{n}_{\eta_{n}(s)}\leq\overline{X}^{n}_{s}$, it follows by
Grönwall’s lemma that, for some constant $C_{p}$ independent of $n\geq 1$,

$\sup_{0\leq t\leq
T}\mathbb{E}\big{[}\big{|}X^{n}_{t}\big{|}^{p}\big{]}~{}\leq~{}\mathbb{E}\big{[}\big{|}\overline{X}_{T}^{n}\big{|}^{p}\big{]}~{}\leq~{}C_{p}.$

$(ii)$ First, we observe that
$\overline{X}^{n}_{t}-\overline{X}^{n}_{s}\leq\sup\limits_{r\in[s,t]}(X_{r}^{n}-X^{n}_{s})$
and
$\widetilde{X}^{n}_{t}-\widetilde{X}^{n}_{s}\leq\sup\limits_{r\in[s,t]}(X_{r}^{n}-X^{n}_{s})$.
Hence it suffices to prove that

$\mathbb{E}[\sup_{r\in[s,t]}(X_{r}^{n}-X^{n}_{s})^{p}]\leq
C\,(1+|V_{0}|^{p})\,(t-s)^{pH}.$ (28)

From the Cauchy-Schwarz inequality and (4)-(5) we get that

$\displaystyle|X_{r}^{n}-X_{s}^{n}|^{p}$
$\displaystyle=\Big{|}V_{0}\,(r-s)+\int_{s}^{r}K(r-\eta_{n}(u))\big{(}\theta\eta_{n}(u)-\lambda\overline{X}^{n}_{\eta_{n}(u)}+\nu
M^{n}_{\eta_{n}(u)}\big{)}~{}du$
$\displaystyle\quad\quad+\int_{0}^{s}\big{(}K(r-\eta_{n}(u))-K(s-\eta_{n}(u))\big{)}\big{(}\theta\eta_{n}(u)-\lambda\overline{X}^{n}_{\eta_{n}(u)}+\nu
M^{n}_{\eta_{n}(u)}\big{)}~{}du\Big{|}^{p}$ $\displaystyle\leq
C|V_{0}|^{p}(r-s)^{p}+C(r-s)^{pH}\Big{|}\int_{s}^{r}\Big{(}\theta\eta_{n}(u)-\lambda\overline{X}^{n}_{\eta_{n}(u)}+\nu
M^{n}_{\eta_{n}(u)}\Big{)}^{2}~{}du\Big{|}^{\frac{p}{2}}$
$\displaystyle~{}~{}~{}+C(r-s)^{pH}\Big{|}\int_{0}^{s}\Big{(}\theta\eta_{n}(u)-\lambda\overline{X}^{n}_{\eta_{n}(u)}+\nu
M^{n}_{\eta_{n}(u)}\Big{)}^{2}~{}du\Big{|}^{\frac{p}{2}}.$

Thus taking the supremum on $[s,t]$ and the expectation, and applying again
the Hölder inequality ($\frac{p}{2}>1$), one gets

$\displaystyle\mathbb{E}\Big{[}\sup_{r\in[s,t]}|X_{r}^{n}-X_{s}^{n}|^{p}\Big{]}$
$\displaystyle\leq
C|V_{0}|^{p}\,(t-s)^{p}+C(t-s)^{pH}\int_{s}^{t}\mathbb{E}\Big{[}\big{|}\theta\eta_{n}(u)-\lambda\overline{X}^{n}_{\eta_{n}(u)}+\nu
M^{n}_{\eta_{n}(u)}\big{|}^{p}\Big{]}~{}du$
$\displaystyle~{}~{}~{}+C(t-s)^{pH}\int_{0}^{s}\mathbb{E}\Big{[}\big{|}\theta\eta_{n}(u)-\lambda\overline{X}^{n}_{\eta_{n}(u)}+\nu
M^{n}_{\eta_{n}(u)}\big{|}^{p}\Big{]}~{}du.$

Hence using (27) and point $(i)$, we deduce from the previous inequality that
(28) holds.

