Question: Given two iid random variables $X$ and $Y$ with same PDF of $f(x) = e^{-x}$, $x>0$, find the PDF of $X+Y$ and $3X+2Y$. Are these two new PDF's independent?
What I've got so far:
First I defined $Z = X + Y$, since $x,y>0$, $z>0$. By applying a convolution on $X$ and $Y$ we get: $f_Z(z) = \int_{0}^{z} f_{XY}(x,z-x) \,dx \rightarrow f_Z(z) = \int_{0}^{z} e^{-x + x - z} \,dx \rightarrow f_Z(z) = ze^{-z}$.
Now for $W = 3X+2Y$, by also applying a convolution, we get: $f_W(w) = \int_{0}^{w/3} f_{XY}(x,\dfrac{w-3x}{2}) \,dx \rightarrow f_W(w) = \int_{0}^{w/3} e^{(x-w)/2} \,dx \rightarrow f_W(w) = 2(e^{-w/3} - e^{-w/2})$.
However, when calculating the joint distribution of $Z$ and $W$ using the jacobian method, we get:
$|J| = 1$, $Y = 3Z-W$ and $X = W-2Z$, and thus:
$f_{WZ}(w,z) = f_{XY}(w-2z,3z-w) \rightarrow f_{WZ}(w,z) = e^{-(w-2z+3z-w)} \rightarrow f_{WZ}(w,z) = e^{-z} $
And so I concluded that the two new PDF's are not independent, because their joint probability is not equal to the product of the marginal probabilities. Are these results correct? Also, how can I get to $f_Z(z)$ from $f_{WZ}(w,z)$? Because I can't seem to figure out the right integration bounds in order to get the right marginal probabilities.
There is a mistake in the pdf of $3X + 2Y$. You are missing factor of $\frac{1}{2}$. It should be,
$ \displaystyle f_W(w) = \frac{1}{2}\int_{0}^{w/3} f_{XY}\left(x,\dfrac{w-3x}{2}\right) \, dx$
$ \displaystyle = \int_{0}^{w/3} e^{(x-w)/2} \,dx = (e^{-w/3} - e^{-w/2})$
You are right that as $ \ f_{WZ} (W, Z) \ne f_W (w) \cdot f_Z (z)$, $W$ and $Z$ are not independent.