00:01
So in this question, we're told that x and y have a joint density.
00:05
F of x and y is c root 1 minus x squared minus y squared for x squared plus y squared less than or equal to 1.
00:13
So first of all, let's find c.
00:18
So we're going to have the double integral from y equals minus root 1 minus x squared to root 1 minus x squared, and from x is minus 1 to 1.
00:32
C root 1 minus x squared minus y squared d x, d x, y, so this is equal to 1, because it's the total probability.
00:45
But what we can also do is express this in polar coordinates.
00:49
So let's use coordinates.
00:51
R is root x squared plus y squared, and we're going to have d x, dy, is r, dr, d theta.
01:04
So we have 1 is the integral.
01:06
Now theta goes between 0 and 2 pi.
01:09
R goes from 0 to 1, and we have c root 1 minus r squared r, dr, d theta.
01:19
So nothing depends on theta.
01:21
So we can put out the c, and we can do the 2 pi integral.
01:24
So we have 2 pi c, and then we're integrating from 0 to 1, r root 1 minus r squared, d r.
01:32
So this gives us 2 pi c.
01:35
Now this is in the form of a quantity and its derivative, so we can just integrate that quantity.
01:41
We get 1 minus r squared to the power of three halves.
01:44
But then remember that when we take a derivative of this, we'll get a three halves and we'll get a minus 2r.
01:50
It's like it's just minus 3r, but we only just want 1r, so we need to divide by the minus 3 and evaluate between 0 and 1.
02:00
So at r equals 1, we don't get anything.
02:02
At r equals 0, we get a third.
02:05
So that's what we get.
02:06
So 1 equals 2 pi c over 3.
02:09
So that tells us that c is 3 over 2 pi.
02:16
So we have part b.
02:20
We want to sketch the joint density.
02:22
So f of x and y is 3 over 2 pi times root 1 minus x squared minus y squared, for x squared plus y squared less than or equal to 1.
02:34
So let's have a look at the region on which this is defined.
02:38
So we've got x, y.
02:40
We are inside the unit disk.
02:44
So we've got 1 minus 1, 1 minus 1.
02:51
And we can see that the density is 0 when x squared plus y squared is equal to 1.
02:57
So it's 0 on the boundary of the disk.
03:00
And when x square plus y squared is equal to 0, it's 3 over 2 pi.
03:04
And it only depends on r squared.
03:06
So we can draw a kind of contour map.
03:12
And it's going like the square root of 1 minus r squared.
03:19
So if we say that, so let's say that z is f of x and y, then we have z squared is 9 over 4 pi squared, 1 minus r squared in polar coordinates, cylindrical polar.
03:38
So that means that z squared plus 9 over 4 pi squared, r squared, is equal to 9 over 4 pi squared, r squared, is equal to 9 over 4 pi squared.
03:52
So this is going to basically look like an ellipse in the rz plane.
04:00
So what we can do is we can draw this out.
04:03
So let's go, x is here, y is here, and f of x and y is up this way.
04:12
Then it's going to look like an ellipse where the distance out in the x, y plane, is 2 pi over 3.
04:22
So when r is zero, z is 3 over 2 pi.
04:33
And then when r goes to 1, z goes to 0.
04:38
So let's just see 3 over 2 pi is less than 1.
04:44
So this is going to be bigger.
04:51
And it's going to kind of look like this elliptical thing.
04:56
It looks like this.
05:22
So that's our joint density.
05:24
And this is what it looks like.
05:25
It's kind of an elliptical shaped hill.
05:28
So now let's find the probability that x squared plus y squared is less than or equal to one half.
05:38
Well, that's the probability that r squared is less than equal to one half, which is equal to the probability that r squared is less than or equal to one over root two, sorry, the probability that r is less than or equal to one over root two, which is the integral from zero to two pi d theta, and from 0 to 1 over root 2, the r of 2 pi of c, which is 3 over 2 pi, and then r root 1 minus r squared.
06:15
So the 2 pi gives us a 3.
06:18
And then integrating this, we get minus 1 3rd, 1 minus r squared to the minus 3 halves, which gets evaluated between 0 and 1 over root 2.
06:32
So that gives us a minus sign.
06:36
And, well, let's actually take out the minus sign and just evaluate the lower limit first...