$(iii)$ From (27), we have

$\mathbb{E}\big{[}\big{|}M^{n}_{t}-M^{n}_{s}\big{|}^{2p}\big{]}\leq
C~{}\mathbb{E}\big{[}\big{|}\overline{X}^{n}_{t}-\overline{X}^{n}_{s}\big{|}^{p}\big{]}.$

Then by $(ii)$ one obtains the result. Similarly, the same holds for
$M^{n,\bot}$. ∎

######  Corollary 4.11.

By passing to a subsequence, one has
$(X^{n},\overline{X}^{n},M^{n},M^{n,\bot})\Longrightarrow(X,\overline{X},M,M^{\bot})$
for some stochastic processes $X,~{}M,~{}M^{\bot}$ in $C([0,T],\mathbb{R})$,
where $\overline{X}_{t}=\sup_{r\in[0,t]}X_{r}$ and $M,~{}M^{\bot}$ are two
martingales in $C([0,T],\mathbb{R})$ with quadratic variation $\langle
M\rangle=\langle M^{\bot}\rangle=\overline{X}$ and covariation $\langle
M,M^{\bot}\rangle=0$.

###### Proof.

In view of Proposition 4.10 and Theorem 2.4.11 of [31], the sequences of
processes $\\{X^{n}\\}_{n\in\mathbb{N}}$,
$\\{\widetilde{X}^{n}\\}_{n\in\mathbb{N}}$,
$\\{\overline{X}^{n}\\}_{n\in\mathbb{N}}$, $\\{M^{n}\\}_{n\in\mathbb{N}}$ and
$\\{M^{n,\bot}\\}_{n\in\mathbb{N}}$ are tight in $C([0,T],\mathbb{R})$. Up to
passing to a subsequence (without loss of generality, we do not rename the
processes), one has
$(X^{n},\widetilde{X}^{n},M^{n},M^{n,\bot})\Longrightarrow(X,Y,M,M^{\bot})$
for some processes $X,Y,M,M^{\bot}$ in $C([0,T],\mathbb{R})$.

For any $s_{1},~{}s_{2},...,s_{k}\in[0,T]$, the mapping

$C([0,T],\mathbb{R})\ni\mathbf{x}~{}\longmapsto~{}\Big{(}\sup\limits_{r\in[0,s_{1}]}\mathbf{x}_{r},\sup\limits_{r\in[0,s_{2}]}\mathbf{x}_{r},\dots,\sup\limits_{r\in[0,s_{k}]}\mathbf{x}_{r}\Big{)}\in\mathbb{R}^{k}$

is continuous. So by the continuous mapping theorem and the convergence in law
of $X^{n}$ to $X$, there is

$\big{(}\widetilde{X}^{n}_{s_{1}},\widetilde{X}^{n}_{s_{2}},\dots,\widetilde{X}^{n}_{s_{k}}\big{)}\Longrightarrow\big{(}\overline{X}_{s_{1}},\overline{X}_{s_{2}},\dots,\overline{X}_{s_{k}}\big{)}.$

Moreover the convergence in law of $\widetilde{X}^{n}$ to $Y$ implies that

$\big{(}\widetilde{X}^{n}_{s_{1}},\widetilde{X}^{n}_{s_{2}},\dots,\widetilde{X}^{n}_{s_{k}}\big{)}\Longrightarrow\big{(}Y_{s_{1}},Y_{s_{2}},...,Y_{s_{k}}\big{)}.$

Thus the finite dimensional distributions of $\overline{X}$ and $Y$ are
identical, and since the class of finite dimensional distributional sets is a
separating class in $C([0,T],\mathbb{R})$, we obtain $Y=\overline{X}$.

We now aim at proving that $\overline{X}^{n}\Longrightarrow\overline{X}$. In
view of the convergence of $\widetilde{X}^{n}$ to $\overline{X}$, it suffices
to prove that
$\sup_{t\in[0,T]}|\widetilde{X}^{n}_{t}-\overline{X}^{n}_{t}|\Longrightarrow
0$.

For $t\in[0,T]$, one has

$\widetilde{X}_{t}^{n}-\overline{X}^{n}_{t}\leq\widetilde{X}^{n}_{t}-\overline{X}^{n}_{\eta_{n}(t)}\leq\sup\limits_{s\in[0,t]}(X^{n}_{s}-X^{n}_{\eta_{n}(s)}).$

From Proposition 4.10 and the Markov inequality, one gets that

$\sup\limits_{t\in[0,T]}|\widetilde{X}^{n}_{t}-\overline{X}^{n}_{t}|\Longrightarrow
0.$

Hence $\overline{X}^{n}\Longrightarrow\overline{X}$.

From
$(M^{n},M^{n,\bot},\overline{X}^{n})\Longrightarrow(M,M^{\bot},\overline{X})$,
we have

$(M^{n})^{2}-\overline{X}^{n}\Longrightarrow M^{2}-\overline{X},$
$(M^{n,\bot})^{2}-\overline{X}^{n}\Longrightarrow(M^{\bot})^{2}-\overline{X},$
$M^{n}M^{n,\bot}\Longrightarrow MM^{\bot}.$

By Proposition IX.1.17 in [28], the processes
$M,~{}M^{\bot},~{}M^{2}-\overline{X},~{}(M^{\bot})^{2}-\overline{X},~{}MM^{\bot}$
are all continuous local martingales w.r.t. the filtration generated by
$(M,M^{\bot},\overline{X},X)$. Therefore, $\langle M\rangle=\langle
M^{\bot}\rangle=\overline{X},~{}\langle M,M^{\bot}\rangle=0$. Moreover, by the
uniform moment estimates on $(M^{n},M^{n,\bot})$ (see Proposition 4.10), it
follows that they are continuous martingales. ∎

As an immediate consequence of Corollary 4.11, we get that up to taking a
subsequence, the following convergence holds:

$(S^{n},Y^{n})\Longrightarrow(S,Y),~{}\mbox{with}~{}S:=\exp(Y)~{}\mbox{and}~{}Y:=\rho
M+\sqrt{1-\rho^{2}}M^{\bot}-\frac{1}{2}\overline{X}.$

We next prove that the limit processes $(X,\overline{X},S,Y,M,M^{\bot})$
satisfy

$\begin{cases}X_{t}=V_{0}t+\int_{0}^{t}K(t,s)(\theta
s-\lambda\overline{X}_{s}+\nu M_{s})~{}ds,\\\ \overline{X}_{t}=\max_{0\leq
s\leq t}X_{s},\\\ S_{t}=S_{0}\exp(\rho
M_{t}+\sqrt{1-\rho^{2}}M^{\bot}_{t}-\frac{1}{2}\overline{X}_{t}),\\\
M,\,M^{\bot}\mbox{ are continuous martingales with }\langle M\rangle=\langle
M^{\bot}\rangle=\overline{X},~{}\langle
M,M^{\bot}\rangle=0~{}\mbox{and}~{}M_{0}=M^{\bot}_{0}=0.\end{cases}$ (29)

######  Proposition 4.12.

Any weak limit $(X,\overline{X},S,M,M^{\bot})$ of
$(X^{n},\overline{X}^{n},S^{n},M^{n},M^{n,\bot})_{n\geq 1}$ in Corollary 4.11
satisfies Equation (29).

###### Proof.

Let
$(\Omega^{n},\mathcal{F}^{n},\mathbb{F}^{n}=(\mathcal{F}^{n}_{t})_{t\in[0,T]}),\mathbb{P}^{n})$
be a filtered probability space in which
$(X^{n},\overline{X}^{n},M^{n},S^{n})$ is defined. Corollary 4.11 gives the
existence of a subsequence $(n_{k})$ such that
$(X^{n_{k}},\overline{X}^{n_{k}},M^{n_{k}},S^{n_{k}})$ converges, and without
loss of generality we simply consider that
$(X^{n},\overline{X}^{n},M^{n},S^{n})\Longrightarrow(X,\overline{X},M,S)$, for
some $X$ in $C([0,T],\mathbb{R})$ and $M$ a martingale with quadratic
variation $\overline{X}$ in a filtered probability space
$(\Omega,\mathcal{F},\mathbb{F}=(\mathcal{F}_{t})_{t\in[0,T]},\mathbb{P})$. By
the Skorokhod representation theorem, one can assume that
$(\Omega^{n},\mathcal{F}^{n},\mathbb{P}^{n})=(\Omega,\mathcal{F},\mathbb{P})$
(but the filtrations $\mathbb{F}^{n}$ and $\mathbb{F}$ are different a priori)
such that
$({X}^{n},{\overline{X}^{n}},{M}^{n},{S}^{n})\longrightarrow({X},{\overline{X}},{M},{S})$
almost surely in $C([0,T],\mathbb{R})^{4}$. From Lemma 4.9, one has

