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We have X ~ geom(p), and given X=n, we generate n number of iid random variables $Y_1...Y_n$ from exp(1) distribution. Denote $Y_{(1)} = min\{Y_1...Y_n\}$, what is $P(Y_{(1)} > t | X=n)$?

My work: $$P(Y_{(1)} > t | X=n) = P(Y_1>t,..., Y_n > t|X=n) = [P(Y>t|X=n)]^n$$ From here I am not sure how to calculate that conditional probability inside. It would be invalid to say it is $1-(1-e^{t})$ right? I also thought I could use definition of conditional probability involving joint pmf, but then I'm not sure how to find the joint pmf, because they are dependent and hence I can't just multiply their individual distributions.

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The statement that $Y$ has $exp(1)$ distribution given $X=n$ means that $P(Y>t|X=n)=e^{-t}$ for all $t >0$. So the answer is $e^{-nt}$.

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  • $\begingroup$ ohh it is as simple as that. It makes sense in terms of words, but is there a way of proving this formally using definition of conditional probability? say $P(Y>t|X=n) = \frac{P(Y>t, X=n)}{P(X=n)}$, from here how can we recover $P(Y>t)$? Clearly X and Y are not independent $\endgroup$
    – MinYoung Kim
    Apr 6, 2020 at 5:26
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    $\begingroup$ @MinYoungKim There is nothing to prove. It is just the meaning of the statement $Y$ has $exp(1)$ distribution given $X=n$. $\endgroup$
    – Kavi Rama Murthy
    Apr 6, 2020 at 5:29
  • $\begingroup$ gotcha, thank you $\endgroup$
    – MinYoung Kim
    Apr 6, 2020 at 5:42

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