Model Specification Testing in Nonparametric and Semiparametric Time Series Econometrics. Jiti Gao

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1 Model Specification Testing in Nonparametric and Semiparametric Time Series Econometrics Jiti Gao Department of Statistics School of Mathematics and Statistics The University of Western Australia Crawley WA 6009, Australia jiti This paper is recent joint work with Maxwell King. Outline: 1. Motivation 2. Estimation and Testing 3. Example of Implementation 4. Discussion 1. Motivation and Examples Example 1.1 (Nonparametric time series regression): Y t = µ(y t 1 ) + σ(y t 1 )e t, (1.1) µ( ) and σ( ) > 0 unknown; and E[e t Y t 1 ] = 0 and E[e 2 t Y t 1 ] = 1. Estimation of µ( ) and σ( ). Univariate testing: H 0 : µ(y) = β 0 (α 0 y) or σ(y) = σ 0 + σ 1 y 2, with finite α 0 and β 0, σ 0 > 0 and σ 1 0. Simultaneous testing: H 0 : µ(y) = β 0 (α 0 y) and σ(y) = σ 0 + σ 1 y 2. 1

2 Example 1.2 (Nonparametric diffusion model): µ( ) Drift function; σ( ) volatility function; and B t standard Brownian motion. Simultaneous testing: An approximate version of model (1.2) is dr t = µ(r t )dt + σ(r t )db t, (1.2) H 0 : µ(r) = µ(r; θ 0 ) and σ 2 (r) = σ 2 (r; θ 0 ). r (t+1) r t = µ(r t ) + σ(r t ) (B (t+1) B t ), (1.3) time between successive observations. In finance, deal with monthly, weekly, daily, or higher frequency observations. Let X t = r (t 1) and Y t = r t r (t 1). Model (2.14) can then be written as e t Gaussian with E[e t ] = 0 and var[e t ] = Estimation and Testing Consider a general nonparametric time series model of the form µ( ) and σ( ) > 0 are unknown functions defined over R d ; both X t and e t can be stationary time series; and {e t } satisfies E[e t X t ] = 0 and E[e 2 t X t ] = 1. Let for x S R d. Y t = µ(x t ) + σ(x t )e t, (1.4) Y t = µ(x t ) + σ(x t )e t, (2.1) m 1 (x) = E(Y t X t = x) = µ(x), (2.2) m 2 (x) = var(y t X t = x) = σ 2 (x) 2

3 Define m(x) = (m 1 (x), m 2 (x)) τ be a bivariate vector; and m θ (x) = (m 1,θ (x), m 2,θ (x)) τ be a vector of some parametric alternatives, θ Θ R q. The interest of this paper is to test for some θ Θ against H 0 : m 1 (x) = m 1θ (x) and m 2 (x) = m 2θ (x) (2.3) H 1 : m 1 (x) = m 1θ (x) + C 1T 1T (x) and m 2 (x) = m 2θ (x) + C 2T 2T (x), (2.4) 1T (x) and 2T (x) are continuous and bounded functions over R d ; and 0 C it 1 and lim T C it = 0 for i = 1, 2. Note that the above hypotheses are equivalent to C T = (C 1T, C 2T ) τ ; and T (x) = ( 1T (x), 2T (x)) τ. H 0 : m(x) = m θ (x) versus H 1 : m(x) = m θ (x) + C T T (x), This contains the parametric case T ( ) 0. Let θ 0 Θ denote the true value of θ if H 0 is true. That is, m(x) = m θ0 (x) for all x S if H 0 is true. Let K be a d-dimensional bounded probability density function with uk(u)du = 0 and uu τ K(u)du = σki 2 d, I d is the d-dimension identity matrix and σ 2 K is a positive constant. Let R(K) = K 2 (x)dx. Let h be a smoothing bandwidth satisfying h 0 and T h d as T. This paper considers only the case of 1 d 3 due to the curse of dimensionality when d 4. 3

