My Math Forum Suppose that $X$ and $Y$ are independent and $Z = X +Y$. Find $f_Z(z)$ if...

 July 28th, 2017, 11:01 AM #1 Member   Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 Suppose that $X$ and $Y$ are independent and $Z = X +Y$. Find $f_Z(z)$ if... >Let $X$ and $Y$ be independent random variables defined on the space $\Omega$, with density functions $f_X(x)$ and $f_Y(y)$, respectively. Suppose $Z = X + Y$. >Find the density $f_Z(z)$ if f_X(x)=f_Y(x)= \begin{cases} \frac{x-3}2 &, & \text{if $3  July 28th, 2017, 11:36 AM #2 Senior Member Joined: Oct 2009 Posts: 428 Thanks: 144 What's the reasoning for the bounds on your integral? Thanks from poopeyey2  July 28th, 2017, 01:09 PM #3 Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591 The integrand is wrong. Should be (z-x+3)(x-3/4. Thanks from poopeyey2  July 29th, 2017, 11:39 AM #4 Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 Why is it (z-x-3)(x-3)/4? July 29th, 2017, 02:40 PM #5 Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591 Quote:  Originally Posted by poopeyey2 Why is it (z-x-3)(x-3)/4? I don't get your comment. You are asking why is (wrong integrand).  July 30th, 2017, 12:11 PM #6 Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 Yeah July 30th, 2017, 12:40 PM #7 Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591 Quote:  Originally Posted by poopeyey2 Yeah The expression I gave is the correct form of the integrand for convolution. Yours is just wrong. The basic equation you need to understand is: z-(x-3)=z-x+3. ; Last edited by mathman; July 31st, 2017 at 12:25 PM.  August 1st, 2017, 01:28 PM #8 Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 Also, are the boundaries correct for the integral?  Tags$fzz$,$x$,$y\$, find, independent, suppose

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