Solution for Assignment 2
Question 1:
AR(2) process can be written as: Zt 1 Zt1 2 Zt2 = at .
Also E (at Zt+k ) = 0 for k 1.
a. For k 1, we have
k = E ( Zt Zt+k ) = 1 E ( Zt1 Zt+k ) + 2 E ( Zt2 Zt+k ) + E (at Zt+k ) = 1 k1 + 2 k2 .
2 + 2
1
Therefore 1 = 1
Solution for Assignment 1
Question 1:
It is given that X0 = 0 and Xi = Xi1 + i with i are independent random errors with
mean 0 and variance 2 . Then, by expressing Xi in terms of j , we have
X0
X1
X2
X3
=0
=1
= 1+
= 2+
2
3
= 2
1
+
2
+
3
a. Using the abov
MATH 4130B / MATH 6633 3.0
Solution to Test 1
Question 1:
a. To be invertible, root of (1 B ) = 0 lies outside unit circle.
| | < 1.
To be stationary, roots of (1 B B 2 ) = 0 les outside unit circle.
This is the same as the stationary conditions required
MATH 4130B / MATH 6633 3.0
Solution to Test 2
Question 1: The following process
(1 B )Zt = (1 0.4B )at (1 B )Zt = (1 B )at
where = 1, = 0.4 and cfw_at is a zero mean white noise.
a. To be stationary, | < 1. In this case, | = 1, therefore, it is not stati
Solution for Assignment 4
Question 1:
a. With the given data, the log likelihood function is
() = 2 log 1.25 ,
the score function is the rst derivative of the log likelihood function, which is
S ( ) =
2
1.25
and the second derivative of the log likelihoo
Solution for Assignment 3
Question 1:
It is simply from the book and I wont type out the detail now. I will add it later this week.
As mentioned in class, this derivation will not be tested in Test 2.
Question 2:
a. Since Zt = at + cat1 + + ca1 where c is