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Let $(X_t)_{t \in \mathbb{N}}$ be a real-valued martingale that is bounded, i.e., there are $a, b \in \mathbb{R}$ such that $a \leq X_t \leq b$ for all $t$.

Define the path length $L$ of $(X_t)_{t \in \mathbb{N}}$ as the distance the martingale covers, $$ L := \sum_{t=0}^\infty | X_t - X_{t+1} |. $$

My questions:

  1. Is $L$ almost surely finite? I.e., can there be a set of positive measure on which $L = \infty$?
  2. If yes to question 1, is $\mathbb{E}[L]$ finite?
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2 Answers 2

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If you have a random walk on $\{0,\ldots,N\}$, it's known that the expected number of steps to hit the boundary starting from the middle is something like $N^2/4$. That is: you travel a distance of $N^2/4$ before hitting the boundary.

You can exploit this to produce a martingale where the expected path length is infinite: Do absorbing random walk on $[0,1]$, where the step sizes are dyadic and depend on time. For example, do $2^{10}$ steps of size $2^{-10}$; then $2^{11}$ steps of size $2^{-11}$ etc. In the $2^{-k}$ phase, your average displacement is something like $2^{-k/2}$, so that it's likely that you'll never hit the boundary.

But any sample path that doesn't hit the boundary has infinite length.

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I like Anthony's answer but here's another construction.

Evaluate a standard Brownian motion on the sequence $t_n = 1- 1/n$. The process has infinite length almost surely by Levy's theorem on the modulus of continuity of Brownian motion.

It's not a bounded martingale but you can stop it the first time it gets larger than $1$ in absolute value. By Doob's stopping theorem this is still a martingale. And it still has infinite length whenever the original Brownian motion remains in $[-1,1]$ up to time $1$.

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