Finding E(max(X,Y)) for random variables X and Y.

Question

We are given two independent exponentially distributed random variables :

$X$ ~ $Exp(\lambda_{1})$ and $Y$ ~ $Exp(\lambda_{2})$ , we need to find $E(max(X,Y))$.

I tried as follows :

Let $Z=max(X,Y)$ , starting off with $P(Z \leq z)$.

$F_Z(z)=P(Z\leq z)=P(max(X,Y) \leq z) = P(X,Y \leq z) = P(X \leq z)P(Y \leq z) $

which turns out to be : $(1-e^{- \lambda_1z})(1-e^{- \lambda_2z})$

Differentiating this gives the density function and hence expectation can be found out. Is this correct ?

Answer 1

which turns out to be : $(1−e^{−λ_1z})(1−e^{−λ_2z})$

Differentiating this gives the density function and hence expectation can be found out. Is this correct ?

Yes, that is viable.   Although to save effort, you might instead employ the fact that for strictly non-negative random variable, $Z$, with a CDF of $F_Z$, then: $$\mathsf E(Z)=\int_0^\infty (1-F_Z(z))\operatorname d z$$