Testing the equality of matrix distributions

Abstract

While matrices are usually used as the basic data structure for experiments with repeated measurements or longitudinal data, testing methods for the equality of two matrix distributions have not been fully discussed in the literature. In this article, we propose three methods to test the equality of two matrix distributions: the likelihood ratio test, the Frobenius norm methods and triangle tests. We present a simulation to compare their performance under the matrix normal distribution. We apply the testing methods to compare the US economy, as measured by closing prices of five market indices, before and after the US stock market crash of 2008.

Appendix

Proof of Lemma 1

The distance between $\mathbf X_1$ and $\mathbf X_2$ is

$$\begin{aligned} ||X_1-X_2||_{Fr}= & {} ||vec(X_1)-vec(X_2)||_{Eu}\\= & {} [{(vec(X_1)-vec(X_2))'(vec(X_1)-vec(X_2))}]^{1/2}. \end{aligned}$$

By definition of $\mathcal {MN}$ distribution, we have $vec(X_1),vec(X_2)\sim N_{dm}(vec(M_1), { \varSigma }_1\otimes \varPsi _1).$ Then we have $vec(X_1)-vec(X_2)\sim N_{dm}(0, 2{ \varSigma }_1 \otimes \varPsi _1 )$. Suppose $\{\lambda _i\},i=1,\ldots ,d$ are the eigenvalues for ${ \varSigma }_1$, and $\{\theta _j\},j=1,\ldots ,m$ are the eigenvalues for $\varPsi _1$, one can show that $\{2\lambda _i\theta _j\},i=1,\ldots ,d,j=1,\ldots ,m$ are the eigenvalues for $2{ \varSigma }_1\otimes \varPsi _1$. Therefore, we have

$$\begin{aligned} ||X_1-X_2||_{Fr}^2 = (vec(X_1) -vec(X_2))'(vec(X_1)-vec(X_2))=2\sum _{i=1}^d\sum _{j=1}^m \lambda _i\theta _j\chi ^2_{(1)}. \end{aligned}$$

Thus $E||X_1-X_2||_{Fr}^2 =2\sum _{i=1}^d\sum _{j=1}^m \lambda _i\theta _j =2tr({ \varSigma }_1 \otimes \varPsi _1)$. It is not difficult to show that $\text {var(}(||X_1-X_2||_{Fr}^2)=8 tr({ \varSigma }_1 \otimes \varPsi _1)^2$. Note that if $\varPsi _1=I_m$, i. e. the columns of the observation matrix are i.i.d., then the expectation reduces to $2m\sum _{i=1}^d \lambda _i$.

Similarly, $E||Y_1-Y_2||_{Fr}^2 =2tr({ \varSigma }_2 \otimes \varPsi _2)$. Consider $\mathbf X_1$ and $Y_1$, we have $vec(X_1)-vec(Y_1)\sim N_{dm}(vec(M_1-M_2), { \varSigma }_1 \otimes \varPsi _1+{ \varSigma }_2 \otimes \varPsi _2 )$. One can show that

$$\begin{aligned} E||X_1-Y_1||_{Fr}^2= & {} E||vec(X_1)-vec(Y_1)||_{Eu}^2\\= & {} tr[{ \varSigma }_1 \otimes \varPsi _1+{ \varSigma }_2 \otimes \varPsi _2]+||M_1-M_2||_{Fr}^2. \end{aligned}$$

Therefore, we have $2E||X_1-Y_1||_{Fr}^2-E||X_1-X_2||_{Fr}^2-E||Y_1-Y_2||_{Fr}^2 =||M_1-M_2||^2_{Fr}\ge 0$. The equality holds if and only if $||M_1-M_2||^2_{Fr}=0$. It follows that $M_1=M_2$. $\square$

Proof of Theorem 1

Let $vec(\cdot )$ be an operator that maps a matrix to a vector by stacking the columns of the matrix on top of one another. Let ${\varvec{u}}_1=vec(X_1)$, ${\varvec{u}}_2=vec(X_2)$, ${\varvec{v}}_1=vec(Y_1)$ and ${\varvec{v}}_2=vec(Y_2)$. Based on Proposition 1, one can show that

$$\begin{aligned} ||{\varvec{u}}_1-{\varvec{u}}_2||_{Eu} = \gamma _{dm} \int _{S^{dm-1}}|a'({\varvec{u}}_1-{\varvec{u}}_2)| d\mu (a), \end{aligned}$$

