Calculate $Ea^{\tau}$ where $\tau=\inf\space[{n: X_1+...+X_n=1}]$

Question

Calculate $Ea^{\tau}$ for any $a\in (0,1)$ where $\tau=\inf\space[{n: X_1+...+X_n=1}]$ $X_i$ are independent and have distrbution as such: $P(X_i=-1)=P(X_i=1)=\frac{1}{2}$ I will add that the first part of the excercise was to find the conditions for a and b such as $Y_n=a^nb^{X_1+...+X_n}$ be a submartingale, the result which I acquired is $ab^2-2b+1>=0$ Thanks for any help, I have litteraly any idea on how to approach it.

Answer 1

If we let $a=\left(\frac{b+b^{-1}}2\right)^{-1}$ for $b>1$, then we can observe that $$ M_n=a^n b^{\sum_{k=1}^n X_k}, \quad n\ge 0 $$ is a martingale from $\Bbb E[M_n|M_{n-1}]=aE[b^{X_n}]M_{n-1}=M_{n-1}$. From this, one can notice that the stopped process $$ M_{n\wedge \tau}=M_\tau 1_{\tau \le n} +M_n1_{\tau> n} $$ is also a non-negative, bounded martingale. By optional sampling theorem, we obtain $$ 1=\Bbb E[M_{\tau\wedge n}]=\Bbb E[a^\tau b1_{\{\tau \le n\}}]+\Bbb E[M_n1_{\tau> n}]. $$ The second term tends to $0$ as $n$ tends to $\infty$, since on $\{\tau>n\}$, $$ 0\le M_n \le a^n b\xrightarrow{n\to\infty} 0, $$ and by Lebesgue's dominated convergence therorem. This gives for all $a\in (0,1)$, $$ \Bbb E[a^\tau 1_{\{\tau<\infty\}}]=\frac1{b}=\frac{1-\sqrt{1-a^2}}{a}. $$ By letting $a\to 1^-$, we get $\Bbb P(\tau <\infty) = 1$, and finally $$ \Bbb E[a^\tau ]=\frac1{b}=\frac{1-\sqrt{1-a^2}}{a}. $$