Testing for diagonal symmetry based on center-outward ranking

Abstract

This paper aims to propose a new class of permutation-invariant tests for diagonal symmetry around a known point based on the center-outward depth ranking. The asymptotic behavior of the proposed tests under the null distribution is derived. The performance of the proposed tests is assessed through a Monte Carlo study. The results show that the tests perform well comparing other procedures in terms of empirical sizes and empirical powers. We demonstrated that the proposed class includes the celebrated Wilcoxon signed-rank test as a special case in the univariate setting. Finally, we apply the tests to a well-known data set to illustrate the method developed in this paper.

Appendix

Proof of Theorem 1

Under the null hypothesis, ${\textbf {X}}_1,\ldots ,{\textbf {X}}_n$ are i.i.d. from F, where F(.) is diagonally symmetric distribution about the origin. Hence, we have

$$\begin{aligned} ({\textbf {X}}_1,\ldots ,{\textbf {X}}_n) \mathop = \limits ^d(\eta _1{\textbf {X}}_1,\ldots ,\eta _n{\textbf {X}}_n). \end{aligned}$$

(3)

where $\eta _i$’s, $i=1,\ldots ,n$ are i.i.d. random variables taking the values 1 and -1 each with probability 1/2. This yields that for any $j=1,\ldots ,p$, $\delta _{1}^j,\ldots ,\delta _n^j$ are independent and identically distributed random variables that take the values 1 and $-1$ with probability 1/2. Let ${\textbf {X}}_{i}=\delta _{i}^j{\textbf {Y}}_{i}^j$ where ${\textbf {Y}}_{i}^j=(Y_{i1}^j,\ldots ,Y_{ip}^j)^T$, $i=1,\ldots ,n$ and $j=1,\ldots ,p$. Thus for ${\textbf {y}}=(y_1,\ldots ,y_p)^T \in \mathbb {R}^p$ and $i=1,\ldots ,n$

$$\begin{aligned}&P_{H_0}\left( {Y_{i1}^j \le y_1 ,\ldots ,Y_{ip}^j \le y_p ,\delta _{i}^j = 1} \right) \\&\quad = P_{H_0}\left( {\delta _{i}^j Y_{i1}^j \le y_1 ,\ldots ,\delta _{i}^j Y_{ip}^j \le y_p ,\delta _{i}^j = 1} \right) \\&\quad = P_{H_0}\left( {X_{i1} \le y_1 ,\ldots , X_{ip} \le y_p ,\delta _{i}^j = 1} \right) \\&\quad = P_{H_0}\left( {X_{i1} \le y_1 ,\ldots ,X_{ip} \le y_p ,X_{ij} > 0} \right) \\&\quad = P_{H_0}\left( {X_{i1} \le y_1 ,\ldots ,0< X_{ij} \le y_j ,\ldots ,X_{ip} \le y_p} \right) \\&\quad = P_{H_0}\left( {-X_{i1} \le y_1 ,\ldots ,0< -X_{ij} \le y_j ,\ldots ,-X_{ip} \le y_p} \right) \\&\quad = P_{H_0}\left( {X_{i1} \ge -y_1 ,\ldots ,-y_j \le X_{ij}< 0 ,\ldots ,X_{ip} \ge -y_p} \right) \\&\quad = P_{H_0}\left( {X_{i1} \ge - y_1 ,\ldots , X_{ip} \ge - y_p ,X_{ij} < 0} \right) \\&\quad = P_{H_0}\left( { -X_{i1} \le y_1 ,\ldots , - X_{ip} \le y_p ,\delta _{i}^j = - 1} \right) \\&\quad = P_{H_0}\left( {\delta _{i}^j X_{i1} \le y_1 ,\ldots ,\delta _{i}^j X_{ip} \le y_p ,\delta _{i}^j = - 1} \right) \\&\quad = P_{H_0}\left( {Y_{i1}^j \le y_1 ,\ldots ,Y_{ip}^j \le y_p ,\delta _{i}^j = - 1} \right) \end{aligned}$$

