00:01
Okay, we are given the joint density function of x and y is equal to c times x times y, where x is between 0 and 18 and y is between 0 and x.
00:09
And we need to find the covariance of x and y.
00:12
So first we need to find the value c, such that this is a valid probability density function.
00:18
And a probability density function is valid when the integral from negative infinity to infinity of this function with respect to y, right, i guess i should say the double integral from negative infinity to infinity of the function of x and y with respect to both y and x is equal to 1.
00:43
And so in this case, that it's going to be equal to, well, anywhere where x is not between 0 and 18, and y is not between 0 and x, the function evaluates to 0.
00:52
So we can ignore those parts of the domains and get a double integral where y reaches from 0 to x, and x reaches from 0 to 18.
01:04
And we have cxy, d, y, d, x.
01:10
And so we're going to get first an integral with respect to y, which is, so c and x are both through just as constants, and we're given 1 over 2 cxy squared, evaluated from 0 to x, which just becomes, it becomes 0 when you plug 0 in, and it becomes x cubed when you plug x in.
01:33
So we get 0 to 18, or sorry, the integral from 0 to 18 of 1 half times c times x cubed dx, which this becomes 1 over 8 times c times x to the 4th, evaluated from 0 to 18, which just becomes 1 over 8 times c times 18 to the 4th.
02:04
Again, we're not subtracting on the second term because it evaluates to 0, and we get the 18 to the 4th divided by 8 is equal to 13 ,122.
02:22
And again, this is equal to 1, which means that c is equal to 1 divided by 13 ,122.
02:29
So i'm going to write this up here.
02:31
C equals 1 over 13 ,122, which we will use for future calculations.
02:43
Now that we can actually or know the value of c, we can find the covariance by saying that the covariance is equal to the expected value of x times y minus the expected value of x times the expected value of y.
02:59
So first let's find the expected value of x, which is just going to be the double integral exactly like we just calculated from 0 to 18 and y goes from 0 to x of of x times f of x.
03:20
So in this case, it's going to be c, x squared, y.
03:25
Again, because we're multiplying an x term to get the expected value of x, d, y, dx, which gives us the integral from 0 to 18 of c over 2 times x squared, y squared, plugging in 0 and x.
03:45
Sorry, there should be d x here, which goes us the integral from 0 to 18 of c, of c, over 2 times x to the 4th, which this becomes c over 10 times x to the 5th, evaluated from 0 to 18, which becomes 18 to the 5th over 10 times 1 over 1300, or 13 ,112, which is equal to 14 .4...