
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
August 9th, 2017, 10:06 AM  #1 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0  Find the support of the density function of the random variable $X + Y$ .
The support of a function $f(x)$ is defined to be the set $\{x : f(x) > 0\}$ . Suppose that $X$ and $Y$ are two continuous random variables with density functions $f_X(x)$ and $f_Y (y)$, respectively, and suppose that the supports of these density functions are the intervals $[a, b]$ and $[c, d]$, respectively. Find the support of the density function of the random variable $X + Y$ . Attempt: I tried to set up and take the integral $$\int_{0}^{z} (zx)(x)\,\mathrm dx$$ and eventually ended getting $\frac{z^3}6$, however this doesnt seem to be right. How would u solve this? 
August 9th, 2017, 11:06 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,169 Thanks: 1141 
The support of a continuous random variable is just the set of values that have nonzero probability. It should be obvious at first glance that the support of $X + Y$ is just $[a+c, b+d]$ The bounds are just the minimum and maximum attainable values of $X+Y$ 
August 10th, 2017, 10:08 AM  #3 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 
Is there a procedure or steps on how you got a+c and b+d, Romsek?


Tags 
density, find, function, random, support, variable 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Let X be a random variable with range [−1,1] and let fX(x) be the density function of  poopeyey2  Advanced Statistics  1  June 15th, 2017 06:02 PM 
function of random variable  mhhojati  Advanced Statistics  3  October 20th, 2015 06:46 PM 
Density of a random variable  alfred_oh  Advanced Statistics  0  April 30th, 2013 06:53 AM 
Probability density function of this variable  lui14  Probability and Statistics  0  February 2nd, 2013 03:56 AM 
function of random variable  benjaminmar8  Advanced Statistics  1  April 10th, 2008 08:14 AM 