and for $k\ne i$

$$\begin{aligned}&P_{H_0}\left( {Y_{i1}^j \le y_1 ,\ldots , Y_{ip}^j \le y_p ,\delta _{k}^j = 1} \right) \\&\quad = P_{H_0}\left( {Y_{i1}^j \le y_1 ,\ldots ,Y_{ip}^j \le y_p ,\delta _{i}^j = 1,\delta _{k}^j = 1} \right) \\&\quad \quad +P_{H_0}\left( {Y_{i1}^j \le y_1 ,\ldots ,Y_{ip}^j \le y_p ,\delta _{i}^j = - 1,\delta _{k}^j = 1} \right) \\&\quad = P_{H_0}\left( {X_{i1} \le y_1 ,\ldots ,X_{ip} \le y_p ,X_{ij}> 0,X_{kj}> 0} \right) \\&\quad \quad +P_{H_0}\left( {X_{i1} \ge - y_1 ,\ldots ,X_{ip} \ge - y_p ,X_{ij}< 0,X_{kj}> 0} \right) \\&\quad = P_{H_0}\left( {X_{i1} \le y_1 ,\ldots ,X_{ip} \le y_p ,X_{ij} > 0,X_{kj}< 0} \right) \\&\quad \quad +P_{H_0}\left( {X_{i1} \ge - y_1 ,\ldots ,X_{ip} \ge - y_p ,X_{ij}< 0,X_{kj} < 0} \right) \\&\quad = P_{H_0}\left( {Y_{i1}^j \le y_1 ,\ldots ,Y_{ip}^j \le y_p ,\delta _{i}^j = 1,\delta _{k}^j = - 1} \right) \\&\quad \quad +P_{H_0}\left( {Y_{i1}^j \le y_1 ,\ldots ,Y_{ip}^j \le y_p ,\delta _{i}^j = - 1,\delta _{k}^j = - 1} \right) \\&\quad = P_{H_0}\left( {Y_{i1}^j \le y_1 ,\ldots ,Y_{ip}^j \le y_p ,\delta _{k}^j = - 1} \right) . \\ \end{aligned}$$

Hence these imply that $\delta _{i}^j$ for $i=1,\ldots ,n$, is independent of ${\textbf {Y}}_{1}^j,\ldots , {\textbf {Y}}_{n}^j$. Now, suppose that $F^s_{{\textbf {X}}_n}$ and $F^s_{{\textbf {Y}}_n^j}$ be the sample distribution functions of $\left\{ { \pm {\textbf {X}}_{1} ,\ldots , \pm {\textbf {X}}_{n} } \right\}$ and $\left\{ { \pm {\textbf {Y}}_{1}^j ,\ldots , \pm {\textbf {Y}}_{n}^j } \right\}$, respectively, $j=1,\ldots ,p$. Since $\left\{ { \pm {\textbf {X}}_{1} ,\ldots , \pm {\textbf {X}}_{n} } \right\} = \left\{ { \pm {\textbf {Y}}_{1}^j ,\ldots , \pm {\textbf {Y}}_{n}^j } \right\}$, it is clear that $F_{{\textbf {X}}_n}^s=F_{{\textbf {Y}}_n^j}^s$, $j=1,\ldots ,p$. This equality along with the affine invariance of depth function concludes that $D\left( {\textbf {X}}_{i},F_{{\textbf {X}}_n}^s\right) =D\left( {\textbf {Y}}_{i}^j,F_{{\textbf {Y}}_n^j}^s\right)$, for all $i=1,\ldots ,n$, $j=1,\ldots ,p$. This shows that $R_{i}$ is a function of ${\textbf {Y}}_{1}^j,\ldots ,{\textbf {Y}}_{n}^j$ and thus is independent of $\delta _{i}^j$, $i=1,2,\ldots ,n$, $j=1,\ldots ,p$. Define ${\textbf {T}}_i=\phi ({i}/(n+1) )\varvec{\delta }_i$. Then we can write that

$$\begin{aligned} \bar{\varvec{V}}_n =\dfrac{1}{n} \sum \limits _{i = 1}^n {{\textbf {V}}_i}=\dfrac{1}{n} \sum \limits _{i = 1}^n \phi \left( {\frac{{ R_i }}{{n + 1}}}\right) \varvec{\delta }_i \,\mathop = \limits ^d \dfrac{1}{n} \sum \limits _{i = 1}^n \phi \left( {\frac{{ i}}{{n + 1}}} \right) \varvec{\delta }_i =\dfrac{1}{n} \sum \limits _{i = 1}^n {{\textbf {T}}_i}= \bar{\varvec{T}}_n . \end{aligned}$$

