Question

The joint density function of two random variables X and Y is given by

Answer 1

Let use the following problem.

The marginal pdf and expectations are as follows:

$p(x) = \int_y p(x,y) \\ = \int^5_1 \frac{xy}{96} dy \\ = \frac{x}{96} \int^5_1 ydy \\ = \frac{x}{96} [\frac{y^2}{2}]^5_1 \\ = \frac{x}{96}[\frac{25}{2}-\frac{1}{2}] \\ = \frac{x}{8} \; 0<x<4$

The expectation of X is as follows:

$E(X) = \int^4_0x \times \frac{x}{8} dx \\ = \frac{1}{8} \int^4_0 x^2 dx \\ =\frac{1}{8}[\frac{x^3}{3}]^4_0 \\ =\frac{1}{8} \times \frac{4^3}{3} \\ = \frac{8}{3} \\ = 2.6667 \\ p(y) = \int_4 p(x,y) \\ = \int^4_0 \frac{xy}{96} dx \\ = \frac{y}{96} \int^4_0 x dx \\ = \frac{y}{96} [\frac{x^2}{2}]^4_0 \\ = \frac{y}{96} [\frac{16}{2}] \\ = \frac{y}{12} \; 1<y<5 \\ E(y) = \int^5_1 y \times \frac{y}{12}dy \\ = \frac{1}{12} \int^5_1 y^2 dx \\ = \frac{1}{12}[\frac{y^3}{3}]^5_1 \\ = \frac{1}{12} \times \frac{5^3 -1}{3} \\ = \frac{31}{9} \\ = 3.4444 \\ p(x) \times p(y) = \frac{x}{8} \times \frac{y}{12} \\ =\frac{xy}{96} \\ =p(x,y)$

Thus, X and Y are independent.

$E(XY)=E(x) \times E(y) \\ = \frac{8}{3} \times \frac{31}{9} \\ = \frac{248}{27} \\ = 9.1852 \\ E(2X+3Y) = 2E(x) +3E(y) \\ = 2 \times \frac{8}{3} + 3 \times \frac{31}{9} \\ = \frac{16}{3} + \frac{31}{3} \\ = \frac{47}{3} \\ = 15.6667$