It was a mid sem exams given to us as a home work so please help me solve it. Thanks!

Question

accepted

Q 1. If X and Y are random variables whose joint density function is the function f given as 

 4xy, 0<x<1,0<y<1 f(x,y) =

0, otherwise

Compute the following:

(a) Expected values of X and Y

(b) Probability P(Y ≥ 1 − X)

H

Answer 1

A standard way to get E(X) on a continuous random variable is to take the integral over the full range of the product of x and the function. The same would go for E(Y). In this case, it would be a double integral. Both x and y run from 0 to 1, so both will be evaluated for the extreme values 0 and 1. I can't put those limits on the integral symbols on this editor, so I'm explaining it here.

∫∫ x · 4xy dy dx = ∫∫ 4x²y dy dx

We do the integral with respect to y first, working from the inside out, and we get

∫ 4x²y²/2 [from 0 to 1] dx

When we evaluate the interval for y = 1 and y = 0 and subtract, we get

∫ 2x² dx [from 0 to 1]

This integral is 2x³/3, evaluated from 0 to 1, which is 2/3.

I could do the one for E(Y), which uses ∫∫ 4xy² dy dx, but you may begin to notice that we're doing exactly the same operations with y as with x, and using the same numbers. So guess what E(Y) will be? That's right! 2/3 also.

Now the one for P(Y ≥ 1 − X) will involve an integral of the given pdf, but the part that is above the y = 1 - x line. But perhaps an easier way is to evaluate the integral below the 1 - x line and then subtract from 1.

So here's roughly what the double integral will look like:

∫ [0 to 1] ∫ [0 to (1 - x)] 4xy dy dx

∫ 4xy²/2 [0 to (1 - x) dx

∫ 2x (1 - x)² dx

∫ (2x³ - 4x² + 2x) dx [0 to 1]

(2x⁴/4 - 4x³/3 + x²) [0 to 1] = 1/6.

This is P(Y < 1 - X), so P(Y > 1 - X) is 1 - 1/6 = 5/6.