3. Suppose X and Y are independent continuous random...

3. Suppose X and Y are independent continuous random variables with PDFs $f_X(x) = \begin{cases} \frac{3}{2}x^2 & x \in [-1, 1] \\ 0 & \text{otherwise} \end{cases}$, and $f_Y(y) = \begin{cases} \frac{1}{2} & y \in [-1, 1] \\ 0 & \text{otherwise} \end{cases}$, respectively. Define a third random variable Z = X² + Y. (a) What is the CDF of Z? (b) What is the joint CDF of Y and Z? (c) What is the joint PDF of X and Z?

Transcript

00:02 So first of all, let me tell you one thing.

00:05 Suppose if we give you a straight line, x is equal to 2y or x is equal to 3y or whatever it may be, x is equal to k.

00:13 In fact, it's a line passing through origin.

00:16 If k is positive, it will be in this way.

00:20 If k is negative, it will be in this way.

00:24 Okay, now let's do the inequality.

00:26 If x is less than equal to ky and k is positive, then it will be above the line, the region above the line.

00:36 So this is the region x is less than or equal to k y.

00:43 If k is negative, if k is negative, it looks like this and x is less than equal to k y for this will be below the line.

00:55 So below the region.

00:57 All right.

00:57 So please keep this in mind because this concept will be helpful to solve this problem.

01:02 Now we are given the distribution function of the two random variables as a constant c, 0 less than x less than 1 and 0 half less than y less than 1 and 0 else how to find c this is the probability density function it should satisfy negative infinity to infinity fxy of x comma y d x dy as 1 but if you can see i can take the limits x from 0 to 1 and y for half to 1 c d x x x x x x is equal to 1 calculating c you can integrate separately with respect to x so c into one and integration of d y is y but upper lower limits is 1 minus half half is equal to 1 so c is equal to 2 so basically your joint distribution function is 2 when 0 less than x less than 1 and half less than y less than 1 and 0 else and let's plot this region this region is actually a rectangle so it looks like this is a rectangle this is the red tending half to one y going from half to one and x going from 0 to 1.

02:13 Now we need to find the cdf of the random variable z is equal to x by y.

02:19 The cdf is defined as f z of u is probability that z less than it equal u probability of z less than it equal u by the definition of probability cumulative distribution function.

02:36 This i can edit as probability of x by y is less than equal u but y i can cross multiply because y is a positive random variable so x is less than is equal to y u now comes the cases if suppose if i take a straight line in this way this is actually half comma one i mean one comma half so what is the equation of this straight line it is is equal to 2 y what will be the region x less than it equal to y it will be the above above the straight line that completely overlaps with this rectangle the completely overlaps with this rectangle.

03:19 So the probability of x less than equal to 2y will be 1 because it is completely overlapping with the region where the joint pdf is defined.

03:31 Now if you increase this to number 2 to 3, that means x is equal to 3y then it will be like this.

03:37 X is equal to 4y it will be like this.

03:40 For all these regions the probability is 1.

03:42 So that means our first case is if the number u is greater than 2.

03:48 Which number? this number u is greater than 2 because all these straight lines inequalities will be above the region that will completely overlap the given rectangle so if u is greater than 2 your fz of z or u is equal to 1 this is only for u is greater than 2 now let's take case 2 case 2 let's take u belongs to 1 to 2 so what you mean by u belongs to 1 to 2? so a typical straight line looks like this this is our region rectangular region half and 1 this is 0 a typical straight line will look like this a typical straight line because i'm taking u from 1 to 2 the last straight line will be when u is equal to 1 it will be the line joining 0 0 in 1 comma 1 now this point where it cuts y is equal to half x is equal to y u where it cuts y is equal to half x coordinate will be u by 2 because substitute y is equal to half in this x is equal to u by 2 and we need to find this region, this region, integrate our own this region.

05:01 That i can do it as 1 minus 1 minus this region.

05:08 So that will be easy for us.

05:10 This region is x is going from u by 2 to 1 and y is going from 0 to the straight line.

05:18 What is that straight line? x by u because when x is y u when x is y u, y is equal to x by u.

05:27 X by u.

05:29 2 divide d x.

05:32 So let's calculate this integration.

05:34 So it is 1 minus x is going from u by 2 to 1...

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