International Macroeconomics and Finance: Theory and Empirical Methods Phần 3 docx

2.7. FILTERING 69 Now let ˜ λ ∼ U[0, π] 29 . Imagine that we take a draw from this distribu- -1.5 -1 -0.5 0 0.5 1 1.5 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 4.4 4.8 5.2 5.6 6 Figure 2.2: π/2Phase shift. Solid: cos(t), Dashed: cos(t + π/2). tion. Let the realization be λ, and form the time-series q t = a co s(ωt + λ). (2.100) Once λ is realized, q t is a deterministic function with periodicity 2π ω and phase shift λ but q t is a random function ex ante. We will need the following two basic trigonometric relations. Two useful trigonometric relations. Let b and c be constants, and i be the imaginary number where i 2 = −1. Then cos(b + c)=cos(b)cos(c) − sin(b)sin(c) (2.101) e ib =cos(b)+i sin(b) (2.102) (2.102) is known as de Moivre’s theorem. You can rearrange it to get cos(b)= (e ib + e −ib ) 2 , and sin(b)= (e ib − e −ib ) 2i . (2.103) 29 Youonlyneedtoworryabouttheinterval[0, π] because the cosine function is symmetric about zero—cos(x)=cos(−x)for0≤ x ≤ π 70 CHAPTER 2. SOME USEFUL TIME-SERIES METHODS Now let b = ωt and c = λ and use (2.101) to represent (2.100) as q t = a cos(ωt + λ) =cos(ωt)[a cos(λ)] − sin(ωt)[a sin(λ)]. Next, build the time-series q t = q 1t + q 2t from the two sub-series q 1t and q 2t ,whereforj =1, 2 q jt =cos(ω j t)[a j cos(λ j )] −sin(ω j t)[a j sin(λ j )], and ω 1 < ω 2 . The result is a periodic function which is displayed on theleftsideofFigure2.3. -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1 6 11 16 21 26 31 36 -30 -20 -10 0 10 20 30 1 6 11 16 21 26 31 36 Figure 2.3: For 0 ≤ ω 1 < ··· < ω N ≤ π, q t = P N j=1 q jt ,whereq jt = cos(ω j t)[a j cos(λ j )] − sin(ω j t)[a j sin(λ j )]. Left panel: N =2. Right panel: N =1000 The composite process with N = 2 is clearly deterministic but if you build up the analogous series with N = 100 of these components, as shown in the right panel of Figure 2.3, the series begins to look like a random process. It turns out that any stationary random process can be arbitrarily well approximated in this fashion letting N →∞. 2.7. FILTERING 71 To summarize at this point, for sufficiently large number N of these underlying periodic components, we can represent a time-series q t as q t = N X j=1 cos(ω j t)u j − sin(ω j t)v j , (2.104) where u j = a j cos(λ j )andv j = a j sin(λ j ), E(u 2 i )=σ 2 i ,E(u i u j )=0, i 6= j,E(v 2 i )=σ 2 i ,E(v i v j )=0,i6= j. Now suppose that E(u i v j ) = 0 for all i, j and let N →∞. 30 You are carving the interval into successively more subintervals and are cramming more ω j into the interval [0, π]. Since each u j and v j is associated with an ω j , in the limit, write u(ω)andv(ω) as functions of ω. For future reference, notice that because cos(−a)=cos(a), we have u(−ω)=u(ω) whereas because sin(−a)=−sin(a), you have v(−ω)=−v(ω). The limit of sums of the areas in these intervals is the integral q t = Z π 0 cos(ωt)du(ω) − sin(ωt)dv(ω). (2.105) Using (2.103), (2.105) can be represented as q t = Z π 0 e iωt + e −iωt 2 du(ω) − Z π 0 e iωt − e −iωt 2i dv(ω) | {z } (a) . (2.106) Let dz(ω)= 1 2 [du(ω)+idv(ω)]. The second integral labeled (a)canbe simpliÞed as ⇐(49) Z π 0 e iωt − e −iωt 2i dv(ω)= Z π 0 e iωt − e −iωt 2i à 2dz(ω) − du(ω) i ! = Z π 0 e −iωt − e iωt 2 (2dz(ω) −du(ω)) = Z π 0 (e −iωt − e iωt )dz(ω)+ Z π 0 e iωt − e −iωt 2 du(ω). Substitute this last result back into (2.106) and cancel terms to get ⇐(50) 30 This is in fact not true because E(u i v i ) 6= 0, but as we let N →∞,the importance of these terms become negligible. 72 CHAPTER 2. SOME USEFUL TIME-SERIES METHODS q t = Z π 0 e −iωt du(ω) | {z } (a) + Z π 0 e iωt dz(ω) | {z } (b) − Z π 0 e −iωt dz(ω) | {z } (c) . (2.107) Since u(−ω )=u(ω), the term labeled (a) in (2.107) can b e written as R π 0 e −iωt du(ω )= R 0 −π e iωt du(ω). The third term labeled (c) in (2.107) is R π 0 e −iωt dz(ω)= 1 2 R π 0 e −iωt du(ω)+ 1 2 R π 0 ie −iωt dv(ω)= 1 2 R 0 −π e iωt du(ω) − 1 2 R 0 −π ie iωt dv(ω). Substituting these results back into (2.107) and can- celing terms you get, q t = 1 2 R 0 −π e iωt [du(ω)+idv(ω)] + R π 0 e iωt dz(ω)(51)⇒ = R π −π e iωt dz(ω). This is known as the Cramer Representation of q t , which we restate as q t = lim N→∞ N X j=1 a j cos(ω j t + λ j )= Z π −π e iωt dz(ω). (2.108) The point of all this is that any time-series can be thought of as be- ing built up from a set of underlying subprocesses whose individual frequency components exhibit cycles of varying frequency. The other side of this argument is that you can, in principle, take any time-series q t and Þgure out what fraction of its variance is generated from those subprocesses that cycle within a given frequency range. The business cycle frequency, which lies between 6 and 32 quarters is of key interest to, of all people, business cycle researchers. Notice that the process dz(ω) is built up from independent incre- ments. Forcoincidentincrements,youcandeÞne the function s(ω)dω to be E[dz(ω) dz(λ)] = ( s(ω)dωλ= ω 0otherwise , (2.109) where an overbar denotes the complex conjugate. 31 Since e iωt e iωt =cos 2 (ωt)+sin 2 (ωt)=1atfrequencyω, it follows that E[e iωt e iωt dz(ω)dz(ω)] = s(ω)dω.Thatis,s(ω)dω is the variance of the ω−frequency component of q t , and is called the spectral density function of q t . Since by (2.108), q t is built up from frequency compo- nents ranging from [−π, π], the total variance of q t must be the integral 31 If a and b are real numbers and z = a + bi is a complex number, the complex conjugate of z is ¯z = a − bi. The product z¯z = a 2 + b 2 is real. 2.7. FILTERING 73 of s(ω). That is 32 E(q 2 t )=E[ Z π −π e iωt dz(ω) Z π −π e iλt dz(λ)] =E[ Z π −π Z π −π e iωt e iλt dz(ω)dz(λ)] = Z π −π E[dz(ω)dz(λ)] = Z π −π s(ω)dω. (2.110) The spectral density and autocovariance g enerating functions.Theau- tocovariance generating function for a time series q t is deÞned to be g(z)= ∞ X j=−∞ γ j z j , where γ j =E(q t q t−j ) is the j-th autocovariance of q t .Ifweletz = e −iω , then 1 2π Z π −π g(e −iω )e iωk dω = 1 2π ∞ X j=−∞ γ j Z π −π e iω(k−j) dω. Let a = k−j.Thene iωa =cos(ωa)+i sin(ωa) and the integral becomes, R π −π cos(ωa)dω +i R π −π sin(ωa )dω =(1/a)sin(aω)| π −π −(i/a)cos(aω)| π −π . The second term is 0 because cos(−aπ)=cos(aπ). The Þrst term is 0 too because the sine of any nonzero integer multiple of π is 0 and a is an integer. Therefore, the only value of a that matters is a = k − j = 0, whic h implies that γ k = 1 2π R π −π g(e −iω )e iωk dω. Setting k =0,youhaveγ 0 =Var(q t )= 1 2π R π −π g(e −iω )dω, but you know that ⇐(52) Var(q t )= R π −π s(ω)dω, so the spectral density function is proportional to the autocovariance generating function with z = e −iω . Notice also, that when you set ω =0,thens(0) = P ∞ j=−∞ γ j . The spectral density function of q t at frequency 0 is the same thing as the long-run variance of q t . It follows that Var(q t )= Z π −π s(ω)dω = 1 2π Z π −π g(e −iω )dω, (2.111) where g(z)= P ∞ j=−∞ γ j z j . ⇐(53) 32 We obtain the last equality because dz(ω) is a process with independent incre- ments so unless λ = ω,Edz(ω)dz(λ)=0. 74 CHAPTER 2. SOME USEFUL TIME-SERIES METHODS Linear Filters You can see how a Þlter changes the character of a time series by com- paring the spectral density function of the original observations with that of the Þltered data. Let the original data q t have the Wold moving-average representa- tion, q t = b(L)² t where b(L)= P ∞ j=0 b j L j and ² t ∼ iid with E(² t )=0 and Var(² t )=σ 2 ² . The k-th autocovariance is γ k =E(q t q t−k )=E[b(L)² t b(L)² t−k ] =E   ∞ X j=0 b j ² t−j ∞ X s=0 b s ² t−s−k   = σ 2 ²   ∞ X j=0 b j b j−k   , and the autocovariance generating function for q t is g(z)= ∞ X k=−∞ γ k z k = ∞ X k=−∞ σ 2 ²   ∞ X j=0 b j b j−k   z k = ∞ X k=−∞   σ 2 ² ∞ X j=0 b j b j−k   z k z j z −j = σ 2 ² ∞ X k=−∞ ∞ X j=0 b j z j b j−k z −(j−k) = σ 2 ²   ∞ X j=0 b j z j ∞ X k=j b j−k z −(j−k)   = σ 2 ² b(z)b(z −1 ). But from (2.111), you know that s(ω)= g(e iω ) 2π . To summarize, these(54)⇒ results, the spectral density of q t can be represen ted as s(ω)= 1 2π g(e −iω )= 1 2π σ 2 ² b(e −iω )b(e iω ). (2.112) Let the transformed (Þltered) data be given by ˜q t = a(L)q t where a(L)= P ∞ j=−∞ a j L j .Then˜q t = a(L)q t = a(L)b(L)² t = ˜ b(L)² t ,where(55)⇒ ˜ b(L)=a(L)b(L). Clearly, the autocovariance generating function of the Þltered data is ˜g(z)=σ 2 ² ˜ b(z) ˜ b(z −1 )=σ 2 ² a(z)b(z)b(z −1 )a(z −1 )= a(z)a(z −1 )g(z), and letting z = e −iω , the sp ectral density function of the Þltered data is ˜s(ω)=a(e −iω )a(e iω )s(ω). (2.113) 2.7. FILTERING 75 The Þlter has the effect of scaling t he spectral density of the original observations by a(e −iω )a(e iω ). Depending on the properties of the Þlter, some frequencies will be magniÞed while others are downweighted. One way to classify Þlters is according to the frequencies that are allowed to pass through and those that are blocked. A high pass Þlter lets through only the high frequency components. A low pass Þlter allows through the trend or growth frequencies. A business cycle pass Þlter allows through frequencies ranging from 6 to 32 quarters. The most popular Þlter used in RBC research is the Hodrick—Prescott Þlter, which we discuss next. The Hodrick—Prescott Filter Hodrick and Prescott [76] assume that the original series q t is generated by the sum of a trend component (τ t ) and a cyclical (c t )component, q t = τ t + c t . The trend is a slow-moving low-frequency component and is in general no t deterministic. The objective is to construct a Þlter to to get rid of τ t from the data. This leaves c t ,whichisthepartof the data to be studied. The problem is that for each observation q t , there ar e two unknowns (τ t and c t ). The question is how to identify the separate components? The cyclical part is just the deviation of the original series from the long-run trend, c t = q t − τ t . Suppose your identiÞcation scheme is to minimize the variance of the cyclical part. You would end up setting its variance to 0 which means setting τ t = q t . This doesn’t help at all—the trend is just as volatile as the original observations. It therefore makes sense to attach a penalty for making τ t too volatile. Do this by minimizing the variance of c t subject to a given amount of prespeciÞed ‘smoothness’ in τ t .Since∆τ t is like the Þrst derivative of the trend and ∆ 2 τ t is like the second derivative of the trend, one way to get a smoothly evolving trend is to force the Þrst derivative of the trend to evolv e smoothly over time by limiting the size of the second derivative. This is what Hodrick and Prescott suggest. Choose a sequence of points {τ t } to minimize T X t=1 (q t − τ t ) 2 + λ T −1 X t=1 (∆ 2 τ t+1 ) 2 , (2.114) 76 CHAPTER 2. SOME USEFUL TIME-SERIES METHODS where λ is the penalty attached to the volatility of the trend component. For quarterly data, research ers typically set λ = 1600. 33 Noting that ∆ 2 τ t+1 = τ t+1 − 2τ t + τ t−1 ,differentiate (2.114) with respect to τ t and re-arrange the Þrst-order conditions to get the Euler equations q 1 − τ 1 = λ[τ 3 − 2 τ 2 + τ 1 ], q 2 − τ 2 = λ[τ 4 − 4 τ 3 +5τ 2 − 2 τ 1 ], . . . . . . q t − τ t = λ[τ t+2 − 4τ t+1 +6τ t − 4 τ t−1 + τ t−2 ],t=3, ,T − 2 . . . . . . q T −1 − τ T −1 = λ[−2τ T +5τ T −1 − 4τ T −2 + τ T −3 ], q T − τ T = λ[τ T − 2τ T −1 + τ T −2 ]. Let c =(c 1 , ,c T ) 0 ,q=(q 1 , ,q T ) 0 ,andτ =(τ 1 , ,τ T ) 0 ,andwrite the Euler equations in matrix form q =(λG + I T )τ, (2.115) where the T × T matrix G is given by G =                     1 −210··· ··· 0 −25−410··· ··· 0 1 −46−410··· ··· 0 01−46−410 . . . . . . . . . . . . 001−46−410 . . .01−46−41 . . .01−45−2 0 ··· ··· 01−21                     . Get the trend component by τ =(λG+I T ) −1 q. The cyclical component follows by subtracting the trend from the original o bservations c = q −τ =[I T − (λG + I T ) −1 ]q. 33 The following derivation of the Þlter follows Pederson [121]. 