I was trying to solve Durrett problem $4.8.8$, where I have to show that
Let $S_n= \xi_1 + \ldots \xi_n$ random walk. Suppose $\varphi(\theta_0)=E[\exp(\theta_0 \xi_1)]$ for some $\theta_0<0$ and $\xi_i$ non constant. Then the exponential martingale $X_n=\exp(\theta_0 S_n)$ is a martingale. Let $\tau= \inf \{ n \in \mathbb{N}: S_n \notin(a,b) \}$ and $Y_n=X_{n \wedge \tau}$. Prove that $E[X_\tau]=1$ and $P(S_{\tau} \leq a) \leq \exp(-\theta_0 a)$
By OST I have that $E[X_{n \wedge \tau}]=E[X_0]=1$.
First, $n \wedge \tau \rightarrow_n \tau$, and $X_{n \wedge \tau} \leq \exp(\theta_0b)$ and by DCT, I have:
\begin{align}
E[X_{\tau}]=E[\lim_n X_{n \wedge \tau} ]= \lim_n E[X_{n \wedge \tau}] = 1
\end{align}
To prove the other point, I sould use that $E[X_{\tau}]=1$. What I tried so far is
\begin{align}
X_{\tau}=\exp(-\theta_0 S_{\tau}) 1_{\tau \leq a} + \exp(-\theta_0 S_{\tau}) 1_{\tau \geq b}
\end{align}
But now I am in trobule because is easy to bound the expectation of the first term, but for the other term I just know that $S_\tau \geq b$ and this is a problem.
How can I move?
Best Answer
Well, it's not actually a problem: The larger $E(\exp(\theta_0 S_{\tau})1_{S_{\tau}\geq \beta})$ is, the worse for the probability that $S_{\tau}\leq a$. Note that your signs in the exponentials aren't entirely consistent. I'm using the ones from your problem statement.
Note that $$ P(S_{\tau}\leq a)\leq \exp(-a \theta_0)E \exp(\theta_0 S_{\tau})1_{S_{\tau}\leq a}, $$
since $t\mapsto \exp(\theta_0 t)$ is decreasing. Now,
$$ \exp(-a \theta_0)E \exp(\theta_0 S_{\tau})1_{S_{\tau}\leq a}\leq \exp(-a\theta_0) E(\exp(\theta_0S_{\tau}))=\exp(-a\theta_0)E(X_{\tau})=\exp(-a\theta_0), $$