Homework4: Pr. 3.18, 3.20, 3.27, 3.31, 3.32 and one more problem. Due on Thursday, Feb 23rd
Solution:
One more question:
(a) The given AR(2) process can be expressed as (B) Yt=Zt, where (z)=1-z-2z2. The roots
of (z)=0 are
and
>1, which reduce to
. By Theo
Homework2: Pr. 1.16, 1.19, 1.20, 1.21, 1.25(b), 1.27, 2.1, 2.8 (a)-(d). Also show that Spencer's
filter
does not distort a cubic polynomial trend. Due on Thursday, Jan 26th
Solutions:
Show that Spencer's filter does not distort a cubic polynomial trend.
Consider the time series
Yt = 0.4Yt1 + 0.45Yt2 + Zt + Zt1 + 0.25Zt2
where Zt W N (0, 2 )
1. Express this equation in terms of the backshift operator B ; determine the order (p, d, q) of this model
2. Determine if the model is causal and/or invertible
3.
Homework 1: Pr. 1.3, 1.4, 1.6, 1.8, 1.9, 1.10, 1.12. Also prove that the autocovariance function of a
stationary process is non-negative definite. Due on Thursday, Jan 19th
Solutions:
Prove that the autocovariance function of a stationary process is non-n
STAT 520
MIDTERM # 2
1. Consider the MA(1) series Xt = Wt + Wt1 where Wt is white noise with mean 0 and
2
variance w . Derive the minimum mean square error one-step forecast based on the rst n
observations and determine the mean square error of this forec
STAT 520
MIDTERM # 1
TOTAL is 20 pt
Feb 27, 2008
1. (6pt) For the model Yt + 0.6Yt2 = t + 1.2t1 determine whether it is causal and invertible
2. (4pt) Show that the autocorrelation function of the stationary second-order AR process
Xt =
1
1
Xt1 + Xt2 + Wt
Changyue Lu Time Series hw 3.18 a. Yule Walker coefficients: 0.4339481 0.4375768 Regression coefficients: 0.4286 0.4418 The two methods yield similar coefficients. b. Yule Walker coefficients standard error: 0.04001303 0.04001303 regression coefficients s
Homework 7: Pr. 4.12, 4.15 (only verify 4.63) and the following two problems Due on Thursday,
Apr 5th
Solution:
5.
(a) f() =
=
=
(b) Solving 1-0.4z+0.7z2=0 gives z=
Since
the process is casual.
The periodogram shows a peak at frequency
meaning that there