Detecting fuzzy periodic patterns in futures spreads

Abstract

This paper extends an optimal frequency domain test for the detection of synchronous patterns in multiple time series to the case of fuzzy patterns, which are not confined to single frequencies or narrow frequency bands. Applying this extension to corn futures with different delivery dates, we obtain significant results only for the spreads between the different contracts but not for the original contracts, which is an indication that spread trading has the advantage of increased predictability.

Appendix A

Under the additional assumption of normality, the statistics $\hat{a}_h (j)$, $\hat{b}_h (j)$, h=1,...,H, j=1,...,k are i.i.d. $N(0,\underline{\sigma }^2)$ and the statistics

$$\begin{aligned} R_{h} ^{*} = \frac{{2\left\{ \tfrac{1}{{\sqrt{k} }}\sum \nolimits _{{j = 1}}^{k} \frac{{\hat{a}_{{h}} (j)}}{{\underline{\sigma } }} \right\} ^{2} + \left\{ \tfrac{1}{{\sqrt{k} }}\sum \nolimits _{{j = 1}}^{k} \frac{{\hat{b}_{{h}} (j)}}{{\underline{\sigma } }} \right\} ^{2}}}{2} \end{aligned}$$

(19)

are i.i.d. $\chi ^{2}$-variables with 2 degrees of freedom, hence

$$\begin{aligned} \frac{1}{\sqrt{H} }\sum \limits _{h=1}^H {\frac{(R_h^*-2)}{\sqrt{4} }} \rightarrow N(0,1). \end{aligned}$$

(20)

Theorem The test statistic $T=\sum _{h=1}^H {R_h }$ has the same limiting distribution as the simpler statistic $T^*=\sum _{h=1}^H {R_h^*}$ , i.e.,

$$\begin{aligned} \frac{1}{\sqrt{H} }\left( {\sum \limits _{h=1}^H {R_h -2H} }\right)\rightarrow N(0,4) \end{aligned}$$

(21)

Proof Using (14), we see that $R_{h}$ can be written as

$$\begin{aligned} R_{h} = \frac{{2X_{2} (h)}}{{\tfrac{1}{{ k }}X_{2k} (h)}}, \end{aligned}$$

(22)

where $X_2 (h)$ and $X_{2k} (h)$ are $\sigma ^{2}$-variables with 2 and 2$k$ degrees of freedom, respectively. We have

$$\begin{aligned} E\left( {R_{h} - R_{h} ^{*} } \right)^{2}&= E\left( {\frac{{2X_{2} (h)}}{{\tfrac{1}{{ k }}X_{{2k}} (h)}} - \frac{{2X_{2} (h)}}{2}} \right)^{2} \\&= 4EX_{2} ^{2} (h)\left( {\frac{k}{{X_{{2k}} (h)}} - \frac{1}{2}} \right)^{2} \\&\le 4\sqrt{EX_{2} ^{4} } \sqrt{E\left( {\frac{k}{{X_{{2k}} (h)}} - \frac{1}{2}} \right)^{4} } \\&= 4{\sqrt{2\left( {2 + 2} \right)\left( {2 + 4} \right)\left( {2 + 6} \right)}\sqrt{\frac{1}{{2^{k} \Gamma \left( k \right)}}\int \limits _{0}^{\infty } {\left( {\frac{k}{x} - \frac{1}{2}} \right)^{4} x^{{k - 1}} } e^{{ - \frac{x}{2}}} dx} } \\&\rightarrow 0, \\ \end{aligned}$$

because

$$\begin{aligned} \begin{array}{l} k^4\frac{1}{2^k\Gamma (k)}\int \limits _0^\infty {x^{k-5}e^{-\textstyle {x \over 2}}dx} =\frac{k^4}{2^4k(k-1)(k-2)(k-3)}\rightarrow \frac{1}{16}, \\ -4\frac{k^3}{2}\frac{1}{2^k\Gamma (k)}\int \limits _0^\infty {x^{k-4}e^{-\textstyle {x \over 2}}dx} =-2\frac{k^3}{2^3k(k-1)(k-2)} \rightarrow -\frac{1}{4}, \\ 6\frac{k^2}{2^2}\frac{1}{2^k\Gamma (k)}\int \limits _0^\infty {x^{k-3}e^{-\textstyle {x \over 2}}dx} =\frac{3}{2}\frac{k^2}{2^2k(k-1)} \rightarrow \frac{3}{8}, \\ -4\frac{k}{2^3}\frac{1}{2^k\Gamma (k)}\int \limits _0^\infty {x^{k-2}e^{-\textstyle {x \over 2}}dx} =-\frac{1}{2}\frac{k}{2k} =-\frac{1}{4}, \\ \frac{1}{2^4}\frac{1}{2^k\Gamma (k)}\int \limits _0^\infty {x^{k-1}e^{-\textstyle {x \over 2}}dx} =\frac{1}{16}. \\ \end{array} \end{aligned}$$

Thus,

$$\begin{aligned} E\left( {\frac{1}{\sqrt{H} }\sum \limits _{h=1}^H {( {R_h -2})-\frac{1}{\sqrt{H} }\sum \limits _{h=1}^H {(R_h^*-2)} } }\right)^2&=E \left( {\frac{1}{\sqrt{H} }\sum \limits _{h=1}^H {R_h -\frac{1}{\sqrt{H} }\sum \limits _{h=1}^H {R_h^*} } }\right)^2 \\&=\frac{1}{H}E\left( {\sum \limits _{h=1}^H {(R_h -R_h^*)} }\right)^2 \\&=\frac{1}{H} \sum \limits _{h=1}^H {E (R_h -R_h^*)^2} \rightarrow 0. \end{aligned}$$

$\square$