A property of the Hodrick-Prescott filter and its application

Transcription

1 A property of the Hodrick-Prescott filter and its application (Job Market Paper #2 ) Neslihan Sakarya Robert M. de Jong November 14, 2016 Abstract This paper explores a remarkable property of the Hodrick-Prescott filter: the cyclical component of the Hodrick-Prescott filter is equal to the trend component of the Hodrick-Prescott filter when applied to the fourth differences, plus an additional term. Our result is remarkable due to its simplicity and its strength in explaining many aspects of the HP filter that have not previously been studied rigorously. We first use our result to analyze the consequences of a deterministic trend break. We find that the effect of a deterministic trend break on the cyclical component is asymptotically negligible for the points that are away from the break point while for the points in Please find the most recent version of the paper at neslihansakarya.weebly.com/research.html Department of Economics, Ohio State University, 371 Arps Hall, Columbus, OH 43210, USA, sakarya.2@osu.edu. Department of Economics, Ohio State University, 444 Arps Hall, Columbus, OH 43210, USA, de-jong.8@osu.edu. 1

2 the neighborhood of the break point, the effect is not negligible even asymptotically and a characterization is provided for it. Second, we apply our result to show that the cyclical component of the Hodrick-Prescott filter when applied to series that are integrated up to order 2 is typically weakly dependent, while the situation for the series that are integrated up to order 3 or more is more subtle. This result contrasts with the conjecture in the literature that the HP filter extracts a stationary series when it is applied to series that are integrated up to order 4. Third, we characterize the behavior of the Hodrick-Prescott filter when applied to deterministic polynomial trends and show that effectively the cyclical component of the filter reduces the order of the polynomial by 4. Finally, we give a characterization of the Hodrick-Prescott filter when applied to an exponential deterministic trend, and this characterization shows that the filter is effectively incapable of dealing with a trend that increase this fast. This result suggests that the HP filter should not be used for series that are measured in nominal terms such as GDP, consumption, investment, house prices, etc. 1 Introduction The Hodrick-Prescott (HP) filter is long-standing standard technique in macroeconomics for separating the long run trend in a data series from short run fluctuations. Introduced initially by Whittaker (1923) and popularized in Econometrics by Hodrick and Prescott (1980,1997), the HP filter is universally used in macroeconomics. The two cited papers by Hodrick and Prescott have thousands of citations; yet, the impact of this work may go beyond that, since the use of the HP filter is universal, and the HP filter is a commonplace tool in macroeconomics. While the HP filter has a long and venerable history, the HP filter 2

3 has recently being analyzed more formally in de Jong and Sakarya (2016), Cornea-Madeira (2016), and Hamilton (2016). These papers analyze the properties of the HP filter rigorously and reconsider its usefulness in the context of macroeconomics. The HP filter calculates the trend of a series y t, t = 1,..., T by minimizing T T 1 (y t τ t ) 2 + λ (τ t+1 2τ t + τ t 1 ) 2, (1) t=1 t=2 over τ = (τ 1,..., τ T ). The parameter λ here is a smoothing parameter that for quarterly data is typically chosen to equal The minimizer, which we will label ˆτ T t, is referred to in the literature as the trend component, while ĉ T t = y t ˆτ T t is referred to as the cyclical component. By writing the minimization problem as a vector differentiation problem, it follows that there exists a unique minimizer. The trend component ˆτ T t and the cyclical component ĉ T t are both weighted averages of {y t } T t=1, and in de Jong and Sakarya (2016), an exact formula for the weights is found. This paper also explores the statistical properties of the cyclical component when the HP filter is applied to a unit root process, and considers adjusting the smoothing parameter for the data frequency. Cornea-Madeira (2016) also provides an exact formula for the weights by using the Sherman-Morrison formula. In this paper, we derive a remarkable property of the HP filter which allows us to derive more general results that are not present in the literature. We first use this elegant result to explore the effect of a deterministic trend break on the cyclical component. It is shown that the cyclical component consists of two terms when the HP filter is applied to a series that has a deterministic trend break at an unknown time point. The first part comprises the residue of the structural break, while the second part is equivalent to the cyclical component 3

4 in the absence of a structural break. Our main result is also applied to the processes that are integrated up to order 4. We show that series that are integrated up to order 2 exhibit weak dependence properties, while for series that are integrated of order 3 or 4 the law of large numbers does not hold for a large class of unbounded functions of the cyclical component. This result shows that the widely known conjecture... the HP filter will render stationary series that are integrated (up to fourth order)... by King and Rebelo (1993) can be formalized and is incorrect in some dimensions. The authors conjecture depends on assuming that the first order condition of the minimization problem in Equation (1) that is valid for t = 3, 4,..., T 2 holds for every t Z. We call this approach the heuristic approach throughout the paper. On the other hand, our representation provides a rigorous analysis of the HP filter. In addition, we give a closed form formula for the cyclical component of a polynomial trend and an exponential deterministic trend by using our new representation. In Section 2 of the paper, we provide an explanation of the heuristic approach. In Section 3 of the paper, we establish our main result. Section 4 explores the consequences of a structural break; we show that a deterministic trend break has asymptotically no effect on the cyclical component ĉ T t if t is away from the structural break point. In Section 5, we show that for series that are integrated up to order 2, the cyclical component possesses weak dependence properties; on the other hand, for the series that are integrated of order 3 or 4, the law of large numbers fails to hold for a large class of unbounded functions of the cyclical component. In Section 6 and 7, we characterize the behavior of the HP filter when applied to a polynomial trend and an exponential deterministic trend, respectively. Section 7 summarizes the findings of the paper. 4

5 2 Explanation of the heuristic approach Letting B and B denote the forward and the backward operators, respectively, the first order conditions of the problem in Equation (1) can be written as ( (1 + λ) 2λ B + λ B2 ) ˆτ T 1 = y 1 (2) ( 2λB + (1 + 5λ) 4λ B + λ B2 ) ˆτ T 2 = y 2 (3) ( 2λ B + (1 + 5λ) 4λB + λb 2 ) ˆτ T,T 1 = y T 1 (4) ( (1 + λ) 2λB + λb 2 ) ˆτ T,T = y T, (5) while for t = 3, 4,..., T 2, ( λ B2 4λ B + (1 + 6λ) 4λB + λb 2) ˆτ T t = y t. (6) By defining 1 B 2 = (1 B)(1 B), the first order condition in Equation (6) can be written as y t = (λ 1 B 4 + 1)ˆτ T t. (7) Analyses of the HP filter based on the first order condition of Equation (7) are for example King and Rebelo (1993), Cogley and Nason (1995), McElroy (2008), Phillips (2010), and Phillips and Jin (2015). Such an analysis cannot be more than a conjecture, since the first order conditions of Equations (2)-(5) are ignored. King and Rebelo (1993) s conjecture (i.e., the HP filter will render stationary series that are integrated up to fourth order) is 5

6 based on a simple manipulation of the first order condition of Equation (7) and the identity y t = ˆτ T t + ĉ T t, which give ĉ T t = (λ 1 B 4 + 1) 1 λ 1 B 4 y t. Thus, one might conjecture that ĉ T t should possess stationarity properties if y t is integrated up to order 4, because 1 B 4 y t = (1 B) 2 (1 B) 2 B 2 B2 y t = 4 y t+2. Conjecturing along these lines, we might also suspect that the HP filter is capable of removing a quadratic trend, since 1 B 4 y t = 1 B 4 t 2 = 0. However, we will show that both conjectures are incorrect in general, since the first order condition of Equation (7) fails to hold for t = 1, 2 and t = T 1, T. After all, it is easy to see (for example, by calculating the cyclical component of a quadratic time trend in a software package) that the cyclical component of the HP filter when applied to a quadratic trend is not equal to zero. Similarly, we will show that the heuristic reasoning needs to be refined when considering the cyclical component of processes that are integrated of order 3 or more, since cyclical component at the end of sample becomes important and determine the rate of convergence of the cyclical component. The results of the aforementioned papers should be interpreted as the results derived from an approximate problem that will likely be valid for values of t away from the begin and end points of the sample and for large values of T. However, such findings cannot render exact results for the HP filter. This paper will seek to derive an exact result for the HP filter that allows us to address those issues formally. Note that the approach taken in this paper towards the analysis of the HP filter is completely different from the approach of de Jong and Sakarya (2016) and Cornea-Madeira (2016). 6

