0:00
Hello everyone.
00:01
So in this question the cross has given is x t is equals to epsilon t plus a1 epsilon t minus 1 plus a 2 epsilon t minus 2.
00:21
Now in part in we have to find the expectation of x t so this will be equals to expectation of epsilon t plus a1 expectation of epsilon t minus 1 plus a2 expectation of epsilon t minus 2.
00:46
Now here we have that epsilon t is normally distributed with mean zero and variance sigma square is iid process.
01:03
So we have expectation of epsilon t is equal to expectation of epsilon t minus 1 is equals to expectation of epsilon t minus 2 and all equals to 0.
01:19
So using this we can write that expectation of xt is equals to 0.
01:28
Solution for part a.
01:30
Let us come to part b.
01:31
In part b we have to find variance of xt.
01:36
So this will be equals to expectation of xt.
01:49
X t square minus expectation of x t whole square.
01:56
This term vanishes so we are only lived with expectation of x t square.
02:03
This term will be vanished using first part of the question part a.
02:10
Now to find the expectation of x t square we have expectation of x t square is equal to expectation of epsilon t square plus a1 square epsilon t minus 1 plus a 2 square epsilon t minus 2 square plus 2 a1 epsilon t x xylem 2 plus a 2 epsilon t x x x minus 1 plus a 2 epsilon t a1, a2, epsilon t minus 1, epsilon t minus 2 plus a2 epsilon t minus 2 plus a2 a2 a2 a1 epsilon t minus 1 epsilon t minus 2.
03:22
Now we'll see clearly here that we are left with all the epsilon 1, epsilon t, epsilon t minus 1 and epsilon t minus 2 terms...