Multivariate tests of uniformity

Abstract

We present tests of multivariate uniformity using data depth, the normal quantiles and the interpoint distances between the observations. We investigate the properties of the interpoint distances among uniform random vectors. We compare the performance of the proposed tests with two existing statistics under the hypothesis of uniformity and obtain their empirical power under various alternatives in a Monte Carlo study.

Appendix

Proof for Theorem 1

Let $e^2_{i, j}=\Vert \mathbf {u}_i - \mathbf {u}_j\Vert ^2$. Consider

$$\begin{aligned}&\displaystyle \bar{\delta }=\frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\sum _{i<j}^n e^2_{i, j}&\nonumber \\&\displaystyle \bar{S}=\frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\sum _{i<j}\big (e^2_{i,j}-\frac{d}{6}\big )^2&\end{aligned}$$

(8)

and

$$\begin{aligned}&\text {Var}(\bar{\delta })=\frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\left[ \text {Var}\left( e^2_{i,j}\right) +2(n-2)\text {Cov}\left( e^2_{i,j},e^2_{i,k}\right) \right] \nonumber \\&\text {Var}(\bar{S})=\frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\left[ \text {Var}\left( \left( e^2_{i,j}-\frac{d}{6}\right) ^2\right) +2(n-2)\text {Cov}\left( \left( e^2_{i,j}-\frac{d}{6}\right) ^2,\left( e^2_{i,k}-\frac{d}{6}\right) ^2\right) \right] . \end{aligned}$$

(9)

From Eq. 9, we have

$$\begin{aligned} \text {Var}\left( \left( e^2_{i,j}-\frac{d}{6}\right) ^2\right) =\text {Var}\ \left( e^4_{i,j}-\frac{d}{3}e^2_{i,j}\right) =E^2\left[ e^4_{i,j}-\frac{d}{3}e^2_{i,j}\right] -E[e^4_{i,j}-\frac{d}{3}e^2_{i,j}]^2. \end{aligned}$$

(10)

Let $\epsilon _{ijkh}=(u_{i,j}-u_{k,h})$. One observes,

$$\begin{aligned} \text {Var}(\bar{\delta })=\frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\left[ d\text {Var}\big (\epsilon _{1121}^2\big )+2d(n-2)\text {Cov}\left( \epsilon _{1121}^2,\epsilon _{1131}^2\right) \right] =\frac{d(2n+3)}{90n(n-1)}. \end{aligned}$$

(11)

Furthermore,

$$\begin{aligned} E[e^4_{i,j}-\frac{d}{3}e^2_{i,j}]&=E\left[ \left( \sum _{i=1}^d\epsilon _{1i2i}^2\right) ^2\right] -\frac{d^2}{18} =E\left[ \sum _{i=1}^d\epsilon _{1i2i}^4+2\sum _{i<j}^d \epsilon _{1i2i}^2\epsilon _{1j2j}^2\right] -\frac{d^2}{18} \nonumber \\&=dE\left[ \epsilon _{1i2i}^4\right] +d(d-1)E\left[ \epsilon _{1i2i}^2\right] ^2-\frac{d^2}{18}=\frac{d}{15}+\frac{d(d-1)}{36}-\frac{d^2}{18}\nonumber \\&=\frac{7d}{180}-\frac{d^2}{36}\quad \end{aligned}$$

(12)

and

$$\begin{aligned} E^2\left[ e^4_{i,j}-\frac{d}{3}e^2_{i,j}\right] =E\left[ e^8_{i,j}-\frac{2d}{3}e^6_{i,j}\right] +\frac{d^2}{9}\left( \frac{d}{15}+\frac{d(d-1)}{36}\right) . \end{aligned}$$

(13)

One can verify that

$$\begin{aligned} E\left[ e^8_{i,j}\right] ={\left\{ \begin{array}{ll}\frac{1}{1296}d^4+\frac{7}{1080}d^3+\frac{2789}{226800}d^2+\frac{101}{37800}d, \quad &{}d\ge 4,\\ \frac{2}{15}+\frac{1}{7}+\frac{2}{25}, \quad &{}d=3, \\ \frac{2}{45}+\frac{2}{75}+\frac{1}{21}, \quad &{}d=2.\end{array}\right. } \end{aligned}$$

