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Introduction Literature Test Statistics Theoretical Results Conclusion

Testing Parametric Conditional Distributions of Dynamic Models Jushan Bai Columbia University, USA

Presented by Xiaojun Song (UC3M) November 2nd, 2010

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Outline

1 Introduction

2 Literature

3 Test Statistics

4 Theoretical Results

5 Conclusion

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Outline

1 Introduction 2 Literature 3 Test Statistics 4 Theoretical Results 5 Conclusion

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Motivation

Classical Kolmogorov test I.I.D. observations. Completely specified null distribution.

But when we have the following problems Data are dependent. The null hypothesis does not completely specify the distribution of the data. Joint distribution of observations is not uniquely determined under the null.

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Motivation



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Motivation



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Motivation



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Motivation



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Motivation



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Motivation



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Objective

Given a sequence of observations (Y1 , X1 ), (Y2 , X2 ), . . . , (Yn , Xn ). Let Ωt = {Xt , Xt−1 , . . . ; Yt−1 , Yt−2 , . . .} represent the information set at time t (not including Yt ). Null Hypothesis of Central Interest H0 : The conditional distribution of Yt conditional on Ωt , is in the parametric family Ft (y|Ωt , θ) for some θ ∈ Θ, where Θ is the parameter space. Example MA(1): Yt = εt + θεt−1 with |θ| < 1 and εtPbeing i.i.d. with cdf F . The ∞ conditional cdf of Yt |Ωt is given by F (y + k=1 (−θ)k Yt−k ) with Ωt = {Yt−1 , Yt−2 , . . .}.

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Outline


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Durbin (1973), the Kolmogorov test is not asymptotically distribution free when parameters are estimated Andrews (1997), conditional Kolmogorov test Diebold, Gunther, and Tay (1998), proposing a framework for evaluating density forecasts and discussing the Kolmogorov test for conditional distributions in time series Koul and Stute (1999), applying Khmaladze’s transformation to marked empirical processes for AR(1) models Zheng (2000), test based on first-order linear expansion of the Kullback-Leibler information function with kernel estimation of the underlying distribution Bai and Ng (2001), constructing a consistent test for conditional symmetry with the transformation method Bai and Chen (2008), testing distributional assumptions in multivariate GARCH models Delgado and Stute (2008), proposing omnibus tests using multiparameter empirical process ,


Description of the Method Examples An Empirical Application

Outline


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Assume that the null hypothesis is true and the true parameter value θ0 is known, Ut = Ft (Yt |Ωt , θ0 ) are i.i.d. and uniform random variables.

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Since θ0 is unknown, the random variables Ut are unobservable. When an estimator θˆ of θ0 is available, ˆ ˆt = Ft (Yt |Ωt , θ) U may be used as an estimator of Ut . The random variables Ut are neither independent nor identically distributed. The unavailability of an infinite history of observations necessitates a truncation of the information sets.

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˜ t = {Xt , Xt−1 , . . . , X1 , 0, 0, . . . , Yt−1 , . . . , Y1 , 0, 0, . . .} be a Let Ω truncated (observable) version of Ωt , we finally have ˆ ˆt = Ft (Yt |Ω ˜ t , θ) U Example For an MA(1) process, Yt = εt + θεt−1 with εt being i.i.d. F , we have ˆt = F (Yt + U

t−1 X

ˆ k Yt−k ) (−θ)

k=1

whereas Ut = F (Yt +

P∞

k k=1 (−θ) Yt−k ).

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ˆ1 , · · · , U ˆn . That is, Let Vˆn (r) be the empirical process based on U n

1 X ˆ Vˆn (r) = √ [I(Ut ≤ r) − r] n t=1

(1)

Under some regularity conditions, √ Vˆn (r) = Vn (r) − g¯(r)0 n(θˆ − θ0 ) + op (1) where

(2)

n

1 X Vn (r) = √ [I(Ut ≤ r) − r] n t=1

(3)

n

1 X ∂Ft g¯(r) = plim (x|Ωt , θ0 )|x=F −1 (r|Ωt ,θ0 ) t n t=1 ∂θ

(4)

