# This does not complicate the application of the

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—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
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 .

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