On a transform of the Vincze-statistic and its exact and asymptotic distribution

TL;DR

This paper introduces a new nonparametric test based on a transform of the Vincze-statistic for two-sample problems, providing exact and asymptotic distributions, with potential applications to one-sided and two-sided tests.

Contribution

It develops a novel transform of the Vincze-statistic and derives its exact and asymptotic distributions, extending the applicability to two-sided tests.

Findings

Derived exact distribution of the transform T

Established asymptotic distribution of T

Proved limit theorem for the two-sided problem

Abstract

We motivate a new nonparametric test for the one-sided two-sample problem, which is based on a transform T of the Vincze-statistic (R,D). The exact and asymptotic distribution of T is derived. The fundamental idea can also be applied to the two-sided problem. Here too, a limit theorem can be proved.

Tables

Table 1. Table 1: mean p-values

$n=m$	$p_V$	$p_S$
10	0.2802	0.2521
15	0.2542	0.2375
20	0.2351	0.2407
25	0.2086	0.2246
30	0.1882	0.2124
35	0.1680	0.2059
40	0.1498	0.1862
50	0.1302	0.1781
60	0.1020	0.1540
70	0.0919	0.1505
80	0.0717	0.1218
100	0.0501	0.0954
200	0.0106	0.0351

Equations

$$\mathbb{P}(\sqrt{n/2}{D}_{nn}\le x)\to 1-\exp (-2x^2)\text{ for all }x\ge 0.$$

$$D_{nm}=\underset{1\le i\le n+m}{\max }\{F_n(Z_{(i)})-G_m(Z_{(i)})\}.$$

$$R_{nm}:=\min \{1\le i\le n+m:F_n(Z_{(i)})-G_m(Z_{(i)})=D_{nm}\}$$

$$\{1,\dots ,n+m\}\times \{\frac{k}{m}:0\le k\le m\}.$$

$$s:=(r+k)\frac{n}{n+m}\in \mathbb{N}\cup \{0\}\text{ and }s\le \min \{r,n\}$$

$$P_{nm}(r,k):=\mathbb{P}(R_{nm}=r,D_{nm}=\frac{k}{m})$$

$$=\frac{\left(\genfrac{}{}{0ex}{}{n+m-r+1}{n-s}\right)}{\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)}\frac{k+1}{n+m-r+1}\sum _{j=0}^{[\frac{n}{n+m}k]}\frac{\frac{m}{n}}{r+k-1-j\frac{n+m}{n}}\left(\genfrac{}{}{0ex}{}{j\frac{n+m}{n}-k}{j}\right)\left(\genfrac{}{}{0ex}{}{r+k-1-j\frac{n+m}{n}}{s-1-j}\right)$$

$$T_{nm}:=\{\begin{array}{ll}\frac{D_{nm}}{\sqrt{R_{nm}(n+m-R_{nm})}}& \text{ if }{R}_{nm}\in \{1,\dots ,n+m-1\},\\ 0& \text{ if }{R}_{nm}=n+m.\end{array}$$

$$\frac{n}{n+m}\to \lambda \in (0,1).$$

$$H_{nm}=\frac{n}{n+m}{F}_n+\frac{m}{n+m}{G}_m\to \lambda F+(1-\lambda )G=:H\text{almost surely (a.s.)}$$

$$(n+m)T_{nm}\to \frac{{\sup }_{x\in \mathbb{R}}\{F(x)-G(x)\}}{\sqrt{H(\tau )(1-H(\tau ))}}a.s.$$

$$D_{nm}\to \underset{x\in \mathbb{R}}{\sup }\{F(x)-G(x)\}a.s.$$

$$F(x):=\left\{\begin{array}[]{l@{\quad,\quad}l}\delta G(x)&x\leq\tau\\ \beta(G(x)-G(\tau))+\delta G(\tau)&x>\tau,\end{array}\right.$$