(14)

Similarly,

$$\begin{aligned} E\left[ e^6_{i,j}\right] ={\left\{ \begin{array}{ll}\frac{1}{216}d^3+\frac{7}{360}d^2+\frac{11}{945}d, \quad &{}d\ge 3,\\ \frac{1}{14}+\frac{1}{15}, \quad &{}d=2.\end{array}\right. } \end{aligned}$$

(15)

It follows from Eqs. 10 to 15,

$$\begin{aligned} \text {Var}\left( \left( e^2_{i,j}-\frac{d}{6}\right) ^2\right) ={\left\{ \begin{array}{ll}\frac{989}{56700}, \quad &{}d=2,\\ \frac{37}{1050}, \quad &{}d=3, \\ \frac{49}{16200}d^2+\frac{101}{37800}d, \quad &{}d\ge 4.\end{array}\right. } \end{aligned}$$

(16)

The covariance term in Eq. 9 can be expanded as

$$\begin{aligned} \text {Cov}\left( \left( e^2_{i,j}-\frac{d}{6}\right) ^2,\left( e^2_{i,k}-\frac{d}{6}\right) ^2\right) =\text {Cov}\left( e^4_{i,j}e^4_{i,k}\right) -\frac{2d}{3}\text {Cov}\left( e^4_{i,j},e^2_{i,k}\right) +\frac{d^3}{1620}. \end{aligned}$$

(17)

One can verify

$$\begin{aligned} E\left[ e^4_{i,j}e^2_{i,k}\right]&=E\left[ \left( \sum _{i=1}^d\epsilon _{1i2i}^2\right) ^2\sum _{j=1}^d\epsilon _{1j3j}^2\right] \nonumber \\&=dE\left[ \epsilon _{1131}^2\sum _{i=1}^d\epsilon _{1i2i}^4\right] +2dE\left[ \epsilon _{1131}^2\sum _{i<k}^d \epsilon _{1i2i}^2\epsilon _{1k2k}^2\right] . \end{aligned}$$

(18)

One observes,

$$\begin{aligned} dE\left[ \epsilon _{1131}^2\sum _{i=1}^d\epsilon _{1i2i}^4\right] =d\left( E\left[ \epsilon _{1131}^2\epsilon _{1121}^4\right] +(d-1)E\left[ \epsilon _{1131}^2\right] E\left[ \epsilon _{1222}^4\right] \right) , \end{aligned}$$

(19)

where $E[\epsilon _{1131}^2\epsilon _{1121}^4]=\frac{19}{1260}$. Similarly,

$$\begin{aligned}&2dE\left[ \epsilon _{1131}^2\sum _{i<k}^d \epsilon _{1i2i}^2\epsilon _{1k2k}^2\right] =2d\left( (d-1)E[\epsilon _{1131}^2\epsilon _{1121}^2]E[\epsilon _{1121}^2]+\left( {\begin{array}{c}d-1\\ 2\end{array}}\right) E[\epsilon _{1121}^2]^3\right) ,\nonumber \\ \end{aligned}$$

(20)

where $E[\epsilon _{1131}^2\epsilon _{1121}^2]=\frac{1}{30}.$ It follows from Eqs. 18 to 20 that

$$\begin{aligned} \text {Cov}\left( e^4_{i,j},e^2_{i,k}\right) =E\left[ e^4_{i,j}e^2_{i,k}\right] -E\left[ e^4_{i,j}\right] E\left[ e^2_{i,k}\right] =\frac{d^2}{540}+\frac{2d}{945}. \end{aligned}$$

(21)

One can also verify

$$\begin{aligned} E\left[ e^4_{i,j}e^4_{i,k}\right]= & {} E\left[ \sum _{i=1}^d\epsilon _{1i2i}^4\sum _{i=1}^d\epsilon _{1i3i}^4+2\sum _{i=1}^d\epsilon _{1i3i}^4\sum _{i<j}^d \epsilon _{1i2i}^2\epsilon _{1j2j}^2 \right. \nonumber \\&\left. +2\sum _{i=1}^d\epsilon _{1i2i}^4\sum _{i<j}^d \epsilon _{1i3i}^2\epsilon _{1j3j}^2+4\sum _{i<j}^d \epsilon _{1i2i}^2\epsilon _{1j2j}^2\sum _{i<j}^d \epsilon _{1i3i}^2\epsilon _{1j3j}^2\right] .\qquad \end{aligned}$$

(22)

Consider the first term in Eq. 22. One observes

$$\begin{aligned} E\left[ \sum _{i=1}^d\epsilon _{1i2i}^4\sum _{i=1}^d\epsilon _{1i3i}^4\right] =dE\left[ \epsilon _{1121}^4\epsilon _{1131}^4\right] +(d-1)E\left[ \epsilon _{1121}^4\right] ^2=\frac{23d}{3150}+\frac{d(d-1)}{225}. \end{aligned}$$

(23)

Similarly,

$$\begin{aligned} E\left[ \sum _{i=1}^d\epsilon _{1i3i}^4\sum _{i<j}^d \epsilon _{1i2i}^2\epsilon _{1j2j}^2\right]&=E\left[ \sum _{i=1}^d\epsilon _{1i2i}^4\sum _{i<j}^d \epsilon _{1i3i}^2\epsilon _{1j3j}^2\right] \nonumber \\&=\frac{19d(d-1)}{7560}+\frac{d(d-1)(d-2)}{1080} \end{aligned}$$

(24)

by symmetry. Consider the last term in Eq. 22. One observes

$$\begin{aligned} E\left[ \sum _{i<j}^d \epsilon _{1i2i}^2\epsilon _{1j2j}^2\sum _{i<j}^d \epsilon _{1i3i}^2\epsilon _{1j3j}^2\right]&=\left( {\begin{array}{c}d\\ 2\end{array}}\right) E\left[ \epsilon _{1121}^2\epsilon _{1222}^2\sum _{i<j}^d \epsilon _{1i3i}^2\epsilon _{1j3j}^2\right] \nonumber \\&=\left( {\begin{array}{c}d\\ 2\end{array}}\right) E\left[ \epsilon _{1121}^2\epsilon _{1222}^2\left\{ \epsilon _{1131}^2\epsilon _{1232}^2+\epsilon _{1131}^2\sum _{2<j}\epsilon _{1j3j}^2\right. \right. \nonumber \\&\quad \left. \left. +\epsilon _{1232}^2\sum _{2<j}\epsilon _{1j3j}^2+\sum _{2<i<j}\epsilon _{1i3i}^2\epsilon _{1j3j}^2\right\} \right] . \end{aligned}$$

(25)

From Eqs. 22 to 25, we have

$$\begin{aligned} \text {Cov}\left( e^4_{i,j},e^4_{i,k}\right)&=E\left[ e^4_{i,j}e^4_{i,k}\right] -E\left[ e^4_{i,j}\right] E\left[ e^4_{i,k}\right] \nonumber \\&={\left\{ \begin{array}{ll}\frac{701}{56700}, \quad &{}d=2,\\ \frac{29}{900}, \quad &{}d=3, \\ \frac{d^3}{1620}+\frac{167d^2}{113400}+\frac{29d}{37800}, \quad &{}d\ge 4,\end{array}\right. } \end{aligned}$$

(26)

Equation 17 becomes

$$\begin{aligned} \text {Cov}\left( \left( e^2_{i,j}-\frac{d}{6}\right) ^2,\left( e^2_{i,k}-\frac{d}{6}\right) ^2\right) ={\left\{ \begin{array}{ll}\frac{101}{56700}, \quad &{}d=2,\\ \frac{1}{350}, \quad &{}d=3, \\ \frac{d^2}{16200}+\frac{29d}{37800}, \quad &{}d\ge 4.\end{array}\right. } \end{aligned}$$

(27)

and

$$\begin{aligned} \text {Var}(\bar{S})={\left\{ \begin{array}{ll}\frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\left( \frac{1}{56700}\left[ 989+202(n-2)\right] \right) , \quad &{}d=2,\\ \frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\left( \frac{1}{1050}\left[ 37+6(n-2)\right] \right) , \quad &{}d=3, \\ \frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\left( \frac{49d^2}{16200}+\frac{101d}{37800}+2(n-2)\left[ \frac{d^2}{16200}+\frac{29d}{37800}\right] \right) , \quad&d\ge 4.\end{array}\right. } \end{aligned}$$

(28)