7 3 Main result 3.1 A property of the HP filter The main result of the paper is presented in the following theorem. Theorem 1. Let ỹ T 1 = 2 y 3, ỹ T 2 = 2 y y 3, ỹ T,T 1 = 2 y T y T, ỹ T,T = 2 y T, and for t = 3, 4,..., T 2, ỹ T t = 4 y t+2 for t = 3, 4,..., T 2. Then ĉ T t (y 1, y 2,..., y T ) = λ ˆτ T t (ỹ T 1, ỹ T 2,..., ỹ T T ). (8) This simple but elegant result provides insights into the structure of the cyclical term. To the best of our knowledge, this property of the cyclical term has not been established before. The result shows that the cyclical component is the trend in the fourth difference of the original series plus an additional term that only affects the first and last two observations, for which the fourth difference of y t+2 is undefined. This property can shed light on the behavior of the cyclical component that is obtained from various data generating processes. The idea behind our result is the following. Using the first order condition in Equation (7) and the identity y t = ˆτ T t + ĉ T t, it follows that ĉ T t (y 1,..., y T ) = λ 1 B 4ˆτ T t (y 1,..., y T ) = λ(1 B) 2 (1 B) 2 B 2 B2ˆτ T t (y 1,..., y T ) = λ(1 B) 4 B2ˆτ T t (y 1,..., y T ). Ignoring the fact that y 1, y 0, y T +1 and y T +2 are undefined, we can now conjecture that the 7

8 last expression is approximately equal to λ(1 B) 4ˆτ T t (y 3,..., y T +2 ), which can be conjectured to approximately equal to λˆτ T t ( 4 y 3,..., 4 y T +2 ). Therefore, the conjecture presents itself that the cyclical component in a series y t is approximately equal to the trend in the fourth difference. Theorem 1 corrects and formalizes the conjecture above by taking into account the first order conditions of Equations (2)-(5) as well. Note that the last two observations ỹ T,T 1 and ỹ T T are important especially if the original series is strongly trended; for example if y t is integrated of order 3 or 4. In that case, those two observations possess the properties of a unit root process or an I(2) process depending on the integration order of the original series. de Jong and Sakarya (2016) gives an analysis of the weak dependence properties of the cyclical component of a unit root process. It is unclear how to extend this analysis to series that are integrated of order 2 or more. Also, the analysis of de Jong and Sakarya (2016) does not give a route for a characterization of structural breaks or deterministic trends (such as polynomial or exponential trends). However, the result that is given in Theorem 1 allows us to give formal results for processes integrated up to order 4 and for deterministic trends in a simple and elegant way by providing a full characterization for the cyclical component of any series. 8

9 4 The effect of a structural break We analyze the effects of the structural breaks to the cyclical component. Specifically, we focus on the intercept break which is assumed to occur at an unknown date in the middle of the sample. The next result formalizes this. Theorem 2. Let u t for t = 1, 2,..., [rt ] y t = µ + u t for t = [rt ] + 1, [rt ] + 2,..., T, where r (0, 1) and [ ] is the floor function. Then for t = 1, 2,..., T ĉ T t (y 1,..., y T ) = λµ 3 w T t,[rt ]+2 + ĉ T t (u 1, u 2,..., u T ), where w T t,[rt ]+3 is defined in Theorem 1 of de Jong and Sakarya (2016), and for k Z lim ĉ T,[rT ]+k(y 1,..., y T ) ĉ T,[rT ]+k (u 1, u 2,..., u T ) = λ µ 3 f λ (k + 1) a.s., T where f λ ( ) is defined in Theorem 3 of de Jong and Sakarya (2016). The first result shows that the presence of the structural break alters the cyclical component ĉ T t by λµ 3 w T t,[rt ]+2. Following Theorem 1 of de Jong and Sakarya (2016), w T ts C t s 3 for t s; therefore, the cyclical component ĉ T t for values of t that are away from [rt ] is not affected much by the structural break. For values of t close to [rt ], the second result shows that the cyclical component is altered by λµ 3 f λ (k + 1) asymptoti- 9

10 cally. It is possible to calculate λ 3 f λ (k + 1) by using the formula for f λ (k + 1) in Theorem 3 of de Jong and Sakarya (2016). In Figure 1, the effect of an intercept break in the cyclical component is illustrated by plotting λµ 3 f λ (k + 1) for λ = 1600 and µ = 1. The figure shows that the structural break mainly impacts ĉ T t s that are 10 time points away from the structural break point.the figure illustrates that the cyclical components ĉ T t are affected by the structural break when t is in the neighborhood of [rt ], but this affect dies out as t gets away from the structural break point. Figure 1: The effect of a deterministic trend break of size 1 when λ =

11 5 The HP filter when applied to integrated processes We consider the weak dependence properties of the cyclical term which is obtained from processes y t that are integrated up to order 4; that is, q y t = u t, for q = 1, 2, 3, or 4, where we assume that u t has some stationarity or weak dependence properties. The cyclical component {ĉ T t } T t=1 is a triangular array, and therefore it cannot be a strictly stationary sequence. On the other hand, it is possible to derive a near epoch dependence type result. The near-epoch dependence idea goes back to Ibragimov (1962) and is formalized by Billingsley (1968), McLeish (1975b), Bierens (1983), Gallant and White (1988), Andrews (1988), and Pötscher and Prucha (1991) with different approaches. The idea behind the near epoch dependence concept is that the process can be approximated arbitrarily well by a function of a finite number of strong mixing variables. Such a function is called the approximator. To formulate our result, for m 1 define the approximator ĉ m T t as T 2 ĉ m T t = λ w T ts ỹ T s I( t s m). (9) s=3 Note that ỹ t = 4 y t+2 for s = 3,..., T 2. Since 4 y t+2 = 4 q u t+2 for q = 1, 2, 3, 4, this approximator depends only on u t 2 m+q,..., u t+2+m. Also, note that the approximator cannot be defined as λ T s=1 w T tsỹ T s I( t s m) because then the approximator would include integrated terms (i.e., ỹ T,T 1 and ỹ T T ) if y t is an I(3) or I(4) process. The approximators such as ĉ m T t are common in the literature on concepts of fading memory of time series that ensures that laws of large numbers hold; see for example, McLeish (1974), McLeish (1975a), McLeish (1975b), Wooldridge (1986), Gallant (1986), and Gallant and White (1988). 11

12 The next result shows that the cyclical component has an approximability property when the HP filter is applied to a process that is integrated of order 4 or less. Theorem 3. Assume that q y t = u t, for q = 1, 2, 3, or 4, where s 1 u s p <. Then for any γ (0, 1/2), there exists constants C 1 > 0 such that for any m 1, T 1 t [γt,(1 γ)t ] ĉ T t w T t1 ỹ T 1 w T t2 ỹ T 2 w T t,t 1 ỹ T,T 1 w T tt ỹ T T ĉ m T t p C 1 m 2 (10) and T 1 t [γt,(1 γ)t ] ĉ T t w T t1 ỹ T 1 w T t2 ỹ T 2 w T t,t 1 ỹ T,T 1 w T tt ỹ T T p <. The assumption that y t is integrated up to order 4 at most implies that, under standard moment assumptions, ỹ T,T 1 p + ỹ T T p = O(T 3/2 ). Therefore, under such assumptions, w T t1 ỹ T 1 + w T t2 ỹ T 2 + w T t,t 1 ỹ T,T 1 + w T tt ỹ T T (11) will be of a small order for t [γt, (1 γ)t ], since w T ts C t s 3 for t, s = 1, 2,..., T and t s by Theorem 1 of de Jong and Sakarya (2016). King and Rebelo (1993) have conjectured, based on considering the first order condition in Equation (7) only, that the cyclical component has weak dependence properties for processes integrated up to order 4. The result above is, to the best of our knowledge, the first formal proof of this. We can now prove the following weak law of large numbers for bounded and continuous functions of the cyclical component: 12

