Stopping time for a martingale 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 $0$.
 A: 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, $\mathbb E(T)=-1/\mu$.
A: Let $S_n' := S_n -1$. Then $$T=\inf\{n \in \mathbb{N}; S_n'=-1\}$$


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*$\mu=0$: 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 $\mathbb{E}T=- \frac{1}{\mu}$.

