I am reading Ramon van Handel's lecture notes on stochastic calculus and came across a confusing claim. Suppose we define the martingale $M_n = M_0 + \sum_{k = 1}^n \xi_k$, where $\{\xi_k\}_{k = 0}^\infty$ are independent Rademacher random variables. Consider the stopping time $\tau = \inf \{n \in \mathbb{N}: M_n \geq 2 M_0\}$. Then, van Handel claims (p. 63) that

$$ M_{n \land \tau} \leq 2 M_0 $$ almost surely for all $n \in \mathbb{N}$. But suppose we set $M_0 = 0.5$. Then $\mathrm{P}(M_1 = 1.5) = \mathrm{P}(M_{1 \land \tau} = 1.5) = 1/2$, so what is going wrong in my thinking? Intuitively, I think the correct bound should be $2M_0 + 1$ in this case.

This does not really matter for the argument he is making (showing that $\mathrm{E}(\tau) = \infty$), but maybe I somehow misunderstood how stopping times (or something else) actually work. Also, I don't think it is a typo, as he writes this multiple times in the text. Thanks in advance!


1 Answer 1


Although he doesn't explicitly state that $M_0$ is an integer, he does write early in the example that "$M_n$ takes only integer values", implying as much. When $M_0$ is an integer, his stated inequality holds.

This is because if $M_0$ is an integer, then so is $M_n$ for all $n$ and $2M_0$. Hence for $M_n$ to be strictly greater than $2M_0$, we must have $M_n\ge2M_0+1$ and hence $\tau<n$, since we must have passed $2M_0$ exactly at some point to arrive at $2M_0+1$.

  • $\begingroup$ I broadly agree, though I think we would have to exclude $M_0 = 0$, right? $\endgroup$ Nov 27, 2022 at 23:41
  • $\begingroup$ I suppose that depends on whether or not you consider $0$ to be a natural number. If so, $\tau\equiv0$ and the inequality holds, but if not, then you would have to exclude it, yes. But in the setting of this being the winnings of a gambler, and $\tau$ being described as the point when they "double their initial bet", it wouldn't really make sense for $M_0$ to be $0$. $\endgroup$ Nov 28, 2022 at 11:16
  • $\begingroup$ Makes sense, thanks! $\endgroup$ Nov 28, 2022 at 11:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .