Sum of Two Continuous Random Variables Let $X$ and $Y$ be two random variables with common PDF.
$f_X(x)=f_Y(y)=e^{-x}$, for all $x, y > 0$
I have set up the integral according to the formula for convolution for two continuous random variables. I am unsure whether the upper bound for the integral is equal to $Z$ itself. The lower bound is obviously zero.
Let $Z=2X+Y$.
Can anyone assist in figuring out how to set up the integrals with the right upper and lower bounds?
 A: We assume that $X$ and $Y$ are independent random variables.
The general idea is that $X$ has a probability $f_X(x)\,\Delta x$ of being between $x$ and $x+\Delta x$, so $2X$ has the same probability of being between $2x$ and $2x+2\Delta x$.  $Y$, meanwhile, has a probability $f_Y(z-2x)\,\Delta x$ of being between $z-2x-\Delta x$ and $z-2x$, so it has a probability $2f_Y(z-2x)\,\Delta x$ of being between $z-2x-2\Delta x$ and $z-2x$ (since the "window" is twice as wide).
We can therefore write, in the limit,
$$
f_Z(z) = \int_{x=0}^{z/2} 2f_X(x) f_Y(z-2x) \, dx
$$
since we must have $X \leq z/2$ in order to permit $Z = z$.
A: This is not an answer, since it does not use a convolution. (The convolution approach has been very well explained by Brian Tung.)
We will find the cdf of $Z$, that is, $\Pr(Z\le z)$. Draw the line $\ell$ with equation $2x+y=z$. Then, for $z\gt 0$, $\Pr(Z\le z)$ is equal to the probability $(X,Y)$ lands in the part of the first quadrant below line $\ell$. Let $T$ be this triangle. Then
$$F_Z(z)=\iint_T e^{-x}e^{-y}\,dx\,dy.$$
Express this integral as the  iterated integral
$$\int_{x=0}^{z/2} e^{-x}\left(\int_{y=0}^{z-2x} e^{-y}\,dy\right)\,dx.$$
For the density, calculate the iterated integral, then differentiate. Alternately, just calculate the inner integral, and differentiate under the integral sign.
