General Jackknife empirical likelihood and its applications

Abstract

This paper is to give a novel Jackknife Empirical Likelihood for both smooth and unsmooth statistical functionals. Specifically, we consider the estimation of unknown statistical functionals under the framework of general delete-d jackknife and a usage of the subsampling method which can reduce the computational burden. Moreover, the corresponding statistical inference issues and asymptotic properties of the proposed method are also investigated. Several related application examples are carefully conducted to check the superiority of our new proposed method. Finally, the finite sample simulation studies are also included.

Appendix

Lemma 1

(Burkholder’s inequality, Theorem 2.10 in Hall and Heyde 2014) If $\left\{ X_i, 1 \leqslant i \leqslant n\right\}$ is a martingale difference sequence and $1<p<\infty$, then there exist constants $C_{1}$ and $C_{2}$ depending only on p such that

$$\begin{aligned} C_{1} \textrm{E}\left| \sum _{i=1}^{n} X_{i}^{2}\right| ^{p / 2} \leqslant \textrm{E}\left| \sum _{i=1}^nX_i\right| ^{p} \leqslant C_{2} \textrm{E}\left| \sum _{i=1}^{n} X_{i}^{2}\right| ^{p / 2}. \end{aligned}$$

Lemma 2

Under conditions of Theorem 1, we have

$$\begin{aligned} \frac{d}{\sqrt{n}f}\sum _{i{=}1}^N(\tilde{\theta }_{{\mathcal {S}}_i}{-}\theta _0){\rightarrow } N(0,\sigma ^2). \end{aligned}$$

Proof

From (4), we have $\sum _{i{=}1}^N(\tilde{\theta }_{{\mathcal {S}}_i}{-}\theta ){=}\frac{N{-}f}{d}\sum _{i{=}1}^n\phi _F(X_i)$${+}o_P(n^{1/2})$. Then, by the Central Limit Theorem, the conclusion follows. $\square$

Lemma 3

Under conditions of Theorem 1, we have

$$\begin{aligned} S= & {} \frac{1}{N}\sum _{i=1}^N(\tilde{\theta }_{{\mathcal {S}}_i}-\theta )^2\\= & {} \frac{1}{Nd^2}\left[ (N-2f+\eta )\left( \sum _{k=1}^n\phi _F(X_k)\right) ^2\right. \\{} & {} \left. +(f-\eta )\sum _{k=1}^n\phi _F^2(X_k)\right] +o_P(n^{-1}). \end{aligned}$$

Proof

Recalling (1), write

$$\begin{aligned}{} & {} \frac{1}{N}\sum _{i=1}^N({\tilde{\theta }}_{{\mathcal {S}}_i}-\theta )^2\\{} & {} \quad =\frac{1}{Nd^2}\sum _{i=1}^N\left[ \sum _{k=1}^n\phi _F(X_k)-\sum _{j\in {\mathcal {S}}_i}\phi _F(X_j)+(nR_n-rR_1)\right] ^2\\{} & {} \quad :=\frac{1}{Nd^2}\sum _{i=1}^N\left( \sum _{k=1}^n\phi _F(X_k)-\sum _{j\in {\mathcal {S}}_i}\phi _F(X_j)\right) ^2+R, \end{aligned}$$

where $R=o(n^{-1})$. Consider the above first sum,

$$\begin{aligned}{} & {} \sum _{i=1}^N\left( \sum _{k=1}^n\phi _F(X_k)-\sum _{j\in {\mathcal {S}}_i}\phi _F(X_j)\right) ^2\\{} & {} \quad =N\left( \sum _{k=1}^n\phi _F(X_k)\right) ^2 -2\sum _{i=1}^N\left( \sum _{k=1}^n\phi _F(X_k)\right) \\{} & {} \qquad \times \left( \sum _{j\in {\mathcal {S}}_i}\phi _F(X_j)\right) +\sum _{i=1}^N\left( \sum _{j\in {\mathcal {S}}_i}\phi _F(X_j)\right) ^2\\{} & {} \quad =(N-2f)\left( \sum _{k=1}^n\phi _F(X_k)\right) ^2 +f\sum _{k=1}^n\phi _F^2(X_k)\\{} & {} \qquad +2\eta \sum _{1\le k<i\le n}\phi _F(X_i)\phi _F(X_k)+O_P(1)\\{} & {} \quad =(N-2f+\eta )\left( \sum _{k=1}^n\phi _F(X_k)\right) ^2\\{} & {} \qquad +(f-\eta )\sum _{k=1}^n\phi _F^2(X_k)+O_P(1), \end{aligned}$$

where in the third equality one applies (5). Hence the conclusion follows. $\square$

Proof of Theorem 1

By the law of large numbers, it is easy to obtain $\frac{1}{n}\sum _{i=1}^n\phi _F^2(X_i)=\sigma ^2+o(1)$. Then combined with the results in Lemma 2 and 3, the conclusion follows. $\square$

Below are the lemmas required for Theorem 2:

Lemma 4

Suppose Y is a random variable with mean zero and nonconstant probability density function $p(\cdot )$ almost everywhere. For $Z=\max (0,Y)$, we have (1). $\textrm{E}[Z]>0$, (2). $\textrm{Var}(Z)\le \textrm{Var}(Y)$.

Proof

For (1), since $p(\cdot )$ is a nonconstant probability density function almost everywhere, $\textrm{Var}(Y)>0$. By $\textrm{E}[Y]=0$, we have ${\textbf{P}}(Y>0)>0$, which in turn implies that $\textrm{E}[Z]>0$. For (2), we observe that

$$\begin{aligned} \textrm{E}[Z^2]= & {} \int _{-\infty }^{\infty }\max (0,y)^2p(y)dy=\int _{0}^{\infty }y^2p(y)dy\\= & {} \int _{0}^{\infty }(y-\textrm{E}[Y])^2p(y)dy<\textrm{Var}(Y), \end{aligned}$$

due to the assumption that ${\mathcal {E}}[Y]=0$ and hence the conclusion follows. $\square$

Lemma 5

Under the conditions of Theorem 2, for the given (9), we have

$$\begin{aligned} {\textbf{P}}\left( \min \limits _{1\le i\le \delta _n,~1\le j\le m_n}(\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)<0<\max \limits _{1\le i\le \delta _n,~1\le j\le m_n}(\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)\right) \rightarrow 1 \end{aligned}$$

Proof

Let $\psi (x)=\max (0,x)$. Without loss of generality, we assume that $\theta _0=0$, and hence

$$\begin{aligned} {\tilde{\theta }}_{{\mathcal {S}}_1}=\underbrace{\frac{1}{d}\sum _{i\in {\mathcal {S}}_1^c}\phi _F(X_i)}_{:=\Delta }+\underbrace{\frac{1}{d}(nR_n-rR_1)}_{:={\tilde{R}}_1}. \end{aligned}$$

We next aim to show that ${\textbf{P}}(\max _{1\le i\le N}{\tilde{\theta }}_{{\mathcal {S}}_i}\le 0)\rightarrow 0$ as $n\rightarrow \infty$. Similarly to Jing et al. (2009), write

$$\begin{aligned}{} & {} {\textbf{P}}(\max _{1\le i\le N}{\tilde{\theta }}_{{\mathcal {S}}_i}\le 0)={\textbf{P}}(\psi ({\tilde{\theta }}_{{\mathcal {S}}_i})=0 \text { for all } 1\le i\le N )\\{} & {} \quad ={\textbf{P}}(\sum _{i=1}^N\psi ({\tilde{\theta }}_{{\mathcal {S}}_i})=0)\\{} & {} \quad \le {\textbf{P}}\left( \left| \sum _{i=1}^N\psi ({\tilde{\theta }}_{{\mathcal {S}}_i})-\textrm{E}\psi ({\tilde{\theta }}_{{\mathcal {S}}_i})\right| \ge N\textrm{E}\psi ({\tilde{\theta }}_{{\mathcal {S}}_1})\right) \\{} & {} \quad \le \frac{N\textrm{Var}(\psi ({\tilde{\theta }}_{{\mathcal {S}}_i}))+N(N-1)\text{ cov }(\psi ({\tilde{\theta }}_{{\mathcal {S}}_1}),\psi ({\tilde{\theta }}_{{\mathcal {S}}_2}))}{N^2(\textrm{E}\psi ({\tilde{\theta }}_{{\mathcal {S}}_i}))^2}. \end{aligned}$$

For $\textrm{Var}(\psi ({\tilde{\theta }}_{{\mathcal {S}}_i}))$, by (2) in Lemma 4, we have $\textrm{Var}(\psi ({\tilde{\theta }}_{{\mathcal {S}}_i}))\le \textrm{Var}(\Delta )+\textrm{Var}({\tilde{R}}_1)+\text{ cov }(\Delta ,{\tilde{R}}_1)\le K/d$. For $\text{ cov }(\psi ({\tilde{\theta }}_{{\mathcal {S}}_1}),\psi ({\tilde{\theta }}_{{\mathcal {S}}_2}))$, by Holder’s inequality, $|\text{ cov }(\psi ({\tilde{\theta }}_{{\mathcal {S}}_1}),\psi ({\tilde{\theta }}_{{\mathcal {S}}_2}))|\le K/d$. From (1) in Lemma 4, $\textrm{E}\psi ({\tilde{\theta }}_{{\mathcal {S}}_1})\ge \frac{1}{d}\sum _{i\in {\mathcal {S}}_1}\textrm{E}\max \{\phi _F(X_i),0\}-o(1)>0$. Therefore, we have ${\textbf{P}}(\max _{1\le i\le N}{\tilde{\theta }}_{{\mathcal {S}}_i}\le 0)\rightarrow 0$ as $n\rightarrow \infty$. Similarly, one can obtain that ${\textbf{P}}(\min _{1\le i\le N}{\tilde{\theta }}_{{\mathcal {S}}_i}\ge 0)\rightarrow 0$ as $n\rightarrow \infty$, and thus the conclusion follows. $\square$

Lemma 6

For the $\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}$ as defined in (8), we have

$$\begin{aligned} \max \limits _{1\le i\le \delta _n,~1\le j\le m}\left| (\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)\right| =O_P((\log n/n)^{1/4}) \end{aligned}$$

Proof

Recall that $\{X_k\}_{k\in {\mathcal {B}}_i}=\{X_k^{(i)}\}_{k=1}^{m_n}$ and (8). Write $\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0=\frac{2}{m_n}\sum _{k\in {\mathcal {S}}_j^c}\phi _F(X_k^{(i)})+R'$, where $R'=o_P(n^{-1/2})$. Note that

$$\begin{aligned}{} & {} {\textbf{P}}\left( \max \limits _{1\le i\le \delta _n,~1\le j\le m}\frac{2}{m_n}\left| \sum _{k\in {\mathcal {S}}_j^c}\phi _F(X_k^{(i)})\right| \ge t\right) \\{} & {} \quad \le n {\textbf{P}}\left( \frac{2}{m_n}\left| \sum _{k\in {\mathcal {S}}_j^c}\phi _F(X_k^{(i)})\right| \ge t\right) . \end{aligned}$$

Using Lemma 1 and Chebyshev’s inequality, and choosing $t=(\log n/n)^{1/4}$, one can conclude the result. $\square$

Lemma 7

$\sum _{i{=}1}^{\delta _n}\!\!\sum _{j{=}1}^{m_n}({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}{-}\theta _0){=}\!\!\sum _{k{=}1}^n\phi _F(X_k){+}o_P(n^{1/2}\!).$

Proof

Following Lemma 2, we have $\sum _{j=1}^{m_n}({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)=\sum _{k\in {\mathcal {B}}_i}\phi _F(X_k)+o_P(m_n^{1/2})$. By the construction of ${\mathcal {B}}_i$’s, we thus have $\sum _{i=1}^{\delta _n}\sum _{j=1}^{m_n}({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)=\sum _{k=1}^n\phi _F(X_k)+o_P(m_n^{1/2}\delta _n^{1/2})$. The conclusion follows. $\square$

Lemma 8

${\tilde{S}}=\frac{1}{n}\sum _{i=1}^{\delta _n}\sum _{j=1}^{m_n}({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)^2=2\sigma ^2/m_n+o_P(1/m_n)$.

Proof

Following Lemma 3, we have

$$\begin{aligned} \frac{1}{m_n}\sum _{j=1}^{m_n}({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)^2= & {} \left( \frac{1}{m_n}\sum _{k\in {\mathcal {B}}_i}\phi _F(X_k)\right) ^2\\{} & {} +\frac{1}{{m_n}^2}\sum _{k\in {\mathcal {B}}_i}\phi _F^2(X_k)+o_P(1/{m_n})\\:= & {} A_j+B_j+o_P(1/{m_n}). \end{aligned}$$

Hence, ${\tilde{S}}=\frac{1}{\delta _n}\sum _{j=1}^{\delta _n}(A_j+B_j)+o_P(1/{m_n})$. Since $A_{j_1}$ and $A_{j_2}$ are constructed from two disjoint sets when $j_1\ne j_2$, by the strong law of large numbers, we have $\frac{1}{\delta _n}\sum _{j=1}^{\delta _n}A_j=\sigma ^2/{m_n}+o(1/{m_n})$ as $\delta _n,m_n\rightarrow \infty$. Moreover, also by the strong law of large numbers, $B_j=\sigma ^2/{m_n}+o_P(1/{m_n})$. We thus obtain the conclusion. $\square$

Proof of Theorem 2.

