# 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}Evaluate the constant k.

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

Evaluate the constant k.

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Given :
$f\left(x,y\right)=\left\{\begin{array}{l}kx\left(x-y\right),\\ 0\end{array}$

To find k
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x,y\right)dxdy=1$
$={\int }_{0}^{2}{\int }_{-x}^{x}kx\left(x-y\right)dxdy=1$
$={\int }_{0}^{2}{\int }_{-x}^{x}k\left({x}^{2}-xy\right)dydx=1$
$={\int }_{0}^{2}k\left[{x}^{2}y-\frac{x{y}^{2}}{2}{\right]}_{-x}^{x}dx=1$
$={\int }_{0}^{2}k\left[{x}^{3}-\frac{{x}^{3}}{2}+{x}^{3}+\frac{{x}^{3}}{2}\right]dx=1$
$={\int }_{0}^{2}k\left[2{x}^{3}\right]dx=1$
$=2k\left[\frac{{x}^{4}}{4}{\right]}_{0}^{2}=1$
=2k(4)=1
$=k=\frac{1}{8}$