Motivation. A Nonparametric test for the existence of moments. Purpose. Outline

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1 A Nonparametric test for the existence of moments Does the k-th moment exist? K. Hitomi 1, Y. Nishiyama 2 and K. Nagai 3 1 Kyoto Institute of Technology, 2 Kyoto University 3 Yokohama National University SWET 2010@ Sapporo Aug 8th In many empirical finance literature, tests on kurtosis and skewness are implemented. We need finite sixth/eighth moment for testing skewness/kurtosis. t 4 = n(m 4 3)/ m 8 ; m k = 1 n n i=1 (X i X) k Almost all the statistic or econometric papers assume that enough moments exist. Question Do such higher order moments really exist? Could we check it from data? Outline Purpose Modified 4 We propose a statistical test for the existence of the second moment. H 0 : H 1 : E X 2 < E X 2 = For general k th moment, we transform Y = X k/2 and test E Y 2 <.

2 Differentiability of C.F. over t 0 Let ψ(t)=ψ 1 (t)+iψ 2 (t) be the characteristic function (C.F.) of a r.v. X. We focus on the real part of the C.F., ψ 1 (t). It is regarded as the C.F. of the symmetrized X or {f(x)+f( x)}/2. Symmetrization does not affect the existence of the 2nd moment. It is well known E X 2 < ψ (0) exist ψ 1 (0) exist. To be precise, the existence of ψ 1 (0) is equivalent to ψ 1 (0)=ψ 1+ (0)=0 and <ψ 1 (0)=ψ 1+(0) 0 where the subscripts + and - indicate the right and left limits, respectively. It is well known that if E X k < then ψ (k) (t) exists for all t. Is ψ(t) k times differentiable at t 0 when E X =? A simple fact: The characteristic function of a Cauchy distribution is infinitely differentiable at t 0, { exp( t), if t >0, ψ(t)=exp( t )= exp(t), if t <0. It is also true for the stable distribution. The characteristic function of the stable distribution is ( ψ(t)=exp (iγt c t α 1 iβ t )) t ω(t,α). When ψ 1+ (0) and ψ 1+ (0) are estimable? Since H 0 is represented in terms of ψ 1+ (0), ψ 1+ (0), ψ 1 (0) and ψ 1 (0), it appears to be possible to test the existence of moments if we can estimate them. In order for this, they must be well defined at least under the null, hopefully under the alternative as well. Then, it is necessary that ψ 1 (t) is sufficiently smooth around the origin, ( ε,0), (0,ε) for ε >0. Is ψ 1 (t) smooth over ( ε,0), (0,ε) in general? Or if not, when is it smooth? Is it a sufficiently wide class of distributions in practice? We use Sato s hyperfunction for treating Fourier transforms of non-integrable functions. Hyperfunctions are a kind of generalized functions and Schwartz s distributions are included in hyperfunctions. Let D be an open subset of the complex z-plain, z =x+iy, which contains a part or the whole of the x-axis. Let F + (z) be an analytic function regular in the upper half-plane y >0 and F (z) be an analytic function regular in the lower half-plane y <0. A hyperfunction f(x) is defined as the following, f(x)= lim ε +0 {F +(x+iε) F (x iε)}. (1) It is a generalization of the concept of boundary values of complex analytic functions. See Isao Imai (1992) Applied Hyperfunction Theory ( (1981) in Japanese).

3 Operations on Hyperfunctions Regularity of the C.F. Two analytic functions F(z)=[F + (z),f (z)] define a hyperfunction f(x) and F(z) is called the generating function of f(x). All operations on hyperfunction f(x) is defined by operations on F(z). Differentiation of hyperfunction f(x) is f (x)= lim ε +0 {F +(x+iε) F (x iε)}. f(x) is differentiable since F(z) is analytic. Integration of f(x) is defined by b f(x)dx = F(z)dz, a where C is an arbitrary closed curve through z =a and z =b. C Theorem Let f(x) be density function of r.v. X. Suppose f(x) is eks α type. Then its characteristic function, ψ(t), of X has at most two singular points and other points on t-axis are regular. Futhermore, if ψ 1 (t) is singular at t =0, it is regular for t 0. An example is f(x)= for some α 1,α 2 >0. { x α 1 (1+o(1)) if x ( x) α 2 (1+o(1)) if x, eks α type function Some Notes of the Theorem Definition A function φ(z) is said to be eks α type if it is regular in the neighbourhood of and (domains such that Re z >R and Re z < R, respectively, with R a positive number), and can be expressed as φ(z)=o[exp{2πikz+ α(±z) s +o(z)}]. Here k,s are real and α complex, with s 1, Re α 0. A domain D in which the above relation holds is called an eks α domain. Note 1: This is a simple corollary of Theorem 6 in Chapter 8 of Imai (1992). Note 2: The assumption of power function approximation might be too strong to get the same results. A weaker sufficient condition is that the generating function of the density function is dominated by an eks α type function with α =0. Note: Power-type hyperfunctions are eks α type with α =0 and their eks α domains are unrestricted (Theorem 1 in Chapter 8 of Imai (1992))

