
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
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. $W=X+Y$ $f_W(w) = \displaystyle \int_{\infty}^\infty f(x)f(wx)~dx = \displaystyle \int_{0}^w f(x)f(wx)~dx =\displaystyle \int_{0}^w 4e^{2x}e^{2(wx)}~dx = 4 w e^{2 w}$ let me take a look at your substitution method a bit and see if I have any comments. 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 
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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389  Quote:
$w = x+y > x = u$ $u < w$ Last edited by romsek; November 11th, 2017 at 11:05 PM.  

Tags 
bivariate, random, sum, transformation, variables 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Transformation of a Random Variable (chisquare)  SGill  Advanced Statistics  3  November 8th, 2014 12:12 PM 
Theory and Problems of Probability, Random Variables, and Random Processes  messaoud2010  Probability and Statistics  1  July 18th, 2014 11:26 AM 
Max Transformation of a normal distributed random variable  hansherman  Advanced Statistics  0  February 6th, 2014 04:25 AM 
PDF of sum of random variables  gama  Advanced Statistics  1  November 3rd, 2013 10:36 AM 
Discrete Bivariate Random Vectors  adamsmc2  Advanced Statistics  1  November 14th, 2012 12:18 PM 