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Let $X_1,X_2,\ldots$ be iid random variables where $X_i\in\{-1,0,1,2,...\}$, $P(X_i=0)<1$ and $E(X_1)=\mu$. Let $S_n=1+X_1+\cdots+X_n$ and $T=\inf \{n:s_n=0\}$. Show that $E(T)=\infty$ if $\mu=0$ and $E(T)<\infty$ if $\mu<0$.

I try to use the Wald's equation but in this case we have $S_0=1$ instead of $0$.

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  • $\begingroup$ The assertion is not correct. Consider for example $X_j := - \frac{2}{3}$ for all $j$. They are clearly iid, $\mu<0$, but $T=\infty$, hence in particular $\mathbb{E}(T)=\infty$. Probably, you want to define $T$ as $T=\inf\{n; S_n \leq 0\}$ instead...? $\endgroup$ – saz Mar 12 '13 at 9:10
  • $\begingroup$ Sorry I forgot to add $X_i\in\{-1,0,1,2,...\} P(X_i=0)<1$ in the problem. $\endgroup$ – BigMike Mar 12 '13 at 17:27
  • $\begingroup$ Unfortunately, the assertion is still not correct. Consider $X_j := -2$. $\endgroup$ – saz Mar 12 '13 at 17:38
  • $\begingroup$ Here $X_i$ is in the set $\{-1,0,1,2,...\}$, so it can not equal to -2. I think I proved the case where $\mu=0$ by making a new martingale $Y_n=S_n-n\mu-1$, but still confused about the case when $\mu<0$ $\endgroup$ – BigMike Mar 12 '13 at 21:32
  • $\begingroup$ Well, since $X_i$ is a random variable, it won't be "in a set". However, so $\mathbb{P}[X_j = j]>0$ for all $j \in \mathbb{Z}$ is another assumption. $\endgroup$ – saz Mar 12 '13 at 21:44
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If $X_1=-1$, $T=1$. Otherwise, $X_1=k$ with $k\geqslant0$ and $T=1+\sum\limits_{i=1}^{k+1}T_i$ where $(T_i)$ is i.i.d. and distributed like $T$ (this is where the hypothesis that the only negative jumps of the random walk are $-1$ is used). Thus, for every $s$ in $(0,1)$, $\mathbb E(s^T)=\sum\limits_{k\geqslant-1}\mathbb P(X=k)s\mathbb E(s^T)^{k+1}$. Introducing $g(t)=\mathbb E(t^X)$, this reads $g(\mathbb E(s^T))=1/s$. Deriving this, one gets $\mathbb E(Ts^{T-1})g'(\mathbb E(s^T))=-1/s^2$. Considering the limit when $s\to1$ yields $\mathbb E(T)g'(1)=-1$, that is, $\mathbb E(T)=-1/\mu$.

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  • $\begingroup$ Thanks for the answer, but I found a easier way using the wald's equation and let $Y_n=S_n-1$. I got the exactly the answer as your. $\endgroup$ – BigMike Mar 13 '13 at 6:15
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Let $S_n' := S_n -1$. Then $$T=\inf\{n \in \mathbb{N}; S_n'=-1\}$$

  • $\mu=0$: Assume $\mathbb{E}(T)<\infty$. Then we have by Wald's equation $$\mathbb{E}S_T' = \mathbb{E}(T) \cdot \mathbb{E}X_1$$ This is clearly a contradiction since $\mathbb{E}X_1=0$ whereas $\mathbb{E}S_T' = -1$.
  • $\mu<0$: By the strong law of large numbers we have $S_n' \to - \infty$ almost surely, thus $\mathbb{P}(T<\infty)$ almost surely (since jumps to the left have magnitude $1$). This implies that $Y_n := S_{n \wedge T}'$ is well-defined. $(Y_n)_n$ is a supermartingale (by optional stopping). By applying Wald's equation to the (bounded) stopping time $T \wedge n$ we obtain $$\mathbb{E}X_1 \cdot \mathbb{E}(T \wedge n) = \mathbb{E}S_{n \wedge T}' \geq \mathbb{E}S_T' = -1$$ i.e. $\mathbb{E}(T \wedge n) \leq - \frac{1}{\mu}$ since $\mu<0$. Monotone convergence theorem yields $\mathbb{E}T \leq - \frac{1}{\mu}< \infty$. By applying Wald's equation (this time to the stopping time $T$), we obtain $\mathbb{E}T=- \frac{1}{\mu}$.
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