13 Theorem 4. Assume that y t satisfies q y t = u t for q = 1, 2, 3, or 4, and assume that u t is strong mixing. In addition, assume that E( ỹ 1T + ỹ 2T + ỹ T 1,T + ỹ T,T ) = O(T 3/2 ). Let g( ) be a function that is bounded and Lipschitz continuous on R. Then T 1 T p (g(ĉ T t ) Eg(ĉ T t )) 0. t=1 Note that in the above result, the term of Equation (11) plays no role asymptotically as long as g( ) is a bounded function. In the case that g( ) is an unbounded function the convergence rate of ỹ T T takes over if y t is integrated of order 3 or more, and Theorem 4 does not hold anymore. 6 The HP filter when applied to deterministic trends 6.1 Deterministic polynomial trends Theorem 1 also allows us to establish the behavior of the HP filter when applied to deterministic polynomial trends. From Theorem 1, a result for the case of a linear trend y t = a + bt immediately follows. After all, for that case, 2 y t = 0, implying that ỹ T t = 0 for t = 1,..., T, which by Theorem 1 implies that ĉ T t = 0. For higher order polynomials, the result is more complex: Theorem 5. Suppose that y t = t p for t = 1, 2,..., T and p N. Then, p 4 ˆτ T t (1, 2 p,..., T p ) = t p λ a pkˆτ T t (1, 2 k,..., T k )I(k p 4) λc T tp + λh T tp, 13

14 where ( p ) ( k 2 p k+1 8 ) if p k is even a pk = 0 if p k is odd, (12) p 2 ( C T tp = c pkˆτ T t 1, 2 k 2, 0,..., 0, (T 3) k 2(T 2) k, (T 2) k) I(k p 2), p 4 ( H T tp = a pkˆτ T t 1, 2 k, 0,..., 0, (T 1) k, T k) I(k p 4), ( ) p c pk = (2 p k 2). (13) k The above result shows that the cyclical component of a polynomial trend of order p is p 4 ĉ T t (1, 2 p,..., T p ) = λ a pkˆτ T t (1, 2 k,..., T k )I(k p 4) + λc T tp λh T tp. For p = 2, ĉ T t = λˆτ T t (1, 1, 0,..., 0, 1, 1) = λ(w T t1 w T t2 w T t,t 1 +w T tt ). This result and Theorem 1 of de Jong and Sakarya (2016) together imply that ĉ T t takes a value close to zero if t is sufficiently away from the begin and end points because w T ts C t s 3 for t s. In the case of a cubic trend, Theorem 5 gives that ĉ T t = 6λˆτ T t (2, 1, 0,..., 0, T, (T 1)) = 6λ(2w T t1 w T t2 T w T t,t 1 + (T 1)w T tt ), which suggests that the cyclical component approaches zero slower than the cyclical component of a quadratic trend in the middle of the sample. The heuristic approach that we explained in Section 2 incorrectly suggests that the cyclical component of a polynomial trend of order 3 or less equals zero. Another implication of our result is that for p = 4, the cyclical component contains a constant, plus some additional terms. This indicates that the cyclical component itself contains terms that 14

15 are defined as trend (i.e., a constant or a linear time trend) when the HP filter is applied to the polynomial trend of order 4 or more. Next, we provide an explicit formula for the cyclical component of a quadratic trend. Theorem 6. Suppose that y t = t 2 for t = 1, 2,..., T, and 0 < λ <. Then, ĉ T t =(C 1T + C 1T ) z 1 t cos(tθ) + i(c 1T C 1T ) z 1 t sin(tθ) (14) + (C 1T + C 1T ) z 1 T t+1 cos((t t + 1)θ) + i(c 1T C 1T ) z 1 T t+1 sin((t t + 1)θ). where i 2 = 1 C 1T = 2λ ( bt ) ( ) + ā T / at bt ā T b T ( ) λ 1 2 z 1 = i 1 2λ λ 1 2 λ ( ( ) ) 1/2 θ = tan 1 2 1/ λ 1 (15) a T = (1 + λ)(z 1 + z T 1 ) 2λ(z z T 1 1 ) + λ(z z T 2 1 ) b T = 2λ(z 1 + z T 1 ) + (1 + 5λ)(z z T 1 1 ) 4λ(z z T 2 1 ) + λ(z z T 3 1 ), and C 1T is the complex conjugate of C 1T. The above result implies that when t is small and T is large, the cyclical component of a quadratic trend is approximately equal to (C 1T + C 1T ) z 1 t cos(tθ) + i(c 1T C 1T ) z 1 t sin(tθ), and when t is close to the end of sample, the last two terms in Equation (15) take over. When t is in the middle of a large sample, ĉ T t takes relatively smaller values since all terms in Equation (15) are small. This is because z 1 < 1 for 0 < λ <. Also, it is easy 15

16 to see that ĉ T t (1, 2 2,..., T 2 ) = ĉ T,T t+1 (1, 2 2,..., T 2 ) for t = 1, 2,..., T. This property of the cyclical term appears only in the quadratic trend case, which highlights the dynamics between the first two terms and the last two terms in Equation (15). Note that it is also possible to derive an expression for the cyclical component of a cubic trend along the lines of Theorem 6: ĉ T t =(C 1T + C 1T ) z 1 t cos(tθ) + i(c 1T + C 1T ) z 1 t sin(tθ) (C 2T + C 2T ) z 1 T t+1 cos((t t + 1)θ) + i(c 2T + C 2T ) z 1 T t+1 sin((t t + 1)θ), where C 1T and C 2T are the terms that depend only on z 1 and T, and C 1T C 2T C 1T. For the sake of brevity, we do not provide the explicit formulas for C 1T and C 2T, but our analysis suggests that C1T and C 2T have a slower rate of convergence than C 1T with respect to T. This explains the fact that the cyclical component of a cubic trend gets closer to zero slower than the cyclical component of a quadratic trend when t is in the middle of a large sample. 6.2 Deterministic exponential trends In this section, we consider exponential deterministic trends in order to exemplify that the HP filter fails to remove the trend of a series when the series increases at a rate that is faster than a certain rate. The following result characterizes an exponential deterministic trend by using the result of Theorem 1. 16

17 Theorem 7. Let y t = exp(t) for t = 1, 2,..., T. Then ĉ T t =C 4 λˆτ T t (exp(1), exp(2),..., exp(t )) + C 2 λˆτ T t ((2 exp( 1)), 1, 0,..., 0, exp(t 1), exp(t 1)(2 exp(1))), where C 2 = exp(2)(1 exp( 1)) 2 and C 4 = exp(2)(1 exp( 1)) 4. The result above suggests that the cyclical component of an exponential deterministic trend contains its HP filter trend multiplied by a constant, plus an additional term. This result illustrates the incapability of the HP filter of removing the trend of a series which increases at a fast rate.in the next result, we compare the convergence rates of the trend and cyclical components of an exponential deterministic trend in order to observe whether the additional term is asymptotically negligible compared to the trend component or not. Theorem 8. Let y t = exp(t) for t = 1, 2,..., T. Then for k 0 ĉ T,T k (exp(1), exp(2),..., exp(t )) lim T ˆτ T,T k (exp(1), exp(2),..., exp(t )) f λ (k 1) + f λ (k + 2) + ξ λ g λ (k + 1)g λ (2) =C 4 λ C 2 λ exp( 1) j=0 (f λ(k j) + f λ (k + j + 1) + ξ λ g λ (k + 1)g λ (j + 1)) exp( j) f λ (k) + f λ (k + 1) + ξ λ g λ (k + 1)g λ (1) + C 2 λ(2 exp( 1) 1) j=0 (f λ(k j) + f λ (k + j + 1) + ξ λ g λ (k + 1)g λ (j + 1)) exp( j) Theorems 7 and 8 together imply that the cyclical component behaves similar to the trend component asymptotically at the end of the sample. This result implies that the HP filter is not capable of removing the deterministic exponential trends. Therefore, it seems unwise to use the HP filter for removing the trends of series measured in nominal terms such 17

