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This seems like a simple question, but I cannot figure out the following. Let $\{M_i\}_{i\geq 0}$ be a martingale adapted to a filtration $\mathcal{F}_i$, with the following conditions:

  • $E((M_{i+1}-M_i)^2|\mathcal{F}_i)\geq \delta$,

  • $M_0=0$,

  • $|M_{i+1}-M_i|\leq h$ almost surely.

Let $\tau=\min\{i~:~|M_i|\geq h\}$. Then $E(M_{\tau\wedge\ell}^2)\leq 4h^2$. (this is what I cannot get).

What I thought was that $\tau\wedge\ell\leq \tau$, so the bound would be $h^2$. But this seems clearly wrong. I think I need to split and condition on something like $\mathcal{F}_\ell$, but I don't seem to be able to get the bound either. Also, I know that $\tau$ is a stopping time, but is $\tau\wedge\ell$ a stopping time?

Thanks for the suggestions!

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For each $\omega$, if $1 \le n \le \tau(\omega)$ then we have $|M_{n-1}(\omega)| \le h$ and so $$|M_n(\omega)| \le |M_{n-1}(\omega)| + |M_n(\omega) - M_{n-1}(\omega)| \le h + h = 2h \quad \text{a.s.} $$ by your third condition. When $\ell = 0$ the result is trivial, and otherwise we have $1 \le \tau \wedge \ell \le \tau$, so applying this with $n=\tau(\omega) \wedge \ell$ we get $|M_{\tau \wedge \ell}| \le 2h$ a.s. Squaring and taking expectations gives the desired result. (You can treat $\ell=0$ as a special case.)

Your first condition seems to be unnecessary for this.

For your last question, yes, $\tau \wedge \ell$ is a stopping time. More generally, it is a good exercise to show that if $\tau, \sigma$ are stopping times then so is $\tau \wedge \sigma$. Constants are also stopping times.

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  • $\begingroup$ Thanks! And thanks for the advice. I don't know why I couldn't see this, this is very clear. $\endgroup$
    – Math-user
    Mar 25, 2015 at 13:39

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