Let x and y be two continuous random variables with joint...

Let X and Y be two continuous random variables with joint density (PDF) fXY(x,y). (a) Show that if X and Y are independent, they are also uncorrelated. (b) Now let the joint PDF of X and Y be given by fXY(x,y) = {A(x^2 + y), -1 <= x <= 1, 0 <= y <= 1, 0, otherwise} and find the constant A and determine if X and Y are independent. (c) Now let X = cosφ and Y = cosφ, where φ is a uniform random variable in the interval (0, π). Show that X and Y are uncorrelated but not independent.

Transcript

00:01 So in this question, we're told this, 8xy for 0 less than equal to x, less than equal to 1, and 0 less than equal to y less than equal to x, and 0 otherwise.

00:20 And let's just look at the region that this covers.

00:23 So x, y.

00:26 Now, x is between 0 and 1, and y is between 0 and x.

00:33 So this is the region that we're covering here, which is going to be important when we come to getting marginal densities, which we're going to do now.

00:45 So the marginal density of x is going to be the integral from 0 to x, dy, 8xy, which is 4xy squared between y equals 0 and y equals x, which is 4x cubed, and this is valid when x is in the interval 0 to 1, and it's 0 otherwise.

01:11 But now when we come to get the marginal density of y, we need to recognize that we're only integrated x from y to 1.

01:19 So we can, so when we're integrating our x, we only integrate from y to 1.

01:24 So f y of x, sorry f y of y is the integral from y to 1 d x 8x y, which is 4x squared y between y and 1, which is going to be 4 y minus 4 y cubed.

01:47 And this is for y in the integral 0 to 1 and it's 0 otherwise.

01:56 So now let's find the conditional density of x and y.

02:01 So f x given y of x given y equals y is going to be the joint density of x and y divided by the marginal density of y, which is equal to.

02:18 So we've got 8 xy divided by 4y 1 minus y squared.

02:30 And since this is only valid when x is between 0 and 1 and y is between 0 and x, we can write that.

02:49 And it's going to be 0 otherwise.

02:52 And we can simplify this a little bit by dividing the top and the bottom by y and then dividing the top and the bottom by 4.

03:00 And we get this.

03:01 Now we can get f of y given x of y given x equals x.

03:07 And this is going to be f of x and y divided by the marginal density of x.

03:16 And this is going to be, so we've got 8xy divided by f of x, which is 4x cubed, and this is valid only in the region in which it has support as well.

03:42 So this is going to be y in the range 0 to 1 and x in the range y to 1.

03:50 So now again we can divide by x here and we can divide by four and we're going to get zero otherwise.

04:15 So okay.

04:17 But now since y is a parameter here, we should probably say that we should probably say that this is in the range y is in the range nought to 1 because y is a parameter and x is in the range y to 1.

04:34 And in this one, because x is the parameter, x can take values between nought and 1.

04:41 So x is in the range nought to 1, and y is in the range 0 to x.

04:50 So what we're doing here is we're letting the parameter take the full range of values because you can't change the parameter it's given to you.

04:58 So yeah, this is how it works.

05:03 So now let's find the conditional expectation...

Question

Please give Ace some feedback