If $X$~$Exp(1)$ and $Y$~$Exp(2)$, what is the pdf of $Z=X+Y$? I am trying to find the following: If $X\sim\mathrm{Exp}(1)$ and $Y\sim\mathrm{Exp}(2)$, what is the pdf of $Z=X+Y$?
I tried to use the convolution formula but am not sure what the limits of the integral are:
$$g(z) = \int_{0}^{z} 1 \cdot 2\cdot\exp(-1\cdot(y-x))\cdot\exp(-2x) \mathrm dx.$$ 
I ended up getting:
$g(z) = 2\cdot(1-e^{-z})\cdot e^{-y}$.
Would this be correct? Thanks!
 A: \begin{equation}
g(z) = 2 \int_{- \infty}^{+ \infty} e^{-(z-x)} e^{-2x} \mathrm{d}x \\ 
=2 \int_{0}^{z} e^{-(z-x)} e^{-2x} \mathrm{d}x \\ 
= 2 e^{-z} \int_{0}^{z} e^{-x} \mathrm{d}x \\
= 2e^{-z} (1 - e^{-z}),
\end{equation}
Note that this result holds if $z>0$. In general, when $p(x) = \alpha e^{-\alpha x}$ and $p(y) = \beta e^{-\beta y}$, the convolution of these two is equal to: 
\begin{equation}
g(z) = \frac{\alpha \beta}{(\alpha - \beta)} (e^{-\beta z} - e^{-\alpha z})
\end{equation}
A: First, are $X$ and $Y$ independent?
(Suppose they are, since you're using the convolution formula.)
It should be easier to figure out the limit if you start with the bivariate transformation.
Let $W=Y$. For the transformation from $(X,Y)$ to $(Z,W)$, the Jacobian is 1, and
$f_{Z,W}(z,w)=f_{X,Y}(z-w,w)|J|=f_{X,Y}(z-w,w)=f_X(z-w)f_Y(w)$.
Then, to obtain the desired pdf $g(z)$ simply integrate out $w$.
Note that $w=y=z-x \le z$, so the limit of integral is $(0,z)$.
$g(z)=\int_{0}^{z} f_{Z,W}(z,w)dw=\int_{0}^{z} f_X(z-w)f_Y(w)dw$,
which is the convolution formula.
Finally,
$g(z)=2\int_{0}^{z} e^{-z+w}e^{-2w}dw=2e^{-z}\int_{0}^{z} e^{-w}dw=2e^{-z}(1-e^{-z})$.
(only for $z\ge0$. If $z<0$, $g(z)=0$.)