$$D(x):=F(x)-G(x)=\left\{\begin{array}[]{l@{\quad,\quad}l}(\delta-\beta)(1-G(\tau))G(x)&x\leq\tau\\ (\delta-\beta)G(\tau)(1-G(x))&x>\tau.\end{array}\right.$$

$$J_{nm}(x)=\mathbb{P}(T_{nm}\le x)=\sum _{r=1}^{n+m-1}\sum _{k=0}^{[\kappa x]\wedge m}{P}_{nm}(r,k)+\frac{m}{(n+m)(n+m-1)},$$

$$\kappa =m\sqrt{r(n+m-r)}.$$

$$\mathbb{P}(T_{nm}\le x)=\sum _{r=1}^{n+m}\sum _{k=0}^m\mathbb{P}(R_{nm}=r,D_{nm}=\frac{k}{m},T_{nm}\le x)$$

$$\displaystyle=\sum_{r=1}^{n+m-1}\sum_{k=0}^{m}\mathbb{P}\Big{(}R_{nm}=r,D_{nm}=\frac{k}{m},\frac{\frac{k}{m}}{\sqrt{r(n+m-r)}}\leq x\Big{)}$$

$$\displaystyle+\sum_{k=0}^{m}\mathbb{P}\Big{(}R_{nm}=n+m,D_{nm}=\frac{k}{m},0\leq x\Big{)}.$$

$$\sum_{k=0}^{m}\mathbb{P}\Big{(}R_{nm}=n+m,D_{nm}=\frac{k}{m},0\leq x\Big{)}=\sum_{k=0}^{m}P_{nm}(n+m,k)=P_{nm}(n+m,0).$$

$$K(x)=2\Phi (x)-\sqrt{\frac{2}{\pi }}x\exp \{-\frac{x^2}{2}\}-1,x\ge 0,$$

$$K_{nm}(x):=\mathbb{P}\Big{(}\sqrt{nm(n+m)}T_{nm}\leq x\Big{)}\rightarrow K(x)$$

$$\sqrt{nm(n+m)}{T}_{nm}:=\{\begin{array}{ll}\frac{\sqrt{\frac{nm}{n+m}}{D}_{nm}}{\sqrt{\frac{R_{nm}}{n+m}(1-\frac{R_{nm}}{n+m})}}& \text{ if }{R}_{nm}\in \{1,\dots ,n+m-1\},\\ 0& \text{ if }{R}_{nm}=n+m.\end{array}$$

$$||H_{nm}-Id||:=\underset{x\in [0,1]}{\sup }|H_{nm}(x)-x|\to \text{ a.s.}$$

$$\bigg{|}\frac{R_{nm}}{n+m}-\tau_{nm}\bigg{|}=|H_{nm}(\tau_{nm})-\tau_{nm}|\leq||H_{nm}-Id||\rightarrow 0\;\text{ a.s.}$$

$${\alpha }_{nm}(x):=\sqrt{\frac{nm}{n+m}}\{F_n(x)-G_m(x)\},x\in [0,1].$$

$$(\sqrt{\frac{nm}{n+m}}{D}_{nm},{\tau }_{nm})=L({\alpha }_n)\stackrel{\mathcal{D}}{\to }L(B)=(M(B),a(B))$$

$$(\sqrt{\frac{nm}{n+m}}{D}_{nm},\frac{R_{nm}}{n+m})\stackrel{\mathcal{D}}{\to }(M(B),a(B)).$$

$$\sqrt{nm(n+m)}{T}_{nm}=h(\sqrt{\frac{nm}{n+m}}{D}_{nm},\frac{R_{nm}}{n+m})\stackrel{\mathcal{D}}{\to }h(M(B),a(B))=:Z.$$

$$L_n(x):=\mathbb{P}(D_{nn}>x)=\frac{\left(\genfrac{}{}{0ex}{}{2n}{n+[nx]+1}\right)}{\left(\genfrac{}{}{0ex}{}{2n}{n}\right)}.$$

$$T_{nm}^{\ast }:=\{\begin{array}{ll}\frac{D_{nm}^{\ast }}{\sqrt{R_{nm}^{\ast }(n+m-R_{nm}^{\ast })}}& \text{ if }{R}_{nm}^{\ast }\in \{1,\dots ,n+m-1\},\\ 0& \text{ if }{R}_{nm}^{\ast }=n+m.\end{array}$$

$$K^{\ast }(x)$$

Full text

MSC Classification]62E15,62E20, 62G10.

