$X_1, X_2, ... $ are independent random variables.
And $P(X_n=1)=P(X_n=-1)=1/2$.
$t=inf (n: X_1+X_2+...+X_n=1)$
Find $E(1/3)^{t}$.
I tried to do it from the definition of expected value:
$E(1/3)^{t}=\sum_{i=1}^{\infty} \quad (1/3)^n \cdot P(t=n)$
$P(t=2n)=0$ but it is harder to calculate
$P(t=2n+1)$
Thanks in advance.
Hints: $X_1+...+X_{2n-1}=1$ iff either $X_1+...+X_{2n-2}=0$ and $X_{2n-1}=1$ or $X_1+...+X_{2n-2}=2$ and $X_{2n-1}=-1$. Also, $X_1+...+X_{2n-2}=0$ iff exactly $n-1$ of $X_i$'s are $+1$ and the rest are $-1$. Similarly, $X_1+...+X_{2n-2}=2$ iff exactly $n$ of the $X_i$'s are $+1$ and the rest are $-1$. Hence $P(X_1+...+X_{2n-1}=1)=\binom {2n-2} {n-1} \frac 1 {2^{2n-1}}+\binom {2n-2} n \frac 1 {2^{2n-1}}$. This gives you $P(t\leq 2n-1)$ and use the fact that $P(t=2n-1)=P(t\leq 2n-1)-P(t\leq 2n-3)$