$\int_{0}^{t}K(t-\eta_{n}(s))(\theta\eta_{n}(s)-\lambda{\overline{X}^{n}}_{\eta_{n}(s)}+\nu{M}^{n}_{\eta_{n}(s)})~{}ds\longrightarrow\int_{0}^{t}K(t-s)(\theta
s-\lambda{\overline{X}}_{s}+\nu{M}_{s})~{}ds$

uniformly in $t\in[0,T]$, almost surely. Therefore, $(X,S,M,M^{\bot})$ solves
(29). ∎

###  4.3 Equivalence and uniqueness of weak solutions

We will adapt the ideas in the proof of [2, Theorem A.1], in order to prove
that the process $V$ in a (weak) solution of (24) is non-negative, and that
the process $X$ in a (weak) solution of (29) is non-decreasing. Consequently,
Equation (24) shares the same unique weak solution as (6), and Equation (29)
shares the same unique weak solution as (7).

Recall that the resolvent $L$ of the first kind of the kernel function $K$ is
defined in (3), together with the definition of the convolution:

$(L^{*}*K)(t)=(K*L^{*})(t):=\int_{[0,t]}K(t-s)L^{*}(ds),~{}~{}\mbox{for a
finite measure}~{}L^{*}~{}\mbox{on}~{}[0,T].$

Let us fix $K(t):=K(T)$ for $t\geq T$, and denote
$\Delta_{h}K(\cdot)=K(h+\cdot)$.

######  Proposition 4.13.

(i) Let $V$ be a (weak) solution to the following equation

$\begin{split}V_{t}~{}=~{}V_{0}+\int_{0}^{t}K(t-s)\big{(}\theta-\lambda(V_{s})_{+}\big{)}~{}ds+\int_{0}^{t}K(t-s)\nu\sqrt{(V_{s})_{+}}~{}dW_{s},\end{split}$
(30)

where $x_{+}=max(x,0)$. Then $V$ is non-negative on $[0,T]$.

(ii) Let $X$ be a (weak) solution to the following equation

$\begin{split}X_{t}~{}=~{}V_{0}t+\int_{0}^{t}K(t-s)\big{(}\theta
s-\lambda\overline{X}_{s}\big{)}~{}ds+\int_{0}^{t}K(t-s)\nu
M_{s}~{}ds,\end{split}$ (31)

where $\overline{X_{t}}=\sup\limits_{s\in[0,t]}X_{s}$, $M_{s}$ is a continuous
martingale with quadratic variation $\langle M\rangle_{s}=\overline{X}_{s}$
and $M_{0}=0$. Then $X$ is non-decreasing on $[0,T]$.

###### Proof.

(i) Define the stopping time $\tau_{n}:=\inf\\{t:V_{t}<-\frac{1}{n}\\}$.
Assume that the set $\\{\tau_{n}<T\\}$ is nonempty, then for every fixed
$\omega\in\\{\tau_{n}<T\\}$, there exists $h>0$ such that

$V_{\tau_{n}+h}<-\frac{1}{n},~{}~{}\mbox{and}~{}V_{s}\leq 0~{}\mbox{for
all}~{}s\in[\tau_{n},\tau_{n}+h].$ (32)

Denote

$Z_{s}:=\int_{0}^{s}b((V_{s})_{+})~{}ds+\int_{0}^{t}\sigma((V_{s})_{+})~{}dW_{s},~{}\mbox{with}~{}b(v):=\theta-\lambda
v_{+},~{}~{}\sigma(x)=\nu\sqrt{x_{+}}.$

It follows that

$V_{t+h}~{}=~{}V_{0}+(K*dZ)_{t+h}~{}=~{}V_{0}+(\Delta_{h}K*dZ)_{t}+Y_{t},$

where

$Y_{t}=\int_{t}^{t+h}K(t+h-s)~{}dZ_{s},~{}t\geq 0.$

By (3.7) of [4], one has for any stopping time $\tau<\infty$,

$Y_{\tau}=\int_{\tau}^{\tau+h}K(\tau+h-s)~{}dZ_{s},~{}\mbox{a.s.}$

Then one has

$V_{\tau_{n}+h}~{}=~{}V_{0}+(\Delta_{h}K*dZ)_{\tau_{n}}+\int_{\tau_{n}}^{\tau_{n}+h}K(\tau_{n}+h-s)~{}dZ_{s}.$

Recall also from [2, Lemma B.2] that for any $F\in L^{1}([0,T])$ such that
$F*L$ is right-continuous and of bounded variation, one has