18 as GDP, investment, consumption, house prices, etc. 7 Conclusion This paper derives a simple but elegant property of the HP filter, which uncovers the behavior of the cyclical component when the HP filter is applied to various processes. Our result is remarkable due to its simplicity and its strength in explaining many aspects of the HP filter which have not been studied rigorously since it is introduced by Hodrick and Prescott (1980,1997). Our main result is used to analyze the effect of a deterministic trend break which is a cvery well possible case in practice. We find that a deterministic trend break affects the cyclical component where this effect has a higher magnitude in the cyclical component that is close to the break point. Next, our main result is applied to the integrated processes of order up to 4. We conclude that the cyclical component of the series that are integrated of order up to 2 possesses weak dependence properties while the situation is more subtle when the HP filter is applied to the processes that are integrated of order 3 or more. This result also reconsiders the conjecture of King and Rebelo (1993) which states that the HP filter extract stationary series when it is applied to the processes that are integrated up to order 4. Lastly, our main result allows us to derive a closed form formula for the deterministic polynomial trends and deterministic exponential trends. 18

19 References Andrews, D. W. (1988): Laws of large numbers for dependent non-identically distributed random variables, Econometric theory, 4, Bierens, H. J. (1983): Uniform consistency of kernel estimators of a regression function under generalized conditions, Journal of the American Statistical Association, 78, Billingsley, P. (1968): Convergence of Probability Measures Wiley Series in Probability and Mathematical Statistics, John Wiley & Sons New York. Cogley, T. and J. M. Nason (1995): Effects of the Hodrick-Prescott filter on trend and difference stationary time series Implications for business cycle research, Journal of Economic Dynamics and control, 19, Cornea-Madeira, A. (2016): The Explicit Formula for the Hodrick-Prescott Filter in Finite Sample, Review of economics and statistics. de Jong, R. M. and N. Sakarya (2016): The econometrics of the Hodrick-Prescott Filter, The Review of Economics and Statistics, 98, Gallant, A. R. (1986): Nonlinear statistical models, New York: Wiley. Gallant, A. R. and H. White (1988): A unified theory of estimation and inference for nonlinear dynamic models, Blackwell. Hamilton, J. D. (2016): Why You Should Never Use the Hodrick-Prescott Filter, working paper. 19

20 Hodrick, R. J. and E. C. Prescott (1980,1997): Postwar US business cycles: an empirical investigation, Carnegie Mellon University discussion paper. Ibragimov, I. A. (1962): Some limit theorems for stationary processes, Theory of Probability & Its Applications, 7, King, R. G. and S. T. Rebelo (1993): Low frequency filtering and real business cycles, Journal of Economic dynamics and Control, 17, McElroy, T. (2008): Exact formulas for the Hodrick-Prescott filter, The Econometrics Journal, 11, McLeish, D. (1974): Dependent central limit theorems and invariance principles, the Annals of Probability, (1975a): Invariance principles for dependent variables, Probability Theory and Related Fields, 32, (1975b): A maximal inequality and dependent strong laws, The Annals of probability, 3, Phillips, P. C. (2010): Two New Zealand pioneer econometricians, New Zealand Economic Papers, 44, Phillips, P. C. and S. Jin (2015): Business Cycles, Trend Elimination, and the HP Filter, Cowles Foundation Discussion Paper. 20

21 Pötscher, B. M. and I. R. Prucha (1991): Basic structure of the asymptotic theory in dynamic nonlineaerco nometric models, part i: consistency and approximation concepts, Econometric reviews, 10, Rudin, W. (1976): Principles of mathematical analysis, vol. 3, McGraw-Hill New York. Whittaker, E. T. (1923): On a new method of graduation, Proceedings of the Edinburgh Mathematical Society, 41, Wooldridge, J. M. (1986): Asymptotic properties of econometric estimators, University of California. Appendix 1: Mathematical proofs Proof of Theorem 1. First, we rewrite the minimization problem of Equation (1) as a minimization problem over c t = y t τ t. We then obtain T T 1 T 1 c 2 t + λ (c t+1 2c t + c t 1 ) 2 2λ (y t+1 2y t + y t 1 )(c t+1 2c t + c t 1 ) t=1 T 1 +λ t=2 (y t+1 2y t + y t 1 ) 2. t=2 t=2 21

22 The last term of the expression above is irrelevant to the minimization problem. Applying summation by parts twice gives (Rudin, 1976, Theorem 3.41 on p. 70) T 1 (y t+1 2y t + y t 1 )(c t+1 2c t + c t 1 ) t=2 2 y t+1 2 c t+1 T 1 = t=2 ( 2 y t 2 y t+1 ) c t + 2 y T 2 c T 2 y 3 c 2 T 1 = t=3 4 y t+2 c t + 2 y 3 c 1 + ( ) 2 y y 3 c2 + ( ) 2 y T y T ct y T c T, T 2 = t=3 and therefore, it suffices to minimize = T T 1 T 2 c 2 t + λ (c t+1 2c t + c t 1 ) 2 2λ 4 y t+2 c t t=1 t=2 t=3 2λ 2 y 3 c 1 2λ ( 2 y y 3 ) c2 2λ ( 2 y T y T ) ct 1 2λ 2 y T c T T T 1 T 2 c 2 t + λ (c t+1 2c t + c t 1 ) 2 2λ ỹ T t c t t=1 t=2 t=3 2λỹ T 1 c 1 2λỹ T 2 c 2 2λỹ T,T 1 c T 1 2λỹ T T c T, 22

23 over (c 1,..., c T ) where {ỹ T t } T t=1 is defined in Theorem 1. The first order conditions for c t for t = 1, 2, T 1, T are ( (1 + λ) 2λ B + λ B2 ) ĉ T 1 = λ 2 y 3 = λỹ T 1, (16) ( 2λB + (1 + 5λ) 4λ B + λ B2 ) ĉ T 2 = λ ( 2 y y 3 ) = λỹt 2, (17) ( 2λ B + (1 + 5λ) 4λB + λb 2 ) ĉ T,T 1 = λ ( 2 y T y T ) = λỹt,t 1, (18) ( (1 + λ) 2λB + λb 2 ) ĉ T T = λ 2 y T = λỹ T T, (19) respectively. Also, the first order condition for c t for t = 3,..., T 2 is ( λ B2 4λ B + (1 + 6λ) 4λB + λb 2) ĉ T t = λ 4 y t+2 = λỹ T t. (20) The analogy between the first order conditions of Equations (17)-(21) and the first conditions of Equations (2)-(6) now reveals that ĉ T t (y 1, y 2,..., y T ) = ˆτ T t (λỹ T 1, λỹ T 2,..., λỹ T T ) = λ ˆτ T t (ỹ T 1, ỹ T 2,..., ỹ T T ). Lemma 1. For r (0, 1) and k, l Z, lim w T,[rT ]+k,[rt ]+l = f λ (k l). T Proof of Lemma 1. We use the definition of weights in Theorem 1 of de Jong and Sakarya 23

24 (2016) and write that lim w T,[rT ]+k,[rt ]+l = lim (f T λ(k l) + f T λ (T )I(2[rT ] + k + l 1 = T )) T T + lim T f T λ(2[rt ] + k + l 1)I(2[rT ] + k + l 1 < T ) + lim T f T λ(2(t [rt ]) k l + 1)I(2[rT ] + k + l 1 > T ) + lim T (ξ T λg T λ ([rt ] + k)g T λ ([rt ] + l) + φ T λ g T λ (T [rt ] k + 1)g T λ ([rt ] + l)) + lim T φ T λg T λ ([rt ] + k)g T λ (T [rt ] l + 1) + lim T ξ T λg T λ (T [rt ] k + 1)g T λ (T [rt ] l + 1) =f λ (k l), since lim T f T λ (k l) = f λ (k l), lim T f T λ (T ) = 0, lim T g T λ (T ) = 0, lim T ξ T λ = ξ λ, and lim T φ T λ = 0 by Theorem 1-3 of de Jong and Sakarya (2016). Proof of Theorem 2. First, we write that y t = µi(t [rt ] + 1) + u t, where we assume that 4 [rt ] T 5. By using the result of Theorem 1, we write that ĉ T t (y 1, y 2,..., y T ) =λˆτ T t ( 2 u 3, 2 u u 3, µ 4 I(5 [rt ] + 1) + 4 u 5,..., µ 4 I(T [rt ] + 1) + 4 u T, 2 u T u T, 2 u T ). 24

