Why the Heston Stochastic Volatility Model Is Markovian

# How can I prove that the solution to the Heston SDE is a Markov process?


# How can I prove that the solution to the Heston SDE is a Markov process?












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.

## Answer by Kevin (score 4)

https://quant.stackexchange.com/a/50954

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.

## Answer by Valometrics.com (score 1)

https://quant.stackexchange.com/a/50934

A stochastic process with independent increments is a markov process. the proof is available in the following document: (Lemma 1.1) http://statweb.stanford.edu/~adembo/math-136/Markov_note.pdf

## Answer by user34971 (score 1)

https://quant.stackexchange.com/a/50989

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.