# Stopped martingale bound in van Handel's notes

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!

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 $$\tau, since we must have passed $$2M_0$$ exactly at some point to arrive at $$2M_0+1$$.
• I broadly agree, though I think we would have to exclude $M_0 = 0$, right? Nov 27, 2022 at 23:41
• 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 $0$. Nov 28, 2022 at 11:16