2.7. FILTERING 77 Properties of the Hodric k—Prescott Filter For t =3, ,T − 2, the Euler equations can be written q t − τ t = λu(L)τ t ,whereu(L)=(1− L) 2 (1 − L −1 ) 2 = P 2 j=−2 u j L j ⇐(56) with u −2 = u 2 =1,u −1 = u 1 = −4, and u 0 = 6. We note for future reference 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 where v(L)=1+λu(L)=1+λ(1 −L) 2 (1 −L −1 ) 2 , so it follows that τ t = v(L) −1 q t = q t 1+λ(1 −L) 2 (1 − L −1 ) 2 . v −1 (L) is the trend Þlter. Once you compute τ t , subtract the result from the data, q t to get c t . This is equivalent to forming c t = δ(L)q t where δ(L)=1−v −1 (L)= λ(1 −L) 2 (1 −L −1 ) 2 1+λ(1 −L) 2 (1 −L −1 ) 2 . Since (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 the cyclical 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), where δ(e −iω )= λ[(1 −e −iω )(1 −e iω )] 2 λ[(1 − e −iω )(1 − e iω )] 2 +1 . From our trigonometric identities, (1−e −iω )(1−e iω )=2(1−cos(ω)), it follows that δ(ω)= 4λ[1−cos(ω)] 2 4λ[1−cos(ω )] 2 +1 . Each frequency of the original series is therefore scaled by |δ(ω)| 2 = h 4λ(1−cos(ω)) 2 4λ(1−cos(ω)) 2 +1 i 2 . This scaling factor is plottedinFigure2.4. 34 This is shown in King and Rebelo (84). 78 CHAPTER 2. SOME USEFUL TIME-SERIES METHODS 0 0.2 0.4 0.6 0.8 1 1.2 -3.1 -2.8 -2.4 -2.1 -1.7 -1.4 -1.0 -0.7 -0.3 0.0 0.4 0.7 1.1 1.4 1.8 2.1 2.5 2.8 Frequency Figure 2.4: Scale factor |δ(ω)| 2 for cyclical component in the Hodric k— Prescott Þlter. [...]... VOLATILITY 89 Table 3. 1: Descriptive statistics for exchange-rate and equity returns, and their fundamentals Mean Std.Dev S&P UKP DEM YEN 2.75 0.41 0.46 0. 73 Div UKP DEM YEN 1 .31 0 0 0 Autocorrelations ρ1 ρ4 ρ8 ρ16 Min Max Returns 5.92 - 13. 34 18 .31 0.24 -0.10 5.50 - 13. 83 16.47 0.12 0. 03 6 .35 - 13. 91 15.74 0.09 0. 23 6.08 -15.00 16.97 0. 13 0.18 Deviation from fundamentals 0 .30 0.49 1.82 1.01 1. 03 0.18 -0.46... formed from CPIs 3. 1 PURCHASING-POWER PARITY 81 US-UK US-Germany 1.2 0.8 0.7 1 0.6 0.8 0.5 0.4 0.6 0 .3 0.2 0.4 0.1 0.2 0 -0.1 0 -0.2 73 75 77 79 81 83 85 87 89 91 93 95 97 73 75 77 79 81 83 85 87 89 91 93 95 97 87 89 91 93 95 97 US-Switzerland US-Japan 1.4 -0.2 1.2 1.2 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 -0.2 -0.2 -0.4 73 75 77 79 81 83 85 87 89 91 93 95 97 73 75 77 79 81 83 85 Figure 3. 1: Log nominal... -20 73 75 77 79 81 83 85 87 89 91 93 95 97 Dollar/Deutschemark 20 73 75 77 79 81 89 91 93 95 97 87 89 91 93 95 97 10 5 87 15 10 85 Dollar/Yen 20 15 83 5 0 0 -5 -5 -10 -10 -15 -15 -20 -20 73 75 77 79 81 83 85 87 89 91 93 95 97 73 75 77 79 81 83 85 Figure 3. 4: Quarterly stock and exchange rate returns (jagged line), 19 73. 1 through 1997.4, with price deviations from the fundamentals (smooth line) 3. 5... 0.975 Switzerland 0.982 Mean 0.991 Median 0.995 ahead 16-quarters ahead p-value U-statistic p-value 0.904 0.864 0.222 0.0 13 0. 837 0. 131 0.424 0.405 0.001 0.074 0.552 0.009 0.912 0.858 0.174 0.527 0.859 0.164 0.155 0.5 83 0.004 0.056 0.518 0.0 03 0.077 0.570 0.012 0.909 1.046 0.594 0.269 0.745 0.016 0.579 0.996 0. 433 0.002 0.486 0.012 0.041 0.7 03 0. 032 0 .38 0 0. 537 0.002 0 .34 1 0.672 0.028 0. 034 0 .37 2 0.001... parity st = pt − p∗ (3. 7) t To simplify the notation, call ∗ ft ≡ (mt − m∗ ) − φ(yt − yt ) t the economic fundamentals Now substitute (3. 