For $(M_t)_{t\geq0}$ to be a martingale w.r.t a filtration $\mathcal{F}_t$, we require
- $M_t \in L^1$
- $E[M_t \mid \mathcal{F}_s] = M_s$, $t\geq s$
If it is known that $M_t \geq 0$, does one have to prove condition 1? Or does it fall out of condition 2 since
$$E[M_t] = E[E[M_t \mid \mathcal{F}_0]] = E[M_0] < \infty?$$
My concern is whether the law of iterated expectations holds if we don't know apriori that $M_t \in L^1$. Relatedly, are there examples of stochastic processes where $M_t \notin L^1$ but $E[M_t \mid \mathcal{F}_s] = M_s$?
Best Answer
It is part of the definition of the conditional expectation that $$E[M_t 1_A] = E[E[M_t \mid \mathcal{F}_0] 1_A]$$ for any $A \in \mathcal{F}_0$. By taking $A = \Omega$ we see that the law of iterated expectations is an immediate consequence of the definition (without assuming that $M_t \in L^1$).
As a result, if $M_0 \in L^1$ and $(M_t)_{t \geq 0}$ satisfies condition $2$ then by your reasoning we have that $M_t$ is a martingale.
It's also fairly easy to see that there exist processes $M_t \not \in L^1$ such that the conditional expectations exist and $E[M_t \mid \mathcal{F}_s] = M_s$. For example, let $X$ be a non-integrable, non-negative random variable and for every $t \geq 0$, let $\mathcal{F}_t = \sigma(X)$ and $M_t = X$. Then we obviously have $$E[M_t \mid \mathcal{F}_s] = E[X \mid \sigma(X)] = X.$$