1 Introduction

Let $X_{1},\ldots,X_{n}$ and $Y_{1},\ldots,Y_{m}$ be two independent samples of i.i.d. real random variables with continuous distribution functions $F$ and $G$, respectively. For deciding wether the hypothesis $H_{0}:F=G$ or the one-sided alternative $H_{1}:F\geq G,F\neq G$ is true the Smirnov-test uses $D_{nm}:=\sup_{x\in\mathbb{R}}\{F_{n}(x)-G_{m}(x)\}$, where $F_{n}$ and $G_{m}$ are the empirical distribution functions of $X_{1},\ldots,X_{n}$ and $Y_{1},\ldots,Y_{m}$, respectively. The distributional behavior of $D_{nm}$ under the hypothesis $H_{0}$ is well-known for a long time. Smirnov [9] shows in case $n=m$ that

[TABLE]

Gnedenko and Korolyuk [4] find the exact distribution of $D_{nn}$ and later on Korolyuk [6] extends the result to $D_{nm}$, when $m$ is an integral multiple of $n$, $m=np$ with $p\in\mathbb{N}$.

Next, let $Z_{1},\ldots,Z_{n+m}$ be the pooled sample, i.e. $Z_{i}:=X_{i},1\leq i\leq n,$ and $Z_{n+i}:=Y_{i},1\leq i\leq m$. If $Z_{(1)}\leq Z_{(2)}\leq\ldots,Z_{(n+m)}$ denote the pertaining order statistics, then

[TABLE]

Thus

[TABLE]

is the smallest index among those order statistics $Z_{(i)}$ at which the two-sample process $\{F_{n}(x)-G_{m}(x):x\in\mathbb{R}\}$ attains its maximal value $D_{nm}$. In the special simple test situation $H_{0}^{*}:F=G=U(0,1)$ against $H_{1}^{a}:F=U(0,1),G=U(a,1)$, for a fixed $a\in(0,1)$, Vinzce [12, 13] uses the pair $(R_{nm},D_{nm})$ instead of the single $D_{nm}$. More precisely, in the case $m=n$ he determines the likelihood ratio of $(D_{nn},R_{nn}/n-D_{nn})$ for $H_{1}^{a}$ and $H_{0}^{*}$ and then uses the corresponding Neyman-Pearson test without randomizing part. As it turns out the Neyman-Pearson test has a significantly superior power in comparison to the Smirnov-test, confer also Koul and Quine [7]. Under $H_{0}$ the exact distribution of the so-called Vincze-statistic $(R_{nm},D_{nm})$ has been computed by Vincze [11] for equal sample sizes $(m=n)$ and for general sample sizes by Steck and Simmons [10]. Gutjahr [5] gives an explicit formula, see Proposition 1 below. Note that $(R_{nm},D_{nm})$ is a discrete random variable with range

[TABLE]

Moreover, it is distribution free under $H_{0}$. Therefore, we can apply Corollary 2.2.7 and Remark 2.2.9 of Gutjahr [5], which yield the following basic result. Here, $[x],x\in\mathbb{R},$ denotes the floor function.

Proposition 1**.**

(Gutjahr) Assume that $F=G$ and that $m=np$ with $p\in\mathbb{N}$. Then for every $r\in\{1,\ldots,n+m\}$ and every $k\in\{0,1,\ldots,m\}$ such that

[TABLE]

the following equation holds:

[TABLE]

If $(r,k)$ in the range (1) does not satisfy (2), then $\mathbb{P}(R_{nm}=r,D_{nm}=\frac{k}{m})=0$.