Proof

From Lemma 5, it follows that with probability tending to one the true $\theta _0$ satisfies $\min \limits _{1\le i\le \delta _n,~1\le j\le m}\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}<\theta _0<\max \limits _{1\le i\le \delta _n,~1\le j\le m}\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}$, and the solution to (12) exists and is unique. By (12) and Lemma 7, we have

$$\begin{aligned} 0= & {} |g(\lambda )|\\= & {} \frac{1}{n}\left| \sum _{i=1}^{\delta _n}\sum _{j=1}^{m_n}({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)} -\theta _0)-\lambda \sum _{i=1}^{\delta _n}\sum _{j=1}^{m_n}\frac{({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)^2}{1+\lambda ({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)}\right| \\\ge & {} \frac{|\lambda |}{n}\sum _{i=1}^{\delta _n}\sum _{j=1}^{m_n}\frac{({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)}{1+\lambda ({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)}-\left| \frac{1}{n}\sum _{i=1}^{\delta _n}\sum _{j=1}^{m_n}({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)\right| \\\ge & {} \frac{|\lambda |{\tilde{S}}}{1+|\lambda |W_n}-O_P(n^{-1/2}), \end{aligned}$$

where $W_n=\max \limits _{1\le i\le \delta _n,~1\le j\le m}\left| (\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)\right|$. From Lemma 8, ${\tilde{S}}=O_P(1/{m_n})$, which combined with Lemma 6 shows that $\lambda =O_P(m_n/n^{1/2})$. Let $\gamma _{ij}=\lambda (\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)$. Applying $\frac{1}{1+x}=1-x+\frac{x^2}{1+x}$ to (12), one can obtain that

$$\begin{aligned} 0=\frac{1}{n}\sum _{i=1}^{\delta _n}\sum _{j=1}^{m_n}({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)\left[ 1-\gamma _{ij}+\frac{\gamma _{ij}^2}{1+\gamma _{ij}}\right] . \end{aligned}$$

By using Lemmas 6, 8 and the fact that $\lambda =O_P(m_n/n^{1/2})$, we have

$$\begin{aligned} \frac{1}{n} \sum _{i=1}^{\delta _n}\sum _{j=1}^{m_n}\lambda ^2(\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)^3=\frac{1}{\delta _n}\left( \frac{\log n}{n}\right) ^{1/4}, \end{aligned}$$

and hence, $\lambda {=}{\tilde{S}}^{{-}1}(\frac{1}{n}\sum _{i{=}1}^{\delta _n}\sum _{j{=}1}^m({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}{-}\theta _0)){+}o_P(m_n/n^{1/2})$. Applying Taylor’s expansion and plugging the expression of $\lambda$ into $-4\log R(\theta _0)/m$, we have

$$\begin{aligned} {-}4\log R(\theta _0)/m&{=}&\frac{4}{m}\left[ \sum _{i{=}1}^{\delta _n}\sum _{j{=}1}^m\gamma _{ij}{-}\frac{1}{2} \sum _{i{=}1}^{\delta _n}\sum _{j{=}1}^m\gamma _{ij}^2\right] {+}o_P(1)\\= & {} \frac{4}{m}\left[ \frac{mn(\frac{1}{n}\sum _{k=1}^n\phi _F(X_k))^2}{4\sigma ^2}\right] \\{} & {} +o_P(1)\overset{d}{\rightarrow }\chi ^2_1. \end{aligned}$$

The conclusion follows.$\square$

Now we consider the estimators satisfying (3). To derive the asymptotic distribution of the proposed statistic, some more lemmas are required. Let $\sigma _h^2=\textrm{E}h^2(X_1,X_2)$.

Lemma 9

Under Condition C1, we have

$$\begin{aligned} \sqrt{\frac{2}{n(n-1)}}\sum _{i<j}h(X_i,X_j)\overset{d}{\rightarrow }N(0,\sigma _h^2). \end{aligned}$$

Proof

Write

$$\begin{aligned} \sum _{1\le i<j\le n}h(X_i,X_j)=\sum _{j=2}^n\sum _{i=1}^jh(X_i,X_j)\equiv \sum _{j=2}^nH_j. \end{aligned}$$

