Let $X$ and $Y$ be two continuous RV jointly distributed with PDF $$ f_{X,Y}(x,y)= 6e^{-(2x+3y)} \text{ for } x,y\geq 0. $$ The PDF takes value $0$ for other values of $x,y$. Clearly, $X$ and $Y$ are independent. I would want to find the probability of $P(X>Y)$.
The usual (easy) way is to take double integrals and to compute $$ \int_0^{\infty}\int_{y}^{\infty}6e^{-(2x+3y)}dxdy=\frac{3}{5}. $$
However, I was thinking another way by letting $Z=X-Y$ and using the convolution integral $$ P(X>Y)=P(Z>0)=\int_{0}^{\infty}\int_{0}^{\infty}f_{X}(x)f_Y(x-z)dxdz=\int_{0}^{\infty}\int_{0}^{\infty}6e^{-(2x+3(x-z))}dxdz. $$ But, the integral above does not converge. I am not clear whether where when wrong in the convolution integral. Would appreciate if anyone can point it out to me. Thanks!
Note that $f_Y(x-z) = 0$ if $x < z$, thus $$ \int_{0}^{\infty}\int_{0}^{\infty}f_{X}(x)f_Y(x-z)dxdz = \int_{0}^{\infty}\int_{z}^{\infty}f_{X}(x)f_Y(x-z)dxdz = \frac{3}{5}. $$