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Let $X_1, X_2....$ be independent exponential(1) distributed variables. Let $N=min[n: X_n>1]$ which I think is the random variable for the minimum number of RVs $X_i$ before getting a value greater than 1. The questions are:

Find $P(X_N>3|N=3)$

and

Find $E(X_N)$

For the first item, I interpreted it is find the probability that $X>3$ given that $X>1$ since this is what is implied by $N=3$ and the $X_i's$ are independent. Is this correct?

For the second question, I'm lost. I was thinking that $N\sim NB(1,e^{-1})$ where N is the trial in which the first success $(X_N>1)$, occurs.

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Your answer for the first part is correct. As for the second, try to find the distribution of $X_N$,

$$ \begin{align} \mathbb{P}(X_N > x) &= \sum\limits_{n=1}^{\infty}\mathbb{P}(X_N > x, N = n) \\ &= \begin{cases} \sum\limits_{n=1}^\infty(1-e^{-1})^{n-1}e^{-1} & 0 \leq x < 1\\ \sum\limits_{n=1}^\infty(1-e^{-1})^{n-1}e^{-x} & x > 1 \end{cases}\\ &= \begin{cases} 1 & 0 \leq x < 1\\ e^{-(x-1)} & x > 1 \end{cases}\end{align}$$

You can find the expectation from here either from direct computation, or noting that $X_N$ has a shifted exponential distribution, $X_N \sim 1 + Exp(1)$.

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  • $\begingroup$ Is a solution generalizing the procedure from the first question also correct? That is, $E(X_N) = E(X|N)$? I found that $Pr(X_N>x|N=n)=e^{-x+1}$ as well. $\endgroup$ – user164144 Aug 3 '17 at 23:07

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