# Bivariate Transformation for Sum of Random Variables

Let $X$ and $Y$ have the joint PDF $f(x,y)=4e^{-2(x+y)}$ on $x>0$ and $y>0$ and zero otherwise. Find the distribution of $W=X+Y$

I make the substitutions $W=X+Y$ and $U=X$ to obtain that the Jacobian is simply $1$.

So then I calculate the joint distribution of $U$ and $W$ to be $f(u,w) = 4e^{-2w}$ on $u>0$ and $w>0$

But when I go to find the marginal distribution function of $W$, $f_W(w)$, I get stuck with a divergent integral. What's going wrong here? Is my support for $u$ and $w$ incorrect? I just don't understand it. Please, somebody explain what's going wrong.

## 1 Answer

You probably did not get the support of $u$ correctly. The density $f_{U,W}$ is zero outside of the set $\{(u,w): 0 \le u \le w\}$. $$f_W(w) = \int_0^w f_{U,W}(u,w) \mathop{du} = \int_0^w 4e^{-2w} \mathop{du} = 4we^{-2w}.$$ This is the gamma distribution with parameters $\alpha=2$ and $\beta=2$.

• I see how that makes a difference, but in general, I do not know how to find the supports of the variables I have substituted in. How exactly do you go about finding the support of u? – JohnTravolski Nov 12 '17 at 6:47
• @JohnTravolski You just have to think carefully about what $(u,w)$ pairs are valid. For example, you cannot have $u=5$ and $w=3$, since that would correspond to $x=5$ and $y=-2$, which violates the $y>0$ constraint. For this case, you can show that the support of $f_{U,W}$ is $\{(u,w) : 0 \le u \le w\}$. – angryavian Nov 12 '17 at 19:05