00:01
For this question, we are asked to compute parts a and b.
00:05
So starting off with part a, it asks us to compute e, y, n, which we have been given that for all i, e, xi equals 3.
00:25
Therefore e y n equals e times xn plus xn minus 1 plus xn minus 2 all over 3.
00:45
So as the constant can be taken out of the expectation, we get eyn equals 133 times eyn equals 1 3rd times.
01:00
Exn plus exn minus 1 plus exn minus 1 plus exn minus 2, which equals 1 3 plus 3, which equals 3.
01:25
Therefore, the expected value of yn is 3.
01:39
Now, for part b, we're asked to compute var times yn.
01:47
So, from the given covariance function for k equals zero, we get co -var times xm xm which equals var times xm xm which equals 8 and also that the covariance for one lag between two variables is 0 .3 and 0 for lag greater than 1...