(4)

where $\mathop = \limits ^d$means equal in distribution. Moreover, for any $j=1,\ldots ,p$, $\delta _{1}^j,\ldots ,\delta _{n}^j$ are independent and identically distributed random variables that take the values 1 and $-1$ with probability 1/2. Then, it is clear that $E({\textbf {T}}_i)=0$. Define

$$\begin{aligned} \varvec{\varphi }_i =cov({\textbf {T}}_i) = \phi ^2 \left( \frac{{i }}{{ {n + 1} }}\right) \varvec{\Omega }, \quad \quad \varvec{\varphi }= \mathop {\lim }\limits _{n \rightarrow \infty } \frac{1}{n}\sum \limits _{i = 1}^n {\varvec{\varphi }_i } = c \varvec{\Omega }\end{aligned}$$

where $\varvec{\Omega }$ is the covariance matrix of the vector $\varvec{\delta }_i$ and c is a constant that does not depend on n. The aim is to establish a multivariate Lindeberg-Feller central limit theorem for the $\bar{\varvec{T}}_n$. For any $\varepsilon >0$ and sufficiently large n, $\phi \left( {\frac{i}{{n + 1}}} \right) \le \frac{{\varepsilon \sqrt{n} }}{{\sqrt{p} }}$ then $\mathop {\lim }\limits _{n \rightarrow \infty } \frac{{\phi \left( {\frac{i}{{n + 1}}} \right) \sqrt{p} }}{{\varepsilon \sqrt{n} }} = 0$ and

$$\begin{aligned} \mathop {\lim }\limits _{n \rightarrow \infty } E\left( {\left\| \phi \left( {\frac{{i }}{{n + 1}}}\right) \varvec{\delta }_i \right\| ^2 1_{\left\{ {\left\| \phi \left( {\frac{{i }}{{n + 1}}}\right) \varvec{\delta }_i \right\| \ge \varepsilon \sqrt{n} } \right\} } } \right) = \mathop {\lim }\limits _{n \rightarrow \infty } p \phi ^2\left( {\frac{{i }}{{n + 1}}}\right) 1_{\left\{ {\phi \left( {\frac{i}{{n + 1}}} \right) \ge \frac{{\varepsilon \sqrt{n} }}{{\sqrt{p} }}} \right\} } = 0 \end{aligned}$$

Consequently, Lindeberg’s condition holds. Now from (4), we conclude that the r.v. $\sqrt{n}\bar{\varvec{V}}_n$ converges to the p-variate normal distribution with mean zero and dispersion matrix $\varvec{\varphi }$. Moreover, we have

$$\begin{aligned} \frac{1}{n}\sum \limits _{i = 1}^n {{\textbf {V}}_i {\textbf {V}}'_i } - \varvec{\varphi } = o_p (1) \end{aligned}$$

(5)

Finally, by applying Slutsky’s theorem it is clear that under the null hypothesis $W_{n,D} {\mathop {\sim }\limits ^{d}} \chi _p^2$. $\square$

Proof of Theorem 2

Let ${\textbf {Z}}_i=c{\textbf {A}}{} {\textbf {X}}_i$, $i=1,\ldots ,n$, and $R_i^*$, $\varvec{\delta }_i^*$ and ${\textbf {V}}^*_i$ are specified in the same way as $R_i$, $\varvec{\delta }_i$ and ${\textbf {V}}_i$ through ${\textbf {Z}}_i$’s instead of ${\textbf {X}}_i$’s. For $i=1,\ldots ,n$, we have

$$\begin{aligned} R_i^*=R_i, \quad \quad and \quad \quad \varvec{\delta }_i^*={\textbf {A}}\varvec{\delta }_i. \end{aligned}$$

Thus

$$\begin{aligned} {\textbf {V}}^*_i={\textbf {A}}{} {\textbf {V}}_i. \end{aligned}$$

Therefore

$$\begin{aligned} W_{n,D} \left( {{\textbf {Z}}_1 ,\ldots ,{\textbf {Z}}_n } \right)&= n\bar{\varvec{V}}^{\prime }_n {\textbf {A}}^{\prime }\left( {\frac{1}{n}\sum \limits _{i = 1}^n {\textbf {A}}{{\textbf {V}}_i {\textbf {V}}^\prime _i }{} {\textbf {A}}^\prime } \right) ^{ - 1} {\textbf {A}}\bar{\varvec{V}}_n \\&= W_{n,D} \left( {{\textbf {X}}_1 ,\ldots ,{\textbf {X}}_n } \right) . \end{aligned}$$