4 A Set of Sufficient Conditions for the Regularity of C.F.s Estimation of the derivatives Modified Theorem Let f(x) be a density function. If, for some (large) R >0, (i) f(x) is regular for x >R, (ii) f(x)=o(exp(o(x))) for x >R, then ψ(t)= exp(itx)f(x)dx is regular except the origin. We can include discrete component without changing the conditions if it possesses a finite variance. Given a random sample of size N, {X 1,,X N }, how to estimate α ψ 1+ (0) and β ψ 1+ (0)? Letting ˆψ 1(t)= 1 N N j=1 cos(tx j) be the real part of the empirical C.F. (ECF), ˆψ 1+ (0) and ˆψ 1+ (0) does not work because they are: ˆψ 1+(0)= d ˆψ 1 (t) lim =0, t +0 dt ˆψ 1+ d 2ˆψ 1 (t) (0)= lim t +0 dt 2 = 1 n n Xj 2. j=1 These quantities do not look useful for the present purpose. Real Part of a C.F. and Test Hypotheses 1 ψ 1 (t) is symmetric, ψ 1 (t)=ψ 1 ( t), since cos(tx)=cos( tx). 2 ψ 1 (t)+ψ 1 ( t)=0 for all t 0, because of symmetry. 3 ψ 1 (t)=ψ 1 ( t) for all t 0, because of symmetry. 4 From 2, E X < ψ 1 (0)= ψ 1+ (0)=0. From 3, E X 2 < <ψ 1 (0) ψ 1+ (0) 0. Our test hypothesis is H 0 :ψ 1+ (0)=0 and <ψ 1+ (0) 0 H 1 :ψ 1+(0) 0 or ψ 1+(0)=. Estimators of ψ 1+ (0) and ψ 1+ (0) Modified We propose to estimate α ψ 1+ (0) and β ψ 1+ (0) from ECF as follows. Dropping the subscript 1 in the sequel, letting h be a small positive real number, and ˆψ(t)= 1 N N j=1 cos(tx j), define and ˆβ = 1 ˆψ(2h) ˆψ(h) { h h Expectations: We have and ˆα = ˆψ(h) 1, (2) h ˆψ(h) 1 }= ˆψ(2h) 2ˆψ(h)+1 h h 2. E(ˆα)={ψ(h) 1}/h, E(ˆβ)={ψ(2h) 2ψ(h)+1}/h 2.

5 Variances of ˆα and ˆβ Modified Modified Variances: using cos 2 (x)={cos(2x)+1}/2 and cos(2x) cos(x) ={cos(3x) + cos(x)}/2, we have and V(ˆα)= 1 nh 2{1 2 ψ(2h)+ 1 2 ψ(h)2 }, V(ˆβ) = 1 nh 4[1 2 ψ(4h)+ 1 2 ψ(2h)2 4{ 1 2 ψ(3h)+ 1 2 ψ(h) ψ(2h)ψ(h)} +4{ 1 2 ψ(h)+1 2 ψ(h)2 }]. Estimators of V(ˆα) and V(ˆβ) Modified The previous expressions yield the following straightforward estimators of V(ˆα) and V(ˆβ); ˆV(ˆα)= 1 nh 2{1 2 ˆψ(2h)+ 1 2 ˆψ(h)2 }, (3) ˆV(ˆβ) = 1 nh 4[1 2 ˆψ(4h)+ 1 2 ˆψ(2h)2 4{ 1 2 ˆψ(3h)+ 1 ˆψ(h) ˆψ(2h)ˆψ(h)} 2 +4{ 1 2 ˆψ(h)+1 2 ˆψ(h)2 }]. Under H 0, α =0, and thus we may consider the t value: t α = ˆα/ ˆV(ˆα). Under H 0, ˆβ E(X p 2 ), which is unknown. We construct two consistent estimates of β with different convergence rates by sample splitting and construct a t value for the difference, ˆβ n1 t β = ˆβ n2 ˆV(ˆβn1 )+ ˆV(ˆβ, n 1 /n 2. ) n2 Question: Do these t values behave as we want? t α and/or t β { d N(0,1), under H0, p or, under H 1. Asymptotics: Behaviour of the C.F. Modified Case 1 The null is true, or E X 2 exists when we can approximate, ψ(t)=1 1 2 σ2 t 2 +R(t), lim t +0 R(t) t 2 0, Case 2 E X < and E X 2 = (H 1 is true), when we can approximate, ψ(t)=1+q(t), lim Q(t) 0, lim t +0 t Q(t) t +0 t 2, Case 3 E X =, when we can approximate, S(t) ψ(t)=1+s(t), lim C( 0). t +0 t