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Martingale Transformation

R1 ˙ )g(τ ˙ )0 dτ . Let g(r) = (r, g¯(r)0 )0 , g(r) ˙ = (1, g¯˙ (r)0 )0 and C(r) = r g(τ Z r Z 1 0 −1 ˆ ˆ ˆ Wn (r) = Vn (r) − g(s) ˙ C (s) g(τ ˙ ) dVn (τ ) ds (5) 0

s

ˆ n (r) converges weakly to a standard Brownian motion W (r). Define W the test statistic as ˆ n (r)| Tn = sup |W 0≤r≤1

We have Tn =⇒ sup |W (r)| 0≤r≤1

Khmaladze’s Approach

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Martingale Transformation (Continued) Martingale transformation is in effect a continuous-time detrending operation, where the trend function is g(r) = (r, g¯(r)0 )0 . √ dVˆn (r) = dVn (r) − g¯˙ (r)0 dr n(θˆ − θ0 ) + op (1) Consider regressing dVˆn (r) on g(r) ˙ over the interval (s, 1], the OLS estimator is given by 1

Z

0

−1 Z

g˙ g˙ dr s

1

g˙ dVˆn = C(s)−1

s

Z

1

g(τ ˙ ) dVˆn (τ )

s

The residual (detrended value) is given by Z dVˆn (s) − g(s) ˙ 0 C −1 (s)

1

g(τ ˙ ) dVˆn (τ ) ds

(6)

s

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Example 1: GARCH(1,1)

0

Yt = Xt δ + εt σt 0

2 σt2 = α + βσt−1 + γ(Yt−1 − Xt−1 δ)2

Under H0 , the conditional distribution of Yt conditional on Ωt is ! 0 y − Xt δ Yt |Ωt ∼ F σt 0

2 ˆ2 σ ˆt2 = α ˆ + βˆσ ˆt−1 + γˆ (Yt−1 − Xt−1 δ)

Define εˆt =

0 Yt −Xt δˆ σ ˆt

ˆt = F (ˆ and U εt ), we have

Vˆn (r) = Vn (r) + f (F −1 (r))pn + f (F −1 (r))F −1 (r)qn + op (1)

(7)

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Example 1: GARCH(1,1)

Martingale transformation is easy to construct. Let g(r) = (g1 , g2 , g3 )0 = (r, f (F −1 (r)), f (F −1 (r))F −1 (r))0

(8)

Therefore, g(r) ˙ = (1, f˙(F −1 (r))/f (F −1 (r)), 1 + g˙ 2 (r)F −1 (r))0 . For testing normality, then g(r) ˙ = (1, −Φ−1 (r), 1 − Φ−1 (r)2 )0 where Φ(r) is the cdf of a standard normal random variable.

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Example 2: ARMA(p, q)

Consider a stationary and invertible ARMA(p, q) process such that Yt = µ + ρ1 Yt−1 + · · · + ρp Yt−p + εt + θ1 εt−1 + · · · + θq εt−q Consider testing the hypothesis that εt are i.i.d. F (· /σ). Let √ θ = (µ, ρ1 , . . . , ρp , θ1 , . . . , θq , σ) and let θˆ be a n-consistent estimator of θ. Given n + p observation Y−p+1 , Y−p+2 , . . . , Y0 , Y1 , . . . , Yn , the residuals can be computed via the recursion εˆt = Yt − µ ˆ − ρˆ1 Yt−1 − · · · − ρˆp Yt−p − θˆ1 εˆt−1 − · · · − θˆq εˆt−q ˆt = F (ˆ Define U εt /ˆ σ ) and Vˆn (r) as in equation (1).

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Example 3: Nonlinear Time Series Consider the general nonlinear time series regressions Yt = h(Ωt , β) + εt

(9)

where Ωt = (Xt , Xt−1 , . . . ; Yt−1 , Yt−2 , . . .). Consider testing the hypothesis that εt has a cdf F (x, λ) with density function f (x, λ), and λ ∈ Rd is vector of unknown parameters. Define ˆ λ), ˆ we have ˆt = F (Yt − h(Ω ˜ t , β), U ∂F (F −1 (r))0 bn + op (1) Vˆn (r) = Vn (r) − f (F −1 (r))an + ∂λ 0 ∂F (F −1 (r))0 g(r) = r, f (F −1 (r)), ∂λ !0 ∂f (F −1 (r))0 f˙(F −1 (r)) −1 g(r) ˙ = 1, , f (F (r)) f (F −1 (r)) ∂λ

(10)

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Data Monthly NYSE equal-weighted returns are fitted with a GARCH(1, 1) process. The range of the data spans from January 1926 to December 1999.

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Testing Conditional Normality

Estimate the following GARCH(1, 1) process from the data: Yt = µ + σt εt 2 + γ(Yt−1 − µ)2 . with σt2 = α + βσt−1 Obtain the parameters by MLE. Compute the residuals according to εˆt = (Yt − µ ˆ)/ˆ σt with 2 σ ˆt2 = α ˆ + βˆσ ˆt−1 + γˆ (Yt−1 − µ ˆ)2 . ˆt = Φ(ˆ Compute U εt )(t = 1, . . . , n) and Vˆn (r).

g(r) ˙ = (1, −Φ−1 (r), 1 − Φ−1 (r)2 )0 . ˆ n (r) and the test statistic Tn . Compute the transformation W Conditional normality is rejected even at the 1% significance level.

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Testing Conditional Normality

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Testing a Conditional t-Distribution

Testing the hypothesis that εt has a Student t-distribution with df = 5, normalized to have a variance of 1. df = 5 is close to the values usually found for asset returns fitted with GARCH models. There is no need to reestimate the model.

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Testing a Conditional t-Distribution Let tν be a Student t random variable with df = ν, and let qν (x) and Qν (x) be the density and cdf of tν , respectively. q ν We have εt ∼ c−1 tν with c = ν−2 . The cdf of εt under the null hypothesis is F (x) = Qν (cx) with f (x) = qν (cx)c. ˆt = Qν (cˆ Define U εt ) and Vˆn as the empirical process based on ˆ ˆ U1 , . . . , Un . −1 −1 0 g(r) = (r, qν (Q−1 ν (r))c, qν (Qν (r))Qν (r)) . We could use −1 −1 0 g(r) = (r, qν (Q−1 ν (r)), qν (Qν (r))Qν (r))

(11)

ˆ n (r) stays well within the 95% confidence band. The process W The hypothesis that innovations to the GARCH process have a conditional t-distribution is not rejected. ,



Testing a Conditional t-Distribution

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General Conditional Distributions Nonlinear Time Series Regressions GARC

Outline


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Lemma If the conditional distribution of Yt conditional on Ft−1 has a continuous cdf Ft (y|Ft−1 ), then the random variables Ut = Ft (Yt |Ft−1 ) are i.i.d. U (0, 1). Proof. Since the conditional cdf of Yt is Ft (y|Ft−1 ), the conditional distribution of Ut = Ft (Yt |Ft−1 ) is U (0, 1). Because the conditional distribution of Ut does not depend on Ft−1 , Ut is independent of Ft−1 . It follows that Ut is independent of Ut−1 , because Ut−1 is Ft−1 -measurable (that is, Ut−1 is a part of Ft−1 ). The latter is true because Ut−1 = F (Yt−1 |Ft−2 ), Yt−1 is Ft−1 -measurable, and Ft−2 ⊂Ft−1 . This implies that Ut is independent of (Ut−1 , Ut−2 , . . .) for all t. This further implies joint independence, because the joint density can be written as product of marginal and conditional densities.

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Assumptions

Let N (θ0 , M ) = {θ : |θ − θ0 | ≤ M n−1/2 }. A1. The cdf Ft (y|Ωt , θ) and its density function ft (y|Ωt , θ) are consistently differentiable with respect to θ; Ft (y|Ωt , θ) is continuous and strictly increasing in y, so that the inverse function Ft−1 is well defined; E sup sup ft (x|Ωt , u) ≤ M1 x

and

u

2

∂Ft

(x|Ωt , u) E sup sup

≤ M1 ∂θ x u

for all t and for some M1 < ∞, where the supremum with respect to u is taken in N (θ0 , M ).

