# Probability of two dependent random variables

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.

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 $$e^{-nt}$$.
• 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 Apr 6, 2020 at 5:26
• @MinYoungKim There is nothing to prove. It is just the meaning of the statement $Y$ has $exp(1)$ distribution given $X=n$. Apr 6, 2020 at 5:29