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Let $(X_n)$ be a martingale. What can be said about the distribution of its maximum over a window of fixed length: $$M_n = \max_{n-10 \leq k \leq n} X_k$$ or about the "range" over a window: $$R_n = \max_{n-10 \leq k \leq n} X_k - \min_{n-10 \leq k \leq n} X_k $$

I know Doob's inequality, but can we give more precise informations about $M_n$ or $R_n$ ? At least when $X_{n+1} - X_n$ has a normal distribution?

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In your latter case ($X_{n+1}-X_n$ normal) you can use a large deviation principle for martingales (if $X_{n+1}-X_n$ is bounded it is called Azuma inequality but there are extensions with worse rates for martingale differences with finite exponential moments) together with a naive union bound to get a good estimate. This will work also for estimating $$\max_{k\in [cn,n]}X_k$$ with $c>0$.

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  • $\begingroup$ Thanks, I'll have a look. In the case $X_n$ is just a simple random walk with independent increments $+1$ or $-1$ with probability $1/2$, is there a well known distribution for $R_n$ ? It looks roughly like a log-normal... $\endgroup$
    – Basj
    Feb 1 '16 at 16:33
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    $\begingroup$ It would be a shame to use a naïve union bound as the Hoeffding-Azuma inequality gives the same bound for the maximum of a martingale as for the last term. Of the many tail bounds for martingales, my personal favorite is [1] which directly applies to the maximum. [1] Freedman, David A. "On tail probabilities for martingales." the Annals of Probability (1975): 100-118. $\endgroup$ Feb 7 '16 at 17:45

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