# Distribution of minimum and sum of two independent exponential random variables

How can I solve this problem? Is there any formula for this problem

Find the distribution of the random variable $Y$ if

1. $Y=\min(X_1,X_2)$
2. $Y=X_1+X_2$

where $X_1$ and $X_2$ are independent exponential random variables with means $\frac 1{\lambda_1}$ and $\frac 1{\lambda_2}$, respectively.

-
Problem 1) look at the complementary cumulative distribution, i.e. $$P(\min(X_1, X_2) > x) = P(X_1 > x)\cdot P(X_2 > x).$$ Substitute what you know about $X_1$ and $X_2$ and you should end up with something familiar.
Problem 2) Use moment-generating functions. Recall that the moment-generating function of a exponential random variable with $t<\lambda$ is $$M_X(t) = \frac{\lambda}{\lambda-t}.$$
@Horizon: what do you mean? An example on how to do the proof? What do you know about the cumulative distribution function? Just multiply the two terms in the expression above and the result is immediate. You will find that the minimum is exponentially distributed with parameter $\lambda = \lambda_1 + \lambda_2$. –  M.B. Nov 3 '12 at 13:24