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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.

Roberto Palermo
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    As quadratic variation is an increasing process that means that it is constant, so what happens to a process which has constant quadratic variation ? – TheBridge Nov 25 '22 at 19:37
  • I'll try, does that mean $[M,M]_t=[M,M]_0=0$ for all $t\ge{0}$? In this case I get $M_t=0$ for all $t\ge{0}$. But why should the quadratic variation be constant? – Roberto Palermo Nov 25 '22 at 19:51
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    Well you can't be increasing and martingale at the same time right unless your constant ? because M^2 and M^2- are martingales then is also a martingale (martingale is linear space) – TheBridge Nov 25 '22 at 20:26
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    You can look at Protter's book there is a theorem that states that a martingale (local?) and its quadratic variations have a.s. the same intervals of "constancy". – TheBridge Nov 25 '22 at 20:29

1 Answers1

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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 $$M_t=\mathbb{E}[M_t]=\mathbb{E}[M_0]=M_0.$$

Nick Halden
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