0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\beta=2$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – JohnTravolski Nov 12 '17 at 6:47
  • $\begingroup$ @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 $\{(u,w) : 0 \le u \le w\}$. $\endgroup$ – angryavian Nov 12 '17 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.