00:01
We have here a continuous random variable x whose probability density function is given by this expression.
00:08
The first part of the problem which asks us to find the expectation value of x is straightforward.
00:14
We simply evaluate this integral for x going from 0 to 1, which is the support of the probability density function.
00:24
It's straightforward to evaluate and gives us 2 over 3 as the expectation.
00:30
Of x.
00:32
For the rest of the problem, we are going to consider a random variable y, which is given by x squared.
00:39
We are going to first find the probability density function for y, and then we are going to find the expectation of y in two different ways.
00:48
In this case, y as a function of x is differentiable and strictly increasing for x between 0 and 1.
00:58
So we can apply the general formula for finding the transformed distributions.
01:06
By finding the inverse function of x as a function of y, which is the square root of y.
01:16
Plugging it into the distribution function of x, we get 2 times square root of y.
01:27
We also need to compute the derivative of the inverse function, which in this case gives us combining these two parts, we get the probability density function for the transformed variable y is simply 1, and that is for y between 0 and 1.
01:58
Otherwise, the density function is just 0.
02:03
Now that we have the pdf of y, we can find its expectation using the standard...