Take the tour ×  Say X and Y are two independent random
variables with exponential density\begin{split} f_{X}(x) = a
e^{-ax}\end{split} and \begin{split} f_{Y}(y) = b
e^{-by}\end{split}, then what is the probability density
function of Z=X-Y? I'm trying to slove this problem, but I have
no sense of the integrating regions. How will they be?  I tried
Shai's hint \begin{split} {\rm P}(X - Y \le z) = \int_0^\infty
{{\rm P}(X - Y \le z|Y = \tau )f_{Y}(\tau ){\rm d}\tau } =
\int_0^\infty {{\rm F_{X}}(z + \tau )f_{Y}(\tau ){\rm d}\tau }
\end{split}  I obtained this \begin{split} {\rm P}(Z \le z) =
F_{Z}(z) = 1 - \frac b {a+b} e^{-a z} \end{split} But it's not
converage to 1 when z is infinite, what's wrong with my
calculation?  share|improve this question You should separate
into the cases $z + \tau < 0$ and $z + \tau > 0$. –  Shai Covo
Apr 11 '11 at 12:05 Also, it does converge to $1$ when $z \to
\infty$; the problem is seen when $z \to -\infty$. –  Shai Covo
Apr 11 '11 at 12:11 One may a priori expect to find $|z