3
$\begingroup$

Let $X_0,X_1,...$ be i.i.d and let $N = \inf \{ n\ge 1:X_n >X_0 \}$.

Prove that $P(N>n) \ge \frac{1}{n+1}$.

My attempt,for $E = \{N>n\} = \{N = n+1 ,n+2,n+3,...\}$ and we know for $\{N = n+k\}$ means for the first $n+k+1$ element the first one i.e. $X_0$ rank exactly $n+k$, and $X_{n+k}$ rank exactly $n+k+1$, which means it has probability $\frac{1}{(n+k+1)(n+k)}$ by symmetric reasoning.Hence the sum for all $N>n$ is $\frac{1}{n+1}$.

What I don't understand is why it may greater than $\frac{1}{n+1}$.

This is the example from Durrett probability page 70.Comic relief

$\endgroup$
3
$\begingroup$

Think of the case where all the $X_i$'s are 0. What is $N$? That should hopefully be enlightening to tell you where the symmetry argument fails, and that it is indeed exact if $P(X_i = X_j) = 0$

$\endgroup$

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.