6 Asymptotics of t α Modified Modified t values Modified Using the approximations of ψ(t), we study the asymptotic behaviour of t α. In fact, we have t α p as N under both H 0 and H 1. The reason is that E(ˆβ α )/ V(ˆβ α ) even under H 0 (too large bias or too small variance). t β also has a similar problem. We need a modification (bias reduction and/or variance inflation). Let n 1, n 2, n 3, n 4 be integers smaller than N and h 1, h 3, h 4, h be bandwidths decaying to zero. We use the following tes: T 1 = ˆα n1 (h 1 ) h 1ˆβn2 (h 1 )/2 +ˆα N (h) hˆβ N (h)/2, ˆV N (ˆα n1 (h 1 ))+h1ˆv 2 N (ˆβ n2 (h 1 ))/4 T 2 = ˆβ n1 (h 3 ) ˆβ n3 (h 4 ) +w{ˆβ N (h) ˆβ n4 (h)}, ˆV N (ˆβ n1 (h 3 ))+ ˆV N (ˆβ n3 (h 4 )) where w is a positive number. Notations Modified Modified t values: remarks on T 1 Modified ˆψ m (t)= 1 m m j=1 cos(tx j ), ˆα m (h)= ˆψ m(h) 1, h ˆβ m (h)= ˆψ m(2h) 2ˆψ m (h)+1 h 2, ˆV M (ˆα m (h))= 1 Mh 2{1 2 ˆψ m(h)+ 1 2 ˆψ m(h) 2 }, ˆV M (ˆβ m (h)) = 1 Mh 4[1 2 ˆψ m(4h)+ 1 2 ˆψ m(2h) 2 4{ 1 2 ˆψ m(3h)+ 1 2 ˆψ m(h) ˆψ m (2h)ˆψ m (h)} +4{ 1 2 ˆψ m(h)+ 1 2 ˆψ m(h) 2 }]. We use bias reduction and sampling splitting. We estimate ˆα n1 (h 1 ) using a smaller number of observations n 1 =o(n 1/2 ) (variance inflation). h 1ˆβn2 (h 1 )/2 estimates the bias of ˆα n1 (h 1 ) (bias reduction). We estimate V(ˆα n1 (h 1 )) by Vˆ N (ˆα n1 (h 1 )) using all of the N observations. We add ˆα N (h) hˆβ N (h)/2 in order to raise the power of the test. We note, { p 0, under case1, ˆα N (h) hˆβ N (h)/2 p, under case3.

7 Modified Asymptotics of modified t values Assumptions Notes Modified Theorem Let f(x) be the density function of X. Suppose (i) n 1 =CN 1/2 ε, n 2 =N n 1, n 3 =Clog(n 1 ), n 4 =Clog(N), n 1 h1 2 C, n 1h 2 γ 3 C, Nh 4+γ 3, n 3 h 2 γ 4 C, Nh, Nh 2 C, where C is a generic positive constant, γ is a small positive constant and ε (0,1/2). (ii) ψ(t) is locally monotone in the neighborhood of the origin. (iii) f(x) is regular for x >R, for some (large) R >0. (iv) f(x)=o(exp(o(x))) for x >R, for some (large) R >0. There exists extremely slowly-growing function such as 0 if 0<x <e 1 if e x <e e 15 H(1/x) = 2 if e e x <e ee if e ee x <e eee We cannot reject H 0 against alternatives of Case 2 with an extremely slowly-varying Q(t)/t 2 =H(t). Asymptotics of T values Theorem Modified Simulation Results: T 1 only Theorem If Case 1 is true, namely E(X 2 ) exists, T 1 d N(0,1) and T 2 d N(0,1). If Case 2 is true with Q(t) satisfying Q(t)/{t 2 log (k) t}, where log (k) t indicates a k-times iterated logarithm and k is a positive finite integer, T 2 p. If Case 2 is true with Q(t) satisfying Q(t)/t 2 δ for some δ (0,1), or Case 3 is true, T 1 p and T 2 p. # of Replications=1000. # of Observations=100, 1000, n=n 0.49, h 1 =n 1/2 H 0 is rejected if T 1 < N N(0, 1) Log Normal U(0,1) Pareto α = Pareto α = Pareto α = Pareto α = Cauchy

8 Simulation Results: T 1 & T 2 # of Replications=1000. # of Observations=100, 1000, n=n 0.49, h 1 =n 1/2, n 1 =10 log(n) 2, n 2 =N n 1, h 2 =N 1/3 H 0 is rejected if T 1 < 1.96 or T 2 < N N(0, 1) Log Normal U(0,1) Pareto α = Pareto α = Pareto α = Pareto α = Cauchy We show a sufficient condition for the smoothness of C.F. on ( ε, 0) and (0, ε) without moment conditions. We construct tes to investigate the existence of moments. Future research The performance depends on the location and scale of the data in small samples (normalization using sample mediun and interquantile range?). Time series expansion (Newey-West typevariance estimation) Consistency against all alternatives (Case 2 with an extremely slowly varying H(t)). Impossible? We apply the method to log-returns (%) of Stock prices of some companies listed in TSE, TDK, Canon, Nissan, Kirin, Osaka gas, Tokyo gas, Tokyo Electric Power Supply, and Osaka Electric Power Supply. We use the daily stock price data for Jan/2000-Sep/2004. N =1170. TDK Canon NSSN KRN OGas TGas Tepco Oep E(R 2 ) E NE E NE E E NE E E(R 4 ) E NE NE NE NE NE NE NE E: Exist, NE: Not exist.