00:01
All right, so for this problem, we're told that x and y have the joint density function of 4x times y, when both x and y are between 0 and 1 and 0 elsewhere.
00:10
So first, let's find the expected value of x.
00:12
And to do this, we need to find the marginal density function of x, which is going to be the derivative from 0 to 1, all values of y, of the joint density function evaluated with respect to y, which is just going to give us 4 ,000.
00:33
Over 2xy squared, evaluated from 0 to 1, which is just going to equal 2x minus 0.
00:43
So we just get 2x.
00:45
And we can show that this is valid by taking the integral of 2x from all values of x, 0 to 1, which gets us x squared, evaluated from 0 to 1, which just gives us 1.
00:55
So because evaluated over or integrated over all possible values of x, this marginal density function gives us an integral of 1, we need.
01:03
Know that this is a valid marginal density function.
01:06
So the expected value of x is just going to equal the integral of x times its density function over all possible values, zero to one, which is going to get us two over three x cubed, evaluated from zero to one, which just gets us two thirds.
01:28
This is our expected value of x.
01:30
And then similarly, we know that this is also equal to the expected value of y, because this function is symmetric with respect to x and y, we know that if we were to find the marginal density function of y, we would essentially do the same operation, just evaluating with respect to x instead of y, which gets us the same result in terms of y.
01:55
And then so as a result, the expected value to y is also going to be two -thirds.
02:03
So the variance of x is going to be equal to the expected value of x minus the expected value of x, which is equal to the integral from 0 to 1 of x minus 2 thirds times the density function of x, which is 2x.
02:24
So we get the integral from 0 to 1 of 2x squared minus 4 thirds x, which is 2 thirds x, which is equal to 2 thirds x, 2 thirds x squared, evaluated from 0 to 1, which gives us a variance of, hold on.
03:01
Okay, i think something went wrong.
03:11
So you're taking x minus 2 thirds times 2x.
03:19
Yeah, so this should give us the integral from 0 to 1 of 2x squared minus 4 thirds x.
03:29
Yeah, so this should give us 2 thirds x.
03:32
Yeah, so this should give us 2 thirds x cubed minus 4 over 6x squared, evaluated from 0 to 1, which is just equal to 2 thirds minus 2 thirds minus 0 plus 0, which gives us a variance of 0, which that does not seem correct.
03:57
Okay, i'm going to pause a video, and let's see what's going on.
04:00
Okay, sorry about that.
04:02
I figured out what the mistake was.
04:04
It's the expected value of x minus e of x.
04:08
Squared.
04:11
I had my formula incorrect.
04:15
And so we're going to get x minus two -thirds squared times two -x, which then simplifies to the integral from zero to one of x squared minus four -thirds x plus 4 over 9 times 2x, valued with respect to x, which then becomes the integral from 0 to 1 of 2x cubed minus 8 over 3x squared plus 8 over 9x, which is equal to 2 thirds, or sorry, two -fourths, x to the fourth, minus 8 over 9 x cubed, plus 4 over 9 x squared, evaluated from 0 to 1, which just gives us 1 half minus 8 over 9 plus 4 over 9, which gets us 1 1� minus 4 over 9, which is equal to 918th minus 818th or 1 .18th.
05:45
This is our variance of x, 1 .18th.
05:52
And so we have our expected value of x is 2 thirds, and our variance of x is 1 .18th.
05:57
All right.
05:58
So now we want to verify that the expected value of x squared times y isn't in fact equal to the expected value of x squared times the expected value.
06:05
Y.
06:06
So first let's find the expected value of x times y or x square times y...