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.
Best Answer
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 $e^{-nt}$.