8 the pdf of continuous random variable y given x are fxx...

8. The PDF of continuous random variable Y given X are $f_X(x) = \begin{cases} 3x^2, & 0 < x \leq 1 \\ 0, & otherwise \end{cases}$ $f_{Y|X}(y|x) = \begin{cases} \frac{2y}{x^2}, & 0 \leq y < x \\ 0, & otherwise \end{cases}$ (a) Find the joint PDF $f_{X,Y}(x,y)$. (b) If Y=1/2, what is the conditional PDF $f_{X|Y}(x|1/2)$? (c) Find $E[Y|X=1/2]$ and $Var[X|Y=1/2]$

Transcript

00:01 Okay, so we have two jointly continuous random variable with the joint pdf, and there's a couple things you asked for.

00:08 First of all, find just one distribution function of just x.

00:14 And remember, x is going from zero to one, so you're taking, even though it starts with the integral from negative infinity to infinity, it turned out just going to be an integral from zero to square root.

00:30 X because you kind of need to sketch out what's the region that you're looking for this is y equal root x and there's one here zero one so the y is actually going from zero to square root of x you zero right here square root of back down there going this direction and so you have this integral from zero to square root of x make sure that you run out the region to see what it is so as the integral of the function for x y d y and we're doing it integral with respect to y.

01:01 So just y here and it's going to be y square over two.

01:05 And that's going to be 2x y square and the bow is going to be 0 n square root x.

01:11 You block it in, you got 2x square.

01:14 So the function going to be 2x square when x is from 0 to 1 and 0 otherwise.

01:20 Okay.

01:21 Part b, find a conditional pdf of x when y equal to little y.

01:28 So the formula is going to be, the whole function f xy but dividing to divided by this guy f the x y of y so you need to find the expectation sorry the distribution f y of y and similarly the distribution f y of y is the integral from negative infinity to infinity of the function in term of x so the x is actually now going from this term right here y square going to one this portion right there again make sure that you have the pictures otherwise you don't know the vision that you're integrating and that integral is just going to be because you're integrating in terms of x so it's for integral x it's going to be x square over two and time y and then you plug in the two bow is going to be one and y square so you're going to get this guy and you can do the integral right here and the time 2y over 1 minus y square.

02:33 Therefore, the conditional expectation is the original for fxy.

02:39 This is fxy of xy, which is 4xy, dividing y the guy that you just found 2y over 1 minus y square, and you cancel the y, so you get 2x over 1 ,000, so you get 2x over 1 my.

02:52 Oops, i'm missing a y.

02:54 Oh, actually, yeah, no, there's no y, that's why i cancel.

02:57 Okay.

02:58 And again, condition is on y square less than equal to x less than equal to one.

03:04 All right.

03:05 Part c, need to find the expectation in the formula for the expectation is going to be x multiply with the pdf...