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Assumptions

A2. There exists a continuously differentiable function g¯(r) such that for every M > 0

n

1 X

∂Ft −1

sup (Ft (r|u)|v) − g¯(r) = op (1)

n ∂θ u,v∈N (θ0 ,M ) t=1 R1 ˙ 2 dr < ∞, and where op (1) is uniform in r ∈ [0, 1]. In addition, 0 kgk R1 0 C(s) = s g˙ g˙ dr is invertible for every s ∈ [0, 1), where g = (r, g¯0 )0 .

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Assumptions

√ A3. The estimator θˆ satisfies n(θˆ − θ0 ) = Op (1). A4. The effect of information truncation satisfies sup u∈N (θ0 ,M )

1 √ n

X n

˜ t , u)|Ωt , θ0 ) − Ft (Ft−1 (r|Ωt , u)|Ωt , θ0 )| = op (1) |Ft (Ft−1 (r|Ω

t=1

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Theorem Under assumptions A1-A4, the asymptotic representations (2), (3), and (4) hold. That is, the following expressions √ Vˆn (r) = Vn (r) − g¯(r)0 n(θˆ − θ0 ) + op (1) where

n

1 X Vn (r) = √ [I(Ut ≤ r) − r] n t=1 n

g¯(r) = plim

1 X ∂Ft (x|Ωt , θ0 )|x=F −1 (r|Ωt ,θ0 ) t n t=1 ∂θ

hold.

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Recall that the martingale transformation is given by Z r Z 1 0 −1 ˆ ˆ ˆ Wn (r) = Vn (r) − g(s) ˙ C (s) g(τ ˙ ) dVn (τ ) ds 0

(12)

s

and the test statistic by ˆ n (r)| Tn = sup |W 0≤r≤1

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Corollary Under the assumptions of Theorem 1, ˆ n (r) =⇒ W (r) W Tn =⇒ sup |W (r)| 0≤r≤1

where W (r) is a standard Brownian motion. Many other tests can be constructed because of the weak convergence. R1 Pn ˆ ˆ 2 2 For example, let Sn = n1 i=1 W n (Ui ) . We have Sn =⇒ 0 W (r) dr.

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Assumptions

Consider nonlinear time series regressions of the form Yt = h(Ωt , β) + εt

(13)

B1. εt are i.i.d. with mean zero, density function f (x, λ), and cdf F (x, λ), where λ ∈ Rd are unknown parameters. The cdf F is strictly increasing and is continuously differentiable with respect to λ. (x,λ) Also,f (x, λ) and ∂F∂λ are bounded for λ in a neighborhood of λ0 and for all x. Furthermore, εt is independent of Ωt . B2. ht (β) is continuously differential in β, and

∂ht (β0 ) 2

≤M E ∂β for some M < ∞.

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Assumptions

ˆ satisfy √n(βˆ − β0 ) = Op (1) and B3. The estimators βˆ and λ √ ˆ n(λ − λ0 ) = Op (1). B4. The effect of information truncation satisfies n

1 X ˜ t , β0 ) − h(Ωt , β0 )| = op (1) √ |h(Ω n t=1

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Theorem Under assumptions B1-B4, equation (10) holds. ∂F (F −1 (r))0 Vˆn (r) = Vn (r) − f (F −1 (r))an + bn + op (1) ∂λ Pn √ ˆ (β0 )0 √ where an = n1 t=1 ∂ht∂β n(βˆ − β0 ), bn = n(λ − λ0 ), and ∂F (F −1 (r)) ∂λ

is equal to

∂F (x,λ0 ) ∂λ

(14)

evaluated at x = F −1 (r, λ0 ).

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Location-Scale Models

When λ is a scale parameter such that F (x, λ) = F 0 (x/λ) for some cdf F 0 , then f (F −1 (r)) = f 0 (F 0−1 (r))λ−1 and ∂F (F −1 (r)) = −f 0 (F 0−1 (r))F 0−1 (r)λ−1 . ∂λ Then the following representation Vˆn (r) = Vn (r) + f 0 (F 0−1 (r))an + f 0 (F 0−1 (r))F 0−1 (r)bn + op (1) is true for all location-scale models.