25 Note that ˆτ T t (x 1 + y 1, x 2 + y 2,..., x T + y T ) = ˆτ T t (x 1, x 2,..., x T ) + ˆτ T t (y 1, y 2,..., y T ), therefore, ĉ T t (y 1, y 2,..., y T ) =λˆτ T t (0, 0, µ 4 I(5 [rt ] + 1),..., µ 4 I(T [rt ] + 1), 0, 0) (21) + λˆτ T t ( 2 u 3, 2 u u 3, 4 u 5,..., 4 u T, 2 u T u T, 2 u T ). Also, note that 4 I(t + 2 [rt ] + 1) = 3 I(t + 2 = [rt ] + 1), thus the first term after the equality in Equation (22) is equivalent to λˆτ T t (0, 0, µ 3 I(5 = [rt ] + 1),..., µ 3 I(T = [rt ] + 1), 0, 0) µ 3 I(s + 2 = [rt ] + 1)w T ts T 2 =λ s=3 T 2 =λµ (I(s + 2 = [rt ] + 1) 3I(s + 1 = [rt ] + 1) + 3I(s = [rt ] + 1) I(s 1 = [rt ] + 1)) w T ts s=3 = λµ 3 w T t,[rt ]+2, where the first equality is obtained by the weighted average representation of the trend given in Theorem 1 of de Jong and Sakarya (2016). The above expression gives that ĉ T t (y 1, y 2,..., y T ) = λµ 3 w T t,[rt ]+2 + λˆτ T t ( 2 u 3, 2 u u 3, 4 u 5,..., 4 u T, 2 u T u T, 2 u T ) = λµ 3 w T t,[rt ]+2 + ĉ T t (u 1, u 2,..., u T ), 25

26 where the last equality follows from Theorem 1. Next, we need to show the second result of the theorem. The first result of the theorem implies that lim ĉ T,[rT ]+k(y 1, y 2,..., y T ) ĉ T,[rT ]+k (u 1, u 2,..., u T ) T = lim T λµ 3 w T,[rT ]+k,[rt ]+2 =λ µ lim T w T,[rT ]+k,[rt ]+2 3w T,[rT ]+k,[rt ]+1 + 3w T,[rT ]+k,[rt ] w T,[rT ]+k,[rt ] 1 =λ µ f λ (k 2) 3f λ (k 1) + 3f λ (k) f λ (k + 1) =λ µ 3 f λ (k + 1), where the third equality follows from Lemma 1. Proof of Theorem 3: By Theorem 1, ĉ T t (y 1, y 2,..., y T ) w T t1 ỹ T 1 w T t2 ỹ T 2 w T t,t 1 ỹ T,T 1 w T tt ỹ T T ĉ m T t p t [γt,(1 γ)t ] = λˆτ T t (ỹ 1, ỹ 2,..., ỹ T ) w T t1 ỹ T 1 w T t2 ỹ T 2 w T t,t 1 ỹ T,T 1 w T tt ỹ T T ĉ m T t p t [γt,(1 γ)t ] = λ T 2 T 2 w T ts ỹ T s w T ts ỹ T s I( t s m) p t [γt,(1 γ)t ] λ t [γt,(1 γ)t ] s=3 s=3 s=3 T 2 w T ts ỹ T s p I( t s > m)i(t m). From the discussion in de Jong and Sakarya (2016) following their Theorem 1, it follows that 26

27 w T ts can be split into eight parts, as 8 w T ts = f T λ (t s) + w j T,t,s (22) j=2 where f T λ (0) 1; for m {1, 2,..., T }, f T λ (m) Cm 3 (23) for some constant C > 0 independent of T ; and T 1 1 s T r [γt,(1 γ)t ] T 3 8 j=2 w j T,[rT ],s <. (24) For m 1 and T m, noting that 3 s T 2 ỹ T s p 16 s 1 u s p if p y t = u t for 27

28 p = 1, 2, 3, or 4, T 2 t [γt,(1 γ)t ] s=3 16 w T ts ỹ T s p I( t s > m)i(t m) T 2 t [γt,(1 γ)t ] s=3 T t [γt,(1 γ)t ] j=m s=3 T 2 32 f T λ (j) s 1 32C s 1 =O(m 2 ) u s p f T λ (t s) I( t s > m) s 1 8 j=m j=2 u s p w j T ts I( t s > m) u s p I(T m) s 1 T 2 u s p +16 s=3 j T 2 s 1 t [γt,(1 γ)t ] u s p 8 j=2 w j T ts s 1 u s p T 1 1 s T t [γt,(1 γ)t ] T 3 8 j=2 w j T ts I(T m) by the results of Equations (23)- (25). This shows the first assertion of the theorem. To show the second assertion, T 1 t [γt,(1 γ)t ] ĉ T t w T t1 ỹ T 1 w T t2 ỹ T 2 w T t,t 1 ỹ T,T 1 w T tt ỹ T T p λ T 1 t [γt,(1 γ)t ] T 2 32λ f T λ (j) s 1 j=0 T 2 w T ts ỹ T s p s=3 T 2 u s p +16λ s=3 t [γt,(1 γ)t ] by a reasoning similar to that of the proof of the first assertion. 8 j=2 w j T ts s 1 u s p = O(1) 28

29 Proof of Theorem 4: We write that T 1 T (g(ĉ T t ) Eg(ĉ T t )) t=1 =T 1 (g(ĉ T t ) Eg(ĉ T t )) + T 1 t {1,...,T },t/ [γt,(1 γ)t ] t {1,...,T },t [γt,(1 γ)t ] (g(ĉ T t ) Eg(ĉ T t )). The first term is bounded in absolute value by 4γ x R g(x) where γ can be chosen arbitrarily small. Therefore, it is sufficient to show that the second term vanishes as T. Let a T t = w T t1 ỹ T 1 + w T t2 ỹ T 2 + w T t,t 1 ỹ T,T 1 + w T tt ỹ T T and note that T 1 T 1 g(ĉ T t ) Eg(ĉ T t ) t {1,...,T },t [γt,(1 γ)t ] t {1,...,T },t [γt,(1 γ)t ] g(ĉ T t ) g(ĉ T t a T t ) (25) + T 1 g(ĉ T t a T t ) Eg(ĉ T t a T t ) (26) t {1,...,T },t [γt,(1 γ)t ] + T 1 Eg(ĉ T t a T t ) Eg(ĉ T t ), (27) t {1,...,T },t [γt,(1 γ)t ] by the triangle inequality. For the expression in Equation (27), it is sufficient to show a weak law of large numbers. This result follows analogously to the proof of Theorem 6 of de Jong and Sakarya (2016). Therefore, we only need to show that the expressions in Equation (26) and (28) vanish in the limit. In order to do that, we first establish the result for the expression in Equation (28) then the result for the expression in Equation (26) will follow. 29

