# Suppose that the random variables X and Y have joint p.d.f.f(x,y)=\begin{cases}kx(x-y),0<x<2,-x<y<x\\0,\ \ \ \ elsewhere\end{cases}Find the marginal p.d.f. of the two random variables.

Suppose that the random variables X and Y have joint p.d.f.
$\left(f\left(x,y\right)=\left\{\begin{array}{l}kx\left(x-y\right)\\ 0\end{array}$Find the marginal p.d.f. of the two random variables.

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Given :
$\left(f\left(x,y\right)=\left\{\begin{array}{l}kx\left(x-y\right)\\ 0\end{array}$
To find marginal P.D.F of x
${f}_{x}\left(x\right)=\frac{1}{8}{\int }_{-x}^{x}x\left(x-y\right)dy$
$\frac{1}{8}{\int }_{-x}^{x}\left({x}^{2}-xy\right)dy$
$=\frac{1}{8}\left[{x}^{2}y-\frac{x{y}^{2}}{2}{\right]}_{-x}^{x}$
$=\frac{1}{8}\left[{x}^{3}-\frac{{x}^{3}}{2}+{x}^{3}+\frac{{x}^{3}}{2}\right]$
$=\frac{1}{8}\left(2{x}^{3}\right)$
$=\frac{{x}^{3}}{4}$
To find marginal P.D.F of y
${f}_{y}\left(y\right)=\frac{1}{8}{\int }_{0}^{2}x\left(x-y\right)dx$
$=\frac{1}{8}{\int }_{0}^{2}\left({x}^{2}-xy\right)dx$
$=\frac{1}{8}\left[\frac{{x}^{3}}{3}-\frac{{x}^{2}y}{2}{\right]}_{0}^{2}$
$=\frac{1}{8}\left[\frac{8}{3}-\frac{4y}{2}-0-0\right]$
$=\frac{1}{8}\left(\frac{8}{3}-y\right)$
$=\frac{1}{3}-\frac{y}{8}$
$=\frac{8-3y}{24}$