where $\mu$ is the uniform distribution on $S^{dm-1}=\{{\varvec{x}}\in \mathfrak {R}^{dm}{:} ||{\varvec{x}}||_{Eu}=1\}$, the surface of the unit sphere in $\mathfrak {R}^{dm}$ and $\gamma _{dm}$ in Proposition 1. Similarly, we have

$$\begin{aligned} ||{\varvec{v}}_1-{\varvec{v}}_2||_{Eu}&= \gamma _{dm} \int _{S^{dm-1}}|a'({\varvec{v}}_1-{\varvec{v}}_2)| d\mu (a),\\ ||{\varvec{u}}_1-{\varvec{v}}_1||_{Eu}&= \gamma _{dm} \int _{S^{dm-1}}|a'({\varvec{u}}_1-{\varvec{v}}_1)| d\mu (a). \end{aligned}$$

Following Baringhaus and Franz (2004), one can show that for each $a\in S^{dm-1}$,

$$\begin{aligned} 2E|a'({\varvec{u}}_1-{\varvec{v}}_1)|-E|a'({\varvec{u}}_1-{\varvec{v}}_2)|-E|a'({\varvec{v}}_1-{\varvec{v}}_2)|\ge 0. \end{aligned}$$

(24)

The equality holds if and only if the distribution of $a'{\varvec{u}}_1$ and $a'{\varvec{v}}_1$ coincide. Integrating with respect to $\mu$ on both sides of inequality (24), we have

$$\begin{aligned} 2E||{\varvec{u}}_1-{\varvec{v}}_1||_{Eu}-E||{\varvec{u}}_1-{\varvec{u}}_2||_{Eu}-E||{\varvec{v}}_1-{\varvec{v}}_2||_{Eu}\ge 0. \end{aligned}$$

(25)

The equality holds if and only if for almost all $a\in S^{dm-1}$, the distribution of $a'{\varvec{u}}_1$ and $a'{\varvec{v}}_1$ coincide. Since for each $t\in \mathfrak {R}$, the function $E(e^{ita' u_1})$ and $E(e^{ita' v_1})$ are continuous, the equality in (25) holds if and only if ${\varvec{u}}_1$ and ${\varvec{v}}_1$ have same Fourier transform, which implies $F=G$. By the definition of the Frobenius and Euclidean norms, we have $\mu _{FF}=E||X_1-X_2||_{Fr}=E||{\varvec{u}}_1-{\varvec{u}}_2||_{Eu}$, $\mu _{GG}=E||Y_1-Y_2||_{Fr}=E||{\varvec{v}}_1-{\varvec{v}}_2||_{Eu}$, $\mu _{FG}=E||X_1-Y_1||_{Fr}=E||{\varvec{u}}_1-{\varvec{v}}_1||_{Eu}$. Therefore, we have $2\mu _{FG}-\mu _{FF}-\mu _{GG}\ge 0$, and the equality holds if and only if $F=G$. $\square$

Proof of Theorem 2

Let $h(X_i,X_j;Y_p,Y_q)$ be a real-valued function such that

$$\begin{aligned} h(X_i,X_j;Y_p,Y_q)=||X_i-Y_p||_{Fr}+||X_j-Y_q||_{Fr}+||X_i-X_j||_{Fr}+||Y_p-Y_q||_{Fr}. \end{aligned}$$

(26)

It is clear that h is symmetric within each argument $(X_i,X_j)$ and $(Y_p,Y_q)$. One can show that EN is the U-statistic corresponding to the kernel function h. That is,

$$\begin{aligned} EN= {n_1\atopwithdelims ()2}^{-1}{n_2\atopwithdelims ()2}^{-1}\sum _{i=1}^{n_1}\sum _{i=1}^{n_1-1}\sum _{j=i+1}^{n_1}\sum _{p=1}^{n_2-1}\sum _{q=p+1}^{n_2}h(X_i,X_j;Y_p,Y_q). \end{aligned}$$

(27)

If the distribution F and G are identical, by Theorem 1, we have $E(h(X_1,X_2;Y_1,$$Y_2)) =0$. In addition, since $E[X_1,X_2;Y_1,Y_2|X_1={\mathbb {X}}_1,Y_1={\mathbb {Y}}_1]=0$ for almost all matrix realization $({\mathbb {X}}_1,{\mathbb {Y}}_1)$, EN is a degenerate kernel U-statistic. The asymptotic distribution of EN can be inferred from the work of Hoeffding and Robbins (1948) for the case $k=1$, which shows that $n\cdot EN$ has a non-degenerate limiting distribution $\sum _{i=1}^\infty \lambda _i (Z_i^2-1)$ where the constants $\lambda _i$ depend on F and $Z_i^2$ are independent $\chi ^2_1$ random variables. $\square$

