My Math Forum Bivariate Transformation for Sum of Random Variables

 November 11th, 2017, 08:29 PM #1 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 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.
 November 11th, 2017, 08:47 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 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. Thanks from John Travolski Last edited by romsek; November 11th, 2017 at 08:59 PM.
 November 11th, 2017, 08:58 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 $f_{UW}(u,w)=4e^{-2w},~w>0$ $f_W(w) = \displaystyle \int_0^w~4e^{-2w}~du$ 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 Thanks from John Travolski
 November 11th, 2017, 09:10 PM #4 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 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 09:30 PM.
November 11th, 2017, 10:48 PM   #5
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Quote:
 Originally Posted by john travolski 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 11th, 2017 at 11:05 PM.

 Tags bivariate, random, sum, transformation, variables

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