00:01
Here on this problem, we are dealing with a continuous random variable with probability distribution function given by 2y, when y is between 0 and 1.
00:18
Now, first here we'd like to find the expected value.
00:25
Now, the expected value of y is going to be the integral from 0 to 1 of y times f of y, the y.
00:33
In order to find the expected value of a random variable, if it is continuous, we take the integral of y times f of y.
00:42
Now here, this would be the integral from 0 to 1 of y times 2y, d .y.
00:55
And so this is the integral from 0 to 1 of 2 y squared, d .y.
01:04
Now, integrating gives us 2 thirds y cubed from y equals 0 to 1.
01:12
And so this gives us 2 over 3.
01:17
And so the expected value of y is 2 over 3.
01:22
Now we also want the variance.
01:23
Now before we can find the variance, we need to find the expected value of y squared.
01:30
And so this is the integral from 0 to 1 of y squared times f of y, d y.
01:38
So this is the integral from 0 to 1 of y squared times 2y, dy, which is the integral from 0 to 1 to 2y cubed, dy.
01:55
Now iterating this gives us 1⁄2 .y to the 4th from y to 1 .4.
02:04
Go 0 to 1 and so it gives us 1 half until e of y squared is 1 half.
02:12
Now for the variance, though, which is what we were really after, the variance is equal to e of y squared minus e of y squared...