4 Define K h (u) = h d K(u/h). The Nadaraya-Watson (NW) estimators of m l (x) for l = 1, 2 are defined by m 1 (x) = m 2 (x) = Tt=1 K h (x X t )Y t Tt=1 K h (x X t ), (2.5) Tt=1 K h (x X t )(Y t m 1 (X t )) 2 Tt=1. K h (x X t ) This paper also considers using the only one smoothing parameter h. One can use two different bandwidth parameters h 1 and h 2 for l = 1 and l = 2 respectively. See Chen and Gao (2003). Similarly, for the parametric forms, one can estimate m l,θ by m l, θ(x) = Tt=1 K h (x X t )m l, θ(x t ) Tt=1 K h (x X t ) (2.6) for l = 1, 2, θ is a consistent estimator of θ under H 0. m(x) = ( m 1 (x), m 2 (x)) τ ; and m θ (x) = ( m 1,θ (x), m 2,θ (x)) τ. ɛ t = Y t m 1 (X t ) and η t = ɛ 2 t m 2 (X t ); [ ] σ ij (x) = E ɛ i tη j t X t = x for i, j = 0, 1, 2; ( σ20 (x) σ 11 (x) Σ 0 (x) = σ 11 (x) σ 02 (x) ). s 0 (x) = Σ 0 (x) 1, A is the determinant of A. f(x) the marginal density of {X t }. Σ(x) = f 1 (x)σ 0 (x) Test Statistic I To construct the first class of our test statistics, we have a look at the following null hypothesis: H 01 : m 1 (x) = m 1θ (x) versus (2.7) H 11 : m 1 (x) = m 1θ (x) + C 1T 1T (x). To test (2.7), Härdle and Mammen (1993) suggested using the following test statistic ( HM T = (T h d ) m1 (x) m 1 θ (x))2 π(x)dx, (2.8) π(x) is a non-negative weight function satisfying π(x)dx = 1 and π 2 (x)dx <. It can be shown that under H 01 as T, HM T = HM T µ 0 σ 0h D N(0, 1) (2.9) 4

5 in which σ 2 (x)π(x) µ 0 = R(K) dx, and f(x) ( σ0h 2 = 2h d K (4) σ 2 ) 2 (x)π(x) (0) dx, f(x) K (4) (0) = ( K(y)K(x + y)dy) 2 dx. To test (2.4), equation (2.9) thus motivates the use of a test statistic of the form N 1T (h) = (T h d ) { m(x) m θ(x)} τ ˆΣ 1 (x){ m(x) m θ(x)}π(x)dx (2.10) provided that ˆΣ 1 (x) exists, ˆΣ 1 (x) = ˆΣ 0 (x) = ˆf(x) = ˆσ ij (x) = 1 ˆf(x)ˆΣ 0 (x), (2.11) ( ) ˆσ20 (x) ˆσ 11 (x), ˆσ 11 (x) ˆσ 02 (x) 1 ( ) x Xt T h d K, h t=1 ( ) Tt=1 K x Xt h ˆɛ i tˆη j t ), Tt=1 K ˆɛ t = Y t m 1 (X t ), ˆη t = ˆɛ 2 t m 2 (X t ). ( x Xt h Theorem 2.1. Under suitable conditions, we have under H 0 as T, L 1T (h) = N 1T (h) 2 σ 1 (h) σ1(h) 2 = 4h d K (4) (0)R 2 (K) D N(0, 1) (2.12) π 2 (x)dx. As can be seen from the construction of L 1T, random denominators are involved in the form. Our experience in finite sample studies suggests that the involvement of such random denominators can reduce the power of the proposed test. This motivates the construction of the following test statistic Test Statistic II To test (2.7), existing studies suggest using N 21T (h) = s=1 t=1, s ( Xs X t Û t p st Û s, with p st = K h 5 ), (2.13)

6 which is a kernel based sample analogue of the following distance function (see Li 1999): [ ] π p (ɛ) = E [ɛe (ɛ X) f(x)] = E E 2 (ɛ X)f(X), (2.14) in which ɛ = Y m 1θ0 (X) and f(x) is the marginal density of X t. To propose a similar test for testing (2.3), we first introduce the following notation: Y = m 1θ0 (X) + ɛ t with ɛ = ɛ 2 = m 2θ0 (X) + η with η = m 2θ0 (X)(e 2 1). m 2θ0 (X)e, (2.15) Obviously, E[ɛ X] = E[η X] = 0 under H 0. Let ξ = ɛ + η. Then E[ξ] = 0 under H 0. Equation (2.15) suggests using a distance function of the form: π v (ξ) = E [ξe (ξ X) f(x)] [( ) ] = E E 2 (ɛ X) + E 2 (η X) f(x) + 2E [E(ɛ η X)f(X)]. (2.16) This would suggest using a kernel based sample analogue of (2.16) of the form Ŵ t = Ût + ˆV t. N 2T (h) = t=1 s=1, t When E[ɛη X] = 0, one may replace N 2T (h) by N 3T (h) = t=1 s=1, t Û s p st Û t + Ŵ s p st Ŵ t, (2.17) t=1 s=1, t ˆV s p st ˆVt, (2.18) (without involving the cross terms Ûs ˆV t and Ût ˆV s ) which is a kernel based sample analogue of the following distance: ψ = (ɛ, η) τ. π w (ψ) = E [ψ τ E (ψ X) f(x)] [( ) ] = E E 2 (ɛ X) + E 2 (η X) f(x), Theorem 2.2. Under suitable conditions, we have under H 0 L 2T = N 2T (h) σ 2 (h) D N(0, 1) (2.19) as T, σ2(h) 2 = 2T 2 h d K 2 (x)dx σ 4 ξ (x)f 2 (x)dx, in which σ 2 ξ (x) = E[ξ2 t X t = x] and ξ t = ɛ t + η t with ɛ t = Y t m 1θ0 (X t ) and η t = ɛ 2 t m 2θ0 (X t ). 6

