Let $(\Omega,\mathcal{F},(\mathcal{F}_t:t\ge{0}),P)$ be a stochastic basis and $M=(M_t:t\ge{0})$ a locally square integrable martingale, which means a stochastic process such that:
- $M_t\in{L^2(\Omega,\mathcal{F_t},P)}$ for all $t\ge{0};$
- $\mathbb{E}[M_t|\mathcal{F}_s]=M_s$ for all $t\ge{s}$.
Show that, if $M^2$ is a locally integrable martingale, then $M_t=M_0$ for all $t\ge{0}$; which means the martingale is constant.
I know that in this case $M^2$ is a submartingale and $M^2-[M,M]$ is a martingale, where $[M,M]$ indicates the quadratic variation, but I don't know how to use this to prove the statement.
Best Answer
Since $M$ and $M^2$ are martingales, for any $t\geq s\geq 0$ holds: $$\mathbb{E}[M_t^2] = \mathbb{E}[M_s^2]=\mathbb{V}[M_s^2]+\mathbb{E}[M_s]^2=\mathbb{V}[M_s^2]+\mathbb{E}[M_t]^2.$$ Substracting $\mathbb{E}[M_t]^2$ from both sides yields $$\mathbb{V}[M_s^2]=0.$$ Thus $$\mathbb{V}[M_s^2]=\mathbb{E}[(M_s-\mathbb{E}[M_s])^2]=0.$$ Since $(M_s-\mathbb{E}[M_s])^2$ is non-negative and its excpeted value is zero, it has to be zero almost everywhere: $$(M_s-\mathbb{E}[M_s])^2=0\quad \text{a.e.},$$ resulting in $$M_s=\mathbb{E}[M_s]\quad \text{a.e.},$$ indicating that $M_s$ has to be constant. And $$M_t=\mathbb{E}[M_t]=\mathbb{E}[M_0]=M_0.$$