Suppose that random variables X and Y vary in accordance with the joint pdf, f_(X,Y)(x,y)=c(x+y), 0<x<y<1. Find c.

Suppose that random variables X and Y vary in accordance with the joint pdf, ${f}_{X,Y}\left(x,y\right)=c\left(x+y\right),0. Find c.
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Ashly Sanford
Given:
0${f}_{X,Y}\left(x,y\right)=c\left(x+y\right)$
If ${f}_{X,Y}$ is a valid pdf, then the integral over all possible values has to be equal to 1:
${\int }_{0}^{1}{\int }_{0}^{y}{f}_{X,Y}$
$\left(x,y\right)dxdy={\int }_{0}^{1}{\int }_{0}^{y}c\left(x+y\right)dxdt=c{\int }_{0}^{1}{\int }_{0}^{y}\left(x+y\right)dxdy$
$=c{\int }_{0}^{1}\left(\frac{{x}^{2}}{2}+xy\right){|}_{0}^{y}dy=c{\int }_{0}^{1}\left(\frac{3}{2}{y}^{2}\right)dy$
$=c\left(\frac{{y}^{3}}{2}\right){|}_{0}^{1}=c\left(\frac{1}{2}\right)=\frac{c}{2}$
This integral has to be equal to 1:
$\frac{c}{2}=1$
Multiply each side by 2:
c=2
Result:
c=2
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