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July 28th, 2017, 12:01 PM  #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 $$f_X(x)=f_Y(x)= \begin{cases} \frac{x3}2 &, & \text{if $3<x<5$,} \\ 0 &, & \text{otherwise.} \end{cases}$$ Attempt: I tried to take the integral $$\int_{z}^{6}\frac{zx3}2 \frac{x3}2\,\mathrm dx$$ then I eventually did the integral to get $(x^3/24)((18x^2)/24)(54x/24)(108/24)$. unfortunately I was unable to figure this out 
July 28th, 2017, 12:36 PM  #2 
Senior Member Joined: Oct 2009 Posts: 232 Thanks: 84 
What's the reasoning for the bounds on your integral?

July 28th, 2017, 02:09 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,438 Thanks: 562 
The integrand is wrong. Should be (zx+3)(x3/4.

July 29th, 2017, 12:39 PM  #4 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 
Why is it (zx3)(x3)/4?

July 29th, 2017, 03:40 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,438 Thanks: 562  
July 30th, 2017, 01:11 PM  #6 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 
Yeah

July 30th, 2017, 01:40 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,438 Thanks: 562  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(x3)=zx+3. ; Last edited by mathman; July 31st, 2017 at 01:25 PM. 
August 1st, 2017, 02:28 PM  #8 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 
Also, are the boundaries correct for the integral?


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