$F(t)=(F*L)(0)K(t)+\big{(}K*(d(F*L))\big{)}(t),~{}\mbox{for
a.e.}~{}t\in[0,T].$

Then for $F=\Delta_{h}K$ and $L$ the resolvent of the first kind of $K$,
$\Delta_{h}K*L$ is right-continuous and of bounded variation (see [2, Remark
B.3.]). Thus one has

$\begin{split}V_{\tau_{n}+h}~{}=~{}&V_{0}+(\Delta_{h}K*L)(0)\big{(}V_{\tau_{n}}-V_{0}\big{)}+\big{(}d(\Delta_{h}K*L)*(V-V_{0})\big{)}_{\tau_{n}}+\int_{\tau_{n}}^{\tau_{n}+h}K(\tau_{n}+h-s)~{}dZ_{s}\\\
~{}=~{}&V_{0}~{}+~{}V_{\tau_{n}}(\Delta_{h}K*L)(0)~{}+\int_{0}^{\tau_{n}}V_{\tau_{n}-s}d\big{(}\Delta_{h}K*L\big{)}(s)~{}-~{}V_{0}~{}\big{(}\Delta_{h}K*L\big{)}(\tau_{n})\\\
&~{}+\int_{\tau_{n}}^{\tau_{n}+h}K(\tau_{n}+h-s)~{}dZ_{s}.\end{split}$

Notice that $V_{\tau_{n}}=-\frac{1}{n}$, $V_{\tau_{n}-s}\geq-\frac{1}{n}$ for
$s\in[0,\tau_{n}]$, and $(\Delta_{h}K*L)$ is non-negative and non-decreasing
(see [2, Remark B.3.]). It follows that

$\displaystyle V_{\tau_{n}+h}~{}\geq~{}$ $\displaystyle
V_{0}~{}-~{}\frac{1}{n}(\Delta_{h}K*L)(0)~{}-~{}\frac{1}{n}(\Delta_{h}K*L)(\tau_{n})~{}+~{}\frac{1}{n}(\Delta_{h}K*L)(0)$
$\displaystyle~{}-~{}V_{0}(\Delta_{h}K*L)(\tau_{n})~{}+~{}\int_{\tau_{n}}^{\tau_{n}+h}\\!\\!\\!K(\tau_{n}+h-s)~{}dZ_{s}.$

Further, notice that for all $t\in[0,T]$,

$\displaystyle(\Delta_{h}K*L)(t)~{}=~{}\int_{0}^{t}K(t+h-s)L(ds)~{}\leq~{}\int_{0}^{t+h}K(t+h-s)L(ds)~{}=~{}(\Delta_{h}K*L)(t+h)~{}=~{}1,$

since $K$ and $L$ are non-negative. Then

$V_{\tau_{n}+h}~{}\geq~{}-\frac{1}{n}(\Delta_{h}K*L)(\tau_{n})+\int_{\tau_{n}}^{\tau_{n}+h}K(\tau_{n}+h-s)~{}dZ_{s}~{}\geq~{}-\frac{1}{n}+\int_{\tau_{n}}^{\tau_{n}+h}K(\tau_{n}+h-s)~{}dZ_{s}.$

Finally, since $b((V_{s})_{+})\geq 0$, $\sigma((V_{s})_{+})=0$ for all
$s\in[\tau_{n},\tau_{n}+h]$, it follows that

$\int_{\tau_{n}}^{\tau_{n}+h}K(\tau_{n}+h-s)~{}dZ_{s}~{}=~{}\int_{\tau_{n}}^{\tau_{n}+h}K(\tau_{n}+h-s)\Big{(}b((V_{s})_{+})~{}ds+\sigma((V_{s})_{+})~{}dW_{s}\Big{)}~{}\geq~{}0,$

and hence $V_{\tau_{n}+h}\geq-\frac{1}{n}$, which is a contradiction to the
fact that $V_{\tau_{n}+h}<-\frac{1}{n}$ in (32). Therefore,
$\mathbb{P}[\tau_{n}<T]=0$. Since $n\geq 1$ is arbitrary, it follows that
$V_{t}\geq 0$ for any $t\in[0,T]$, a.s.

