Need help determining PDF, please.

Question

Given are two random variables X and Y for which the joint probability density function fX,Y (x, y) is given by:

fX,Y (x, y) = 0.5xy for 0 ≤ x ≤ y ≤ 2

fX,Y (x, y) = 0 otherwise.

Let Z = Y − X. Determine the probability density function fZ(z) for 0 ≤ z ≤ 2 and the expectation E(Z) of the random variable Z.

I was trying to solve this by using a double integral, and then taking the derivative to obtaint the PDF. But this did not seem to work. Is there a way to solve this using a double integral? If so what would the limits have to be? Could anyone please show me how to solve it I have been stuck for hours on this question.

Answer 1

Z ranges from 0 to 2. To have z <= a, we must have y-x <= a, are x >= y - a. For y in [0,a], we have all x <= y.

For y in [a, 2], x is in [y-a,y]

Thus, we have two double integrals. Use I[a,b] for the integral from a to b and E[a,b] for evaluation from a to b

I[0,a]I[0,y] 1/2 xy dx dy + I[a,2]I[y-a,y] 1/2 xy dx dy

I[0,a]1/4 x^2y E[0,y] dy + I[a,2]1/4 x^2y E[y-a,y] dy = as y^2 - (y-a)^2 = 2ay - a^2

I[0,a] 1/4y^3 dy + I[a,2] 1/2ay^2 - 1/4a^2y dy =

1/16y^4 E[0,a] + 1/6ay^3 - 1/8a^2y^2 E[a,2] =

1/16a^4 + 4/3a - 1/2a^2 -1/6a^4 + 1/8a^4 =

1/48a^4 - 1/2a^2 + 4/3a

F'(a) = 1/12a^3 - a + 4/3

f(z) = 1/12z^3 - z + 4/3

Answer 2

First of all, are you sure this is a probability distribution. Do you get a volume of 1?