With a view to Vincze’s idea, our approach is to use a transform $T_{nm}=h(R_{nm},D_{nm})$ of the Vincze-statistic $(R_{nm},D_{nm})$ as follows:

[TABLE]

By Remark 1 the event $\{R_{nm}=n+m\}$ has positive probability $\frac{m}{(n+m)(n+m-1)}$ and therefore cannot be ignored.

The paper is structured as follows: In section 2 we motivate the upper $T_{nm}$-test (called $V$-test) and determine the exact null-distribution of $T_{nm}$ in case that $m$ is an integral multiple of $n$. In section 3 it is shown that $\sqrt{nm(n+m)}T_{nm}$ converges in distribution to a random variable $Z$, which follows the Maxwell-Boltzmann law. Moreover, we report on the results of a small Monte Carlo simulation study. Here, the $V$-test proves to be superior to the Smirnov-test ($S$-test). In the last section, the basic idea is carried over to the two-sided alternative $H_{1}^{*}:F\neq G$. For the corresponding transformation, we also determine the limit distribution.

2 Exact null-distribution of $T_{nm}$

To motivate our test-statistic we investigate $T_{nm}$ under the usual limiting regime, that is $\min\{n,m\}\rightarrow\infty$ and

[TABLE]

Let $H_{nm}$ be the empirical distribution function pertaining to the pooled sample $Z_{1},\ldots,Z_{n+m}$. Observe that

[TABLE]

uniformly on $\mathbb{R}$ by the Glivenko-Cantelli Theorem. Assume that $F$ and $G$ in the alternative are such that $D=F-G$ has a unique maximizing point $\tau$ (as for instance when $F$ arises from $G$ as in Example 1 below). According to Lemma 1 in the appendix $H(\tau)\in(0,1)$ and

[TABLE]

whereas

[TABLE]

Notice that the limit in (5) is positive. On the other hand, $(n+m)T_{nm}\rightarrow 0$ in probability under the hypothesis, see Lemma 2 in the appendix. Therefore, given a level of significance $\alpha\in(0,1)$ our test rejects the hypothesis $H_{0}$, if $\sqrt{nm(n+m)}T_{nm}>c_{nm}(\alpha)$, where the exact critical value $c_{nm}(\alpha)$ can be computed by Theorem 2, where we give an explicit formula for the exact distribution of $T_{nm}$ under $H_{0}$. If $n$ and $m$ are large the exact critical value can be replaced by the $(1-\alpha)$-quantile of the Maxwell-Boltzmann distribution as a consequence of Theorem 3, which gives the limit distribution of $\sqrt{nm(n+m)}T_{nm}$.

Notice that the limit in (5) is larger than the limit in (6) by the factor $1/\sqrt{H(\tau)(1-H(\tau))}\geq 2$. Now, if for instance $F$ differs from $G$ only slightly in the left tails of $G$, then $H(\tau)$ is small, because $H(\tau)\in(G(\tau),F(\tau)).$ Consequently, the factor becomes significantly greater than $2$. As a numerical example assume that $G(\tau)<F(\tau)=0.02$, then the factor is greater than $7$. For that reason we strongly expect that the V-test is much more likely to detect the alternative than the Smirnov-test.

Example 1**.**

Let $\tau\in\mathbb{R}$ and $\delta>1$. Define $F=F_{\tau,\delta}$ by

[TABLE]

If $\delta G(\tau)<1$ and $\beta>0$ satisfies () $\delta G(\tau)+\beta(1-G(\tau))=1$, then $F$ is a continuous distribution function. Further, one verifies that*

[TABLE]

Since $\delta>1$ and therefore by () $\beta<1$ we see that $\delta-\beta$ is positive and hence $D=F-G\geq 0$. Moreover, it follows that $\tau$ is a maximizing point of $D$ with $D(\tau)>0$. This shows that $F$ and $G$ lie in the alternative $H_{1}$. If in addition $G$ is strictly increasing in a neighborhood of $\tau$, then $\tau$ is the unique maximizer.*

In the following theorem we use the notation $a\wedge b:=\min\{a,b\}$ for reals $a$ and $b$.

