pdf of combination of two distinct exponential random variables

Question

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?

Answer 1

Hint: $P(X - Y \le z) = \int_0^\infty {P(X \le z + y)f_Y (y)dy}$, $z \in \mathbb{R}$.

Answer 2

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.