$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


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 $P(t=2n-1)=P(t\leq 2n-1)-P(t\leq 2n-3)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.