Theorem 2**.**

Assume that $F=G$ and that $m=np$ with $p\in\mathbb{N}$. Then for each $x\geq 0$ the distribution function $J_{nm}$ of $T_{nm}$ is given by

[TABLE]

where

[TABLE]

For $x<0$ the probability is equal to zero.

Proof.

Since the set in (1) is the range of $(R_{nm},D_{nm})$, a decomposition yields:

[TABLE]

Since $\frac{\frac{k}{m}}{\sqrt{r(n+m-r)}}\leq x\Leftrightarrow k\leq[m\sqrt{r(n+m-r)}\;x]$, it follows that the sum in (8) is equal to the first summand in (7). As to the sum in (9) observe that for $r=n+m$ the second requirement in (2) is equivalent to $k\leq 0$. Consequently, $P_{nm}(n+m,k)=0$ for all $k\in\{1,\ldots,m\}$ and therefore (by $x\geq 0$),

[TABLE]

The pair $(n+m,0)$ with $s=n$ satisfies condition (2), whence after some simple algebra one obtains that $P_{nm}(n+m,0)=\frac{m}{(n+m)(n+m-1)}$, which is the second summand in (7). Finally, since $T_{nm}\geq 0$, it follows that $\mathbb{P}(T_{nm}\leq x)=0$ for all negative $x$. ∎

Remark 1**.**

Notice that the left side in (10) is equal to $\mathbb{P}(R_{nm}=n+m)$. So, the last part of the above proof shows that $\mathbb{P}(R_{nm}=n+m)=\frac{m}{(n+m)(n+m-1)}.$

Remark 2**.**

The source code in Mathematica [14] that implements Gutjahr’s formula is only a few lines long. According to Theorem 2, the distribution function $J_{nm}$ of $T_{nm}$ can therefore be calculated quickley without any problems.

3 Asymptotic null-distribution

We derive the asymptotic distribution of $T_{nm}$ under the hypothesis $H_{0}$. It turns out that $\sqrt{nm(n+m)}T_{nm}$ converges to the Maxwell-Boltzmann distribution, which has distribution function

[TABLE]

and $K(x)=0$ for all negative $x$. Here, $\Phi$ is the distribution function of the standard normal law $N(0,1)$.

Theorem 3**.**

Assume that $H_{0}$ holds. Then under the limiting regime (3),

[TABLE]

for all $x\geq 0$. For negative $x$ the probabilities $K_{nm}(x),n,m\in\mathbb{N},$ are equal to zero.

Proof.

Recall that $(R_{nm},D_{nm})$ is distribution free under $H_{0}$, whence we can assume that $F$ and $G$ both are equal to the distribution function of $U(0,1)$, the uniform distribution on the unit interval $(0,1)$. So, $F$ is equal to the identity map $Id$ on $[0,1]$. We start with the representation

[TABLE]

Note that $D_{nm}=\sup_{x\in[0,1]}\{F_{n}(x)-G_{m}(x)\}$ a.s. Observe that by (4)

[TABLE]

Introduce $\tau_{nm}:=Z_{(R_{nm})}$. Since $\frac{R_{nm}}{n+m}=H_{nm}(Z_{(R_{nm})})=H_{nm}(\tau_{nm})$, it follows from (12) that

[TABLE]

As a consequence the sequences $(\sqrt{\frac{nm}{n+m}}D_{nm},\frac{R_{nm}}{n+m})$ and $(\sqrt{\frac{nm}{n+m}}D_{nm},\tau_{nm})$ are asymptotical stochastically equivalent.

