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An explicit formula for the smoother weights of the Hodrick–Prescott filter

  • Hiroshi Yamada EMAIL logo und Fatima Tuj Jahra
Veröffentlicht/Copyright: 23. Januar 2019
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Abstract

By applying the Sherman–Morrison–Woodbury (SMW) formula and a discrete cosine transformation matrix, De Jong and Sakarya [De Jong, R. M., and N. Sakarya. 2016. “The Econometrics of the Hodrick–Prescott Filter.” Review of Economics and Statistics 98 (2): 310–317] recently derived an explicit formula for the smoother weights of the Hodrick–Prescott filter. More recently, by applying the SMW formula and the spectral decomposition of a symmetric tridiagonal Toeplitz matrix, Cornea-Madeira [Cornea-Madeira, A. 2017. “The Explicit Formula for the Hodrick–Prescott Filter in Finite Sample.” Review of Economics and Statistics 99: 314–318] provided a simpler formula. This paper provides an alternative simpler formula for it and explains the reason why our approach leads to a simpler formula.

MSC 2010: 62G05
JEL Classification: C22

Award Identifier / Grant number: 16H03606

Funding statement: The Japan Society for the Promotion of Science supported this work through KAKENHI Grant Number 16H03606.

Acknowledgments

We appreciate two anonymous referees for their valuable suggestions and comments. An earlier draft entitled “An Alternative Explicit Formula for the Hodrick-Prescott Filter in Finite Sample” was presented at the 26th Annual Symposium of the Society for Nonlinear Dynamics and Econometrics held at Keio University in Japan. Our thanks to the participants for their useful comments. The usual caveat applies.

A Appendix

A.1 Proof of (5)

From (4), γ1γ2 is

γ1γ2=[1T2Tcos((21)(10.5)πT)2Tcos((T1)(10.5)πT)][1T2Tcos((21)(20.5)πT)2Tcos((T1)(20.5)πT)].

Let βj=π(j1)/(2T) for j = 2, …, T. Then, it follows that

2Tcos((j1)(10.5)πT)2Tcos((j1)(20.5)πT)=2T[cos(βj)cos(3βj)]=32Tsin2(βj)cos(βj).

The last equality follows from cos(βj)cos(3βj)=4sin2(βj)cos(βj).

A.2 Proof of (6)

From (4), γTγT1 is

γTγT1=[1T2Tcos((21)(T0.5)πT)2Tcos((T1)(T0.5)πT)][1T2Tcos((21)(T10.5)πT)2Tcos((T1)(T10.5)πT)]

Let βj=π(j1)/(2T) and κj=2Tβj=π(j1) for j = 2, …, T. Then, it follows that

2Tcos((j1)(T0.5)πT)2Tcos((j1)(T10.5)πT)=2T[cos(βj(2T1))cos(βj(2T3))]=2T[cos(κjβj)cos(κj3βj)].

Here, since sin(κj)=0, it follows that

cos(κjβj)cos(κj3βj)=cos(κj)cos(βj)+sin(κj)sin(βj)cos(κj)cos(3βj)sin(κj)sin(3βj)=cos(κj)[cos(βj)cos(3βj)]+sin(κj)[sin(βj)sin(3βj)]=cos(κj)[4sin2(βj)cos(βj)]+sin(κj)[4sin3(βj)2sin(βj)]=4sin2(βj)[cos(κj)cos(βj)+sin(κj)sin(βj)]2sin(κj)sin(βj)=4sin2(βj)cos(κjβj),

where

κjβj=2Tβjβj=βj(2T1)=π(j1)(2T1)2T=π(j1)(T0.5)T.

A.3 Application of the SMW formula to (BαVV)−1

As in Cornea-Madeira (2017), by applying the SMW formula to (Bαq1q1αqTqT)1, we obtain the following results:

(27)(Bαq1q1αqTqT)1=(Bαq1q1)1+α(Bαq1q1)1qTqT(Bαq1q1)11αqT(Bαq1q1)1qT,

where

(28)(Bαq1q1)1=B1+αB1q1q1B11αq1B1q1.

On the other hand, by applying the SMW formula to (BαVV)1, we obtain

(29)(BαVV)1=B1B1V(VB1Vα1I2)1VB1=B1[B1q1,B1qT][q1B1q1α1q1B1qTqTB1q1qTB1qTα1]1[q1B1qTB1].

By comparing (29) with (27) and (28), it is observable that (29) is preferable to (27), mainly because (29) is symmetric with respect to q1 and qT.

A.4 A MATLAB/GNU Octave function to calculate (IT + α DD)−1 in (1) based on (23)

function HP_hat_matrix = calc_HP_hat_matrix(T, alpha) % T: sample size % alpha: smoothing parameter Lam = diag( 4*(sin((1:T-2)*pi/(2*(T-1))).^2) ); G = zeros(T-2,T-2); for i = 1:T-2 for j = 1:T-2 G(i,j) = sqrt(2/(T-1))*sin(i*j*pi/(T-1)); end end invC = zeros(T-2,T-2); for i = 1:T-2 for j = 1:T-2 s = 0; for k = 1:T-2 s = s + G(i,k)*G(j,k)/( (1/alpha)+Lam(k,k)^2 ); end invC(i,j) = s; end end Xi = zeros(T,T); DG = diff([zeros(2,T-2);G;zeros(2,T-2)],2); for i = 1:T for j = 1:T s = 0; for k = 1:T-2 s = s+DG(i,k)*DG(j,k)/((1/alpha)+Lam(k,k)^2); end Xi(i,j) = s; end end Up1 = zeros(T,1); Up2 = zeros(T,1); for i=1:T s1 = 0; s2 = 0; for k = 1:T-2 s1 = s1+DG(i,k)*G(1,k)/((1/alpha)+Lam(k,k)^2); s2 = s2+DG(i,k)*G(end,k)/((1/alpha)+Lam(k,k)^2); end Up1(i) = s1; Up2(i) = s2; end c11 = invC(1,1); c22 = invC(end,end); c12 = invC(1,end); c21 = invC(end,1); den = (1+c11)*(1+c22)-c12*c21; Tau = zeros(T,T); for i=1:T for j=1:T num = (1+c22)*Up1(i)*Up1(j)-c12*Up1(i)*Up2(j)-c21*Up2(i)*Up1(j) ... +(1+c11)*Up2(i)*Up2(j); Tau(i,j) = num/den; end end Z = zeros(T,T); I = eye(T); for i=1:T for j=1:T Z(i,j) = I(i,j)-Xi(i,j)+Tau(i,j); end end HP_hat_matrix = Z; end

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Published Online: 2019-01-23

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Heruntergeladen am 9.9.2025 von https://www.degruyterbrill.com/document/doi/10.1515/snde-2018-0035/pdf
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