(ii) For equation (31), denote $Z_{s}:=\theta s-\lambda\overline{X}_{s}+\nu
M_{s}$, where $M$ is a continuous martingale with quadratic variation $\langle
M\rangle_{s}=\overline{X}_{s},~{}M_{0}=0$. We follow the proof of [1, Lemma
2.1] and apply the stochastic Fubini theorem (see e.g. [34, p.175]) to obtain
that

$\begin{split}X_{t}&~{}=~{}V_{0}t+\int_{0}^{t}K(t-s)Z_{s}~{}ds~{}=~{}V_{0}t+\int_{0}^{t}K(s)\Big{(}\int_{0}^{t-s}~{}dZ_{r}\Big{)}~{}ds\\\
&~{}=~{}V_{0}t+\int_{0}^{t}\Big{(}\int_{0}^{t-r}K(s)~{}ds\Big{)}~{}dZ_{r}~{}=~{}V_{0}t+\int_{0}^{t}\Big{(}\int_{0}^{t}K(s-r)\textbf{1}_{r\leq
s}~{}ds\Big{)}~{}dZ_{r}\\\
&~{}=~{}V_{0}t+\int_{0}^{t}\Big{(}\int_{0}^{s}K(s-r)~{}dZ_{r}\Big{)}~{}ds.\end{split}$

Then $X$ is absolutely continuous in $t$ so that both $\frac{d{X}_{t}}{dt}$
and $\frac{d\overline{X}_{t}}{dt}$ are well defined, and

$X_{t}~{}=~{}\int_{0}^{t}\Big{(}V_{0}+\int_{0}^{s}K(s-r)b\big{(}\frac{d\overline{X}_{r}}{dr}\big{)}~{}dr+\int_{0}^{s}K(s-r)\nu\sqrt{\frac{d\overline{X}_{r}}{dr}}~{}d\hat{W}_{r}\Big{)}~{}ds,$

with $b(v):=\theta-\lambda v_{+}$ and for some Brownian motion $\hat{W}$. Let
us define $V_{t}:=\frac{d{X}_{t}}{dt}$, then $V$ satisfies

$V_{t}~{}=~{}V_{0}+\int_{0}^{t}K(t-s)b\big{(}\frac{d\overline{X}_{s}}{ds}\big{)}~{}ds+\int_{0}^{t}K(t-s)\nu\sqrt{\frac{d\overline{X}_{s}}{ds}}~{}d\hat{W}_{s}.$

Notice that $V_{t}<0$ implies that $\frac{d\overline{X}_{t}}{dt}=0$, then we
can apply the same arguments as in (i) to deduce that $V_{t}\geq 0$ for any
$t\in[0,T]$. Since $X_{t}=\int_{0}^{t}V_{s}~{}ds$, this proves that all
solutions to equation (31) are non-decreasing over [0,T]. ∎

###  4.4 Proofs of Theorems 2.2 and 2.3

In view of Proposition 4.13, any weak solution $V$ to (30) is non-negative.
Therefore, a weak solution to (24) is also a weak solution to (6), which is
unique (see Remark 2.3). One then proves Theorem 2.2 by Proposition 4.7 and
Remark 4.8.

Again by Proposition 4.13, any weak solution $X$ to (31) is non-decreasing.
Therefore, a weak solution to (29) is also a weak solution to (7), which is
unique (see Remark 2.5). One can then prove Theorem 2.3 by Proposition 4.12. ∎

