00:01
So in this question we're told that a joint pdf is given by, sorry, xy, 8xy for x between 0 and 1 and y between 0 and x and 0 otherwise.
00:22
So first of all, we want the expectation of x given y is equal to y.
00:32
So this is going to be the integral over all values x from...
00:40
So if x can be between 0 and 1, so this expectation is going to be a function of y.
00:47
So this is a function of y.
00:50
That means we need to have a look at this region and find out what values of x we need to integrate out.
00:57
So we know that x is between 0 and 1 and y is between 0 and x.
01:01
So here's the line y equals x.
01:08
So here's our region, our allowed region.
01:12
But there's also, so y between 0 and 1, sorry, x between 0 and 1 and y between 0 and x, is equivalent to the region y between 0 and 1 and x between y and 1.
01:36
So that's the integration we need to do.
01:39
So our conditional distribution is going to be the joint distribution divided by the marginal distribution of y.
01:54
So this is going to be 8xy divided by the integral from x equals y to 1, 8xy d x.
02:03
So that's 8xy divided by 4x squared y between y and 1, x equals y and x equals 1.
02:12
So this is 8xy over 4y minus 4y cubed, which is 2x over 1 minus y squared.
02:28
So our conditional distribution of x given y equals y is equal to, so x given y equals y, is 2x over 1 minus y squared for y between 0 and 1.
02:48
And x between y and 1, and 0 otherwise.
02:57
So that means that the expectation value of x given y equals y is going to be the integral from y to 1, 2x over 1 minus y squared the x, which is x squared over 1 minus y squared between y and 1, x equals y and x equals 1.
03:19
So this is 1 minus y squared over 1 minus y squared, which is just one.
03:30
Oh no, sorry, there should be an extra x in here, because it's the expectation value of x squared...