# Ornstein-Uhlenbeck process: Markov, but not martingale?

I'm puzzled about properties of the Ornstein-Uhlenbeck process, given by the Itō integral $$X_t = x e^{-\lambda t} + \sigma \int_0^t e^{-\lambda(t-s)} d W_s \,.$$

1. I compute that $\{X_t\}$ is not a martingale process: $$E[X_{t'} | \mathcal{F}_t] = X_t e^{-\lambda {t'}} \neq X_t , \qquad 0 \leq t < t'.$$ by splitting the integral at $s=t$; $\mathcal{F}_t$ is the history of the Wiener process $\{W_t\}$ up to time t.

2. On the other hand, the process is a time-homogeneous Itō diffusion as it is of the form $d X_t = -\lambda X_t\,d t + \sigma \, dW_t$. And as such it fulfills the Markov property.

I had expected that any Markov process is also a martingale, but not vice versa. This is also expressed in Martingale that is not a Markov process

So where I am wrong? Perhaps the OU process is actually a martingale and my mistake is in the computation of the conditional expectation?

• Are you sure that it is not $\frac{dX_t}{X_t}=-\lambda dt+\sigma dW_t$? If it is, then you may want to change measure, such that $dW_t=\phi(t)dt+dW^*_t$. Under some technicalities, you can apply Girsanov theorem, and $W^*_t$ is a Wiener process. Then, you can reconduce the $X$-dynamic under the new measure, posing that the drift is null, in order to find $\phi(t)$. – 7raiden7 Mar 28 '14 at 9:57
• No, the given form is correct. A physical interpretation of the process is Brownian motion with a harmonic restoring force. – Erwin411 Mar 28 '14 at 10:00

You are right; the Ornstein-Uhlenbeck process is a Markov process but not a martingale. It is simply not correct that any Markov process is a martingale (and vica versa).

An easier counterexample is the following: Let $(B_t)_{t \geq 0}$ a Brownian motion and $$X_t := B_t +a \cdot t, \qquad t \geq 0$$ for some $a \in \mathbb{R}$, $a \neq 0$. Then $(X_t)_{t \geq 0}$ is not a martingale since, as $(B_t)_{t \geq 0}$ is a martingale,

$$\mathbb{E}(X_t \mid \mathcal{F}_s) = B_s +a \cdot t \neq X_s.$$

On the other hand,

\begin{align*} \mathbb{E}(f(X_t) \mid \mathcal{F}_s) &= \mathbb{E}(f(B_t-B_s+B_s+at) \mid \mathcal{F}_s) \\ &= \mathbb{E}(f(B_{t-s}+y+at)) \big|_{y=B_s} \\ &= \mathbb{E}(f(B_{t-s}+y+a(t-s)))) \big|_{y=X_s}, \end{align*}

and this shows that $(X_t)_{t \geq 0}$ is a Markov process.

• Thanks a lot for the clarification and for the detailed counterexample. – Erwin411 Mar 28 '14 at 15:24