7 This section then suggests using H T is a set of bandwidths satisfying certain conditions. L 2 = max h H T L 2T (h), (2.20) Simulation Scheme: We now discuss how to obtain a critical value for L 2. The exact α level critical value, lα (0 < α < 1) is the 1 α quantile of the exact finite-sample distribution of L 2. Because θ 0 is unknown, lα cannot be evaluated in practice. We therefore suggest choosing a simulated α level critical value, l α, by using the following simulation procedure: 1. For each t = 1, 2,..., T, generate Yt = m 1 θ(x t ) + m 2 θ(x t )ɛ t, {ɛ t } is sampled independently from a specified distribution with E [ɛ t ] = 0 and E [ ɛ 2 ] t = 1. In Example 3.1 below, we consider the case the specified distribution is either the standard Normal distribution N(0, 1) or the normalized exponential distribution Exp(1) 1. It should be noted that under H 0, m 1 (X t ) and m 2 (X t ) can be specified simultaneously. As a result, m 2 (X t ) can be specified parametrically as m 2θ (X t ) under H 0. When only m 1 (X t ) is specified, one usually needs to replace m 2 (X t ) by a nonparametric estimate. 2. Use the data set {Yt, X t : t = 1, 2,..., T } to estimate θ. Denote the resulting estimate by ˆθ. Compute the statistic ˆL 2 that is obtained by replacing Y t and θ with Yt and ˆθ on the right hand side of (2.20). 3. Repeat the above steps M times and produce M versions of ˆL 2 denoted by ˆL 2m for m = 1, 2,..., M. Use the M values of ˆL 2m to construct their empirical bootstrap distribution function: F (u) = 1 Mm=1 M I(ˆL 2m u). Use the empirical bootstrap distribution function to estimate the asymptotic critical value, l α. We now state the following result. Theorem 2.3. Under suitable conditions, we have under H 0 lim P T (L 2 > l α ) = α. This result shows that l α is an asymptotically correct α level critical value under any model in H 0. Let ρ(h 0, H 1 ) define the distance between H 0 and H 1. The following result shows that a consistent test will reject a false H 0 with probability approaching one as T. Theorem 2.4. Assume that the conditions of Theorem 2.3 hold. C 0 > 0 such that lim T P (ρ(h 0, H 1 ) C 0 ) = 1 holds, then lim P T (L 2 > l α ) = 1. In addition, if there is some 3. An example of implementation Example 3.1. Consider a nonlinear time series model of the form Y t = α + βx t + σ Xt 2 e t, X t = 0.5X t 1 + ɛ t, 1 t T, (3.1) 7

