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2.7. FILTERING 77Properties of the Hodrick—Prescott FilterFor t = 3,...,T − 2, the Euler equations can be writtenq t − τ t = λu(L)τ t ,whereu(L) =(1− L) 2 (1 − L −1 ) 2 = P 2j=−2 u j L j ⇐(56)with u −2 = u 2 =1,u −1 = u 1 = −4, and u 0 = 6. We note for futurereference that c t = q t − τ t implies that c t = λu(L)τ t .You’ve already determined that q t =(λu(L)+1)τ t = v(L)τ t wherev(L) =1+λu(L) =1+λ(1 − L) 2 (1 − L −1 ) 2 , so it follows thatτ t = v(L) −1 q t =q t1+λ(1 − L) 2 (1 − L −1 ) 2 .v −1 (L) is the trend Þlter. Once you compute τ t , subtract the resultfrom the data, q t to get c t . This is equivalent to forming c t = δ(L)q twhereδ(L) =1− v −1 (L) = λ(1 − L)2 (1 − L −1 ) 21+λ(1 − L) 2 (1 − L −1 ) . 2Since (1 − L) 2 (1 − L −1 )=L −2 (1 − L) 4 ,theÞlter is equivalent to Þrst ⇐(57)applying (1 − L) 4 on q t , and then applying λL −2 v −1 (L) ontheresult. 34⇐(58)This means the Hodrick-Prescott Þlter can induce stationary into thecyclical component from a process that is I(4).The spectral density function of the cyclical component is s c (ω) =δ(e −iω )δ(e iω )s q (ω), whereδ(e −iω )= λ[(1 − e−iω )(1 − e iω )] 2λ[(1 − e −iω )(1 − e iω )] 2 +1 .From our trigonometric identities, (1−e −iω )(1−e iω )=2(1−cos(ω)), itfollows that δ(ω) =4λ[1−cos(ω)]2 . Each frequency of the original series4λ[1−cos(ω)] 2 +1is therefore scaled by |δ(ω)| 2 = h i4λ(1−cos(ω)) 2. 24λ(1−cos(ω)) 2 +1 This scaling factor isplottedinFigure2.4.34 This is shown in King and Rebelo (84).