Question

# 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.

Random variables

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

Evaluate the constant k.

2021-05-30

Given :
$$f(x,y)=\begin{cases}kx(x-y),\\0\end{cases}$$

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