Hoeffding-Blum-Kiefer-Rosenblatt independence test statistic on partly not identically distributed data

Abstract

The established Hoeffding-Blum-Kiefer-Rosenblatt independence test statistic is investigated for partly not identically distributed data. Surprisingly, it turns out that the statistic has the well-known distribution-free limiting null distribution of the classical criterion under standard regularity conditions. An application is testing goodness-of-fit for the regression function in a non parametric random effects meta-regression model, where the consistency is obtained as well. Simulations investigate size and power of the approach for small and moderate sample sizes. A real data example based on clinical trials illustrates how the test can be used in applications.

Keywords:

• Brownian pillow
• Hoeffding-Blum-Kiefer-Rosenblatt independence test
• not identically distributed
• random effects meta-regression model

2000 MSC:

• 62G10
• 62G08

Acknowledgments

The author wishes to thank the two referees for very helpful comments and suggestions.

Appendix A: proofs

Proof of Proposition 3.1.

First, let us consider the stochastic process $V_n=(V_n(x,y),(x,y)\in {\overline{\mathbb{R}}}^2),$ defined by $$V_n(x,y)=\sqrt{n}({\widehat{F}}_{(X,Y),n}(x,y)-F_{(X,Y),n}(x,y)), (x,y)\in {\overline{\mathbb{R}}}^2,$$ where $$F_{(X,Y),n}(x,y)=\sum _{i=1}^k\frac{n_i}{n}{F}_{(X,Y_i)}(x,y), (x,y)\in {\overline{\mathbb{R}}}^2.$$

The crucial point is that under the null hypothesis of independence $\mathcal{H}$ $$\begin{array}{c}{F}_{(X,Y_i)}(x,y)=F_X(x)F_{Y_i}(r), (x,y)\in {\overline{\mathbb{R}}}^2, i=1,\dots ,k,\hfill \end{array}$$ and so $$F_{(X,Y),n}(x,r)=F_X(x)F_{Y,n}(y), (x,y)\in {\overline{\mathbb{R}}}^2,$$ in the null hypothesis case, where $$F_{Y,n}(y)=\sum _{i=1}^k\frac{n_i}{n}{F}_{Y_i}(y), y\in \overline{\mathbb{R}}.$$

Under the null hypothesis $\mathcal{H},$ it follows that the stochastic process U_n can be rewritten as $U_n=U_{1,n}+U_{2,n},$ where the stochastic process $U_{1,n}=(U_{1,n}(x,y),(x,y)\in {\overline{\mathbb{R}}}^2)$ is given by $$U_{1,n}(x,y)=V_n(x,y)-F_X(x)V_n(\infty ,y)-F_{Y,n}(y)V_n(x,\infty ), (x,y)\in {\overline{\mathbb{R}}}^2,$$ and the stochastic process $U_{2,n}=(U_{2,n}(x,y),(x,y)\in {\overline{\mathbb{R}}}^2)$ is defined by $$U_{2,n}(x,y)=-\frac{1}{\sqrt{n}}{V}_n(x,\infty )V_n(\infty ,y), (x,y)\in {\overline{\mathbb{R}}}^2.$$

The process V_n is an empirical process based on not necessarily identically distributed random vectors. Limiting results are available by Ziegler (1997). In our situation, the conditions in 4.2 in Ziegler (1997) can be easily verified and the convergence in distribution $$V_n⟿V$$ follows, where $V=(V(x,y),(x,y)\in {\overline{\mathbb{R}}}^2)$ is a centered Gaussian process with a.s. uniformly d₂-continuous sample paths and covariance function $$\begin{array}{c}v(x,y,r,s)=F_{(X,Y)}(x\wedge r,y\wedge s)-F_{(X,Y,X,Y)}(x,y,r,s), (x,y),(r,s)\in {\overline{\mathbb{R}}}^2,\hfill \end{array}$$ and $(X,Y,X,Y)$ denotes a random vector with distribution function $$F_{(X,Y,X,Y)}(x,y,r,s)=\sum _i{\rho }_i{F}_{(X,Y_i)}(x,y)F_{(X,Y_i)}(r,s), (x,y,r,s)\in {\overline{\mathbb{R}}}^4.$$

Note that in general $F_{(X,Y,X,Y)}(x,y,r,s)=F_{(X,Y)}(x,y)F_{(X,Y)}(r,s),(x,y,r,s)\in {\overline{\mathbb{R}}}^4,$ does not hold, and the same applies to $v(x,y,r,s)=F_{(X,Y)}(x\wedge r,y\wedge s)-F_{(X,Y)}(x,y)F_{(X,Y)}(r,s),(x,y,r,s)\in {\overline{\mathbb{R}}}^4.$ For that reason, in contrast to the classical situation of identically distributed data, the process V has in general not the structure of a bivariate Brownian bridge.

