# stopping time a.s. bounded?

let $$(X_n)_n$$ be a sequence of independent random variables and identically distributed $$B(1,\frac{1}{2})$$ (Bernoulli distribution). We set $$Y_n=\sum_{k=1}^n(2X_k-1)$$ and $$\mathcal{F_n}=\sigma(X_0,...,X_n).$$ We notice that $$(Y_n)_n$$ is a martingale for $$(\mathcal{F_n})_n$$.

Let $$k \in \mathbb{N^*}$$ and $$T=\inf(n \in \mathbb{N^*};Y_n=k).$$

$$T$$ is a stopping time for $$(\mathcal{F_n})_n.$$

My question is the following : is $$T$$ bounded a.s. ? ($$\exists p>0;T \leq p \ \ a.s$$ ?)

Thanks

For $$k \neq 0$$, no. First suppose $$k \neq 1$$. Then with probability $$(1/2)^n$$ the first $$n$$ steps of the walk are $$1, 0, 1, 0, 1, 0, 1, 0, \ldots 1$$ (or 0 depending on $$n$$), meaning with probability at least $$(1/2)^n$$ we have $$T \geq n$$. Now $$(1/2)^n$$ gets very small as $$n$$ gets big, but it's never zero. So $$T$$ can't be almost surely bounded. Similar argument if $$k = 1$$ with $$0, -1, 0, -1$$ instead.
• can we say that, since $$\forall n \in \mathbb{N^*}, P(T>n )=P(\bigcap_{p=1}^{n}{\left\{Y_p\neq k \right\}})\geq P(X_1=0,...,X_n=0)=\frac{1}{2^n}>0$$ then we can deduce it is not bounded a.s. Apr 1, 2019 at 5:30