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 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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post SGill Advanced Statistics 3 November 8th, 2014 12:12 PM messaoud2010 Probability and Statistics 1 July 18th, 2014 11:26 AM hansherman Advanced Statistics 0 February 6th, 2014 04:25 AM gama Advanced Statistics 1 November 3rd, 2013 10:36 AM adamsmc2 Advanced Statistics 1 November 14th, 2012 12:18 PM

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