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Find the distribution of U=min(X,Y) where X and Y are independent random variables and both exponentially distributed with parameters lambda and mu respectively.

The only headway I have made is that P(U< u)= P(X< u)P(Y< u) by considering the joint distribution of X and Y and the fact X and Y are independent; is this right? If it is then U isn't a 'named' distribution function so is just stating the cdf enough to answer the question?

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You have the rough idea, but it's not quite correct: for example, $\Pr[U \le 1]$ only means that the lesser of $X$ and $Y$ is at most $1$. So for example, you could have $X = 1/2$, $Y = 5$, and yet $U = \min(X,Y) = 1/2 \le 1$.

Instead, consider $$\Pr[U > u] = \Pr[\min(X,Y) > u] = \Pr[(X > u) \cap (Y > u)] \overset{\text{ind}}{=} \Pr[X > u]\Pr[Y > u].$$ This way works because $U$ is greater than $u$ if and only if both $X$ and $Y$ are also greater than $u$. Now recall that $$\Pr[X > x] = S_X(x) = 1 - F_X(x) = e^{-\lambda x}$$ and $$\Pr[Y > y] = S_Y(y) = e^{-\mu y},$$ (assuming that your exponential distributions are parametrized by rate and not by scale (thus $\operatorname{E}[X] = 1/\lambda$ and $\operatorname{E}[Y] = 1/\mu$). I leave the rest to you.


An additional note: If we had defined $W = \max(X,Y)$, then the distribution of $W$ could be found by considering $\Pr[W \le w]$; but with $U = \min(X,Y)$, we need to use $\Pr[U > u]$ as shown above. But for an extra challenge, what would we need to do if we had, say, $X_i = \operatorname{Exponential}(\lambda_i)$, $i = 1, 2, 3$, and we wanted the distribution of $X_{(2)}$, the middle order statistic, rather than the minimum or maximum?

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  • $\begingroup$ I am not sure, maybe use indicator function somehow? $\endgroup$ – guest10923 Mar 31 '15 at 11:04

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