4), (3. 5), and (3. 6) into (3. 7) to get (3. 8) st = ft + λ(Et st+1 − st ), and solving for st gives st = γft + ψEt st+1 , (3. 9) where γ ≡ 1/(1 + λ), ψ ≡ λγ = λ/(1 + λ) (3. 9) is the basic Þrst-order stochastic difference equation of the monetary model and serves the same... parity, and purchasing-power parity 3. 3 THE MONETARY MODEL UNDER FLEXIBLE EXCHANGE RATES85 Under ßexible exchange rates, the money stock is exogenous Equilibrium in the domestic and foreign money markets are given by mt − pt = φyt − λit , ∗ m∗ − p∗ = φyt − λi∗ , t t t (3. 4) (3. 5) where 0 < φ < 1 is the income elasticity of money demand, and λ > 0 is the interest rate semi-elasticity of money demand Money... seminal contributions to this literature are Leroy and Porter [90] and Shiller [127] 92 CHAPTER 3 THE MONETARY MODEL = ∞ X ψ j Et ∆ft+j (3. 18) j=1 (3. 17) and (3. 18) allow you to represent the deviation of the exchange rate from the fundamental as the present value of future fundamentals growth ζt = st − ft = ∞ X ψ j Et ∆ft+j (3. 19) j=1 Since st and ft are cointegrated they can be represented by... 3. 3 displays a realization of a st for 200 time ˆ periods where ψ = 0.99 and the fundamentals follow a driftless random walk with innovation variance 0. 035 2 Early on, the exchange rate seems to return to the fundamentals but the exchange rate diverges as time goes on 88 CHAPTER 3 THE MONETARY MODEL 25 20 15 exchange rate with bubble 10 5 0 fundamentals -5 -10 1 26 51 76 101 126 151 176 Figure 3. 3:... element and zeros elsewhere and let e2 be a (1 × 2p) row vector with a 1 as the p + 1−th element and zeros elsewhere e1 = (1, 0, , 0), e2 = (0, , 0, 1, 0, , 0) ⇐(65) These are selection vectors that give e1 Y t = ∆ft , e 2 Y t = ζt Now the k-step ahead forecast of ft is conveniently expressed as E(∆ft+j |Ht ) = e1 E(Y t+j |Ht ) = e1 Bj Y t (3. 23) 94 CHAPTER 3 THE MONETARY MODEL Substitute (3. 23) ... credible, market participants expect no change in the exchange rate and UIP implies that the interest rate it = i∗ is given by the exogenous foreign interest rate Assume that the t money market is continuously in equilibrium by equating md in (3. 2) t to mt in (3. 1) and rearranging to get θrt = s + p∗ + φyt − λi∗ − (1 − θ)dt + ²t ¯ t t (3. 3) (3. 3) embodies the central insights of the monetary approach to . substitute (3. 4), (3. 5), and (3. 6) into (3. 7) to get s t = f t + λ(E t s t+1 − s t ), (3. 8) and solving for s t gives s t = γf t + ψE t s t+1 , (3. 9) where γ ≡ 1/(1 + λ), ψ ≡ λγ = λ/(1 + λ). (3. 9) is. 97 US-Germany -0.2 -0.1 0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0.8 73 75 77 79 81 83 85 87 89 91 93 95 97 US-Japan -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 73 75 77 79 81 83 85 87 89 91 93 95 97 US-Swit zerland -0.2 0 0.2 0.4 0.6 0.8 1 1.2 73. on theleftsideofFigure2 .3. -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1 6 11 16 21 26 31 36 -30 -20 -10 0 10 20 30 1 6 11 16 21 26 31 36 Figure 2 .3: For 0 ≤ ω 1 < ··· < ω N ≤ π, q t = P N j=1 q jt ,whereq jt = cos(ω j t)[a j cos(λ j )]