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GARCH(1,1) Assumptions

C1. The εt are i.i.d. random variables with zero mean and unit variance. The density of εt is f (x), and the cdf is F (x). The latter is continuous and strictly increasing. In addition, E|εt |2+τ < ∞ for some τ > 0, and εt is independent of Xs for s ≤ t. Pn 0 C2. n1 t=1 Xt Xt converges to a nonrandom and positive definite matrix. √ C3. n(θˆ − θ) = Op (1), where θ = (δ 0 , α, β, γ)0 .

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Theorem Under assumptions C1-C3, Vˆn (r) = Vn (r) + f (F −1 (r))pn + f (F −1 (r))F −1 (r)qn + op (1) where pn and qn are are stochastically bounded and are given by pn =

√ n 1 X Xt n(δˆ − δ) n t=1 σ ˆt

and qn =

+

t n X √ 1 X 1 √ n(ˆ α − α) βˆj + n(ˆ σ02 − σ02 )βˆt 2 2n t=1 σ ˆt j=0

√

n(βˆ − β)

t−1 X

2 βˆj σt−1−j

j=0

+

√

n(ˆ γ − γ)

t−1 X

! ˆj

0

ˆ β (Yt−1−j − Xt−1−j δ)

2

j=0 ,



In general, the function g(r) ˙ depends on the unknown parameter θ0 , so that g(r) ˙ = g(r, ˙ θ0 ). A natural solution is to replace θ0 by a √ ˆ Assuming g˙ is continuously differentiable with n-consistent estimator θ. √ respect to θ, we shall have a pointwise n-consistent estimate of g, ˙ because √ ∂ g(r, ˙ θ∗ ) √ ˆ ˆ − g(r, n[g(r, ˙ θ) ˙ θ0 )] = n(θ − θ0 ) (15) ∂θ where θ∗ is between θˆ and θ0 .

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Assumption

D1. Let g˙ n (r) be an estimator of g(r), ˙ either parametric or nonparametric, such that Z 1 2 kg˙ n (r) − g(r)k ˙ = op (1)

(16)

0

and Z

1

[g˙ n (r) − g(r)] ˙ dVn (r) = op (1)

(17)

s

uniformly in s ∈ [0, 1].

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Consider the transformed process based on g˙ n , Z r Z 1 ˜ n (r) = Vˆn (r) − W g˙ n (s)0 Cn−1 (s) g˙ n (τ ) dVˆn (τ ) ds 0

where Cn (s) =

R1 s

(18)

s

0

g˙ n g˙ n dr. The test statistic is defined as Tn, =

sup

˜ n (r)| |W

0≤r≤1−

where > 0 is a small number.

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Theorem Under assumptions A1-A4 and D1, we have for every ∈ (0, 1), in the space D[0, 1 − ], ˜ n (r) =⇒ W (r) W Tn, =⇒

sup

|W (r)|

0≤r≤1−

Note that Tn∗ = √

1 Tn, =⇒ sup |W (s)| 1− 0≤s≤1

1 sup0≤s≤1− |W (s)| and sup0≤s≤1 |W (s)| have the same since √1− distribution. Hence the same set of critical values for Tn are applicable for Tn, after a simple rescaling.

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How to verify D1?

√ Assumption D1 does not require n-consistency of g˙ n as in equation (15). Suppose g˙ n (r) has the following representation: g˙ n (r) − g(r) ˙ = κn (r)an where κn (r) is a matrix of (random) functions and an = op (1). ∗ ˙ ) For example, in (15), κn (r) = ∂ g(r,θ and an = θˆ − θ. In this case, ∂θ −1/2 an = Op (n ), which is more than necessary. If we assume R1 2 kκn (r)k dr = Op (1), then equation (16) holds because an = op (1). 0

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If

R1

Z s

s 1


κn (r) dVn (r) is stochastically bounded, that is, n

1 X κn (r) dVn (r) = √ I(Ut > s)κn (Ut ) − n t=1

Z

!

1

κn (r) dr

= Op (1)

s

(19) then equation (17) holds. Equation (19) is generally a consequence of the uniform central limit ∗ ˙ ) theorem. For example, with κn (r) = κ(r, θn∗ ) = ∂ g(r,θ , the left side of ∂θ equation (19) is bounded by

n

X 1

√ sup [I(Ui > s)κ(Ui , λ) − E{I(Ui > s)κ(Ui , λ)}] = Op (1)

n λ∈N (θ0 ) i=1 (20) where N (θ0 ) is a (shrinking) neighborhood of θ0 .

