ex2_sol - TIME SERIES Example Sheet 2 Solutions Version...

This preview shows page 1 - 3 out of 11 pages.

We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
A Transition to Advanced Mathematics
The document you are viewing contains questions related to this textbook.
Chapter 6 / Exercise 3
A Transition to Advanced Mathematics
Smith/Eggen
Expert Verified
TIME SERIES Part III Example Sheet 2 - Solutions YC/Lent 2015 Version dated on 25 February 2015 . Comments and corrections should be emailed to [email protected] 1. Find the Yule–Walker equations for the AR(2) process X t = 1 / 3 X t - 1 + 2 / 9 X t - 2 + t , t i.i.d. N (0 , σ 2 ) . Show that it has autocorrelation function ρ h = 16 21 2 3 | h | + 5 21 - 1 3 | h | , h Z .
2. Suppose that X t is weakly stationary with zero mean and recall the definition of the PACF. That is, for a positive integer h , let u t = X t - a 1 X t - 1 - · · · - a h - 1 X t - h +1 v t - h = X t - h - b 1 X t - h +1 - · · · - b h - 1 X t - 1 be the two residuals where { a 1 , . . . , a h - 1 } and { b 1 , . . . , b h - 1 } are chosen so that they minimise the mean-squared errors Eu 2 t and Ev 2 t - h . The PACF at lag h is defined as the correlation between u t and v t - h , i.e., φ hh = E ( u t v t - h ) / q Eu 2 t Ev 2 t - h . Let ρ h be the ACF of { X t } . Let R h be the h × h matrix with elements ρ ( i - j ) , i, j = 1 , . . . , h , and let ρ h = ( ρ 1 , . . . , ρ h ) T be the vector of lagged autocorrelations. In addition, let ˜ ρ h = ( ρ h , . . . , ρ 1 ) T be the reversed vector, and let - X h t denote the best linear predictor of X t given { X t - 1 , . . . , X t - h } , i.e., - X h t = α h 1 X t - 1 + . . . + α hh X t - h , where { α h 1 , . . . , α hh } are picked to minimise the mean-squared error E ( X t - - X h t ) 2 . Here we denote the minimum mean-squared error by P h . (a) Prove that a i = b i for i = 1 , . . . , h - 1. (b) ( ? ) Prove that φ hh = ρ h - ˜ ρ T h - 1 R - 1 h - 1 ρ h - 1 1 - ˜ ρ T h - 1 R - 1 h - 1 ˜ ρ h - 1 = α hh . 1
We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
A Transition to Advanced Mathematics
The document you are viewing contains questions related to this textbook.
Chapter 6 / Exercise 3
A Transition to Advanced Mathematics
Smith/Eggen
Expert Verified
(c) ( ? ) Show that given the ACF, α ji ( i j ) can be solved iteratively as follows: α 00 = 0 , P 0 = γ 0 , where γ 0 is the variance of the series. For h 1, α hh = ρ h - h - 1 k =1 α h - 1 ,k ρ ( h - k ) 1 - h - 1 k =1 α h - 1 ,k ρ k , P h = P h - 1 (1 - α 2 hh ) , where, for h 2, α hk = α h - 1 ,k - α hh α h - 1 ,h - k , k = 1 , 2 , . . . , h - 1 . This procedure is also known as the Durbin–Levinson algorithm. Solution.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture