On change-points tests based on two-samples U-Statistics for weakly dependent observations

Abstract

We study change-points tests based on U-statistics for absolutely regular observations. Our method avoids some technical assumptions on the data and the kernel. The asymptotic properties of the U-statistics are studied under the null hypothesis, under fixed alternatives and under a sequence of local alternatives. The asymptotic distributions of the test statistics under the null hypothesis and under the local alternatives are given explicitly and the tests are shown to be consistent. A small set of simulations is done for evaluating the performance of the tests in detecting changes in the mean, variance and autocorrelation of some simple time series.

Appendix: proofs of the results

1.1 Preliminary results

In this subsection, we prove some preliminary results necessary to the proofs of Theorems 1 and 2 .

Proposition 1

Under the conditions of Theorem 1, we have, in probability

$$\begin{aligned} n^{-3/2}\sup _{0\le \lambda \le 1}\bigg | \sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},X_{j})\bigg | \longrightarrow _{n\rightarrow \infty }0. \end{aligned}$$

Under the conditions of Theorem 2, we have, in probability

$$\begin{aligned} n^{-3/2}\sup _{0\le \lambda \le 1}\bigg | \sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_{G^{(n)}}(X_{i},Y_{nj})\bigg | \longrightarrow _{n\rightarrow \infty }0. \end{aligned}$$

Proof

We only prove the first part. This needs two lemmas that we first state and prove.

Lemma 1

Under the conditions of Theorem 1, there exists a Constant $Cst>0$ such that

$$\begin{aligned} {\mathbb {E}}\left\{ \left[ \sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},X_{j})\right] ^{2} \right\} \le Cst[n\lambda ](n-[n\lambda ]). \end{aligned}$$

Proof of Lemma 1

We can write

$$\begin{aligned} {\mathbb {E}}\left\{ \left[ \sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},X_{j})\right] ^{2} \right\}\le & {} \sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n} {\mathbb {E}}\left\{ \left[ g_F(X_{i},X_{j})\right] ^{2} \right\} \\&+2\sum _{1\le i_{1}<i_{2}\le [n\lambda ]}\sum _{[n\lambda ]+1\le j_{1}<j_{2}\le n} H\left[ (i_{1},j_{1}),(i_{2},j_{2}) \right] \\= & {} A_{n}(\lambda )+2B_{n}(\lambda ). \end{aligned}$$

where

$$\begin{aligned} H\left[ (i_{1},j_{1}),(i_{2},j_{2}) \right]= & {} {\mathbb {E}}\bigg \{ \left[ g_F(X_{i_{1},}X_{j_{1}})-h_{F,1}(X_{i_{1}})-h_{F,2}(X_{j_{1}})+\theta (F,F) \right] \\&\times \left[ g_F(X_{i_{2}}X_{j_{2}})-h_{F,1}(X_{i_{2}})-h_{F,2}(X_{j_{2}})+\theta (F,F)\right] \bigg \}. \end{aligned}$$

From the integrability condition, we have

$$\begin{aligned} \sup _{i,j\in {\mathbb {N}}} {\mathbb {E}}\left\{ \left[ g_F(X_{i},X_{j})\right] ^{2} \right\} \le Cst, \end{aligned}$$

then

$$\begin{aligned} A_{n}(\lambda )\le Cst[n\lambda ](n-[n\lambda ]). \end{aligned}$$

Since

$$\begin{aligned} \int _{{\mathbb {R}}}\left[ g_F(x,y)-h_{F,1}(x)-h_{F,2}(y)+\theta (F,F)\right] dF(x)=0 \end{aligned}$$

so from Lemma 1 of Yoshihara (1976), we have the following inequalities:

(a) If $1\le i_{1}<j_{1}\le [n\lambda ], \ [n\lambda ]+1\le i_{2}<j_{2}\le n$ and $i_{2}-i_{1}\ge j_{2}-j_{1}$, then

$$\begin{aligned} H\left[ (i_{1},j_{1}),(i_{2},j_{2}) \right] \le Cst \beta ^{\frac{\delta }{2+\delta }}(i_{2}-i_{1}). \end{aligned}$$

Then we deduce

$$\begin{aligned}&\sum _{1\le i_{1}<i_{2}\le [n\lambda ]}\sum _{[n\lambda ]+1\le j_{1}<j_{2}\le n} H\left[ (i_{1},j_{1}),(i_{2},j_{2}) \right] \le Cst [n\lambda ](n-[n\lambda ]) \sum _{k=1}^{n}k\beta ^{\frac{\delta }{2+\delta }}(k) \\&\quad Cst[n\lambda ](n-[n\lambda ])\sum _{k=1}^{n}k\beta ^{\frac{\delta }{2+\delta }}(k)\le Cst[n\lambda ](n-[n\lambda ]), \end{aligned}$$

where $k=j_{2}-i_{2}$.

Suppose k fixed, we have $[n\lambda ]$ ways to choose $i_{1},$ once $i_{1}$ is chosen we have one way to choose $i_{2}=i_{1}+k$. For $j_{1}$ we have $n-[n\lambda ]$ ways to choose $j_{1}$ and then for each $j_{1}$, $j_{2}$ need to be in the interval $[j_{1},j_{1}+k]$ and there are exactly $\ k$ integers in such interval.

(b) Similarly, if $1\le i_{1}<j_{1}\le [n\lambda ], \ [n\lambda ]+1\le i_{2}<j_{2}\le n$ and $i_{2}-i_{1}\le j_{2}-j_{1}$, then

$$\begin{aligned} H\left[ (i_{1},j_{1}),(i_{2},j_{2}) \right] \le M_{0}\beta ^{\frac{\delta }{2+\delta }}(j_{2}-j_{1}). \end{aligned}$$

Thus, we deduce that

$$\begin{aligned} B_{n}(\lambda )\le Cst[n\lambda ](n-[n\lambda ]) \end{aligned}$$

and Lemma 1 is proved. $\square$

We now define the process ${\mathcal {G}}_{n}(\lambda ),0\le \lambda \le 1$ by

$$\begin{aligned} {\mathcal {G}}_{n}(\lambda )=n^{-3/2}\sum _{i=1}^{[n\lambda ]} \sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},X_{j}),0\le \lambda \le 1. \end{aligned}$$

Lemma 2

Under the conditions of Theorem 1, we have

$$\begin{aligned} {\mathbb {E}}\Big (\Big | {\mathcal {G}}_{n}(\lambda )-{\mathcal {G}}_{n}(\lambda ^{\prime })\Big |^{2}\Big )\le \frac{Cst}{n}(\lambda -\lambda ^{\prime }), \text{ for } \text{ all } \ 0\le \lambda ^{\prime }\le \lambda \le 1. \end{aligned}$$

Proof of Lemma 2

We can write

$$\begin{aligned} {\mathbb {E}}\big (\left| {\mathcal {G}}_{n}(\lambda )-{\mathcal {G}}_{n}(\lambda ^{\prime })\right| ^{2}\big )\le & {} 2n^{-3}{\mathbb {E}}\left\{ \left[ \sum _{i=1}^{[n\lambda ^{\prime }]}\sum _{j=[n\lambda ^{\prime }]+1}^{[n\lambda ]}g_F(X_{i},X_{j})\right] ^{2} \right\} \\&+2n^{-3}{\mathbb {E}}\left\{ \left[ \sum _{i=1+[n\lambda ^{\prime }]}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},X_{j})\right] ^{2} \right\} . \end{aligned}$$

From Lemma 1, we deduce that

$$\begin{aligned} {\mathbb {E}}\left( \left| {\mathcal {G}}_{n}(\lambda )-{\mathcal {G}}_{n}(\lambda ^{\prime })\right| ^{2}\right)\le & {} \frac{Cst}{n^{3}}\big \{[n\lambda ^{\prime }]([n\lambda ]-[n\lambda ^{\prime }]) +([n\lambda ]-[n\lambda ^{\prime }])(n-[n\lambda ^{\prime }])\big \}\\&\times \frac{Cst}{n}(\lambda -\lambda ^{\prime }) \end{aligned}$$

and Lemma 2 is proved. $\square$

From Lemma 2, we deduce that

$$\begin{aligned} {\mathbb {P}}\left( \left| {\mathcal {G}}_{n}(\lambda )-{\mathcal {G}}_{n}(\lambda ^{\prime })\right| \ge \epsilon \right) \le \frac{Cst}{\epsilon ^{2}n}(\lambda -\lambda ^{\prime }) \end{aligned}$$

for all $\epsilon >0.$ It implies for $0\le l_{1}\le l_{2}\le n$ with $l_{1},l_{2},n\in {\mathbb {N}}$,

$$\begin{aligned} {\mathbb {P}}\left( \left| {\mathcal {G}}_{n}\Big (\frac{l_{2}}{n}\Big )-{\mathcal {G}}_{n}\Big (\frac{l_{1}}{n}\Big )\right| \ge \epsilon \right) \le \frac{Cst}{\epsilon ^{2}n}(l_{2}-l_{1})\le \frac{Cst}{\epsilon ^{2}n^{\frac{5}{3}}}(l_{2}-l_{1})^{\frac{4}{3}}. \end{aligned}$$

Consider the partial sum process defined by $S_{0}=0$ and $S_{i}=\sum _{j=1}^{i}A_{j}$ where $A_{j}={\mathcal {G}}_{n}(\frac{j}{n})-{\mathcal {G}}_{n}(\frac{j-1}{n})$ if $1\le j\le n-1$ and 0 otherwise. It results that $S_{i}={\mathcal {G}}_{n}(\frac{i}{n})$.

The last inequality is equivalent to

$$\begin{aligned} {\mathbb {P}}\big (\left| S_{l_{2}}-S_{l_{1}}\right| \ge \epsilon \big )\le \frac{Cst}{\epsilon ^{2}}\left( \frac{l_{2}-l_{1}}{n^{\frac{5}{4}}}\right) ^{\frac{4}{3}}. \end{aligned}$$

From Theorem 10.2 of Billingsley (1999), we easily deduce that

$$\begin{aligned} {\mathbb {P}}\Big (\max _{1\le i\le n-1}\left| S_{i}\right| \ge \epsilon \Big )\le \frac{Cst}{\epsilon ^{2}}\left( \frac{ l_{2}-l_{1}}{n^{\frac{5}{4}}}\right) ^{\frac{4}{3}}\le \frac{Cst}{\epsilon ^{2}n^{\frac{1}{3}}} \end{aligned}$$

which implies that, in probability,

$$\begin{aligned} n^{-3/2}\sup _{0\le \lambda \le 1}\bigg | \sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},X_{j})\bigg | \longrightarrow _{_{n\rightarrow \infty }}0. \end{aligned}$$

This completes the proof of Proposition 1. $\square$

We need the following result proved by Oodaira and Yoshihara (1972).

Let $\xi _1,\xi _2,\ldots ,\xi _n,\ldots$ be a strictly stationary sequence of zero-mean random variables, and let

$$\begin{aligned} \sigma _{*}^{2}={\mathbb {E}}(\xi _1^{2})+2\sum _{i=1}^{\infty } {\mathbb {E}} \left( \xi _1\xi _{i+1} \right) . \end{aligned}$$

Proposition 2

Assume ${\mathbb {E}} \left( \left| \xi _{i}\right| ^{2+\delta } \right) <\infty$ for some positive $\delta$ and $\xi _1,\xi _2,\ldots ,\xi _n,\ldots$ is $\alpha$-mixing with $\alpha$-rate satisfying

$$\begin{aligned} \sum _{i=1}^{\infty }[\alpha (i)]^{\frac{\delta }{2+\delta }}<\infty . \end{aligned}$$

Then $\sigma _{*}^{2}<\infty .$

If $\sigma _{*}>0$, then the sequence of processes

$$\begin{aligned} S_{n}(\lambda )=\frac{1}{\sigma _{*}\sqrt{n}}\sum _{i=1}^{[n\lambda ]}\xi _{i}, \lambda \in [0,1] \end{aligned}$$

converges weakly to a Wiener measure on $(D,\mathcal {D)}$, where ${\mathcal {D}}$ is the $\sigma$-fields of Borel sets for the Skorohod topology.

Proof

See the proof of Theorem 2 of Oodaira and Yoshihara (1972).

Proposition 3

Under the conditions of Theorem 1, we have

$$\begin{aligned} \left\{ \frac{1}{\sqrt{n}}\sum _{i=1}^{[n\lambda ]} \begin{pmatrix} h_{F,1}(X_{i}) \\ h_{F,2}(X_{i}) \end{pmatrix} \right\} _{0\le \lambda \le 1}\longrightarrow _{n\rightarrow \infty } \left\{ \begin{pmatrix} W_{1}(\lambda ) \\ W_{2}(\lambda ) \end{pmatrix} \right\} _{0\le \lambda \le 1}. \end{aligned}$$

(10)

Under the conditions of Theorem 2, we have

$$\begin{aligned} \left\{ \frac{1}{\sqrt{n}}\sum _{i=1}^{[n\lambda ]} \begin{pmatrix} h_{F,1}(X_{i}) \\ h_{G^{(n)},2}(Y_{ni}) \end{pmatrix} \right\} _{0\le \lambda \le 1}\longrightarrow _{n\rightarrow \infty } \left\{ \begin{pmatrix} W_{1}(\lambda ) \\ W_{2}(\lambda ) \end{pmatrix} \right\} _{0\le \lambda \le 1}. \end{aligned}$$

(11)

Proof

We only prove (11). The part relating to (10) involves a series which is not a triangular array. It is more easier to handle than (11).

For establishing (11), we need to establish a finite-dimensional convergence and a tightness results.

Starting by the finite-dimensional convergence, by the Cramér-Wold device it suffices to show that for any $k\in {\mathbb {N}}^{*}$, any $a_{j,}b_{j},\lambda _{j}\in {\mathbb {R}}$, $a_{1}<\ldots <a_{k}$, $b_{1}<\ldots <b_{k}$, $0=\lambda _{0}<\lambda _{1}<\ldots <\lambda _{k}=1$

$$\begin{aligned} \sum _{j=1}^{k}\frac{1}{\sqrt{n}}\sum _{i=[n\lambda _{j-1}]+1}^{[n\lambda _{j}]}\left[ a_{j}h_{F,1}(X_{i})+b_{j}h_{G^{(n)},2}(Y_{ni}) \right] \end{aligned}$$

converges in distribution to a Gaussian random variable.

For that, we need the following lemma.

Lemma 3

(Harel and Puri 1989) Let $\{X_{ni}\}$ be a sequence of zero-mean absolutely regular random variables (rv)’s with rates satisfying

$$\begin{aligned} \sum _{n\ge }\big [\beta (n)\big ]^{\delta /(2+\delta )} <\infty \ \text{ for } \text{ some } \ \delta > 0. \end{aligned}$$

(12)

Suppose that for any $\kappa$, there exists a sequence $\{Y^\kappa _{ni}\}$ of rv’s satisfying (12) such that

$$\begin{aligned} \sup _{n\in {\mathbb {N}}}\max _{0\le i\le n}|Y^\kappa _{ni}|\le B_\kappa < \infty , \end{aligned}$$

(13)

where $B_\kappa$ is some positive constant

$$\begin{aligned}&\sup _{n\in {\mathbb {N}}}\max _{0\le i\le n} {\mathbb {E}}\big (|X^*_{ni}-Y^\kappa _{ni}|^{2+\delta }\big )\longrightarrow 0 \ \text{ as } \ \kappa \rightarrow \infty \end{aligned}$$

(14)

$$\begin{aligned}&\frac{1}{n} {\mathbb {E}}\bigg [\Big (\sum ^n_{i=1}X^*_{ni}\Big )^2\bigg ]\longrightarrow c \ \text{ as } \ n\rightarrow \infty , \end{aligned}$$

(15)

where c is some positive constant

$$\begin{aligned} \frac{1}{n} {\mathbb {E}}\bigg [\Big (\sum ^n_{i=1}Y^\kappa _{ni}-{\mathbb {E}}(Y^\kappa _{ni})\Big )^2\bigg ]\longrightarrow c_\kappa \ \text{ as } \ n\rightarrow \infty , \end{aligned}$$

(16)

where $c_\kappa$ is some constant $> 0$

$$\begin{aligned} c_\kappa \longrightarrow c \ \text{ as } \ \kappa \rightarrow \infty . \end{aligned}$$

(17)

Then

$$\begin{aligned} \frac{1}{n}\sum ^n_{i=1}X^*_{ni} \end{aligned}$$

converges in distribution to the normal distribution with mean 0 and variance c.

Without loss of generality, we take $k=2$ and $0=\lambda _{0}<\lambda _{1}<\lambda _{2}=1$, $a_{1}<a_{2}$, $b_{1}<b_{2}$.

The assumption (12) readily holds from (5).

Define, for $j=1,2$,

$$\begin{aligned} \psi _{ni}^{(j)}=a_{j}h_{F,1}(X_{i})+b_{j}h_{G^{(n)},2}(Y_{ni}). \end{aligned}$$

For establishing (15), we need proving that, as n tends to infinity,

$$\begin{aligned} {\mathbb {E}} \left[ \left( \frac{1}{\sqrt{n}}\left\{ \sum _{i=1}^{[n\lambda _{1}]} \psi _{ni}^{(1)} +\sum _{i=[n\lambda _{1}]+1}^{n} \psi _{ni}^{(2)}\right\} \right) ^{2} \right] \end{aligned}$$

tends to some positive constant c.

We have

$$\begin{aligned}&{\mathbb {E}}\left\{ \left[ \frac{1}{\sqrt{n}}\left( \sum _{i=1}^{[n\lambda _{1}]}\psi _{ni}^{(1)}+\sum _{i=[n\lambda _{1}]+1}^{n}\psi _{ni}^{(2)}\right) \right] ^{2} \right\} = \frac{1}{n}\bigg \{{\mathbb {E}}\left[ \left( \sum _{i=1}^{[n\lambda _{1}]}\psi _{ni}^{(1)}\right) ^{2} \right] \\&\ \ +{\mathbb {E}}\left[ \left( \sum _{i=1}^{[n\lambda _{1}]} \psi _{ni}^{(1)}\right) \left( \sum _{i=[n\lambda _{1}]+1}^{n}\psi _{ni}^{(2)}\right) \right] +{\mathbb {E}}\left[ \left( \sum _{i=[n\lambda _{1}]+1}^{n}\psi _{ni}^{(2)}\right) ^{2}\right) \bigg \}. \end{aligned}$$

Since the random variables $\psi _{ni}^{(1)}$ and $\psi _{ni}^{(2)}$ are centered, we obtain

$$\begin{aligned} {\mathbb {E}}\left[ \bigg (\sum _{i=1}^{[n\lambda _{1}]}\psi _{ni}^{(1)}\bigg )^{2} \right] =[n\lambda _{1}] {\mathbb {E}}\left[ \big (\psi _{n1}^{(1)}\big )^{2} \right] +2\sum _{i=1}^{[n\lambda _{1}]} \sum _{j=1}^{[n\lambda _{1}]-i}{\mathbb {E}}\left[ \big (\psi _{ni}^{(1)}\psi _{n,i+j}^{(1)}\big ) \right] . \end{aligned}$$

From the condition of Theorem 2, we deduce that ${\mathbb {E}}\left[ \left( \psi _{n1}^{(1)}\right) ^{2+\delta }\right] <\infty$, which implies that

$$\begin{aligned} \sup _{n, i,j\ge 1}\left| {\mathbb {E}}\left( \psi _{ni}^{(1)}\psi _{n,i+j}^{(1)}\right) \right| \le \beta ^{\frac{\delta }{2+\delta }}(j)\big \{{\mathbb {E}}\left( \big |\psi _{ni}^{(1)}\big |^{2+\delta }\right) \big \}^{\frac{1}{2+\delta }} \big \{{\mathbb {E}}\left( \big |\psi _{n,i+j}^{(1)}\big |^{2+\delta }\right) \big \}^{\frac{1}{2+\delta }}. \end{aligned}$$

We get

$$\begin{aligned} {\mathbb {E}}\left[ \bigg (\sum _{i=1}^{[n\lambda _{1}]}\psi _{ni}^{(1)}\bigg )^{2} \right] \le [n\lambda _{1}] {\mathbb {E}}\left[ \left( \psi _{n1}^{(1)}\right) ^{2} \right] +2[n\lambda _{1}]\sum _{j=1}^{[n\lambda _{1}]}\beta ^{\frac{\delta }{2+\delta }}(j)M^{2}, \end{aligned}$$

where $M=\sup _{n \ge 1}\left\{ {\mathbb {E}}\left[ \left( \psi _{n1}^{(1)}\right) ^{2+\delta }\right] \right\} ^{\frac{1}{2+\delta }}$.

It results that

$$\begin{aligned}&\lim _{n\rightarrow \infty }\frac{1}{n}\bigg \{[n\lambda _{1}]{\mathbb {E}}\left[ \big (\psi _{n1}^{(1)}\big )^{2} \right] +2\sum _{i=1}^{[n\lambda _{1}]}\sum _{j=1}^{[n\lambda _{1}]-i}\Big |{\mathbb {E}}\big (\psi _{ni}^{(1)}\psi _{n,i+j}^{(1)}\big )\Big |\bigg \} \nonumber \\&\quad \le \lambda _1\Big \{{\mathbb {E}}\left[ \big (\psi _{n1}^{(1)}\big )^{2}\right] +2\sum _{j=1}^{\infty }\beta ^{\frac{\delta }{2+\delta }}(j)M^{2}\Big \}. \end{aligned}$$

(18)

We also have

$$\begin{aligned} {\mathbb {E}}\left[ \left( \sum _{i=1}^{[n\lambda _{1}]}\psi _{ni}^{(1)}\right) \left( \sum _{i=[n\lambda _{1}]+1}^{n}\psi _{ni}^{(2)}\right) \right] =\sum _{i=1}^{[n\lambda _{1}]}\sum _{j=[n\lambda _{1}]+1}^{n}{\mathbb {E}}\big (\psi _{ni}^{(1)}\psi _{nj}^{(2)}\big ). \end{aligned}$$

From

$$\begin{aligned} \sup _{n \ge 1}\sup _{i,j\ge 1}\left| {\mathbb {E}}\big (\psi _{ni}^{(1)}\psi _{nj}^{(2)}\big )\right| \le \beta ^{\frac{\delta }{2+\delta }}(j-i)MM^{*}, \end{aligned}$$

where $M^{*}=\sup _{n \ge 1}\left\{ {\mathbb {E}}\left[ \left( \psi _{n1}^{(2)}\right) ^{2+\delta } \right] \right\} ^{\frac{1}{2+\delta }}$, it results that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{i=1}^{[n\lambda _{1}]}\sum _{j=[n\lambda _{1}]+1}^{n}\left| {\mathbb {E}}\big (\psi _{ni}^{(1)}\psi _{nj}^{(2)}\big )\right| \le \lambda _{1}\sum _{j=1}^{\infty }\beta ^{\frac{\delta }{2+\delta }}(j)MM^{*}. \end{aligned}$$

(19)

Similarly, we get

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}{\mathbb {E}}\left[ \bigg (\sum _{i=[n\lambda _{1}]+1}^{n}\psi _{ni}^{(2)}\bigg )^{2}\right] \le (1-\lambda _{1})\Big \{{\mathbb {E}}\left[ \big (\psi _{n1}^{(1)}\big )^2 \right] +2\sum _{j=1}^{\infty }\beta ^{\frac{\delta }{2+\delta }}(j)(M^{*})^{2}\Big \}. \end{aligned}$$

(20)

From (18)-(20), we deduce (15).

Now, we turn to proving (14). For all $i\ge 1$, and for any $\kappa >0$, define

$$\begin{aligned} \psi _{ni}^{(j),\kappa }= \left\{ \begin{array}{ll} \psi _{ni}^{(j)} &{} \text{ if } \big |\psi _{ni}^{(j)}\big | \le \kappa \\ 0 &{} \text{ if } \big |\psi _{ni}^{(j)}\big | \ge \kappa , \ j=1,2. \end{array} \right. \end{aligned}$$

It is immediate that

$$\begin{aligned} \sup _{n\ge 1}\sup _{i\ge 1}\big | \psi _{ni}^{(j)}\big | \le \kappa <\infty . \end{aligned}$$

It results from the integrability condition in Theorem 2 that the sequences $\{\psi _{ni}^{(j)}; \ i\ge 1, \ j=1,2\}$ are uniformly integrable.

Whence

$$\begin{aligned} \sup _{i\ge 1}{\mathbb {E}}\left( \Big | \psi _{ni}^{(j)}-\psi _{ni}^{(j),\kappa }\Big |^{2+\delta }\right) \longrightarrow 0 \ \text { as } \ \kappa \rightarrow \infty , \ j=1,2 \end{aligned}$$

and (14) is proved.

The proof of (16), that is

$$\begin{aligned} \lim _{n\rightarrow \infty } {\mathbb {E}}\left[ \left( \frac{1}{\sqrt{n}}\left\{ \sum _{i=1}^{[n\lambda _{1}]}\left[ \psi _{ni}^{(1),\kappa }-{\mathbb {E}}\left( \psi _{ni}^{(1),\kappa }\right) \right] +\sum _{i=[n\lambda _{1}]+1}^{n}\left[ \psi _{ni}^{(2),\kappa }-{\mathbb {E}}\left( \psi _{ni}^{(2),\kappa }\right) \right] \right\} \right) ^{2} \right] =c_\kappa , \end{aligned}$$

where $c_\kappa$ is some positive constant, is similar to that of (15).

It remains to prove (17).

For any $i,j=1,2$, denote by $\psi _{i}^{(j),\kappa }$ the counterpart of $\psi _{ni}^{(j),\kappa }$ obtained by substituting the $Y_{ni}$’s for the $X_i$’s.

We have

$$\begin{aligned} c_\kappa= & {} \lambda _{1} {\mathbb {E}}\left\{ \left[ \psi _{1}^{(1),\kappa }- {\mathbb {E}}\big (\psi _{1}^{(1),\kappa }\big )\right] ^{2} \right\} +2\lambda _{1}\sum _{i=1}^{\infty } {\mathbb {E}}\left\{ \left[ \psi _{1}^{(1),\kappa }-{\mathbb {E}}\big (\psi _{1}^{(1),\kappa }\big )\right] \left[ \psi _{i+1}^{(1),\kappa }-{\mathbb {E}}\big (\psi _{i+1}^{(1),\kappa }\big )\right] \right\} \\&+\lambda _{1}\sum _{i=1}^{\infty }{\mathbb {E}}\left\{ \left[ \psi _{1}^{(1),\kappa }-{\mathbb {E}}\big (\psi _{1}^{(1),\kappa }\big )\right] \left[ \psi _{i+1}^{(2),\kappa }-{\mathbb {E}}\big (\psi _{i+1}^{(2),\kappa }\big )\right] \right\} \\&+(1-\lambda _{1}) {\mathbb {E}}\left\{ \left[ \psi _{1}^{(2),\kappa }-{\mathbb {E}}\big (\psi _{1}^{(2),\kappa }\big )\right] ^{2} \right\} \\&+2(1-\lambda _{1})\sum _{i=1}^{\infty }{\mathbb {E}}\left\{ \left[ \psi _{1}^{(2),\kappa }-{\mathbb {E}}\big (\psi _{1}^{(2),\kappa }\big )\right] \left[ \psi _{i+1}^{(1),\kappa }-{\mathbb {E}}\big (\psi _{i+1}^{(2),\kappa }\big )\right] \right\} . \end{aligned}$$

By the Lebesgue dominated convergence theorem, one obtains

$$\begin{aligned}&{\mathbb {E}}\left\{ \left[ \psi _{1}^{(1),\kappa }-{\mathbb {E}}\left( \psi _{1}^{(1),\kappa }\right) \right] ^{2} \right\} \longrightarrow {\mathbb {E}}\left[ \left( \psi _{1}^{(1)}\right) ^{2} \right] \text{ as } \kappa \rightarrow \infty , \\&{\mathbb {E}}\left\{ \left[ \psi _{1}^{(1),\kappa }-{\mathbb {E}}\big (\psi _{1}^{(1),\kappa } \right] \left[ (\psi _{i+1}^{(1),\kappa }-{\mathbb {E}}\big (\psi _{i+1}^{(1),\kappa }\big )\right] \right\} \longrightarrow {\mathbb {E}} \left( \psi _{1}^{(1)}\psi _{i+1}^{(1)}\right) \text{ as } \kappa \rightarrow \infty ,\\&{\mathbb {E}}\left\{ \left[ \psi _{1}^{(1),\kappa }-{\mathbb {E}}\big (\psi _{1}^{(1),\kappa }\big )\right] \left[ \psi _{i+1}^{(2),\kappa }-{\mathbb {E}}\big (\psi _{i+1}^{(2),\kappa }\big )\right] \right\} \longrightarrow {\mathbb {E}}\left( \psi _{1}^{(2)}\psi _{i+1}^{(2)}\right) \text{ as } \kappa \rightarrow \infty , \\&{\mathbb {E}}\left\{ \left[ \psi _{1}^{(2),\kappa }-{\mathbb {E}}\big (\psi _{1}^{(2),\kappa }\big )\right] ^{2}\right\} \longrightarrow {\mathbb {E}}\left[ \left( \psi _{1}^{(2)}\right) ^{2} \right] \text{ as } \kappa \rightarrow \infty \end{aligned}$$

and

$$\begin{aligned} {\mathbb {E}}\left\{ \left[ \psi _{1}^{(2),\kappa }-{\mathbb {E}}\big (\psi _{1}^{(2),\kappa }\big )\right] \left[ \psi _{i+1}^{(2),\kappa } -{\mathbb {E}}\big (\psi _{i+1}^{(2),\kappa }\big )\right] \right\} \longrightarrow {\mathbb {E}}\left( \psi _{1}^{(2)} \psi _{i+1}^{(2)} \right) \text{ as } \kappa \rightarrow \infty . \end{aligned}$$

Therefore

$$\begin{aligned} \lim _{\kappa \rightarrow \infty }c_\kappa =c \end{aligned}$$

and (17) is proved. Whence, the finite dimensional convergence is established.

For proving the tightness, we need the following Lemma.

Lemma 4

(Phillips and Durlauf 1986) Probability measures on a product space are tight iff all the marginal probability measures are tight on the component spaces.

It results from this lemma that it suffices to prove the tightness of each component of the sequence of processes in (11). It is immediate from Proposition 2 that the first is tight. For the second, define

$$\begin{aligned} {\mathcal {M}}_n(\lambda )=\sigma _{22}^{-1/2}\frac{1}{\sqrt{n}}\sum _{i=1}^{[n\lambda ]} h_{G^{(n)},2}(Y_{ni}). \end{aligned}$$

If $\lambda _1\le \lambda \le \lambda _2$, from the integral conditions and condition (5), there exists a constant C such that

$$\begin{aligned} {\mathbb {E}}\left( |{\mathcal {M}}_n(\lambda )-{\mathcal {M}}_n(\lambda _1)|^2|{\mathcal {M}}_n(\lambda _2)-{\mathcal {M}}_n(\lambda )|^2 \right)\le & {} C{1 \over n^2}([n\lambda ]-[n\lambda _1])([n\lambda _2]-[n\lambda ])\\\le & {} C{1 \over n^2}([n\lambda _1]-[n\lambda ])([n\lambda ]-[n\lambda _2])\\\le & {} C{1 \over n^2}([n\lambda _2]-[n\lambda _1])^2 \\\le & {} C(\lambda _2-\lambda _1)^2. \end{aligned}$$

If $\lambda _2-\lambda _1\ge 1/n$ the last inequality follows and if $\lambda _2-\lambda _1<1/n$, then either $\lambda _1$ and $\lambda$ lie in the same subinterval $[(i-1)/n,i/n]$ or else $\lambda$ and $\lambda _2$ do. In either of these cases the left hand of last inequality vanishes. From Theorem 13.5 of Billingsley (1999), the process ${\mathcal {M}}_n$ is tight. This ends the proof of Proposition 3. $\square$

1.2 Proof of Theorem 1

Using the Hoeffding decomposition, we can write $Z_{n}(\lambda )$ as

$$\begin{aligned} Z_{n}(\lambda )= & {} n^{-3/2}\sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n} \left[ h_{F,1}(X_{i})+h_{F,2}(X_{j})+g_F(X_{i},X_{j})\right] \nonumber \\= & {} n^{-3/2}\left[ (n-[n\lambda ])\sum _{i=1}^{[n\lambda ]}h_{F,1}(X_{i})+[n\lambda ] \sum _{j=[n\lambda ]+1}^{n}h_{F,2}(X_{j})\right] \nonumber \\&+n^{-3/2}\sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},X_{j}). \end{aligned}$$

(21)

From Proposition 1, we have

$$\begin{aligned} n^{-3/2}\sup _{0\le \lambda \le 1}\bigg |\sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},X_{j})\bigg | \longrightarrow _{_{n\rightarrow \infty }}0 \end{aligned}$$

in probability.

Thus, by Slutsky’s lemma, it suffices to show that the sum of the first two terms

$$\begin{aligned} \Big \{n^{-3/2}(n-[n\lambda ])\sum _{i=1}^{[n\lambda ]} h_{F,1}(X_{i})+ n^{-3/2} [n\lambda ] \sum _{j=[n\lambda ]+1}^{n} h_{F,2}(X_{j})\Big \}_{0\le \lambda \le 1} \end{aligned}$$

converges in distribution to the desired limit process.

It results from Proposition 2 that the process

$$\begin{aligned} \Big \{\frac{1}{\sqrt{n}}\sum _{i=1}^{[n\lambda ]}h_{F,1}(X_{i})\Big \}_{0\le \lambda \le 1} \end{aligned}$$

converges weakly to a Brownian motion $\{W(\lambda )\}_{0\le \lambda \le 1}$.

Proposition 3 yields

$$\begin{aligned} \Big \{\frac{1}{\sqrt{n}}\sum _{i=1}^{[n\lambda ]} \begin{pmatrix} h_{F,1}(X_{i}) \\ h_{F,2}(X_{i}) \end{pmatrix} \Big \}_{0\le \lambda \le 1}\longrightarrow _{n\rightarrow \infty } \Big \{\begin{pmatrix} W_{1}(\lambda ) \\ W_{2}(\lambda ) \end{pmatrix} \Big \}_{0\le \lambda \le 1} \end{aligned}$$

in distribution on the space $(D[0,1])^{2}$ to $(D[0,1])^{2}.$

Now, we consider the mapping defined by

$$\begin{aligned} \begin{pmatrix} x_{1}(\lambda ) \\ x_{2}(\lambda ) \end{pmatrix} \mapsto (1-\lambda )x_{1}(\lambda )+\lambda (x_{2}(1)-x_{2}(\lambda )), \ 0\le \lambda \le 1. \end{aligned}$$

This is a continuous mapping from $(D[0,1])^{2}$ to D[0, 1]. Whence,

$$\begin{aligned}&\left\{ n^{-3/2}(n-[n\lambda ])\displaystyle \sum _{i=1}^{[n\lambda ]}h_{F,1}(X_{i})+n^{-3/2}[n\lambda ] \displaystyle \sum _{j=[n\lambda ]+1}^{n}h_{F,2}(X_{j})\right\} _{0\le \lambda \le 1} \longrightarrow _{_{n\rightarrow \infty }} \left\{ Z(\lambda )\right\} _{0\le \lambda \le 1}, \end{aligned}$$

where for any $\lambda \in [0,1]$,

$$\begin{aligned} Z(\lambda )= (1-\lambda )W_{1}(\lambda )+\lambda [W_{2}(1)-W_{2}(\lambda )]. \end{aligned}$$

Whence, Theorem 1 is proved. $\square$

1.3 Proof of Theorem 2

Now we prove Theorem 2. Under the conditions of Theorem 2, we have the following equality

$$\begin{aligned} Z_{n}(\lambda )= & {} Z_{n}^{*}(\lambda )-n^{-3/2}[n\lambda ](n-[n\lambda ])\theta (F,F) \\= & {} n^{-3/2}[n\lambda ](n-[n\lambda ])\theta (F,G^{(n)})-n^{-3/2}[n\lambda ](n-[n\lambda ])\theta (F,F) \\&+\frac{[n\lambda ](n-[n\lambda ])}{n^{3/2}}\frac{1}{[n\lambda ]}\sum _{i=1}^{[n\lambda ]}h_{F,1}(X_{i}) \\&+\frac{[n\lambda ](n-[n\lambda ])}{n^{3/2}}\frac{1}{(n-[n\lambda ])}\sum _{j=[n\lambda ]+1}^{n}h_{G^{(n)},2}(Y_{nj}) \\&+\frac{1}{n^{3/2}}\sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},Y_{nj}). \end{aligned}$$

From Proposition 1, we deduce that

$$\begin{aligned} n^{-3/2}\sup _{0\le \lambda \le 1}\bigg | \sum _{i=1}^{[n\lambda ]}\sum _{j=[n\lambda ]+1}^{n}g_F(X_{i},Y_{nj})\bigg | \longrightarrow _{n\rightarrow \infty }0 \end{aligned}$$

in probability.

From Proposition 2, we deduce that

$$\begin{aligned} n^{-1/2}\sum _{i=1}^{[n\lambda ]}h_{F,1}(X_{i}) \end{aligned}$$

converges weakly to the Brownian process $\{W_1(\lambda )\}_{_{0\le \lambda \le 1}}$ and

$$\begin{aligned} n^{-1/2}\sum _{j=[n\lambda ]+1}^{n}h_{G^{(n)},2}(Y_{nj}) \end{aligned}$$

converges weakly to the Brownian process $\{W_2(1)-W_2(\lambda )\}_{_{0\le \lambda \le 1}}$.

We have also from the ${\mathcal {H}}_{1,n}$

$$\begin{aligned} \lim _{n\rightarrow \infty } n^{-3/2}[n\lambda ](n-[n\lambda ])\theta (F,G^{(n)})-n^{-3/2}[n\lambda ](n-[n\lambda ])\theta (F,F)=\lambda (1-\lambda )A. \end{aligned}$$

From Proposition 3, we obtain that

$$\begin{aligned}&\left\{ n^{-3/2}(n-[n\lambda ])\displaystyle \sum _{i=1}^{[n\lambda ]}h_{F,1}(X_{i})+n^{-3/2}[n\lambda ] \displaystyle \sum _{j=[n\lambda ]+1}^{n}h_{G^{(n)},2}(Y_{nj})\right\} _{0\le \lambda \le 1}\\&\quad \longrightarrow _{_{n\rightarrow \infty }} \left\{ {{\widetilde{Z}}}(\lambda )\right\} _{0\le \lambda \le 1}, \end{aligned}$$

where for any $\lambda \in [0,1]$,

$$\begin{aligned} {{\widetilde{Z}}}(\lambda )=(1-\lambda )W_{1}(\lambda )+\lambda (W_{2}(1)-W_{2}(\lambda )). \end{aligned}$$

This establishes Theorem 2.

1.4 Proof of Theorem 3

Let $1\le [(n+1)t]\le [n\lambda _0]$, then

$$\begin{aligned} Z_n^*(t)= & {} n^{-3/2}\sum _{1\le i<j\le [n\lambda _0]}h(X_i,X_j) + n^{-3/2}\sum _{i=1}^{[n\lambda _0]}\sum _{j=[n\lambda _0]+1}^nh(X_i,X_j)\\&-n^{-3/2}\bigg [\sum _{1\le i<j\le [(n+1)t]}h(X_i,X_j) +\sum _{[(n+1)t] +1\le i<j\le [n\lambda _0]}h(X_i,X_j)\\&+\sum _{[(n+1)t]+1\le i\le [n\lambda _0]}\sum _{[n\lambda _0]+1\le j\le n}h(X_i,X_j)\bigg ] \\= & {} R_n^{(1)} + R_n^{(2)} - \left( R_n^{(3)} + R_n^{(4)} + R_n^{(5)}\right) . \end{aligned}$$

First we prove that

$$\begin{aligned} n^{-1/2}R_n^{(1)}{\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty } \lambda ^2_0\theta (F,F)/2. \end{aligned}$$

From the Hoeffding decomposition (4), we have

$$\begin{aligned} n^{-1/2}R_n^{(1)}= & {} \frac{1}{2n^{2}}U_{[n\lambda _0]}\nonumber \\= & {} \frac{[n\lambda _0]([n\lambda _0]-1)}{2n^2}\theta (F,F)+ \frac{[n\lambda _0]([n\lambda _0]-1)}{n^2}\frac{2}{[n\lambda _0]}\sum _{i=1}^{[n\lambda _0]}h_{F,1}(X_i)\nonumber \\&+ \frac{[n\lambda _0]([n\lambda _0]-1)}{2n^2} U_{[n\lambda _0]}^{(2)}. \end{aligned}$$

