An intuitive skewness-based symmetry test applicable to stationary time series data

Luke Hartigan

doi:10.1515/snde-2017-0031

Abstract

I propose a simple skewness-based test of symmetry suitable for a stationary time series. The test is based on the difference between the squared deviation of a process above its median with that below it. The test has many attractive features: it is applicable to weakly dependent processes, it has a familiar form, it can be implemented using regression, and it has a standard Gaussian limiting distribution under the null hypothesis of symmetry. The finite sample properties of the test statistic are examined via Monte Carlo simulation and suggest that it has better size-adjusted power compared to competing tests in the literature when examining moderately persistence processes. I apply the test to a range of US economic and financial data and find stronger support for asymmetry in financial series compared to economic series.

Keywords: hypothesis testing; Monte Carlo simulation; skewness; symmetry

Acknowledgement

This paper is based on the fourth chapter of my Ph.D. dissertation. I am grateful to Yunjong Eo, Jiti Gao, James Morley, Timothy Neal, and Philip Rothman as well as the editor Bruce Mizrach and two anonymous referees for helpful comments and suggestions which greatly improved the quality of this paper. I also thank William Dunsmuir for useful comments on preliminary work related to this paper. All remaining errors are solely my own.

Appendix

Proof of Proposition 1

The proof of Proposition 1 follows Feunou, Jahan-Parvar, and Tédongap (2016) very closely with a few small changes.

P1: For any a > 0 and b, the median of ax_t + b is equal to am + b. The upside squared deviation of ax_t + b is given by:

E[(axt+b−E[axt+b∣axt+b≥am+b])2∣axt+b≥am+b]=a2σu2

which holds because:

a2σu2=a2E[(xt−E[xt∣xt≥m])2∣xt≥m]=E[(axt−E[axt∣xt≥m])2∣xt≥m]=E[(axt+b−E[axt+b∣xt≥m])2∣xt≥m]=E[(axt+b−E[axt+b∣axt+b≥am+b])2∣axt+b≥am+b]

Similarly, the downside squared deviation of ax_t + b is given by:

E[(axt+b−E[axt+b∣axt+b<am+b])2∣axt+b<am+b]=a2σd2

which holds since:

a2σd2=a2E[(xt−E[xt∣xt<m])2∣xt<m]=E[(axt−E[axt∣xt<m])2∣xt<m]=E[(axt+b−E[axt+b∣xt<m])2∣xt<m]=E[(axt+b−E[axt+b∣axt+b<am+b])2∣axt+b<am+b]

Hence, the skewness-based measure of symmetry for ax_t + b is given by:

δ(axt+b)=a2σu2−a2σd2=a2(σu2−σd2)=a2δ

Which satisfies P1 up to a multiplicative constant (this is similar to Feunou, Jahan-Parvar and Tédongap, 2016, see footnote 5 on page 1266).

P2: Suppose that x_t is from a symmetric and unimodal distribution, then we know that the median is equal to the mean for the distribution.

As a result, x_t − m is symmetric and unimodal with zero mean. As a consequence, x_t − m and its opposite, m − x_t have the same distribution. The upside squared deviation of x_t − m is equal to σu2 and the downside squared deviation of x_t − m is equal to σd2. However:

σd2=E[(xt−m−E[xt−m∣xt−m<0])2∣xt−m<0]=E[(m−xt−E[m−xt∣xt−m<0])2∣xt−m<0]=E[(m−xt−E[m−xt∣m−xt<0])2∣m−xt<0]

So, σd2 is also the upside squared deviation of m − x_t. Hence x_t − m and m − x_t have the same distribution, which means σu2=σd2 and as a result δ(xt)=0. This shows that the skewness-based measure of symmetry satisfies P2.

P3: Note that the median of −x_t is just −m. The upside squared deviation of −x_t is therefore the downside squared deviation of x_t:

E[(−xt−E[−xt∣−xt≥−m])2∣−xt≥−m]=E[(xt−E[xt∣−xt≥−m])2∣−xt≥−m]=E[(xt−E[xt∣xt<m])2∣xt<m]=σd2

By the same reasoning it can be shown that the downside squared deviation of −x_t is equal to σu2, the upside squared deviation of x_t. As a result, we have:

δ(−xt)=E[(−xt−E[−xt∣−xt≥−m])2∣−xt≥−m]−E[(−xt−E[−xt∣−xt<−m])2∣−xt<−m]=σd2−σu2=σu2−σd2=−δ(xt)

Hence, the skewness-based measure of symmetry satisfies P3.

Proof of Proposition 2

The proof of Proposition 2 closely follows the logic of the proof of Theorem 4 from Giacomini and White (2006). I separately show that under ℍ₀, N(δ¯/σδ)⟶dN(0,σδ2), where σδ2=var[Nδ¯] and that σ^δ−σδ⟶p0, from which the result follows.

The variable σδ2 is finite by (ii) and it is positive for all N sufficiently large by (iii). By writing N(δ¯/σδ)=N1/2∑n=1Nσδ−1δ^n, it can be shown that the sequence {σδ−1δ^n} satisfies the conditions of Wooldridge and White (1988)’s Central Limit Theorem for mixing processes. First, {σδ−1δ^n} is mixing of the same size as x_t. Furthermore, by (ii) E|σδ−1δ^n|2+ϵ<∞ for some small ϵ > 0. Hence, {σδ−1δ^n} satisfies the conditions of Corollary 3.1 of Wooldridge and White (1988), which implies that N(σδ−1δ¯)⟶dN(0,σδ2).

Based on a similar line of reasoning, {δ^n} is mixing of the same size as x_t which implies that {δ^n} is also mixing with ϕ of size −r/(r−1) or α of size −2r/(r−2). This, together with assumption (ii) and with the fact that E[δ^n]=0 under ℍ₀, implies that the conditions of Theorem 6.20 of White (2000) are satisfied, and therefore σ^δ−σδ⟶p0.

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Published Online: 2019-03-15

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