00:02
All right, so the objective here is to find the distribution between xn plus 1 minus x of n.
00:10
So in this case, let's denote it as d being equal to the difference between those two.
00:22
It's going to be x of n plus 1 minus x of n.
00:31
So in this case, since both variables, x of n and x of n plus 1 are taken from.
00:41
From the or they follow the normal distribution of some mean and one, that means that their expected values should be equivalent to each other, right? and then that their variance values should just be equal to one, right? so since they are both from the normal distribution, then their covariance value should be zero.
01:34
So the covariance, of x n plus 1 and x of n should be equal to 0.
01:56
But then it also tells us that the covariance of x of n plus 1 and the x n bar will also be 0.
02:13
So knowing that then we know that the difference between those, if one's find the expected value, of d, well, that's just going to be equal to the expected value of each one of these, which we know this is just the mean times the mean from this portion over here...