8 α, β and σ > 0 are unknown parameters; {ɛ t : t 1} and {e t : t 1} are i.i.d. and independent of X 0 ; ɛ t U( 0.5, 0.5), X 0 U( 1, 1); and {e t } is either the standard N(0, 1) or the normalized exponential Exp(1) 1 error, which has mean zero and variance one. In the following detailed simulation, we consider using a class of alternatives of the form Y t = α + βx t + 1 ψ φ(x t/ψ) ( + σ Xt ) ψ φ(x t/ψ) e t, (3.2) φ(x) = 1 2π e x2 2 and ψ 0 is to be chosen. The vector of unknown parameters, θ = (α, β, σ), involved in (3.1) was then estimated using the pseudo maximum likelihood method, which is quite common in the estimation of parametric ARCH models. We choose the following weight function and the kernel function given by and K(x) = π(x) = { 1 2 if x [ 1, 1] 0 otherwise { (1 x2 ) 2 if x [ 1, 1] 0 otherwise. (3.3) (3.4) With the choice of π( ) and K( ) in (3.3) and (3.4), the constant C(K, π) involved in L 1T (h) is In order to calculate both L 1T (h) and L 2T (h), one needs to find suitable h values, which are chosen by the following simulation procedure: For the simulation, we start with some initial values for θ 0 and X 0. For each t = 1, 2,..., T, generate the data (X t, Y t ) from (3.2). Use the data set {(Y t, X t ) : t = 1, 2,..., T } to estimate θ. Denote the resulting estimate by θ. For each fixed h, compute the resulting function of h given by L 1 (h) = N 1T (h) h 5 Repeat the above steps M = 1000 times and produce M versions of ˆL 1 (h) denoted by ˆL 1m (h) for m = 1, 2,..., M. Use the M functions of h, ˆL 1m (h) for m = 1, 2,..., M, to construct their empirical bootstrap distribution function, that is, F 1h (u) = 1 M I(ˆL 1m (h) u), M m=1 I(U u) is the usual indicator function. 8

9 For the given empirical value l 0.05 = 1.65, one can calculate the following power function φ 1 (h) = 1 F 1h (l 0.05 ). Find approximately at which h value the power function φ 1 (h) is maximized. Denote the maximizer by h. Similarly, one can find the maximizer, h, of the corresponding power function φ 2 (h) for ( Tt=1 Ts=1, t ) p st Ŵ s Ŵt L 2 (h) =, ˆσ 2 (h) ˆσ 2 2(h) = 2 T Ts=1 t=1 p 2 stŵ t 2 Ŵs 2. We now can calculate the following test statistic L 1 = L 1 (h ). (3.5) Similarly, we can compute L 2 = L 2 (h ). (3.6) We suggest choosing two simulated 5% level critical values, l 1,0.05 and l 2,0.05, by using the following simulation procedure: For the simulation, we start with some initial values θ 0 and X 0. For each t = 1, 2,..., T, generate the data (X t, Y t ) from model (3.1). Use the data set {(Y t, X t ) : t = 1, 2,..., T } to estimate θ. Denote the resulting estimate by θ. For the chosen H T, compute the statistics L 1 and L 2 given by (3.5) and (3.6). Repeat steps 2 3 M = 1000 times and produce M versions of L 1 and L 2 denoted by L 1m and L 2m for m = 1, 2,..., M. Use the M values of L 1m and L 2m to construct their empirical bootstrap distribution functions, that is, Fi (u) = 1 Mm=1 M I(L im u) for i = 1, 2. Use the empirical bootstrap distribution functions to calculate the two bootstrap simulated critical values, l 1,0.05 and l 2,0.05. For each case both ψ and T are chosen, we can compute the rejection rates. The number of simulations in producing Tables 3.1 and 3.2 below was The detailed results are given in Tables 3.1 and 3.2 below. Table 3.1. Rejection Rates for the Simultaneous Tests at the 5% level 9

10 Normal Error Distribution Truncation Observation Null Hypothesis Is True T L 1 L Truncation Observation Null Hypothesis Is False ψ T L 1 L Normalized Exponential Error Distribution Truncation Observation Null Hypothesis Is True T L 1 L Truncation Observation Null Hypothesis Is False ψ T L 1 L Remark 3.1. As can be seen from Table 3.1, for the standard Normal error the power can be close to one when T = 500 and the value of ψ 1 is between 4% and 10%. This may show that L 2 is not only asymptotically optimal but also practically applicable to both the small and medium sample cases, since the differences between H 0 and H 1 were made deliberately close. We also computed the power of the tests for the case ψ = 1 or 0.25, our small sample results showed that the power of L 2 was already 100% even when T = 250. In Table 3.2, we have provided some small sample results for the case the error is the normalized exponential random variable. The results show that the power of L 2 is uniformly higher than that for the standard N(0, 1) case. Table 3.1 also shows that L 2 is more powerful than L 1. To test H 01 : m 1 (x) = m 1θ0 (x), existing studies suggest using L 21T (h) = Ts=1 Tt=1, s p st Û t Û s S 21T, (3.7) 10