(22)

As $(h_{1}^{(1)}(X_{i}))_{1\le i\le n}$ is stationary and ergodic, we have

$$\begin{aligned} \frac{1}{[n\lambda _0]}\sum _{i=1}^{[n\lambda _0]}h_{F,1}(X_i){\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty } 0. \end{aligned}$$

For any $\varepsilon >0$, put

$$\begin{aligned} A_{[n\lambda _0]}={\mathbb {P}}\Big (\Big |U_{[n\lambda _0]}^{(2)}\Big |>\varepsilon \Big ). \end{aligned}$$

One has from Markov inequality and Lemma 2 of Yoshihara (1976)

$$\begin{aligned} {\mathbb {P}}\Big (\Big |U_{[n\lambda _0]}^{(2)}\Big |>\varepsilon \Big )\le & {} \frac{1}{\varepsilon ^2} {\mathbb {E}}\left[ \left( U_{[n\lambda _0]}^{(2)}\right) ^2 \right] \\= & {} {\mathcal {O}}([n\lambda _0]^{-1-\gamma }), \ \gamma >0, \end{aligned}$$

which implies

$$\begin{aligned} \sum _{i=1}^{[n\lambda _0]}A_{[i\lambda _0]}<\infty . \end{aligned}$$

Then from Borel–Cantelli lemma

$$\begin{aligned} \frac{[n\lambda _0]([n\lambda _0]-1)}{2n^2} U_{[n\lambda _0]}^{(2)}{\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty } 0. \end{aligned}$$

Then from (22), we have

$$\begin{aligned} n^{-1/2}R_n^{(1)}{\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty } \lambda ^2_0\theta (F,F)/2. \end{aligned}$$

Similarly, we prove

$$\begin{aligned} n^{-1/2}R_n^{(3)}{\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty } t^2\theta (F,F)/2 \end{aligned}$$

and

$$\begin{aligned} n^{-1/2}R_n^{(4)}{\mathop {=}\limits ^{ {\mathcal {D}} }}\sum _{1\le i < j \le [n\lambda _0] -[(n+1)t]}h(X_i,X_j){\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty }(t-\lambda _0)^2\theta (F,F)/2. \end{aligned}$$

Now, we establish that

$$\begin{aligned} n^{-1/2}R_n^{(2)}{\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty } \lambda _0(1-\lambda _0)\theta (F,G). \end{aligned}$$

From (21), we have

$$\begin{aligned} n^{-1/2}R_n^{(2)}= & {} \frac{1}{n^2}\sum _{i=1}^{[n\lambda _0]}\sum _{j=[n\lambda _0]+1}^n\{\theta (F,G)+ h_{G,1}(X_i)+ h_{G,2}(Y_j)+ g_G(X_i,Y_j)\}\nonumber \\= & {} \frac{[n\lambda _0](n-[n\lambda _0])}{n^{2}}\theta (F,G) +\frac{[n\lambda _0](n-[n\lambda _0])}{n^{2}}\frac{1}{[n\lambda _0]} \sum _{i=1}^{[n\lambda _0]}h_{F,1}(X_{i})\nonumber \\&+\frac{[n\lambda _0](n-[n\lambda _0])}{n^{2}}\frac{1}{(n-[n\lambda _0])} \sum _{j=[n\lambda _0]+1}^{n}h_{G,2}(Y_{j})\nonumber \\&+\frac{1}{n^{2}}\sum _{i=1}^{[n\lambda _0]}\sum _{j=[n\lambda _0]+1}^{n}g_F(X_{i},Y_{j}), \end{aligned}$$

(23)

where we recall that the $Y_{j}$’s are random variables with cumulative distribution function G and satisfy (5).

From the ergodic theorem, we have that

$$\begin{aligned} \frac{1}{[n\lambda _0]}\sum _{i=1}^{[n\lambda _0]}h_{F,1}(X_i){\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty } 0. \end{aligned}$$

and

$$\begin{aligned} \frac{1}{(n-[n\lambda _0])}\sum _{j=[n\lambda _0]+1}^{n}h_{G,2}(Y_{j}){\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty }0. \end{aligned}$$

From Lemma 2, we deduce that

$$\begin{aligned} {\mathbb {E}}\left\{ \left[ \frac{1}{n^{2}}\sum _{i=1}^{[n\lambda _0]}\sum _{j=[n\lambda _0]+1}^{n}g_F(X_{i},Y_{j})\right] ^{2} \right\} \le Cst[n\lambda _0](n-[n\lambda _0])n^{-4}. \end{aligned}$$

From Markov inequality, we deduce for any $\epsilon >0$ that

$$\begin{aligned} {\mathbb {P}}\left( \left| \frac{1}{n^{2}}\sum _{i=1}^{[n\lambda _0]}\sum _{j=[n\lambda _0]+1}^{n}g_F(X_{i},Y_{j})\right| >\epsilon \right) ={\mathcal {O}}(n^{-2}). \end{aligned}$$

Also, by Borel–Cantelli Lemma one has

$$\begin{aligned} \frac{1}{n^{2}}\sum _{i=1}^{[n\lambda _0]}\sum _{j=[n\lambda _0]+1}^{n}g_F(X_{i},Y_{j}){\mathop {\longrightarrow }\limits ^{ a.s. }}_{n\rightarrow \infty }0. \end{aligned}$$

Similarly, we prove that

$$\begin{aligned} n^{-1/2}R_n^{(5)}{\mathop {\longrightarrow }\limits ^{{\mathbb {P}} }}(\lambda _0-t)(1-\lambda _0)\theta (F,G). \end{aligned}$$

These observations clearly imply the first part of (7). The proof of its second part is similar. $\square$