It is easily seen that $F_{Y,n}$ converges uniformly to F_Y. Applying Slutsky’s theorem, that is Example 1.4.7 in van der Vaart and Wellner (1996), yields $$(V_n,F_{Y,n})⟿(V,F_Y),$$ and from the continuous mapping theorem, that is Theorem 1.3.6 in van der Vaart and Wellner (1996), it follows that $$U_{1,n}⟿U.$$

Moreover, it follows from Lemma 1.10.2. (iii) in van der Vaart and Wellner (1996) that $$-\frac{1}{\sqrt{n}}{V}_n\to 0 \text{in}\mathrm{ }\text{probability}$$ in the uniformly sense. The latter combined with Slutsky’s theorem yields $$(V_n,-\frac{1}{\sqrt{n}}{V}_n)⟿(V,0),$$ and it follows from the continuous mapping theorem that $$U_{2,n}\to 0 \text{in}\mathrm{ }\text{probability},$$ where the convergence holds uniformly. From a further application of Slutsky’s theorem, we deduce $$(U_{1,n},U_{2,n})⟿(U,0),$$ and finally $$U_n⟿U$$ from the continuous mapping theorem again. It is clear that $U$ is a centered Gaussian process with a.s. uniformly d₂-continuous sample paths. The covariance function u turns out by simple calculation.

Note that, form the latter one it follows that the stochastic process U has the structure of the well-known Brownian pillow analogous to the classical situation of identically distributed data. Regarding that, in contrast to the classical situation of identically distributed data, V has in general not the structure of a multivariate Brownian bridge, this finding surprises. It is a special feature of the process U. □

Proof of Lemma 3.1.

The function ${\widehat{F}}_{(X,Y),n}$ is an empirical distribution function based on not necessarily identically distributed random vectors. Limiting results are available by Gänßler and Ziegler (1994). Corollary 4.1 (i) in Gänßler and Ziegler (1994) implies $${\widehat{F}}_{(X,Y),n}-F_{(X,Y),n}\to 0 \text{in}\mathrm{ }\text{probability}$$ where the convergence holds uniformly. It is easily seen that $F_{(X,Y),n}$ converges uniformly to $F_{(X,Y)}.$ In all, we obtain $$\begin{array}{c}\underset{(x,y)\in {\overline{\mathbb{R}}}^2}{\sup }|{\widehat{F}}_{(X,Y),n}(x,y)-F_{(X,Y)}(x,y)|\le \underset{(x,y)\in {\overline{\mathbb{R}}}^2}{\sup }|{\widehat{F}}_{(X,Y),n}(x,y)-F_{(X,Y),n}(x,y)|\hfill \\ +\underset{(x,y)\in {\overline{\mathbb{R}}}^2}{\sup }|F_{(X,Y),n}(x,y)-F_{(X,Y)}(x,y)|\to 0 \text{in}\mathrm{ }\text{probability}.\hfill \end{array}$$

This implies the statement. □

Proof of Theorem 3.1.

Because the null hypothesis $\mathcal{H}$ is true, the convergence in distribution of ${\text{HBKR}}_n$ to $\text{HBKR}=\int \int U{(x,y)}^2\mathrm{d}{F}_X(x)\mathrm{d}{F}_Y(y)$ follows immediately from Proposition 3.1 and Lemma 3.1 combined with the continuous mapping theorem. At first, we consider the special case that k = 1 is fixed. In this situation, ${\text{HBKR}}_n$ based on independent and identically distributed pairs of bivariate random vectors with underlying distribution function $F_{(X,Y)},$ where the only restriction on $F_{(X,Y)}$ is that the related marginal distributions are continuous. Under the stated null hypothesis of independence, ${\text{HBKR}}_n$ has the distribution-free distribution of the classical Hoeffding-Blum-Kiefer-Rosenblatt test statistic based on independent and identically distributed bivariate random vectors with continuous marginal distributions. On the one hand, it follows from the results in Blum, Kiefer, and Rosenblatt (1961) that ${\text{HBKR}}_n$ converges in distribution to a real-valued random variable $\widetilde{\text{HBKR}},$ say, where $\widetilde{\text{HBKR}}$ is distribution-free, has a continuous and strictly increasing distribution function, and the characteristic function $${\phi }_{\widetilde{\text{HBKR}}}(t)=\prod _{j,\mathcal{\ell }=1}^{\infty }(1-\frac{2it}{{\pi }^4{j}^2{\mathcal{\ell }}^2}{)}^{-\frac{1}{2}}, t\in \mathbb{R}.$$

On the other hand, it is already shown that ${\text{HBKR}}_n$ converges in distribution to $\text{HBKR}.$ In all, $\widetilde{\text{HBKR}}$ and $\text{HBKR}$ have the same distribution. Because $F_{(X,Y)}$ is an arbitrary uniformly continuous distribution function with continuous marginal distributions, the latter finding is also valid for general k. □

Proof of Lemma 6.1.

