Zero mean but not a martingale

Question

I am looking for a simple stochastic process which has zero mean for all $t\geq0$ but it is not a martingale. I been looking in to local martingales but having trouble keeping the mean zero for all t. Any advise on how I should think?

Answer 1

A martingale is not a process that has zero mean for all time. The defining feature of a martingale process $M_t$ is:

$E(M_t|M_1,...,M_{t-1})=M_{t-1}$

Therefore, any stationary process will violate this property. Take a stationary gaussian process, which is about as simple as you can get:

Let $X_t \sim \mathcal{N}(0,1)$

Therefore, $E(X_t|X_1,...,X_{t-1})=0\neq X_{t-1}$

Answer 2

It is actually very easy to construct a process $X$, such that $E(X_t)=0$, for all $t\geq 0$, and which is not a martingale. For example, let $\bigl(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq 0},P\bigr)$ be a filtered probability space, let $\xi$ be an $\mathcal{F}_0$-measurable Bernoulli random variable with $P(\xi=1)=P(\xi=-1)=\frac{1}{2}$, and define $X$ by setting $X_t:=t\xi$, for all $t\geq 0$. Then $E(X_t)=tE(\xi)=0$, for all $t\geq 0$. However, $X$ is not a martingale, since $E(X_t\,|\,\mathcal{F}_s)=t\xi\neq s\xi=X_s$, for all $t>s\geq 0$.

A more interesting question is whether there exists a strict local martingale (i.e. a local martingale that is not a martingale) $L$, such that $E(L_t)=0$, for all $t\geq 0$. The answer is yes; for example, set $L:=X^{-1}-Y^{-1}$, where $X$ and $Y$ are independent Bessel processes of dimension three.