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November 11th, 2017, 09:29 PM   #1
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Exclamation 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.
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November 11th, 2017, 09:47 PM   #2
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well... I wouldn't approach this problem the way you have. I'll show you what I'd do and that might shed some light on the divergent integral.

$f(x,y) = (2e^{-2x})(2e^{-2y})$, i.e. $X,~Y$ are i.i.d.

The sum of independent rvs is distributed as the convolution of the two individual distributions.

$W=X+Y$

$f_W(w) = \displaystyle \int_{-\infty}^\infty f(x)f(w-x)~dx = \displaystyle \int_{0}^w f(x)f(w-x)~dx =\displaystyle \int_{0}^w 4e^{-2x}e^{-2(w-x)}~dx = 4 w e^{-2 w}$

let me take a look at your substitution method a bit and see if I have any comments.
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Last edited by romsek; November 11th, 2017 at 09:59 PM.
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November 11th, 2017, 09:58 PM   #3
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$f_{UW}(u,w)=4e^{-2w},~w>0$

$f_W(w) = \displaystyle \int_0^w~4e^{-2w}~du$

because $0 \leq u \leq w$

$f_W(w) = 4 w e^{-2w}$

which is identical to the answer I got in the previous post
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November 11th, 2017, 10:10 PM   #4
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I'm sorry for not seeing this, but how are we obtaining that $u ≤ w$? That's the only reason I'm struggling with this. I thought the support of $U$ was $u>0$.

Last edited by John Travolski; November 11th, 2017 at 10:30 PM.
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November 11th, 2017, 11:48 PM   #5
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Quote:
Originally Posted by john travolski View Post
i'm sorry for not seeing this, but how are we obtaining that $u ≤ w$? That's the only reason i'm struggling with this. I thought the support of $u$ was $u>0$.
$x,y > 0$

$w = x+y > x = u$

$u < w$

Last edited by romsek; November 12th, 2017 at 12:05 AM.
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