Some hypothesis tests based on random projection

Abstract

Two new non-parametric tests are proposed based on continuous one-dimensional random projections. The first one addresses central symmetry and the second addresses independence. These tests are implemented for finite and infinite dimensional (functional) data sets. Both tests are distribution-free and universally consistent. Additionally, different techniques are proposed to improve the power of the tests. Promising results have been obtained by comparing the new tests with existing ones using simulation study. Real data in Banach spaces have been used to develop an application.

Appendix.

Proof of Theorem 3

Let M and Q be the probability measures induced by the random elements $\mathbf{X }$ and $-\mathbf{X }$. Then

$$\begin{aligned} \mathcal {E}(M,Q)= & {} \{ h \in E^* / Q_{h}=M_{h} \} \\= & {} \{ h \in E^* / f(\mathbf{X }) \text {and} -f(\mathbf{X }) \text {have the same distribution} \}\\= & {} \mathcal {E}(\mathbf{X }). \end{aligned}$$

From Theorem 2, because $\mathcal {E}(M,Q)$ has a positive $\mu$-measure, it can be concluded that $Q=M$, namely, $\mathbf{X }$ and $-\mathbf{X }$ have the same distribution, so $\mathbf{X }$ is a centrally symmetric random element.

Proof of Theorem 4

Given two probability measures M and Q on $\textit{E} \times \textit{E}$ over the determinant class of the half-spaces $A=\{ (x,y) \in E \times E / h(x,y) \le t, h \in (\textit{E} \times \textit{E})^* \}$,

$$\begin{aligned} \begin{array}{l} M(A) = \int \mathbf 1 _{ \{ h(x,y) \le t \}} dP_{\mathbf{X }}(x) dP_{\mathbf{Y }}(y), \quad \forall h \in (\textit{E} \times \textit{E})^*, \quad \forall t \in \mathbb {R}, \\ Q(A) = \int \mathbf 1 _{ \{ h(x,y) \le t \}} dP_{(\mathbf{X },\mathbf{Y })}(x,y), \quad \forall h \in (\textit{E} \times \textit{E})^*, \quad \forall t \in \mathbb {R}. \\ \end{array} \end{aligned}$$

(23)

The maximum is used as the norm on $\textit{E}\times \textit{E}$, namely

$$\begin{aligned} \Vert (x,y) \Vert = \max \{ \Vert x \Vert _{E} ,\Vert y \Vert _{\textit{E}} \}, \end{aligned}$$

(24)

therefore,

$$\begin{aligned} m_M(n)= & {} \int \Vert (x,y) \Vert ^n dP_{\mathbf{X }}(x) dP_{\mathbf{Y }}(y) \\\le & {} \left[ \underbrace{\int \Vert x \Vert ^n dP_{\mathbf{X }}(x)}_{m_{\mathbf{X }}(n)} + \underbrace{\int \Vert y \Vert ^n dP_{\mathbf{Y }}(y)}_{m_{\mathbf{Y }}(n)} \right] \\\le & {} 2 \max \{ m_{\mathbf{X }}(n), m_{\mathbf{Y }}(n)\}. \end{aligned}$$

So,

$$\begin{aligned} m^{-1/n}_M(n)\ge & {} 2^{-1/n} \left[ \max \{ m_{\mathbf{X }}(n), m_{\mathbf{Y }}(n)\} \right] ^{-1/n} \\\ge & {} 2^{-1/n} \left[ \min \{ m^{-1/n}_{\mathbf{X }}(n), m^{-1/n}_{\mathbf{Y }}(n)\} \right] . \end{aligned}$$

Thus, the moments are finite, and Carleman’s condition for the measure M holds.

Let the set $\mathcal {E}(M,Q) = \{ h \in (E \times E)^* / Q_{h}=M_{h} \}$. Let $h \in \mathcal {E}(\mathbf{X },\mathbf{Y })$. Now, setting $f(\mathbf{X })= h(\mathbf{X },0)$ and $g(\mathbf{Y })=h(0,\mathbf{Y })$, one has that $h(\mathbf{X },\mathbf{Y })=f(\mathbf{X })+g(\mathbf{Y })$. Then,

$$\begin{aligned} Q_{h} \left( (-\infty ,t] \right)= & {} P \left( f(\mathbf{X })+g(\mathbf{Y }) \le t \right) \\= & {} \int _{E \times E} \mathbf 1 _{ \{ f(x) + g(y) \le t \}} dP_{(\mathbf{X },\mathbf{Y })}(x,y)\\= & {} \int _{\mathbb {R} \times \mathbb {R} } \mathbf 1 _{ \{ u + v \le t \}} dP_{(f(\mathbf{X }),g(\mathbf{Y }))}(u,v) \\= & {} \int _{\mathbb {R} \times \mathbb {R} } \mathbf 1 _{ \{ u + v \le t \}} dP_{f(\mathbf{X })}(u) dP_{g(\mathbf{Y })}(v)\\= & {} M_{h} \left( (-\infty ,t] \right) . \end{aligned}$$

Then, $h \in \mathcal {E}(M,Q)$, implying that $\mathcal {E}(M,Q)$ has a positive $\mu$-measure. By Theorem 2, it follows that $M=Q$, and thus $\mathbf{X }$ and $\mathbf{Y }$ are independent.$\square$

