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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

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1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – mathex
    Apr 1, 2019 at 5:30
  • $\begingroup$ Yep, that works $\endgroup$
    – bitesizebo
    Apr 1, 2019 at 16:02

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