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The test based on martingale transformation has nontrivial power against root-n local alternatives. Consider the following local alternatives: for δ > 0 and 1 > √δn , δ δ Gnt (y|Ωt , θ0 ) = (1 − √ )Ft (y|Ωt , θ0 ) + √ Ht (y|Ωt , θ0 ) n n

(21)

where both Ft and Ht are conditional distribution functions. Assume Ft and Ht are different, so that n

1X Ht (Ft−1 (r|Ωt , θ0 )|Ωt , θ0 ) − r 6= 0 k(r) = plim n t=1

(22)

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If Ht = Ft , then Gnt is identical to Ft , and moreover, Ht (Ft−1 (r)) = r and k(r)=0. Under the alternative hypothesis, the random variables Ut = Ft (Yt |Ωt , θ0 ) are no longer uniform random variables and not necessarily independent. Rather, Ut∗ = Gnt (Yt |Ωt , θ0 ) are i.i.d. uniformly distributed random variables.

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Theorem Under the local alternative hypothesis, we have √ Vˆn (r) = Vn∗ (r) − g¯(r)0 n(θˆ − θ0 ) + δk(r) + op (1) where k(r) is defined in equation (22), g is given in equation (4), and n

Vn∗ (r)

1 X =√ [I(Ut∗ ≤ r) − r] n t=1

In addition, ˆ n (r) =⇒ W (r) + δk(r) − δφg (k)(r) W Rr R1 where φg (k)(r) = 0 [g(s) ˙ 0 C(s)−1 s g(k) ˙ dk] ds.

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An Interesting Question

What kind of function k(r) satisfies the following integral equation? k(r) − φg (k)(r) ≡ 0

(23)

Lemma A function k(r) satisfies the integral equation (23) if and only if k(r) = a0 g(r) for some constant vector a.

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An Application Consider the local power of the test for GARCH models. Let εt be i.i.d. with cdf δ δ Gn (x) = (1 − √ )F (x) + √ H(x) n n where F and H are distribution functions. Here k(r) = H(F −1 (r)) − r, the integral equation (23) is equivalent to H(F −1 (r)) − r = a1 r + a2 f (F −1 (r)) + a3 f (F −1 (r))F −1 (r) for some a = (a1 , a2 , a3 )0 6= 0. Applying a change in variable such that x = F −1 (r), we have H(x) − F (x) = a1 F (x) + a2 f (x) + a3 f (x)x

(24)

Under the assumption that x3 f (x) → 0 for |x| → ∞, it can be shown that the only distribution function H(x) satisfying equation (24) is F (x) itself, and in this case, ai = 0, i = 1, 2, 3. ,



Size

Random variables xt are generated from normal and tν with ν = 5 distributions. Estimate the mean and variance parameters, µ ˆ and σ ˆ. xt −ˆ µ Compute the residuals as εˆt = σˆ . For xt normal, test εt as having a standard normal distribution based on residuals εˆt =. q ν For xt being tν , test εt ν−2 as having a tν -distribution. Results are obtained from 1,000 repetitions.

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Size

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Power

Generate data xt from tν and χ2v distributions with ν = 5. Compute the residuals as εˆt = xtσˆ−ˆµ . Test εt = xtσ−µ to have a standard normal distribution based on the residuals εˆt . Results are obtained from 1,000 simulations. The power of the test should decrease as v increases.

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Power

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Outline


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A nonparametric test for conditional distributions of dynamic models is proposed. Some weak-convergence results for empirical distribution functions under parameter estimation and information truncation are established. Dimension reduction in the transformation can be achieved in conditional-mean and conditional-variance models. The consistency property of the test is explored. An empirical application demonstrates the usefulness of the test procedure. Possible developments?

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The martingale approach of Khmaladze (1981) effectively transforms a nonmartingale process to a martingale one. Let V (r) be a standard Brownian bridge on [0, 1]. Then Z r V (s) ds W (r) = V (r) + 0 1−s is a standard Brownian motion on [0, 1]. Here W (r) is a martingale transformation of the Brownian bridge. Back

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Muchas Gracias!

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