11 S 2 21T = 2 T t=1 Ts=1 p 2 stû 2 s Û 2 t. We also propose testing H 02 : m 2 (x) = m 2θ0 (x) by L 22T (h) = Tt=1 Ts=1, t p st ˆVs ˆVt S 22T, (3.8) S 2 22T = 2 T t=1 Ts=1 p 2 st ˆV 2 s ˆV 2 t. When there is a model specification problem with only either the conditional mean or the conditional variance, but it is difficult to determine which one may have caused the model specification problem, it would be interesting to know whether there would be any significant reduction of the power when using L 2. Table 3.2. Rejection Rates for Testing the Conditional Mean at the 5% level Normal Error Distribution Truncation Observation Null Hypothesis Is True T L 2 L Truncation Observation Null Hypothesis Is False ψ T L 2 L Normalized Exponential Error Distribution Truncation Observation Null Hypothesis Is True T L 2 L Truncation Observation Null Hypothesis Is False ψ T L 2 L

12 4. Future Work Table 3.3. Rejection Rates for Testing the Conditional Variance at the 5% level Extension to nonstationarity. Normal Error Distribution Truncation Observation Null Hypothesis Is True T L 2 L Truncation Observation Null Hypothesis Is False ψ T L 2 L Normalized Exponential Error Distribution Truncation Observation Null Hypothesis Is True T L 2 L Truncation Observation Null Hypothesis Is False ψ T L 2 L Nonstationary cases are much more difficult but important. For example, a nonlinear random walk of the form Y t = Y t 1 + g(y t 1 ) + σ(y t 1 )ɛ t (4.1) is nonstationary. See Granger, et al (1997): J. Econometrics. It would be interesting to test whether H 0 : g( ) = 0 and σ( ) = σ 0. In other words, before using model (4.1) one should test whether a random walk model of the form Y t = Y t 1 + e t would still be useful in modelling some nonlinear and nonstationary time series. Consider a nonparametric cointegration of the form Y t = m(x t ) + σ(x t )e t, (4.2) X t = X t 1 + u t, m( ) and σ( ) > 0 are both unknown to be specified. 12

13 It would then be interesting to test H 0 : m(x t ) = m(x t, θ 0 ) and σ(x t ) = σ(x t, θ 0 ). 5. By products Lemma A.1. Let ξ t be a r-dimensional strictly stationary and α-mixing stochastic process. Let φ(, ) be a symmetric Borel function defined on R r R r. Assume that for any fixed x R r, E[φ(ξ 1, x)] = 0 and E[φ(ξ i, ξ j ) Ω j 1 0 ] = 0 for any i < j, Ω j i denotes the σ field generated by {ξ s : i s j}. Let φ st = φ(ξ s, ξ t ), σst 2 = var(φ st ) and σt 2 = σst. 2 For some small constant 0 < δ < 1, let M T 1 = M T 2 = 1 s<t T { max max E φ ik φ jk 1+δ}, 1 i<j<k T { max max E φ ik φ jk 2(1+δ)}. 1 i<j<k T Assume that all the M T s are finite. Let { M T = max T 2 M 1 1+δ T 1 } and N T = max {T 32 1 M 2(1+δ) T 2 }. If as T then as T σt 2 T 0 and max{m T, N T } σt 2 0, 1 σ T 1 s<t T φ(ξ s, ξ t ) D N(0, 1). Case study: As an application, one can show that as T L T = eτ P e E[e τ P e] var(e τ P e) N(0, 1), e is a random vector and P is a random matrix. When P = X(X τ X) 1 X τ is a non-random matrix with tr(p ) = q as T and e N(0, I), one can have L T = χ2 q q N(0, 1) 2q as T. Lemma B.1. (i) Let ψ(,, ) be a symmetric Borel function defined on R r R r R r. Let the process ξ i be defined as in Lemma A.1. Assume that for any fixed x, y R r, E[ψ(ξ 1, x, y)] = 0. Then E 1 i<j<k T ψ(ξ i, ξ j, ξ k ) 13 2 CT 3 M 1 1+δ 1,

14 0 < δ < 1 is a small constant, C > 0 is a constant independent of T and the function ψ, and { M 1 = max max E ψ(ξ 1, ξ i, ξ j ) 2(1+δ)}. 1<i<j T (ii) Let φ(, ) be a symmetric Borel function defined on R r R r. Let the process ξ i be defined as in Lemma B.1. Assume that for any fixed x R r, E[φ(ξ 1, x)] = 0. Then E 1 i<j T φ(ξ i, ξ j ) 2 CT 2 M 1 1+δ 2, δ > 0 is a constant, C > 0 is a constant independent of T and the function φ, and { M 2 = max max E φ(ξ 1, ξ i ) 2(1+δ)}. 1<i<j T 14

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