In the null hypothesis case, it is f = g. This implies $Y=\eta +\epsilon .$ Because X, η, and ε are independent, it follows that X and Y are independent. Now, we consider alternatives. Let $f\ne g$ be true. Well, suppose that X and Y are independent for a moment. It is well-known that the distribution of a random vector W with values in ${\mathbb{R}}^d,d\in \mathbb{N},$ is uniquely determined by its characteristic function ${\phi }_W(t)=E({\mathrm{e}}^{\mathrm{i}{t}^{\top }W}),t\in {\mathbb{R}}^d.$ The independence assumption implies that for all $t=(t_1,t_2)\in {\mathbb{R}}^2$ $${\phi }_{(X,Y)}(t)={\phi }_X(t_1){\phi }_Y(t_2).$$

Because X, η, and ε are independent, the latter is equivalent to $${\phi }_{(X,f(X)-g(X))}(t)={\phi }_X(t_1){\phi }_{f(X)-g(X)}(t_2)$$ for all $t=(t_1,t_2)\in {\mathbb{R}}^2.$ We obtain the independence of X and $f(X)-g(X).$ Because I is the support of the distribution of X, it follows that the map $x\mapsto f(x)-g(x),x\in I,$ is constant. From $f(x_0)=0=g(x_0),$ we deduce f = g, that contradicts the assumption $f\ne g$ and completes the proof. □

Proof of Theorem 6.1.

We define the stochastic process $W_n=(W_n(x,y),(x,y)\in {\overline{\mathbb{R}}}^2)$ by $$W_n(x,r)={\widehat{F}}_{(X,Y),n}(x,y)-{\widehat{F}}_{X,n}(x){\widehat{F}}_{Y,n}(y), (x,y)\in {\overline{\mathbb{R}}}^2,$$ and the map $$W(x,y)=F_{(X,Y)}(x,y)-F_X(x)F_Y(y), (x,y)\in {\overline{\mathbb{R}}}^2.$$

It follows from Lemma 3.1 that $$\begin{array}{c}\underset{(x,y)\in {\overline{\mathbb{R}}}^2}{\sup }|W_n(x,y)-W(x,y)|\le 3\underset{(x,y)\in {\overline{\mathbb{R}}}^2}{\sup }|{\widehat{F}}_{(X,Y),n}(x,y)-F_{(X,Y)}(x,y)|\to 0\hfill \end{array}$$ in probability. The latter combined with Lemma 3.1 again yields $$(W_n,{\widehat{F}}_{(X,Y),n})\to (W,F_{(X,Y)}) \text{in}\mathrm{ }\text{probability}$$ in the uniformly sense. Regarding the expressions $$\begin{array}{c}\frac{1}{n}{\text{HBKR}}_n=\int W_n{(x,y)}^2\mathrm{d}{\widehat{F}}_{(X,Y),n}(x,y) \text{and} \Delta =\int W{(x,y)}^2\mathrm{d}{F}_{(X,Y)}(x,y)\hfill \end{array}$$ the continuous mapping theorem implies the convergence of ${\text{HBKR}}_n/n$ to Δ in probability. It remains to show that $\Delta >0.$ Denote by ε a real-valued random variable with distribution function $$F_{\epsilon }(e)=\sum _i{\rho }_i{F}_{{\epsilon }_i}(e), e\in \overline{\mathbb{R}},$$ such that X, η, and ε are independent. Setting $$Y=f(X)-g(X)+\eta +\epsilon ,$$ we have $$\begin{array}{c}{F}_{(X,Y)}(x,y)=P(X\le x,Y\le r), (x,y)\in {\overline{\mathbb{R}}}^2,\hfill \end{array}$$ as well as $$\begin{array}{c}{F}_Y(y)=P(Y\le y), y\in \overline{\mathbb{R}},\hfill \end{array}$$ and under the alternative it follows from Lemma 6.1 that X and Y are not independent. In addition, it is easy to see that $F_{(X,Y)}$ is absolutely continuous. In all, $\Delta >0$ follows from Yanagimoto (1970). □