This does not complicate the application of the

  • No School
  • AA 1
  • 60

This preview shows page 41 - 43 out of 60 pages.

—this does not complicate the application of the Kalman filter, since the Bayesian estimation that we use takes turns to simulate { z t } T t =1 , then { x t } T t =1 , and then { s t } T t =1 . For example, during the step that simulates { z t } T t =0 , the value s t μ 0 +(1 - s t ) μ 1 + s t x t is fixed and therefore treated as a constant. Equations ( A.2 ), ( A.7 )–( A.12 ) summarize the model that we estimate, and the model is summarized by parameter vector θ : θ = n p, q, μ 0 , μ 1 , σ v , σ f , γ 1 , ..., γ Q + M , { ψ 1 , 1 , ..., ψ 1 ,P } , ..., { ψ Q + M, 1 , ..., ψ Q + M,P } , σ 1 , ..., σ Q + M o . A.2 Bayesian estimation Let Y = { y t } T t =0 be the observed data that consists of quarterly and monthly growth rates normalized by their standard deviations. Let Z i = { z i t } T t =0 , X i = { X i t } T t =0 , and S i = { S i t } T t =0 be the i -th simulation of the unobserved variables, and let θ i be the i -th simulation of the parameter vector. The Bayesian procedure follows ( Carter and Kohn , 1994 ) algorithm, iterating upon Z , X , S , and θ . 1. Given Y , S i , X i , and θ i , and using the dynamics equation ( A.10 ) and the measurement equation ( A.7 ), simulate Z i +1 using Carter and Kohn ( 1994 ): a. Initialize z 1 | 0 as a vector with the first element equal to μ 0 s i 1 + μ 1 (1 - s i 1 ) + s i 1 x i 1 and the remaining elements equal to zero, and P 1 | 0 = I . 15 We restrict the individual components for the quarterly series to be white noises in the model, so that the first row of Ψ i is zero for i Q . ECB Working Paper Series No 2381 / March 2020 40
Image of page 41
b. Run standard Kalman filter to evaluate { z t | t , P t | t } T t =0 : Account for missing observations in vector y t : define y * t by eliminating the miss- ing observations, define H * t by eliminating the rows that correspond to the miss- ing observations, and define R * t by eliminating the rows and the columns that correspond to the missing observations. For t = 1 , ..., T , compute the following: Forecast: Ω t | t = H * t P t | t - 1 ( H * t ) 0 + R * t ; ν t = y t - H * t z t | t - 1 ; Update: z t | t = z t | t - 1 + P t | t - 1 ( H * t ) 0 t | t - 1 ) - 1 ν t ; P t | t = P t | t - 1 - P t | t - 1 ( H * t ) 0 t | t - 1 ) - 1 H * t P t | t - 1 ; Prediction: z t +1 | t = Fz t | t + h s i t +1 ( μ 0 + x i t +1 ) + (1 - s i t +1 ) μ 1 , 0 , ..., 0 i 0 ; P t +1 | t = FP t | t F 0 + Q. c. Using the output from the Kalman filter, run the smoothing filter backwards in order to obtain { z t | T , P t | T } T t =0 : Account for the fact that in the dynamics equation ( A.10 ), shocks in vector ε t do not affect all the state variables contemporaneously, so that the matrix Q is sin- gular, and Ω t | t - 1 is potentially non-invertible. Define ˆ z t by eliminating the rows that correspond to zero elements in the vector ε t : according to equation ( A.11 ), this means discarding all elements but the first, the fifth, etc. Similarly, de- fine ˆ F by eliminating the corresponding rows, and define ˆ Q by eliminating the corresponding rows and columns. For t = T , Kalman filter has already delivered z t | t = z t | T and P t | t = P t | T in step (b). We can use this information to simulate z i +1 T ∼ N ( z T | T , P T | T ), and then eliminate part of its elements, as described above, to get ˆ z i +1 T .
Image of page 42
Image of page 43

You've reached the end of your free preview.

Want to read all 60 pages?

  • Fall '19
  • Economics, Recession, Late-2000s recession, GWI

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors