Let $S_n = \xi_1 + \cdots + \xi_n$ be a random walk and suppose that $\phi(\theta_o) = \mathbb{E}\exp{\theta_o \xi_1} = 1$ for some $\theta_o <0$ and $\xi_i$ is not constant. Suppose $\xi_i$ are integer valued with $P(\xi_i < -1) = 0$, $P(\xi_i = -1) > 0 $ and $\mathbb{E}\xi_i > 0 $. Let $ T = \inf\{n: S_n = a\}$ with $a<0$. Use the martingale $X_n = \exp\{\theta_o S_n\}$ to conclude that $P(T< \infty) = \exp\{-\theta_o a\}$.
My attempt until now :
Since $X_n$ is a martingale, $X_{n \wedge T}$ is also a martingale. Thus, $\mathbb{E}[X_{n \wedge T}] = 1$. Next, $\mathbb{E}[\exp\{\theta_o (S_{n \wedge T} - a)\}]=\exp\{-\theta_0 a\}$
$\exp\{-\theta_0 a\} = \mathbb{E}[\mathbb{1}_{T<\infty}\exp\{\theta_o (S_{n \wedge T} - a)\}] + \mathbb{E}[\mathbb{1}_{T=\infty}\exp\{\theta_o (S_{n \wedge T} - a)\}] $
The first term in the right hand term goes to $\mathbb{P}(T<\infty)$ by the dominated convergence theorem since $\mathbb{1}_{T<\infty}\exp\{\theta_o (S_{n \wedge T} - a)\}$ goes to $\mathbb{1}_{T<\infty}\exp\{\theta_o (S_{T} - a)\} = \mathbb{1}_{T<\infty}$ almost surely and $\mathbb{1}_{T<\infty}\exp\{\theta_o (S_{n \wedge T} - a)\} < 1$ by def of $T$.
To conclude I would like to show that $\mathbb{E}[\mathbb{1}_{T=\infty}\exp\{\theta_o (S_{n \wedge T} - a)\}] = \mathbb{E}[\mathbb{1}_{T=\infty}\exp\{\theta_o (S_{n} - a)\}]$ goes to zero as $n$ goes go infinity but I don't know how to proceede.
Do you see any other approach to conclude the exercise ?
By the strong law of large numbers, we have
$$\frac{S_n}{n} \to \mathbb{E}\xi_1>0$$
almost surely, and so
$$S_n -a = n \left( \frac{S_n}{n} - \frac{a}{n} \right) \xrightarrow[]{n \to \infty} \infty$$
almost surely. As $\theta_0<0$ this implies
$$\exp(\theta_0 (S_n-a)) \xrightarrow[]{n \to \infty} 0 \tag{1}$$
almost surely. On the other hand, $S_{n \wedge T} \geq a$ and $\theta_0<0$ give
$$\exp(\theta_0(S_{n \wedge T}-a)) \leq 1$$
Combining both considerations, we conclude that we may apply the dominated convergence theorem to obtain that
$$\mathbb{E}(1_{\{T=\infty\}} \exp(\theta_0(S_n-a))) \xrightarrow[]{n \to \infty} 0.$$