Proof of Theorem 5

Under $H_0$, since $\mathbf{X }_1$ is symmetric for any $h \in \textit{E}^*$, Theorem 1 implies that $h(\mathbf{X }_1)\in \mathbb {R}$ is also symmetric and therefore it fulfills the condition given in (11) for the null assumption in the one dimensional case. Let $h(\mathbf{Y }_1) \ge h(\mathbf{Y }_2) \ge \ldots \ge h(\mathbf{Y }_n)$ be the order statistics (sorted from the largest to the smallest) of the absolute values $\left| {h(\mathbf{X }_1} \right| , \left| {h(\mathbf{X }_2} \right| , \ldots ,\left| {h(\mathbf{X }_n} \right|$. Let $t^{h}_{n,i}= F_{n}(-h(\mathbf{Y }_{i}))$. Then, $0 \le t^{h}_{n,1} \le t^{h}_{n,2} \le \ldots \le t^{h}_{n,n} \le F(0)= 1/2$. Because F is continuous, ties not occur a.s. and therefore,

$$\begin{aligned} 0< t^{h}_{n,1}< t^{h}_{n,2}< \ldots< t^{h}_{n,n}< F(0)< 1/2 \quad \text {a.s.} \end{aligned}$$

Via the canonical transformation, one may define $V^{h}_n(t)= n^{1/2}[G^{h}_n(t)-t]$ with $0<t<1$, where $G^{h}_n(t)= \frac{1}{n}\sum _{i=1}^{n} \mathbf 1 _{[-\infty , t )} \left( F^{h}(X_i) \right)$, and define

$$\begin{aligned} \tilde{V}^{h}_n(t)= V^{h}_n(t^{-}) + V^{h}_n(1-t) , \quad 0 \le t \le 1/2. \end{aligned}$$

(25)

Then, $\tilde{V}^{h}_n(t)$ is a stochastic process defined on (0, 1 / 2) having n jumps of 1 or $-1$ at the points $t^{h}_{n,1},t^{h}_{n,2}, \ldots , t^{h}_{n,n}$. Let $p_{i,j}= P \left( h(\mathbf{Y }_{n-i+1})= \vert h(\mathbf{X }_j \vert \right)$. Then

$$\begin{aligned} {\begin{matrix} P \left( h(\mathbf{Y }_{n-i+1}) \quad \text { matches with a positive value of} \quad h(\mathbf{X }_j) \right) \\ = \sum _{j=1}^{n} p_{ij} P \left( h(\mathbf{X }_j) >0 \Big / Y ^{h}_{n-i+1}= \vert h(\mathbf{X }_j) \vert \right) = 1/2 \sum _{j=1}^{n} p_{i,j}= 1/2. \end{matrix}} \end{aligned}$$

(26)

Let sg(.) stand for the sign function, and $\vert \cdot \vert$ for the absolute value function. It is well known that the vectors

$$\begin{aligned} \left( sg \left( h(\mathbf{X }_1) \right) , \ldots , sg \left( h(\mathbf{X }_n) \right) \right) \ \text{ and } \left( \left| { h(\mathbf{X }_1)} \right| ,\ldots ,\left| { h(\mathbf{X }_n)} \right| \right) \end{aligned}$$

are independent under the null assumption.

Then, the jumps of $n ^{1/2} \tilde{V}^{h}_n(t)$ at $t^{h}_{n,1},t^{h}_{n,2}, \ldots , t^{h}_{n,n}$ are independent. Therefore, under $H_0$, the distributions of the statistics $D^h_-(n)$ and $D^h_+(n)$ follow the distribution of the maximum of a symmetric random walk of n steps from the origin, and $n D^{h}(n)$ follows the distribution of the maximum of the absolute values of the random walk as can be seen in Takács (1967).$\square$

Proof of Theorem 6

Write P for the distribution of X and Q for the distribution of $-X$. The set $\mathcal {E}(P,Q)$ has H-measure zero in $\mathbb {R}^d$, since if the H-measure were positive, then by Theorem 3 X and $-X$ would have the same distribution, which contradicts the hypothesis.

For each $h \in \mathcal {E}^{c}(P,Q)$, by Theorem 1, $h(\mathbf{X })$ is not symmetric, so if we define

$$\begin{aligned} \delta (F^{h})= \sup _{x \ge 0} \vert F^{h}(x)+ F^{h}(-x) -1 \vert , \end{aligned}$$

it holds that

$$\begin{aligned} \delta (F^{h}) = \left\{ \begin{array}{l@{\quad }l} 0 &{} \text {if } F ^{h}\in \mathcal {F}^{h}_0\\ >0 &{} \text {if } F ^{h}\in \mathcal {F}^{h}_1. \end{array} \right. \end{aligned}$$

(27)

There exists, under $H_1$, $t_h \in \mathbb {R}$ such that

$$\begin{aligned} P \left( x \in \mathbb {R}^d / \langle x, h \rangle \le t_h \right) \ne Q \left( x \in \mathbb {R}^d / \langle x, h \rangle \le t_h \right) , \end{aligned}$$

(28)

namely,

$$\begin{aligned} F ^{h}(t_h)+ F ^{h}(-t_h) - 1 \ne 0. \end{aligned}$$

(29)

By the Glivenko–Cantelli theorem, $\sup _{x } \vert F^{h}_n(x)- F^{h}(x) \vert \mathop {\longrightarrow }\limits ^{a.s}0$, which entails

$$\begin{aligned} D^{h}_n&\ge \vert F_n^{h}(t_h)+ F_n^{h}(-t_h) - 1 \vert \\&\ge \vert F ^{h}(t_h)+ F ^{h}(-t_h) - 1 \vert - \vert F^{h}_n(t_h)- F^{h}(t_h) \vert - \vert F^{h}_n(-t_h)- F^{h}(-t_h)\vert \\&\ge \frac{\delta (F^{h})}{2}, \end{aligned}$$

almost surely when $n \rightarrow +\infty$.