$\square$

Proof of Proposition 1

It is clear that

$$\begin{aligned} I_{\left| {X_j } \right|< \left| {X_i } \right| }&= I_{X_i> X_j } I_{X_i> - X_j } + I_{X_i< X_j } I_{X_i < - X_j } \\&= 2I_{X_i> X_j } I_{X_i> - X_j } - I_{X_i> X_j } - I_{X_i> - X_j } + 1\\&=1-\left| {I_{X_i> X_j } - I_{X_i > - X_j } } \right| . \end{aligned}$$

Therefore

$$\begin{aligned} r_i = \sum \limits _{j = 1}^n {I_{\left| {X_j } \right| < \left| {X_i } \right| } } + 1 = n + 1 - \sum \limits _{j = 1}^n {\left| {I_{X_i> X_j } - I_{X_i > - X_j } } \right| }. \end{aligned}$$

Define

$$\begin{aligned} S_{i}&=\sum \limits _{j = 1}^n {\left( {I_{\left( {X_j \le X_i} \right) } + I_{\left( { - X_j \le X_i} \right) } } \right) },\\ L_{i}&=\sum \limits _{j = 1}^n {\left( {I_{\left( {X_j \ge X_i} \right) } + I_{\left( { - X_j \ge X_i} \right) } } \right) }. \end{aligned}$$

For $p=1$, $HD({\textbf {x}},F) = \inf \left\{ {F\left( x \right) ,1 - F\left( x^- \right) } \right\}$. Thus

$$\begin{aligned} HD\left( {X_i ,F_n^s } \right)&= \frac{1}{{2n}}\min \left\{ {S_i ,L_i } \right\} \\&= \frac{1}{{4n}}\left\{ {2\left( {n + 1} \right) - \left| {S_i - L_i } \right| } \right\} \\&= \frac{1}{{2n}}\left\{ {\left( {n + 1} \right) - \left| {S_i - \left( {n + 1} \right) } \right| } \right\} \end{aligned}$$

Analoguos to $r_i$, we can write $R_i$, $i=1,\ldots ,n$ defined in (1) as

$$\begin{aligned} R_i&= \sum \limits _{j= 1}^n {I_{\left\{ {HD\left( {X_j ,F_n^s } \right)> HD\left( {X_i ,F_n^s } \right) } \right\} } } + 1 = \sum \limits _{j = 1}^n {I_{\left\{ {\left| {S_{j} - \left( {n + 1} \right) } \right| < \left| {S_{i} - \left( {n + 1} \right) } \right| } \right\} } } + 1 \\&= n + 1 - \sum \limits _{j = 1}^n {\left| {I_{S_{i}> S_{j} } - I_{S_{i}> \left\{ {2\left( {n + 1} \right) - S_{j} } \right\} } } \right| } = n + 1 - \sum \limits _{j = 1}^n {\left| {I_{S_{i}> S_{j} } - I_{S_{i} > L_{j} } } \right| } \end{aligned}$$

Now, for arbitrary $i, j=1,\ldots ,n$ assume that ${\left| {I_{X_i> X_j } - I_{X_i > - X_j } } \right| }=1$. Then two cases occurs. Either $X_i > X_j$ and $X_i \le - X_j$ or $X_i \le X_j$ and $X_i > - X_j$. First if $X_i > X_j$ and $X_i \le - X_j$ then

$$\begin{aligned} I_{\left\{ {S_{i}> S_{j} } \right\} } = 1, \quad I_{\left\{ {S_{i} > L_{j} } \right\} } = 0 \end{aligned}$$

and for $X_i \le X_j$ and $X_i > - X_j$ then

$$\begin{aligned} I_{\left\{ {S_{i}> S_{j} } \right\} } = 0, \quad I_{\left\{ {S_{i} > L_{j} } \right\} } = 1. \end{aligned}$$

Then, in both cases

$$\begin{aligned} \left| {I_{S_{i}> S_{j} } - I_{S_{i} > L_{j} } } \right| = 1 \end{aligned}$$

Similarly, for ${\left| {I_{X_i> X_j } - I_{X_i > - X_j } } \right| }=0$, we have

$$\begin{aligned} \left| {I_{S_{i}> S_{j} } - I_{S_{i} > L_{j} } } \right| = 0. \end{aligned}$$

By a similar argument, the above relations can be concluded in the opposite direction. Therefore, it is shown that $r_i=R_i$, $i=1,\ldots ,n$. $\square$