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Consider $(X_n)_{n\ge 1}$ iid with $P(X_1=1)=P(X_1=-1)=\frac{1}{2}$. Then there is the sumprocess $S_n:=\sum_{i=1}^n X_i$ and the stopping times $$T_{a,b}:=\min\{n\ge 1:S_n\in\{a,b\}\},$$ $$T_{a}:=\min\{n\ge 1:S_n=a\},$$ of first moment when $S_n\in\{a,b\}$, in the second case, when $S_n=a$.

I already was able to show, that $$P(T_{a,b}>n(b-a))\le(1-\frac{1}{2^{b-a}})^n$$ by induction,

$$E[T_{a,b}]<\infty$$ by using estimations with the probability above, $$P(S_{T_{a,b}}=a)=\frac{b}{b-a}$$ by using Wald's equation.

Now, by using this last result, I want to show that

$$P(T_a<\infty)=1$$ and

$$E[T_a]=\infty.$$

It may be again useful to use Wald's equation (maybe by contradiction?), but I do not really see how $T_{a,b}$ and $T_a$ are connected.

saz
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user408858
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1 Answers1

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First of all, note that for any $k \in \mathbb{N}$

$$\left\{S_{T_{a,k}}=a \right\} \subseteq \{T_a < \infty\}. \tag{1}$$

Indeed: If $\omega \in \{S_{T_{a,k}}=a\}$, then $S_{T_{a,k}}(\omega)=a$, and in particular there exists $n \in \mathbb{N}$ such that $S_n(\omega)=a$ (namely, $n=T_{a,k}(\omega)$).

It follows from $(1)$ that

$$\mathbb{P}(T_a<\infty) \geq \mathbb{P} \left( S_{T_{a,k}}=a \right) = \frac{k}{k-a}. \tag{2}$$

As $k \in \mathbb{N}$ is arbitrary, this implies

$$\mathbb{P}(T_a<\infty) \geq \sup_{k \in \mathbb{N}} \frac{k}{k-a}=1.$$

This proves the first assertion. To prove that $\mathbb{E}(T_a)=\infty$ we can indeed use a proof by contradiction. If $\mathbb{E}(T_a)$ was finite, then Wald's equation would imply $\mathbb{E}(S_{T_a})=\mathbb{E}(S_0)=0$; this is clearly wrong as $S_{T_a}=a$ a.s. implies $\mathbb{E}(S_{T_a})=a$.

saz
  • 123,507