Answer to Question #252320 in Statistics and Probability for pas

Question

The random variables X and Y have joint p.d.f given by

f(x,y)=k, 0<x<y<2 zero otherwise

i) Obtain the value of the constant k

ii)Find the conditional variance of Y given X=x

Answer 1

i)

$$\displaystyle\int_{-\infin}^{\infin}\displaystyle\int_{-\infin}^{\infin}f(x, y)dydx=1$$

$$\displaystyle\int_{0}^{2}\displaystyle\int_{x}^{2}kdydx=1$$

$$k\displaystyle\int_{0}^{2}[y]dx\begin{matrix} 2 \\ x \end{matrix}=1$$

$$k[2x-\dfrac{x^2}{2}]\begin{matrix} 2 \\ 0 \end{matrix}=1$$

$$k=\dfrac{1}{2}$$

ii)

$$f_X(x)=\displaystyle\int_{x}^2\dfrac{1}{2}dy=1-\dfrac{1}{2}x, 0\leq x\leq 2$$

Thus for $0\leq x\leq 2$ we obtain

$$f_{Y|X}(y|X)=\dfrac{f_{XY}(x,y)}{f_X(x)}=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}x}=\dfrac{1}{2-x}$$

$$E(Y|x)=\displaystyle\int_{x}^2y(\dfrac{1}{2-x})dy=\dfrac{4-x^2}{2(2-x)}=\dfrac{2+x}{2}$$

$$E(Y^2|x)=\displaystyle\int_{x}^2y^2(\dfrac{1}{2-x})dy=\dfrac{8-x^3}{3(2-x)}=\dfrac{4+x+x^2}{3}$$

$$Var(Y|X=x)=E(Y^2|x)-(E(Y^2|x))^2$$

$$=\dfrac{4+x+x^2}{3}-(\dfrac{2+x}{2})^2$$

$$=\dfrac{16+4x+4x^2-12-12x-3x^2}{12}$$

$$=\dfrac{x^2-8x+4}{12}$$