30 We write that T 1 T 1 Eg(ĉ T t a T t ) Eg(ĉ T t ) t {1,...,T },t [γt,(1 γ)t ] t {1,...,T },t [γt,(1 γ)t ] E g(ĉ T t a T t ) g(ĉ T t ). Since g( ) is Lipschitz continuous, we have T 1 E g(ĉ T t ) g(ĉ T t a T t ) LT 1 E a T t, (28) t {1,...,T },t [γt,(1 γ)t ] t {1,...,T },t [γt,(1 γ)t ] and by the definition of a T t and Theorem 1 of de Jong and Sakarya (2016), we have E a T t w T t1 E ỹ T 1 + w T t2 E ỹ T 2 + w T t,t 1 E ỹ T,T 1 + w T tt E ỹ T T ( f T λ (t 1) + 8 wt k t1 )E ỹ T 1 + ( f T λ (t 2) + k=2 + ( f T λ (t T + 1) + s {1,2,T 1,T } + s {1,2,T 1,T } C 1 T 3/2 8 wt k t2 )E ỹ T 2 k=2 8 wt k t,t 1 )E ỹ T,T 1 + ( f T λ (t T ) + k=2 f T λ (t s) E( ỹ T 1 + ỹ T 2 + ỹ T,T 1 + ỹ T T ) 8 wt k ts E( ỹ T 1 + ỹ T 2 + ỹ T,T 1 + ỹ T T ) k=2 s {1,2,T 1,T } f T λ (t s) + C 2 T 3/2 s {1,2,T 1,T } 8 wt k tt )E ỹ T T k=2 8 wt k ts (29) k=2 where the last line follows from the fact that E( ỹ T 1 + ỹ T 2 + ỹ T,T 1 + ỹ T T ) = O(T 3/2 ) under standard moment assumptions if y t is integrated up to order 4. Therefore, we use the 30

31 upper bound for E a T t in Equation (30) to rewrite for the expression in Equation (29) as follows E g(ĉ T t ) g(ĉ T t a T t ) (30) t {1,...,T },t [γt,(1 γ)t ] C 1 LT 3/2 T 1 s {1,2,T 1,T } t {1,...,T },t [γt,(1 γ)t ] + C 2 LT 3/2 T 1 s {1,2,T 1,T } t {1,...,T },t [γt,(1 γ)t ] f T λ (t s) 8 wt k ts. k=2 Note that lim T 3/2 T 1 T lim T T 3/2 T 1 s {1,2,T 1,T } t {1,...,T },t [γt,(1 γ)t ] r [γ,(1 γ)] s {1,2,T 1,T } t {1,...,T },t [γt,(1 γ)t ] = lim T 3/2 T 1 ((1 2γ)T + 1) T C 3 lim T 3/2 T 1 ((1 2γ)T + 1) T =0 f T λ (t s) (31) r [γ,(1 γ)] s {1,2,T 1,T } r [γ,(1 γ)] s {1,2,T 1,T } f T λ ([rt ] s) f T λ ([rt ] s) T 3 [rt ] s 3 where the last inequality follows from the fact that f T λ (m) C m 3 for m {1, 2,..., T } 31

32 (de Jong and Sakarya, 2016, Theorem 1). Similarly, we write that lim T 3/2 T 1 T lim T T 3/2 T 1 s {1,2,T 1,T } t {1,...,T },t [γt,(1 γ)t ] 8 wt k ts (32) k=2 r [γ,(1 γ)] s {1,2,T 1,T } t {1,...,T },t [γt,(1 γ)t ] = lim T 3/2 T 1 ((1 2γ)T + 1) T =0, r [γ,(1 γ)] s {1,2,T 1,T } 8 k=2 T 3 8 k=2 w k T,[rT ],s w k T,[rT ],s since T 1 1 s T T 3 8 k=2 w T,[rT ],s < for any r (0, 1) (de Jong and Sakarya, 2016, Equation (18)). Therefore, we have shown that the expression in Equation (26) vanishes as T. Lastly, we need to show that the expression in Equation (26) vanishes in the limit, which is implied by lim T 1 T t {1,...,T },t [γt,(1 γ)t ] E g(ĉ T t a T t ) g(c T t ) = 0. Lemma 2. 2 (t + 2) p = p 2 c pkt k I(k p 2) for t = 1, 2,..., T, where c pk is defined in Equation (14). 32

33 Proof of Lemma 2. The Binomial Theorem gives the following equality (t + m) p = p ( ) p t k m p k. k By the Binomial Theorem, 2 (t + 2) p = (t + 2) p 2(t + 1) p + t p p ( ) p p = t k 2 p k 2 k p 1 ( ) p t k + t p k = c pk t k I(k p 1) + (t p 2t p + t p ), p 2 = c pk t k I(k p 2), where c pk = ( p k) (2 p k 2). The last equality follows from the fact that c p,p 1 = 0. Lemma 3. 4 (t + 2) p = p 4 a pkt k I(k p 4) for t = 1, 2,..., T, where a pk is defined in Equation (13). 33

34 Proof of Lemma 3. By the Binomial Theorem, we have where 4 (t + 2) p = (t + 2) p 4(t + 1) p + 6t p 4(t 1) p + (t 2) p p ( ) p p ( ) p p ( ) p = t k 2 p k 4 t k + 6t p 4 t k ( 1) p k + k k k p 1 = a p,k t k I(k p 1) + (t p 4t p + 6t p 4t p + t p ) p 1 = a pk t k I(k p 1), p ( ) p t k ( 2) p k k a pk = ( ) p (2 p k 4 4( 1) p k + ( 2) p k). k This implies that ( p ) ( k 2 p k+1 8 ) if p k is even a pk = 0 if p k is odd. Note that a pk = 0 for k = p 1 and k = p 3, since p k is odd in both cases. It is also easy to see that a p,p 2 = 0. Thus, we conclude that p 4 4 (t + 2) p = a p,k t k I(k p 4). 34

35 Proof of Theorem 5. Let y t = t p for t = 1, 2,..., T. Theorem 1 implies that ĉ T t (1, 2 p,..., T p ) = λˆτ T t (ỹ T 1, ỹ T 2,..., ỹ T T ), where p 2 ỹ T 1 = 3 p 2 p = c pk I(k p 2), (33) p 2 ỹ T 2 = (4 p 2 3 p + 2 p ) 2(3 p 2 p+1 + 1) = c pk (2 k 2)I(k p 2), (34) ỹ T,T 1 = ((T 1) p 2(T 2) p + (T 3) p ) 2(T p 2(T 1) p + (T 2) p ) (35) p 2 = c pk ((T 3) k 2(T 2) k )I(k p 2), (36) p 2 ỹ T T = T p 2(T 1) p + (T 2) p = c pk (T 2) k I(k p 2), (37) by Lemma 2, and for t = 3, 4,..., T 2, p 4 ỹ T t = 4 (t + 2) p = a pk t k I(k p 4), (38) by Lemma 3. Also, note that ˆτ T t (x 1 + y 1, x 2 + y 2,..., x T + y T ) = ˆτ T t (x 1, x 2,..., x T ) + ˆτ T t (y 1, y 2,..., y T ), (39) 35

36 therefore, ĉ T t (1, 2 p,..., T p ) = λˆτ T t (0, 0, ỹ T 3,..., ỹ T,T 2, 0, 0) + λˆτ T t (ỹ T 1, ỹ T 2, 0,..., 0, ỹ T,T 1, ỹ T T ). By replacing ỹ T t with the expression in Equation (39) for t = 3, 4,..., T 2, we obtain p 4 p 4 ˆτ T t (0, 0, ỹ T 3,..., ỹ T,T 2, 0, 0) = ˆτ T t (0, 0, a pk 3 k I(k p 4),..., a pk (T 2) k I(k p 4), 0, 0) p 4 = a pkˆτ T t (0, 0, 3 k,..., (T 2) k, 0, 0)I(k p 4), where the last equality follows from the linearity of ˆτ T t (de Jong and Sakarya, 2016, Theorem 1). Similarly, by replacing ỹ T 1, ỹ T 2, ỹ T,T 1 and ỹ T T with the expressions in Equations (34)- (38) and by the linearity of ˆτ T t, we obtain p 2 ˆτ T t (ỹ T 1, ỹ T 2, 0,..., 0, ỹ T,T 1, ỹ T T ) = c pkˆτ T t (1, 2 k 2, 0..., 0, (T 3) k 2(T 2) k, (T 2) k )I(k p 2). Thus, we have p 4 ĉ T t (1, 2 p,..., T p ) =λ a pkˆτ T t (0, 0, 3 k,..., (T 2) k, 0, 0)I(k p 4) p 2 + λ c pkˆτ T t (1, 2 k 2, 0..., 0, (T 3) k 2(T 2) k, (T 2) k )I(k p 2). By using the identity in equation (40), we write that ˆτ T t (0, 0, 3 k,..., (T 2) k, 0, 0) = 36

37 ˆτ T t (1, 2 k,..., T k ) ˆτ T t (1, 2 k, 0,..., 0, (T 1) k, T k ), which gives p 4 ĉ T t (1, 2 p,..., T p ) =λ a pkˆτ T t (1, 2 k,..., T k )I(k p 4) p 2 + λ c pkˆτ T t (1, 2 k 2, 0..., 0, (T 3) k 2(T 2) k, (T 2) k )I(k p 2) p 4 λ a pkˆτ T t (1, 2 k, 0,..., 0, (T 1) k, T k )I(k p 4). In order to conclude the proof, we write that ˆτ T t (1, 2 p,..., T p ) =t p ĉ T t (1, 2 p,..., T p ) p 4 =t p λ a p,kˆτ T t (1, 2 k,..., T k )I(k p 4) p 2 λ c pkˆτ T t (1, 2 k 2, 0..., 0, (T 3) k 2(T 2) k, (T 2) k )I(k p 2) p 4 + λ a pkˆτ T t (1, 2 k, 0,..., 0, (T 1) k, T k )I(k p 4). Lemma 4. λz 4 4λz 3 + (1 + 6λ)z 2 4λz + λ = 0 has four roots z 1, z 2, z 3 and z 4 where ( ) λ 1 2 z 1 = i 1 2λ λ 1 2 λ (40) and z 2 = z 1 1, z 3 = z 1, z 4 = z 1 1. Proof of Lemma 4. Note that λz 4 4λz 3 + (1 + 6λ)z 2 4λz + λ = λ(z 1) 4 + z 2. Therefore, 37

38 we write that ( ( ) λ(z 1) 4 + z 2 = λ(z 1) 2 + iz) λ(z 1) 2 iz (41) = 0. First, we consider λ(z 1) 2 +iz = 0, which is equivalent to λz 2 +(i 2 λ)z+ λ = 0. A tedious calculation shows that this quadratic equation is equivalent to λ(z z1 )(z z 2 ) = 0, where z 1 is defined in Equation (41) and z 2 = z 1 1. By the fundamental theorem of algebra, the polynomial equation in Equation (42) has four complex roots where two of the roots are the complex conjugate of the other two. Therefore, λ(z 1) 4 + z 2 = (z z 1 )(z z 1 1 )(z z 3 )(z z 4 ), where z 3 = z 1 and z 4 = z 1 1. Proof of Theorem 6. For t = 3, 4,..., T 1, the first order condition for ĉ T t given in equation (21) as λĉ T,t+2 4λĉ T,t+1 + (1 + 6λ)ĉ T t 4λĉ T,t 1 + λĉ T,t 2 = 0. The above equation is a fourth order difference equation, which has four roots z 1, z 2, z 3 and 38

39 z 4 given in Lemma 4. It is possible to write that ĉ T t = C 1T z t 1 + C 2 z t 1 + C 3 z t 1 + C 4 z t 1, by using the fact that z 2 = z 1 1, z 3 = z 1, and z 4 = z 1 1, which are implied by Lemma 4. Note that the HP filter always produces trends and cyclical components that are real as long as the original series is real. Thus, ĉ T t = ĉ T t C 1T z t 1 + C 2 z t 1 + C 3 z t 1 + C 4 z t 1 = C 1T z t 1 + C 2 z t 1 + C 3 z t 1 + C 4 z t 1, which implies that C 3 = C 1T and C 4 = C 2. Furthermore, Theorem 1 implies that ĉ T t (1, 2 2,..., T 2 ) = λˆτ T t (1, 1, 0,..., 0, 1, 1) and ĉ T,T t+1 (1, 2 2,..., T 2 ) = λˆτ T,T t+1 (1, 1, 0,..., 0, 1, 1). Equation (20) of de Jong and Sakarya (2016) implies that ĉ T t (1, 2 2,..., T 2 ) = ĉ T,T t+1 (1, 2 2,..., T 2 ). 1 Therefore, we write that ĉ T t (1, 2 2,..., T 2 ) = ĉ T,T t+1 (1, 2 2,..., T 2 ) (T +1) C 1T z1 t + C 2 z1 t + C 1T z 1 t + C 2 z 1 t = C 1T z1 T +1 z1 t + C 2 z1 z1 t + C 1T z T +1 1 z t (T +1) 1 + C 2 z 1 z 1, t which implies that C 2 = C 1T z T This identity allows us to write that ĉ T t (1, 2 2,..., T 2 ) = C 1T (z t 1 + z T t+1 1 ) + C 1T ( z t 1 + z T t+1 1 ). (42) 1 Equation (20) of de Jong and Sakarya (2016) states that ˆτ T t (y 1, y 2,..., y T ) = ˆτ T,T t+1 (y T, y T 1,..., y 2, y 1 ). 39

40 Next, we use the first order conditions for t = 1 and 2, given in Equations (17) and (18), to solve for constant C 1T. Let a T = (1 + λ)(z 1 + z T 1 ) 2λ(z z T 1 1 ) + λ(z z T 2 1 ) b T = 2λ(z 1 + z T 1 ) + (1 + 5λ)(z z T 1 1 ) 4λ(z z T 2 1 ) + λ(z z T 3 1 ), then Equations (17) and (18) are equivalent to C 1T a T + C 1T ā T = 2λ and C 1T b T + C 1T bt = 2λ, respectively, when y t = t 2 for t = 1, 2,..., T. and ĉ T t = C 1T (z t 1 + z T t+1 1 ) + C 1T ( z t 1 + z T t+1 1 ). These imply that C 1T = 2λ( b T + ā T )/(a T bt ā T b T ). By using the polar coordinate form, we write that z 1 = z 1 (cos(θ) + i sin(θ)) where θ = tan 1 ( 2 1/2 ( λ 1) 1/2 ). Then, when we replace z 1 with its polar coordinate form in Equation (43), we obtain ĉ T t =(C 1T + C 1T ) z 1 t cos(tθ) + i(c 1T C 1T ) z 1 t sin(tθ) + (C 1T + C 1T ) z 1 T t+1 cos((t t + 1)θ) + i(c 1T C 1T ) z 1 T t+1 sin((t t + 1)θ). Lemma 5. 2 exp(t+2) = C 2 exp(t) and 4 exp(t+2) = C 4 exp(t) where C 2 = exp(2) (1 exp( 1)) 2 and C 4 = exp(2) (1 exp( 1)) 4. 40

41 Proof of Lemma 5. Note that 2 exp(t + 2) = exp(t + 2) 2 exp(t + 1) + exp(t) = exp(t + 2) (1 2 exp( 1) + exp( 1)) = exp(t + 2) (1 exp( 1)) 2 = C 2 exp(t), where C 2 = exp(2) (1 exp( 1)) 2. Similarly, we can write that 4 exp(t + 2) = exp(t + 2) 4 exp(t + 1) + 6 exp(t) 4 exp(t 1) + exp(t 2) = exp(t + 2) (1 4 exp( 1) + 6 exp( 2) 4 exp( 3) + exp( 4)) = exp(t + 2) (1 exp( 1)) 4 = C 4 exp(t), where C 4 = exp(2) (1 exp( 1)) 4. Proof of Theorem 7. Let y t = exp(t) for t = 1, 2,..., T. By Theorem 1, we write that ĉ T t (exp(1), exp(2),..., exp(t )) = λˆτ T t (ỹ T 1, ỹ T 2,..., ỹ T T ), 41

42 where by Lemma 5 ỹ T 1 = exp(3) 2 exp(2) + exp(1) = C 2 exp(1) ỹ T 2 = (exp(4) 2 exp(3) + exp(2)) 2(exp(3) 2 exp(2) + exp(1)) = C 2 (exp(2) 2 exp(1)) ỹ T,T 1 = (exp(t 1) 2 exp(t 2) + exp(t 3)) 2(exp(T ) 2 exp(t 1) + exp(t 2)) = C 2 (exp(t 3) 2 exp(t 2)) ỹ T T = exp(t ) 2 exp(t 1) + exp(t 2) = C 2 exp(t 2), and for t = 3, 4,..., T 2, ỹ T t = exp(t) 4 exp(t 1) + 6 exp(t 2) 4 exp(t 3) + exp(t 4) = C 4 exp(t). Since, ˆτ T t (ỹ T 1, ỹ T 2,..., ỹ T T ) = ˆτ T t (0, 0, ỹ T 3, ỹ T 4,..., ỹ T,T 2, 0, 0) + ˆτ T t (ỹ T 1, ỹ T 2, 0,..., 0, ỹ T,T 1, ỹ T T ), 42

43 we write that ĉ T t (exp(1), exp(2),..., exp(t )) =C 4 λˆτ T t (0, 0, exp(3), exp(4),..., exp(t 2), 0, 0) + C 2 λˆτ T t (exp(1), (exp(2) 2 exp(1)), 0,..., 0, (exp(t 3) 2 exp(t 2)), exp(t 2)) =C 4 λˆτ T t (exp(1), exp(2),..., exp(t )) + C 2 λˆτ T t (exp(1), (exp(2) 2 exp(1)), 0,..., 0, (exp(t 3) 2 exp(t 2)), exp(t 2)) C 4 λˆτ T t (exp(1), exp(2), 0..., 0, exp(t 1), exp(t )), where the first equality is due to the linearity of ˆτ T t (de Jong and Sakarya, 2016, Theorem 1), and the last line follows from ˆτ T t (0, 0, exp(3), exp(4),..., exp(t 2), 0, 0) =ˆτ T t (exp(1), exp(2),..., exp(t )) ˆτ T t (exp(1), exp(2), 0..., 0, exp(t 1), exp(t )). Note that C 4 = C 2 (1 exp( 1)) 2, therefore (C 2 C 4 ) exp(1) = C 2 ( 1 (1 exp( 1)) 2 ) exp(1) = C 2 (2 exp( 1)), 43

44 C 2 (exp(2) 2 exp(1)) C 4 exp(2) = (C 2 C 4 ) exp(2) 2C 2 exp(1) = C 2 (2 exp(1) 1) 2C 2 exp(1) = C 2, C 2 (exp(t 3) 2 exp(t 2)) C 4 exp(t 1) =C 2 (exp(t 3) 2 exp(t 2)) C 2 (1 exp( 1)) 2 exp(t 1) = C 2 exp(t 1), and C 2 exp(t 2) C 4 exp(t ) = C 2 exp(t 2) C 2 (1 exp( 1)) 2 exp(t ) = C 2 exp(t 1)(2 exp(1)). By using the above equalities, we can write that C 2 λˆτ T t (exp(1), (exp(2) 2 exp(1)), 0,..., 0, (exp(t 3) 2 exp(t 2)), exp(t 2)) C 4 λˆτ T t (exp(1), exp(2), 0..., 0, exp(t 1), exp(t )) =C 2 λˆτ T t ((2 exp( 1)), 1, 0,..., 0, exp(t 1), exp(t 1)(2 exp(1))). 44

45 Therefore, we can conclude that ĉ T t (exp(1), exp(2),..., exp(t )) =C 4 λˆτ T t (exp(1), exp(2),..., exp(t )) + C 2 λˆτ T t ((2 exp( 1)), 1, 0,..., 0, exp(t 1), exp(t 1)(2 exp(1))). Lemma 6. For k, j 0, lim w T,T k,j+1 = 0 T lim w T,T k,t j = f λ (k j) + f λ (k + j + 1) + ξ λ g λ (k + 1)g λ (j + 1) T Proof of Lemma 6. We use the definition of weights in Theorem 1 of de Jong and Sakarya (2016) and write that lim w T,T k,j+1 = lim (f T λ(t k j 1) + f T λ (T )I(T k + j = T )) T T =0, + lim T (f T λ(t k + j)i(t k + j < T ) + f T λ (T + k j)i(t k + j > T )) + lim T (ξ T λg T λ (T k)g T λ (j + 1) + φ T λ g T λ (k + 1)g T λ (j + 1)) + lim T (φ T λg T λ (T k)g T λ (T j) + ξ T λ g T λ (k + 1)g T λ (T j)) since lim T f T λ (T ) = 0, lim T g T λ (T ) = 0, lim T g T λ (j + 1) = g λ (j + 1), lim T ξ T λ = 45

46 ξ λ, and lim T φ T λ = 0 by Theorems 1 and 2 of de Jong and Sakarya (2016). Similarly, we write that lim w T,T k,t j = lim (f T λ(k j) + f T λ (T )I(2T k j 1 = T )) T T + lim T (f T λ(2t k j 1)I(2T k j 1 < T ) + f T λ (k + j + 1)I(2T k j 1 > T )) + lim T (ξ T λg T λ (T k)g T λ (T j) + φ T λ g T λ (k + 1)g T λ (T j)) + lim T (φ T λg T λ (T k)g T λ (j + 1) + ξ T λ g T λ (k + 1)g T λ (j + 1)) =f λ (k j) + f λ (k + j + 1) + ξ λ g λ (k + 1)g λ (j + 1), since lim T f T λ (m) = f λ (m) and lim T g T λ (m) = g λ (m) for all λ 0 and m Z, and lim T ξ T λ = ξ λ by Theorems 2 and 3 of de Jong and Sakarya (2016). Proof of Theorem 8. By Theorem 7, we have ĉ T,T k (exp(1), exp(2),..., exp(t )) lim T ˆτ T,T k (exp(1), exp(2),..., exp(t )) =C 4 λ + C 2 λ lim T ˆτ T,T k ((2 exp( 1)), 1, 0,..., 0, exp(t 1), exp(t 1)(2 exp(1))). ˆτ T,T k (exp(1), exp(2),..., exp(t )) Note that ˆτ T,T k ((2 exp( 1)), 1, 0,..., 0, exp(t 1), exp(t 1)(2 exp(1))) lim T ˆτ T,T k (exp(1), exp(2),..., exp(t )) =(2 exp( 1)) lim T w T,T k,1 T s=1 w T,T k,s exp(s) lim T w T,T k,2 T s=1 w T,T k,s exp(s) (43) exp(t 1)w T,T k,t 1 exp(t 1)w T,T k,t lim T T s=1 w + (2 exp(1)) lim T,T k,s exp(s) T T s=1 w T,T k,s exp(s). 46

47 The first term of equation (44) can be written as exp( T )w T,T k,1 (2 exp( 1)) lim T T s=1 w T,T k,s exp(s T ) = (2 exp( 1)) lim T exp( T )w T,T k,1 T 1 j=0 w T,T k,t j exp( j) = 0, since lim T w T,T k,1 = 0, and lim T T 1 j=0 w T,T k,t j exp( j) = j=0 (f λ(k j)+f λ (k + j + 1) + ξ λ g λ (k + 1)g λ (j + 1)) exp( j) by Lemma 6. The second term of equation (44) is zero by the argument similar to the one above. The third term of equation (44) is equivalent to w T,T k,t 1 exp( 1) lim T T 1 j=0 w T,T k,t j exp( j) f λ (k 1) + f λ (k + 2) + ξ λ g λ (k + 1)g λ (2) = exp( 1) j=0 (f λ(k j) + f λ (k + j + 1) + ξ λ g λ (k + 1)g λ (j + 1)) exp( j), by Lemma 6. The last term of equation (44) is equivalent to w T,T k,t (2 exp( 1) 1) lim T T 1 j=0 w T,T k,t j exp( j) f λ (k) + f λ (k + 1) + ξ λ g λ (k + 1)g λ (1) =(2 exp( 1) 1) j=0 (f λ(k j) + f λ (k + j + 1) + ξ λ g λ (k + 1)g λ (j + 1)) exp( j). 47