Question

Suppose X and Y are continuous random variables with joint density function 1 + xy 9 fx,y(2, y) = 4 [2] < 1, [y] < 1 otherwise 0, (1) (4 pts) Find the marginal density function for X and Y separately. (2) (2 pts) Are X and Y independent? Verify your answer. (3) (9 pts) Are X2 and Y2 independent? Verify your answer.

Answer 1

Q1) The marginal PDFs for X and Y here are obtained as:

$f_x(x) = \int_{-1}^{1} \frac{1 + xy}{4} \ dy = \frac{2 + 0.5x(1 - 1)}{4} = \frac{2}{4} = 0.5$

Therefore, the PDF for X is given as:

Due to symmetry between X and Y here, the pdf for Y would also be the same given as:

b) f(x)f(y) = 0.5*0.5 = 0.25 which is not equal to the given joint PDF f(x, y)

Therefore X and Y are not independent variables here.

c) We have here:

The expected value of X²Y² is computed here as:

$E(X^2Y^2) =\int_{-1}^{1}\int_{-1}^{1} \frac{x^2y^2 + x^3y^3}{4} \ dy \ dx = \int_{-1}^{1} 0.25*(\frac{2x^2}{3} ) \ dx$

$E(X^2Y^2) = \int_{-1}^{1} (\frac{x^2}{6} ) \ dx = \frac{1 + 1}{6*3} = \frac{1}{9}$

Now getting the expected value of X² here as:

$E(X^2) = \int_{-1}^{1}0.5x^2 \ dx = \frac{1}{3}$

Also due to symmetry,
E(Y²) = 1/3

Therefore, E(X²Y²) = E(X²)E(Y²) = 1/9

therefore X² and Y² are independent here.