Consider the Heston model expressed as \begin{align} dS_t &= \mu S_t dt + S_t \sqrt{V_t} \big(\rho dW_t^{(1)}+\sqrt{1-\rho^2}dW_t^{(2)} \big); \tag*{(1)} \\ dV_t &= \kappa(\theta - V_t)dt + \sigma \sqrt{V_t}dW_t^{(1)}, \tag*{(2)} \end{align} where $(W^{(1)},W^{(2)})$ is a two-dimensional standard Brownian motion (under the probability measure $P$) and $\mu, \rho, \kappa, \theta$ and $\sigma$ are constants. We assume that the Feller condition is satisfied, i.e. $$2 \kappa \theta > \sigma^2,$$ which ensures that $V_t >0.$

In Shreve's book, I read that the solution $(S_t,V_t)_{0 \leq t \leq T}$ to the two-dimensional SDE above is a Markov process but he doesn't prove it. I have already checked a couple of books and I only have found a sufficient condition, which requires the coefficients (drift and diffusion functions) to satisfy the Lipschitz and linear growth conditions. This is not the case for this SDE, so I don't know how to proceed. Any ideas?

Edit: I see in the comments asking for the definition of a Markov process. Any definition is fine as long as I can get a rigorous proof. For example:

The solution $(X_t,V_t)_{0 \leq t \leq T}$ of the above SDE is a Markov process if for any bounded Borel measurable function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ and for all $0 \leq s \leq t \leq \infty,$ we have $$E[f(X_t,V_t) | \mathscr{F}_s]=[E[f(X_t,V_t) |(X_s,V_s) ],$$ or we could also use the transition probability function of the Markov process.


3 Answers 3


I am not providing a full proof but a reference for you to read up the details. The key step is mentioned below.

Most models used in finance are Markovian which is kind of in line with the efficient market hypothesis. The key step of of seeing that the Heston process is Markovian is the following theorem.

Let $f$ be a bounded Borel function from $\mathbb{R}^n$ to $\mathbb{R}$. Then for $t,h>0$, $$\mathbb{E}^x[f(X_{t+h})\mid\mathcal{F}_t^{(m)}](\omega)=\mathbb{E}^{X_t(\omega)}[f(X_h)],$$ where $\mathcal{F}_t^{(m)}$ is the $\sigma$-algebra generated by $\{B_s;s\leq t\}$.

The statement above is from Øksendal (2003, Theorem 7.1.2), a great recourse on SDEs. In his setting, $X_t$ is the solution to a SDE. Shreve (2004, Theorem 6.3.1) covers essentially the same theorem. Øksendal gives a proof, Shreve merely outlines it but highlights the intuition. The latter follows up with the corollary

Solutions to stochastic differential equations are Markov processes.

As you see, this corollary helps you to see that awful lot of models in finance are indeed Markovian.

  • $\begingroup$ Thank you for your answer. As you said the theorem from Oksendal deals with the solution to the SDE. The problem is that the Heston SDE doesn't satisfy the conditions of existence and uniqueness because the function $f(x)=\sqrt{x}$ is not Lipschitz. So we don’t know if there exists a unique solution that fits Oskendal theorem. I would have to prove that first in order to use Oksendal theorem. $\endgroup$
    – user39119
    Commented Feb 1, 2020 at 18:34
  • $\begingroup$ A possible strategy could be to treat the 2-dimenensional SDE as two 1-dimensional SDEs. Then show existence and uniqueness for the variance SDE. After that, solve the SDE for the stock and prove that is unique. Then show that this solution is a two-dimensional Ito process, as required in Oksendal theorem, and use Oksendal theorem. But not sure... $\endgroup$
    – user39119
    Commented Feb 1, 2020 at 18:35
  • $\begingroup$ No, it doesn't work. In definition 7.1.1 (Ito diffusion), he requires the coefficients to satisfy the Lipschitz (and linear growth) conditions as in Theorem 5.2.1. And as I said before, Heston SDE coefficients don't satisfy this requirement. $\endgroup$
    – user39119
    Commented Feb 1, 2020 at 18:50
  • $\begingroup$ Yeah, I just read that up, sorry. I jumped straight to the theorem instead of looking at the previous page... That theorem doesn't help either $\endgroup$
    – Kevin
    Commented Feb 1, 2020 at 18:52
  • 1
    $\begingroup$ @user1987 The SDEs Shreve considers are given by $\mathrm{d}X_t=\mu(t,X_t)\mathrm{d}t+\sigma(t,X_t)\mathrm{d}W_t$. So you're example doesn't apply. $\endgroup$
    – Kevin
    Commented Feb 1, 2020 at 19:22

In this (extremely technical) paper by Duffie et al it is shown that a Markov process is infinitely decomposable if and only if it is a regular affine process. So their results establishes a correspondence between Markov processes and regular affine processes.

Okay, that (the paper) is too technical for me, but if I look at the characteristic function of the Heston model (and other affine (jump) diffusions) I see that it depends on $S_t$ and $V_t$ only, which looks very Markov process to me.

Although this may not be a full answer to your question I believe it will point you in the right direction if you really want to get into the weeds.

  • $\begingroup$ I'll have a look at it. Thank you! $\endgroup$
    – user39119
    Commented Feb 4, 2020 at 11:19
  • $\begingroup$ @UBM You're welcome. I think though that to note that the characteristic function does not depend on the history but only of the current price of the spot and vol process is sufficient since given the characteristic function you can price general (path-independent) claims which therefore will also not depend on the history of the processes. $\endgroup$
    – user34971
    Commented Feb 4, 2020 at 11:29

A stochastic process with independent increments is a markov process. the proof is available in the following document: (Lemma 1.1)

  • $\begingroup$ Could you elaborate a bit, we prefer answers to be self contained. In this case the proof of the lemma depends on Definition 6.1.10 but for me it's not obvious where to find that. $\endgroup$
    – Bob Jansen
    Commented Jan 31, 2020 at 15:45
  • $\begingroup$ Could you give me your definition of a markov process? i will show you how to find the definition 6.1.10 $\endgroup$ Commented Jan 31, 2020 at 15:53
  • $\begingroup$ @user1987 Thank you. How do we prove that the solution to the Heston SDE has independent increments? $\endgroup$
    – user39119
    Commented Jan 31, 2020 at 18:08
  • $\begingroup$ the increment of S and V over a small time interval is determined by the Brownian increment which does not depend on the history prior to time t. $\endgroup$ Commented Jan 31, 2020 at 18:17
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    $\begingroup$ @user1987 Are you quite sure that this theorem applies to the Heston model? The mean-reversion in the variance prevents you from having independent increments doesnt it? To quote Gatheral (2006) The Heston process is very path-dependent; increments are from independent and it is not a Levy process? $\endgroup$
    – Kevin
    Commented Feb 1, 2020 at 17:11

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