Proof of Theorem 3

If $\mu _{FF}= \mu _{GG}= \mu _{FG}$, we have $2\mu _{FG}-\mu _{FF}-\mu _{GG}= 0$, which implies $F=G$ by Theorem 1. Suppose $F=G$, the distributions of $\mathbf{X}_1, \mathbf{X}_2,\mathbf{Y}_1$ and $\mathbf{Y}_2$ are equal. Therefore, the distributions of $||\mathbf{X}_1-\mathbf{X}_2||_{Fr}, ||\mathbf{Y}_1-\mathbf{Y}_2||_{Fr}$ and $||\mathbf{X}_1-\mathbf{Y}_1||_{Fr}$ are also equal, implying the fact that $\mu _{FF}= \mu _{GG}= \mu _{FG}$. $\square$

Proof of Theorem 4

Following Biswas and Ghosh (2014) for vector distributions, we express $n\cdot BG(\mathscr {A},\mathscr {B})$ as

$$\begin{aligned} n\cdot BG(\mathscr {A},\mathscr {B})=\frac{1}{2}([\sqrt{n}({\hat{\mu }}_{FF} -{\hat{\mu }}_{GG} )]^2+[\sqrt{n}\cdot EN(\mathscr {A},\mathscr {B})]^2), \end{aligned}$$

where ${\hat{\mu }}_{FF}$ and ${\hat{\mu }}_{GG}$ are given in Eqs. (8)–(10). From Theorem 2, we have ${n}\cdot EN(\mathscr {A},\mathscr {B})=O_p(1)$, and hence $\sqrt{n}\cdot EN(\mathscr {A},\mathscr {B}) \overset{p}{\rightarrow }0$, as $n\rightarrow \infty$.

Let $\mu _{FF}=E||\mathbf{X}_1-\mathbf{X}_2 ||_{Fr}$ and $\mu _{GG}=E||\mathbf{Y}_{1}-\mathbf{Y}_{2} ||_{Fr}$. Under null hypothesis, we have $\mu _{FF}=\mu _{GG}$, and hence $\sqrt{n}({\hat{\mu }}_{FF} -{\hat{\mu }}_{GG} )=\sqrt{n}[({\hat{\mu }}_{FF}-\mu _{FF})-({\hat{\mu }}_{GG} -\mu _{GG})].$ Note that

$$\begin{aligned} {\hat{\mu }}_{FF} -\mu _{FF}={n_1\atopwithdelims ()2}^{-1}\sum _{i=1}^{n_1-1}\sum _{j=i+1}^{n_1}(||\mathbf{X}_i-\mathbf{X}_j||_{Fr}-\mu _{FF}) \end{aligned}$$

is a U-statistic with symmetric kernel function $h(\mathbf{X}_i,\mathbf{X}_j)=||\mathbf{X}_i-\mathbf{X}_j||_{Fr}-\mu _{FF}$. Therefore, we have $R_1=\sqrt{n_1}({\hat{\mu }}_{FF} -\mu _{FF})\overset{d}{\rightarrow }N(0,4{\sigma }_0)$, where ${\sigma }_0=Var[E(||\mathbf{X}_1-\mathbf{X}_2||_{Fr}|\mathbf{X}_1)]$. Similarly, we have $R_2=\sqrt{n_2}({\hat{\mu }}_{GG} -\mu _{GG})\overset{d}{\rightarrow }N(0,4{\sigma }_0)$. Since ${\hat{\mu }}_{FF}$ and ${\hat{\mu }}_{GG}$ are independent, one can show

$$\begin{aligned} \sqrt{n}({\hat{\mu }}_{FF} -{\hat{\mu }}_{GG} )=&\sqrt{n/n_1}R_1-\sqrt{n/n_2}R_2\overset{d}{\rightarrow }N\left( 0,\left( \frac{1}{\lambda }+\frac{1}{1-\lambda }\right) 4{\sigma }_0\right) . \end{aligned}$$

Therefore, as $min(n_1,n_2)\rightarrow \infty$, we obtain

$$\begin{aligned} n\cdot BG(\mathscr {A},\mathscr {B}) =\frac{1}{2}([\sqrt{n}({\hat{\mu }}_{FF} -{\hat{\mu }}_{GG} )]^2+[\sqrt{n}\cdot EN(\mathscr {A},\mathscr {B})]^2) \overset{d}{\rightarrow }\frac{ 2{\sigma }_0}{\lambda (1-\lambda )}\chi _1^2. \end{aligned}$$

$\square$