Tests for aggregated dispersion: Van Valen’s test and a new competitor

Abstract

Van Valen’s test is usually applied as a two sample test for equality of dispersion for multivariate data. Motivated by a comment of Manly (Van Valen’s test. Encyclopedia of Statistical Sciences, 2006) that “Little is known about the properties of Van Valen’s test” we develop an alternative test and compare the Van Valen test with our alternative robust test in an extensive simulation study. We find that Van Valen’s test does not actually test for equality of variance sums; however, for that null hypothesis it still performs well in terms of closeness to the nominal significance level. Due to testing the correct null hypothesis and the excellent adherence to the nominal significance level, we recommend the use of the robust test as a permutation test.

Appendix: Sum of squares of the robust test statistic

The robust analogue is based on the variable sample variances $S^2_{jk} = \frac{\sum _{i=1}^{n_j} \left( X_{ijk} - {\bar{X}}_{\bullet jk} \right) ^2 }{n_j - 1}$ and in particular on the sum of these, $\sum _{k=1}^{p} S^2_{jk}$. It is important here to not confuse $S_j^2$ and $S_{jk}^2$. The numerator in the robust analogue is the square of $\sum _{k=1}^{p} S^2_{1k} - \sum _{k=1}^{p} S^2_{2k}$. Now

$$\begin{aligned} \sum _{k=1}^{p} S^2_{jk} = \frac{\sum _{i=1}^{n_j} \left( X_{ijk} - {\bar{X}}_{\bullet jk} \right) ^2 }{n_j - 1} = \frac{\sum _{i=1}^{n_j} D_{ij}}{n_j - 1} = \frac{n_j {\bar{D}}_j}{n_j - 1}, \end{aligned}$$

and hence

$$\begin{aligned} \sum _{k=1}^{p} S^2_{1k} - \sum _{k=1}^{p} S^2_{2k} = \frac{n_1 {\bar{D}}_1}{n_1 - 1} - \frac{n_2 {\bar{D}}_2}{n_2 - 1}. \end{aligned}$$

The denominator of the robust analogue is equivalent to ${\varvec{1}}_p^T \hat{\varvec{\varOmega }}_1 {\varvec{1}}_p + {\varvec{1}}_p^T \hat{\varvec{\varOmega }}_2 {\varvec{1}}_p$ with

$$\begin{aligned} {n_j} {\varvec{1}}_p^T \hat{\varvec{\varOmega }}_j {\varvec{1}}_p = \sum _{k}^{p} \sum _{l}^{p} \left( m^{22}_{jkl} - m^{20}_{jkl} m^{02}_{jkl}\right) , \end{aligned}$$

where, as before, ${n_j} m^{rs}_{jkl} = \sum _{i=1}^{n_j} \left( X_{ijk} - {\bar{X}}_{\bullet jk} \right) ^r \left( X_{ijl} - {\bar{X}}_{\bullet jl} \right) ^s$. We seek to express these in terms of $D_{ij}$. Now $\sum _{i=1}^{n_j} \left( D_{ij} - {\bar{D}}_{ij}\right) ^2 = \sum _{i=1}^{n_j} D_{ij}^2 - n_j {\bar{D}}_j^2$. First

$$\begin{aligned} \sum _{i=1}^{n_j} D_{ij}^2&= \sum _{i=1}^{n_j} \left[ \sum _{k=1}^{p} \left( X_{ijk} - {\bar{X}}_{\bullet jk} \right) ^2 \right] ^2 \\&= \sum _{i=1}^{n_j} \sum _{k=1}^{p} \sum _{l=1}^{p} \left( X_{ijk} - {\bar{X}}_{\bullet jk} \right) ^2 \left( X_{ijl} - {\bar{X}}_{\bullet jk} \right) ^2 \\&= \sum _{k=1}^{p} \sum _{l=1}^{p} n_j m_{jkl}^{22}. \end{aligned}$$

Next,

$$\begin{aligned} n_j^2 {\bar{D}}^2_j = \left[ \sum _{i=1}^{n_j} D_ij \right] ^2&= \sum _{i=1}^{n_j} \sum _{u=1}^{n_j} D_{ij} D_{uj} \\&= \sum _{i=1}^{n_j} \sum _{u=1}^{n_j} \left[ \sum _{k=1}^{p} \left( X_{ijk} - {\bar{X}}_{\bullet jk} \right) ^2 \right] \left[ \sum _{l=1}^{p} \left( X_{ujl} - {\bar{X}}_{\bullet jl} \right) ^2 \right] \\&= \sum _{i=1}^{n_j} \sum _{u=1}^{n_j} \sum _{k=1}^{p} \sum _{l=1}^{p} \left( X_{ijk} - {\bar{X}}_{\bullet jk} \right) ^2 \left( X_{ujl} - {\bar{X}}_{\bullet jl} \right) ^2 \\&= \sum _{k=1}^{p} \sum _{l=1}^{p} \left[ \sum _{i=1}^{n_j} \left( X_{ijk} - {\bar{X}}_{\bullet jk} \right) ^2 \right] \left[ \sum _{u=1}^{n_j} \left( X_{ujl} - {\bar{X}}_{\bullet jl} \right) ^2 \right] \\&= n_j^2 \sum _{k=1}^{p} \sum _{l=1}^{p} m_{jk}^{20} m_{jl}^{02}. \end{aligned}$$

Therefore

$$\begin{aligned} \sum _{i=1}^{n_j} \left( D_{ij} - {\bar{D}}_j\right) ^2&= n_j \sum _{k=1}^{p} \sum _{l=1}^{p} \left( m_{jkl}^{22} - m_{jk}^{20} m_{jl}^{02} \right) \\&= {n_j}^2 {\varvec{1}}_p^T \hat{\varvec{\varOmega }}_j {\varvec{1}}_p \end{aligned}$$

and $S_j^2 = \frac{n_j^2 {\varvec{1}}_p^T \hat{\varvec{\Omega }}_j {\varvec{1}}_p }{n_j - 1}$, so ${\varvec{1}}_p^T \hat{\varvec{\varOmega }}_j {\varvec{1}}_p = \frac{\left( n_j - 1\right) S_j^2}{n_j^2}$. Thus the denominator in the robust analogue is ${\varvec{1}}_p^T \hat{\varvec{\varOmega }}_1 {\varvec{1}}_p + {\varvec{1}}_p^T \hat{\varvec{\varOmega }}_2 {\varvec{1}}_p = \frac{\left( n_1 - 1\right) S_1^2}{n_1^2} + \frac{\left( n_2 - 1\right) S_2^2}{n_2^2}$. Combining these the robust analogue is

$$\begin{aligned} T^2_{\text {R}} = \frac{\left[ \frac{n_1 {\bar{D}}_1}{n_1 - 1} - \frac{n_2 {\bar{D}}_2}{n_2 - 1}\right] ^2}{\frac{\left( n_1 - 1\right) S_1^2}{n_1^2} + \frac{\left( n_2 - 1\right) S_2^2}{n_2^2}}. \end{aligned}$$