Suppose $X_1$ is $\mathrm{Exp}(\lambda_1)$ and $X_2$ is $\mathrm{Exp}(\lambda_2)$. $X_1$ and $X_2$ are independent.
Let $Y = \min (X_1, X_2)$ and $Z = \max (X_1, X_2)$ and $W = ZY$ . Compute the distribution function and/or the density function of $W$.
The way I'm approaching this problem is as follows:
Probability (W <= x) = $ \int_{0}^{\infty }Pr(W\leqslant x | \min(X_{1},X_{2})=y)(\lambda _{1}+\lambda _{2})e^{(\lambda _{1}+\lambda _{2})y}dy$
= $\int_{0}^{\infty }Pr(\max(X_{1},X_{2})\leqslant \frac{x}{y} | \min(X_{1},X_{2})=y)(\lambda _{1}+\lambda _{2})e^{(\lambda _{1}+\lambda _{2})y}dy$
But I'm getting lost while doing the actual integration. Can anyone tell me whether there is a better way to solve this?
Hint:
Note that if $Y=X_1$, then $Z=X_2$; similarly, if $Y=X_2$, then $Z=X_1$. So, no matter what, $W=YZ=X_1X_2$. So, you need only compute the distribution/density of $X_1X_2$, which is (hopefully) easier to deal with!
In this case: to get that $X_1X_2\leq x$, $X_1X_2\leq x$ is equivalent to saying that either $X_1=0$, or $X_1\neq0$ but $X_2\leq\frac{x}{X_1}$. So, $$ P(X_1X_2\leq x)=\int_0^{\infty}\int_0^{x/x_1}f_{X_1}(x_1)f_{X_2}(x_2)\,dx_2\,dx_1, $$ where $f_{X_1}$ and $f_{X_2}$ are the densities of $X_1$ and $X_2$.