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July 28th, 2017, 11:01 AM   #1
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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 $$f_X(x)=f_Y(x)=
\begin{cases}
\frac{x-3}2 &, & \text{if $3<x<5$,} \\
0 &, & \text{otherwise.}
\end{cases}$$

Attempt: I tried to take the integral $$\int_{z}^{6}\frac{z-x-3}2 \frac{x-3}2\,\mathrm dx$$ 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
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July 28th, 2017, 11:36 AM   #2
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What's the reasoning for the bounds on your integral?
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July 28th, 2017, 01:09 PM   #3
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The integrand is wrong. Should be (z-x+3)(x-3/4.
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July 29th, 2017, 11:39 AM   #4
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Why is it (z-x-3)(x-3)/4?
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July 29th, 2017, 02:40 PM   #5
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Quote:
Originally Posted by poopeyey2 View Post
Why is it (z-x-3)(x-3)/4?
I don't get your comment. You are asking why is (wrong integrand).
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July 30th, 2017, 12:11 PM   #6
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Yeah
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July 30th, 2017, 12:40 PM   #7
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Quote:
Originally Posted by poopeyey2 View Post
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.








;
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Last edited by mathman; July 31st, 2017 at 12:25 PM.
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August 1st, 2017, 01:28 PM   #8
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Also, are the boundaries correct for the integral?
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