##  Appendix A Tables of simulation results

| Mean Value | Stat. Error | Comp. Time | Mean Value | Stat. Error | Comp. Time
---|---|---|---|---|---|---
Ref. | 0.056832 | - | - | - | - | -
n=4 | 0.059642 | 0.000245 | 7.856192 | 0.065483 | 0.000279 | 7.050599
n=10 | 0.058905 | 0.000238 | 17.419916 | 0.059996 | 0.000244 | 16.307883
n=20 | 0.058630 | 0.000234 | 35.061742 | 0.058635 | 0.000234 | 31.698260
n=40 | 0.058344 | 0.000232 | 70.921679 | 0.057363 | 0.000228 | 60.035371
n=80 | 0.058280 | 0.000232 | 129.678460 | 0.056967 | 0.000225 | 123.884151
n=160 | 0.057965 | 0.000230 | 266.368216 | 0.056905 | 0.000225 | 247.920679
n=320 | 0.057858 | 0.000229 | 582.703205 | 0.056897 | 0.000225 | 482.391579
Table 1: European call $(S_{T}-K)_{+}$ option price estimation with Scheme (8) (left) and Scheme (9) (right). The computation time is in second. | Mean Value | Stat. Error | Comp. Time | Mean Value | Stat.Error | Comp. Time
---|---|---|---|---|---|---
Ref. | - | - | - | - | - | -
n=10 | 0.036363 | 0.000145 | 18.126627 | 0.038373 | 0.000155 | 16.334577
n=20 | 0.034653 | 0.000136 | 34.257258 | 0.034992 | 0.000138 | 31.231823
n=40 | 0.033647 | 0.000132 | 68.872156 | 0.033387 | 0.000130 | 61.623905
n=80 | 0.033266 | 0.000130 | 144.862723 | 0.032474 | 0.000126 | 121.671897
n=160 | 0.032915 | 0.000128 | 292.438511 | 0.032230 | 0.000125 | 239.012198
n=320 | 0.032479 | 0.000127 | 563.534848 | 0.031907 | 0.000124 | 450.094275
Table 2: Asian option $(A_{T}-K)_{+}$ price estimation with Scheme (8) (left) and Scheme (9) (right). The computation time is in second. | Mean Value | Stat. Error | Comp. Time | Mean Value | Stat. Error | Comp. Time
---|---|---|---|---|---|---
Ref. | - | - | - | - | - | -
n=10 | 0.081591 | 0.000234 | 18.394217 | 0.076769 | 0.000238 | 17.260538
n=20 | 0.085409 | 0.000228 | 33.485885 | 0.079742 | 0.000226 | 33.581574
n=40 | 0.088502 | 0.000226 | 66.453965 | 0.083297 | 0.000222 | 64.918632
n=80 | 0.090927 | 0.000225 | 130.933455 | 0.086043 | 0.000219 | 124.431435
n=160 | 0.092118 | 0.000223 | 290.638306 | 0.088685 | 0.000219 | 246.881069
n=320 | 0.092904 | 0.000222 | 558.946791 | 0.090338 | 0.000219 | 492.821703
Table 3: Lookback option $(M_{T}-K)_{+}$ price estimation with Scheme (8) (left) and Scheme (9) (right). The computation time is in second. | Mean Value | Stat. Error | Comp. Time | Mean Value | Stat. Error | Comp. Time
---|---|---|---|---|---|---
Ref. | 0.028295 | - | - | - | - | -
n=4 | 0.033967 | 0.000080 | 8.389909 | 0.033565 | 0.000090 | 8.291969
n=10 | 0.030781 | 0.000085 | 17.706663 | 0.030522 | 0.000098 | 17.645600
n=20 | 0.029736 | 0.000088 | 33.061506 | 0.029387 | 0.000098 | 31.253372
n=40 | 0.029218 | 0.000091 | 69.696746 | 0.028756 | 0.000098 | 59.500006
n=80 | 0.029165 | 0.000093 | 132.443006 | 0.028527 | 0.000097 | 123.055661
n=160 | 0.028875 | 0.000095 | 267.132785 | 0.028477 | 0.000098 | 248.295026
n=320 | 0.028685 | 0.000094 | 575.724998 | 0.028328 | 0.000098 | 486.069715
Table 4: Estimation of $\mathbb{E}[X_{T}]$ with Scheme (8) (left) and Scheme (9) (right). The computation time is in second. | Mean Value | Stat. Error | Comp. Time | Mean Value | Stat. Error | Comp. Time
---|---|---|---|---|---|---
Ref. | 0.013517 | - | - | - | - | -
n=4 | 0.016940 | 0.000074 | 8.425556 | 0.017096 | 0.000081 | 7.537352
n=10 | 0.014542 | 0.000076 | 18.414490 | 0.015131 | 0.000088 | 15.891882
n=20 | 0.014043 | 0.000079 | 36.577253 | 0.014254 | 0.000088 | 30.900766
n=40 | 0.013752 | 0.000082 | 70.948186 | 0.013905 | 0.000088 | 59.677628
n=80 | 0.013841 | 0.000084 | 144.526883 | 0.013770 | 0.000088 | 118.789620
n=160 | 0.013705 | 0.000085 | 273.132594 | 0.013775 | 0.000088 | 230.445799
n=320 | 0.013641 | 0.000085 | 585.570704 | 0.013600 | 0.000088 | 471.164119
Table 5: Estimation of $\mathbb{E}[((X_{T}-V_{0})_{+}]$ with Scheme (8) (left)
and Scheme (9) (right). The computation time is in second.

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