# pdf of combination of two distinct exponential random variables

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?

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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 $|z|$ in the answer. – Shai Covo Apr 11 '11 at 12:13
Thank you! sorry for the mistake ,but how to obtain |z| in the answer? – Zrst Apr 11 '11 at 12:27
Have you calculated the distribution function for all $z \in \mathbb{R}$? Note that $Z$ takes values in $(-\infty,\infty)$. You can use $|z|$ to shorten the final answer. – Shai Covo Apr 11 '11 at 12:53

Hint: $P(X - Y \le z) = \int_0^\infty {P(X \le z + y)f_Y (y)dy}$, $z \in \mathbb{R}$.
Where exponential random variables are concerned, one can prefer considering $P(X\ge t)$ rather than $P(X\le t)$, in which case the quantity to introduce would be $P(X-Y\ge z)$ rather than $P(X-Y\le z)$. – Did Apr 11 '11 at 11:43
I believe that it's eventually the same. But you can also do this $$P(X-Y\le z)=\iint_{\{x-y\le z\}}f_{X,Y}(x,y)dxdy$$ where the joint PDF is given by $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ because of independence.