Square-Integrable Martingale, Submartingale, and Supermartingale

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},P)$ be a filtered probability space and let $M=(M_t)_{t\ge0}$ be a stochastic process on $(\Omega,\mathcal{F},P)$. Write $\mathbf{1}_{A}$ for the function equal to $1$ on $A$ and $0$ off $A$.

The process $M$ is a square-integrable martingale with respect to $(\mathcal{F}_t)_{t\ge0}$ if:

(i) $M$ is adapted to $(\mathcal{F}_t)_{t\ge0}$;

(ii) $M_t$ is square-integrable for every $t\ge0$;

(iii) for all real $0\le s\le t$, the random variable $M_s$ is a conditional expectation of $M_t$ given $\mathcal{F}_s$.

Under conditions (i) and (ii), the variable $M_s$ is automatically $\mathcal{F}_s$-measurable and square-integrable, so by Conditional Expectation of a Square-Integrable Random Variable condition (iii) is equivalent to the martingale property in averaged form:

and, by the uniqueness assertion of Existence and Uniqueness of Conditional Expectation for Square-Integrable Random Variables, also to the requirement that for all $0\le s\le t$ every conditional expectation $Y$ of $M_t$ given $\mathcal{F}_s$ satisfies $P(Y=M_s)=1$. Taking $A=\Omega$: a square-integrable martingale has constant expectation, $\mathbb{E}[M_t]=\mathbb{E}[M_0]$ for all $t\ge0$.

The process $M$ satisfying (i) and (ii) is a square-integrable submartingale if instead of the displayed identity one requires

and a square-integrable supermartingale if the reverse inequality holds for all such $s,t,A$. A process is a square-integrable martingale if and only if it is both a square-integrable submartingale and a square-integrable supermartingale, since the two inequalities together give the averaged identity.