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Suppose we have $X$ and $Y$ are jointly distributed with density of $2$ if $0 < x < y < 1$ (and $0$ otherwise). I'd like to find the PDF of the sum, $X + Y$.

I read the Jacobian method here: pdf of sum of two dependent random variables

and set $Z = X + Y, W = X$ to apply that method.

However, I am having trouble writing the bounds for the joint density function. I know that $0 < x < y < 1$ and thus $2x < z < 1 + x$, and since $w = x$ we know that $2w < z < 1 + w$ and $0 < w < z$.

Finally when I integrate the density function $2$ from $0$ to $z$ to find PDF of $Z$ the solution is simply...$2z$ for (what are the bounds?) How can I finish up the problem?

1 Answers1

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The joint density is

$$f_{WZ}(w,z)=2$$

because tha jacobian is 1. Thus it is still uniform. The only difficulty is to derive the joint support of $(W;Z)$

considering that you have

$$0<w<z-w<1$$

that is

$$0<2w<z<1+w$$

the following region is the joint support $(W;Z)$

enter image description here

in order to get $f_Z(z)$ you have to integrate wrt $w$ obtaining

When $0<z<1$

$$f_Z(z)=\int_0^{z/2}2dw=z$$

and when $1<z<2$

$$f_Z(z)=\int_{z-1}^{z/2}2dw=2-z$$

tommik
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  • Just to make sure I am concluding correctly, would the density of $Z$ specifically be $2$ for $0 < z < 2$? Since $0 < w < 1$? –  Dec 09 '21 at 16:28
  • @freshprinter68 : obviously no. $f_Z(z)$ is a triangular density – tommik Dec 09 '21 at 16:31
  • so how should I obtain $f_Z(z)$? Sorry I'm not sure the picture is helping me find the method. I took $\int_0^1 f_{WZ}(w,z),dw$ which yields $2$? –  Dec 09 '21 at 16:33