You have been captured and blindfolded by pirates, then places somewhere on a five-meter-long wooden plank. Disoeriented, each step you take is a meter long but in a random direction - either towards the sharks waiting at one end or eventual freedom at the other. If x (integer) is the distance in meters you start from the safe end, determine the probability of your survival as a function of x.

My attempt, $M_n$ be position at time $n$, so that $M_0 = x$ and clearly is a martingale.

Let $\tau = min\{n: M_n = 0 \space\text{and} \space M_{n-1},...,M_0 \neq 5\}$. (Case $M_0 = 5$ is trivial.)

Then by optional stopping theorem we have $\mathbb{E}M_{\tau} = \mathbb{E}M_0 = x = 0 * \mathbb{P}(\text{reaching safety before sharks})$

But clearly I'm not able to get the probability out of this.

Could please someone explain what is wrong?

  • $\begingroup$ Your definition of $\tau$ doesn't make sense. What is the smallest $n$ such that $M_n=0$ "before" $M_n=5$? $\endgroup$ – Jack M Sep 25 '18 at 18:04
  • $\begingroup$ You right, it should be Let $\tau = min\{n: M_n = 0 \space\text{and} \space M_{n-1},...,M_0 \neq 5\}$. $\endgroup$ – Paul Sep 25 '18 at 18:07
  • $\begingroup$ Well, sure, but a time like that might never happen. What if you start at $1$ and just go $1,2,3,4,5$? $\endgroup$ – Jack M Sep 25 '18 at 18:13
  • $\begingroup$ If that's the case, then $\tau$ is unbounded on non-countable set and therefore you can not apply optional stopping theorem (not with my definition of $\tau$ that is)? $\endgroup$ – Paul Sep 25 '18 at 18:19
  • $\begingroup$ $\tau$ is simply not defined, it would only be unbounded if you arbitrarily set $\tau=+\infty$ when $5$ comes up first. But yes, it's undefined on a non-negligible (uncountable is neither here nor there) set and therefore you presumably can't apply optional stopping. $\endgroup$ – Jack M Sep 25 '18 at 18:22

What you really want is the stopping time $T=\min\{n\mid M_n=0\text{ or }M_n=5\}$. This stopping time is well-defined everywhere.

After that, you had the right idea... indeed, we have $E(M_T)=E(M_0)=x$. And normally you'd be right, in general the expected value of a random variable tells you almost nothing about the probability distribution of the variable. But this is a very special case, and it happens that in this case, you can in fact deduce the probability you want from the expected value.

Here's a hint: what values can the variable $M_T$ take?

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  • $\begingroup$ Thank you! Got the answer I was looking for. $\endgroup$ – Paul Sep 25 '18 at 18:40

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