Our basic idea is to write $(\sqrt{\frac{nm}{n+m}}D_{nm},\tau_{nm})$ as a functional of the two-sample empirical process

[TABLE]

To that let $(D[0,1],s)$ be the Shorokhod-space. For every $f\in D[0,1]$ define $M(f):=\sup_{t\in[0,1]}f(t)$ and $A(f):=\{t\in[0,1]:\max\{f(t),f(t-)\}=M(f)\}$ with the convention $f(0-):=f(0)$. By Lemma A.2 in Ferger [3] $A(f)$ is a non-empty compact subset of [0,1], whence the argsup-functional $a(f):=\min A(f)$ is well defined. By definition of $R_{nm}$ the random variable $\tau_{nm}$ is the smallest maximizing point of $F_{n}-G_{m}$, i.e. $\tau_{nm}=a(F_{n}-G_{m})$. Since $a(cf)=a(f)$ for every positive constant $c$, it follows that $\tau_{nm}=a(\alpha_{nm})$. Similarly, $M(cf)=cM(f)$ and thus $\sqrt{\frac{nm}{n+m}}D_{nm}=M(\alpha_{nm})$. The functional $L:=(M,a):(D[0,1],s)\rightarrow\mathbb{R}^{2}$ is Borel-measurable by Lemma A.3 in Ferger [3] and continuous on the subset $C_{u}:=\{f\in C[0,1]:f\text{ has a unique maximizing point}\}\subseteq D[0,1]$. To see this note that $M$ is continuous on $C[0,1]\supseteq C_{u}$ and $a$ is continuous on $C_{u}$ by Lemma A.4 in Ferger [3]. It is well-known that $\alpha_{nm}\stackrel{{\scriptstyle\mathcal{D}}}{{\rightarrow}}B$ in $(D[0,1],s)$, where $B$ is a Brownian bridge. Now, $B\in C_{u}$ almost surely, whence an application of the Continuous Mapping Theorem (CMT) yields that

[TABLE]

and by the asymptotical stochastical equivalence we obtain that

[TABLE]

Let $h:\mathbb{R}\times(0,1]\rightarrow\mathbb{R}$ be defined by $h(x,y):=x/\sqrt{y(1-y)}$ for $(x,y)\in\mathbb{R}\times(0,1)=:G$ and $h(x,y):=0$ otherwise. Obviously $h$ is continuous on $G$. Moreover, $a(B)\in(0,1)$ almost surely. (Indeed, $a(B)$ is uniformly distributed on $(0,1)$.) Thus (11) and another application of the CMT gives

[TABLE]

Now the assertion follows from Theorem 1.1 of [2], which says that the distribution of $Z$ is equal to the Maxwell-Boltzmann distribution. ∎

Figure 1 illustrates the convergence of $K_{nm}$ to $K$.

In a small simulation study with $F(x)=\Phi(x+0.4)$ and $G(x)=\Phi(x)$ (normal location model) and equal subsample sizes $n=m$ we computed $l=1000$ Monte-Carlo replicates $T_{nn}^{(1)},\ldots,T_{nn}^{(l)}$ and $D_{nn}^{(1)},\ldots,D_{nn}^{(l)}$ of $T_{nn}$ and $D_{nn}$, respectively. They yielded the corresponding $p$-values $p_{V}^{(j)}=1-J_{nn}(T_{nn}^{(j)})$ and $p_{S}^{(j)}=L_{n}(D_{nn}^{(j)}),1\leq j\leq l,$ where

[TABLE]

The last equation can be found, for example, in Shorack and Wellner [8] on p.401. Table 1 below shows the mean $p$-values $p_{V}=l^{-1}\sum_{j=1}^{l}p_{V}^{(j)}$ and $p_{S}=l^{-1}\sum_{j=1}^{l}p_{S}^{(j)}$.

