Answer to Question #190372 in Statistics and Probability for muhammad amir

Question

The joint pdf of random variables ‘X’ and ‘Y’ is

 

Find

a. The marginal pdfs, f_X(x) and f_Y(y).

b. The conditional pdfs, f_X/Y(x/y) and f_Y/X(y/x)

c. The E(X/Y=1)

d. Are ‘X’ and ‘Y’ statistically independent?

Answer 1

Since joint pdf is not given in the question hence we can assume that

Joint pdf of variables X and Y is

$f_{X,Y}(x,y) = 6xy$ when $0 \le x \le 1, 0\le y \le \sqrt{x}$

Otherwise

a.) $f_{X}(x) = \int_{-\infty}^{\infty} f_{XY}(x,y)dy$

$= \int_{0}^{\sqrt{x}}6xy dy$

$= 3x^2$

Thus, $f_{X}(x) = 3x^2 \hspace{3mm}0 \le x\le 1$

0 Otherwise

To find $f_{Y}(y) \hspace{2mm}for \hspace{2mm} 0 \le y \le 1$ , we can write,

$f_{Y}(y) = \int_{-\infty}^{\infty} f_{XY}(x,y)dx$

$= \int_{y^2}^{1} 6xydx$

$= 3y(1-y^4)$

$f_{Y}(y ) = 3y(1-y^4)$ $0 \le y \le 1$

$= 0$ Otherwise

b.) The conditional pdf,

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

$= \dfrac{6xy}{3x^2} = \dfrac{2y}{x}$

$f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(x)}$

$= \dfrac{6xy}{3y(1-y^4)} = \dfrac{2x}{(1-y^4)}$

c.) The $E(X/Y = y)$

$= \int_{-\infty}^{\infty} xf_{X|Y}(x|y)dx$

$= \int_{y^2}^{1} x \dfrac{2x}{1-y^4}dx$

$= \dfrac{2(1-y^6)}{3(1-y^4)}$

d.) Yes X and Y are statistically independent.