Note that, with the exception of $n\in\{10,15\}$, the V-test performs significantly better than the Smirnov test

4 Two-sided alternatives

Consider the test problem $H_{0}:F=G$ against the alternative $H_{1}^{*}:F\neq G.$ Here, the Kolmogorov-Smirnov test rejects the hypothesis in the case of large values of $D_{nm}^{*}=\sup_{x\in\mathbb{R}}|F_{n}(x)-G_{m}(x)|.$ Since $D_{nm}^{*}=\max_{1\leq i\leq n+m}|F_{n}(Z_{(i)})-G_{m}(Z_{(i)})|$, the random variable $R_{nm}^{*}:=\min\{1\leq i\leq n+m:|F_{n}(Z_{(i)})-G_{m}(Z_{(i)})|=D_{nm}^{*}\}$ is the smallest index among those order statistics $Z_{(i)}$ at which the reflected two-sample process $\{|F_{n}(x)-G_{m}(x)|:x\in\mathbb{R}\}$ attains its maximal value. Now, the counterpart $T_{nm}^{*}$ of $T_{nm}$ is given by:

[TABLE]

Theorem 4**.**

If $F=G$, then $\mathbb{P}(\sqrt{nm(n+m)}T_{nm}^{*}\leq x)\rightarrow K^{*}(x)$, where

[TABLE]

Here, $\alpha_{j}=2j+1,j\in\mathbb{N}_{0},$ and $\varphi$ denotes the density of $N(0,1)$.

Proof.

One shows completely analogously to the proof of Theorem 3 that

[TABLE]

Thus the assertion follows from Theorem 1.1 of [2], which gives the distribution of $Z^{*}$. ∎

5 Appendix

In this short appendix we describe the asymptotic behavior of $T_{nm}$ under (special) alternatives (Lemma 1) and under the hypothesis (Lemma 2).

Lemma 1**.**

Let $F$ and $G$ be two different continuous distribution functions such that $F-G\geq 0$ has a unique minimizing point $\tau\in\mathbb{R}$. Then $F(\tau)>G(\tau)$ and $H(\tau)\in(0,1)$. Moreover,

[TABLE]

Proof.

First, notice that

[TABLE]

Let \tau_{nm}:=\raisebox{-5.12149pt}{{{\displaystyle\rm arg,!max}\atop{\scriptstyle x\in\mathbb{R}}}}(F_{n}(x)-G_{m}(x)) be the (smallest) maximizing point of $F_{n}-G_{m}$. Recall from the proof of Theorem 3 that $\frac{R_{nm}}{n+m}=H_{nm}(\tau_{nm}).$ Thus we obtain

[TABLE]

It follows from Theorem 3.3 in Ferger [1] that $\tau_{nm}\rightarrow\tau$, whence (4), (15) and continuity of $H$ yield that $R_{nm}/({n+m})\rightarrow H(\tau)\;a.s.$ Conclude with (6) that

[TABLE]

We know that $F(\tau)-G(\tau)>0$ is positive, because otherwise it followed that $F=G$ in contradiction to our assumption. Moreover, $H(\tau)\in(G(\tau),F(\tau))$, because $\lambda\in(0,1).$ Consequently, $H(\tau)\in(0,1)$, whence the limit in (16) lies in $G=\mathbb{R}\times(0,1).$ Recall the function $h$ in the proof of Theorem 3, which is continuous on $G$. By (14) and (16) it follows that

[TABLE]

which gives the desired result (13). ∎

Lemma 2**.**

Assume $H_{0}$ holds. Then

[TABLE]

Proof.

Notice that $(n+m)T_{nm}=\sqrt{\frac{n+m}{nm}}Z_{nm}$, where $Z_{nm}=\sqrt{nm(n+m)}T_{nm}\stackrel{{\scriptstyle\mathcal{D}}}{{\rightarrow}}Z$ by Theorem 3. Now, the assertion follows, because the deterministic factor $\sqrt{\frac{n+m}{nm}}$ converges to zero. ∎