Let

$$\begin{aligned} M_r=\sum _{j=2}^r\sum _{i=1}^{j-1}h(X_i,X_j), \end{aligned}$$

and define the filtration ${\mathcal {F}}_r=\sigma \{Z_1,\ldots ,Z_r\}$. Note that from the condition that

$$\begin{aligned} \textrm{Var}(\textrm{E}[h(X_i,X_j)|X_j])=0, \end{aligned}$$

we have $\textrm{E}[h(X_i,X_j)|X_j]=0$ with probability one. So $\textrm{E}[M_r]=0$ and

$$\begin{aligned} \textrm{E}[M_{r'}|{\mathcal {F}}_r]= & {} M_r+\textrm{E}\left[ \sum _{j=r+1}^{r'}\sum _{i=1}^{j-1} h(X_i,X_j)\Big |{\mathcal {F}}_r\right] \\= & {} M_r\text { for }r'>r, \end{aligned}$$

and hence $\{M_r\}_{r=2}^n$ forms a mean-zero martingale sequence. Define $V_j=\sum _{i=1}^{j-1}h(X_i,X_j)$. Following Corollary 3.1 in Hall and Heyde (2014), we will prove that

$$\begin{aligned} \sum _{j=1}^nB^{-2}\textrm{E}[V_j^2{\textbf{1}}\{|V_j|>\epsilon B\}|{\mathcal {F}}_{j-1}]\overset{p}{\rightarrow }0 \end{aligned}$$

(23)

for B such that

$$\begin{aligned} \sum _{j=1}^n\textrm{E}[V_j^2|{\mathcal {F}}_{j-1}]/B^2 \overset{p}{\rightarrow }C>0. \end{aligned}$$

(24)

Note that

$$\begin{aligned} \sum _{j=1}^n\textrm{E}[V_j^2|{\mathcal {F}}_{j-1}]= & {} \sum _{i,k=1}^{j-1}\textrm{E}\left[ h(X_i,X_j)h(X_k,X_j)\Big |{\mathcal {F}}_{j-1}\right] \\\equiv & {} \sum _{i,k=1}^{j-1}H(X_i,X_k) \end{aligned}$$

and

$$\begin{aligned} \sum _{j=1}^n\textrm{E}[V_j^2]= & {} \sum _{j=1}^n\sum _{i,k=1}^{j-1}\textrm{E}\left[ h(X_i,X_j)h(X_k,X_j)\right] \\= & {} \frac{n(n-1)}{2}\textrm{E}[h(X_1,X_2)^2]\equiv \frac{n(n-1)}{2}\sigma _h^2. \end{aligned}$$

To prove (24), by setting $C{=}1$ and $B{=}\frac{q(q{-}1)}{2}\textrm{E}[h(X_1,X_2)^2]$, we consider $\textrm{Var}\left( \sum _{j{=}1}^q\textrm{E}[V_j^2|{\mathcal {F}}_{j-1}]\right)$. For $j_1\ge j_2$, write

$$\begin{aligned}{} & {} \text{ cov }\left( \textrm{E}[V_{j_1}^2|{\mathcal {F}}_{{j_1}-1}],\textrm{E}[V_{j_2}^2|{\mathcal {F}}_{{j_2}-1}]\right) \\{} & {} \quad =\sum _{i_1,k_1=1}^{j_1-1}\sum _{i_2,k_2=1}^{j_2-1}\text{ cov }\left( H(X_{i_1},X_{k_1}),H(X_{i_2},X_{k_2})\right) \\{} & {} \quad =(j_2-1)\textrm{Var}(H(X_1,X_1))\\{} & {} \qquad +2(j_2-1)(j_2-2)\textrm{E}[H(X_1,X_2)^2]. \end{aligned}$$

Notice that

$$\begin{aligned}{} & {} \sum _{j_1,j_2=1}^q\text{ cov }\left( \textrm{E}[V_{j_1}^2|{\mathcal {F}}_{{j_1}-1}],\textrm{E}[V_{j_2}^2|{\mathcal {F}}_{{j_2}-1}]\right) \\{} & {} \quad =C_1q^3 \textrm{Var}(H(X_1,X_1)) +C_2q^4\textrm{E}[H(X_1,X_2)^2]. \end{aligned}$$

Hence, under Condition C1, combining Chebyshev’s inequality, one can obtain (24). To show (23), it suffices to prove that

$$\begin{aligned} B^{-4}\sum _{j=1}^q\textrm{E}[|V_j|^4]\rightarrow 0. \end{aligned}$$

Note that

$$\begin{aligned} \sum _{j=1}^q\textrm{E}[|V_j|^4]= & {} \sum _{j=1}^q\sum _{i_1,i_2,i_3,i_4=1}^{j-1}\textrm{E}\left[ h(X_{i_1},X_j)h(X_{i_2},X_j)\right. \\{} & {} \left. h(X_{i_3},X_j)h(X_{i_4},X_j)\right] \\= & {} \frac{q(q-1)}{2}\textrm{E}[h(X_1,X_2)^4]\\{} & {} +3\sum _{j=2}^q\sum _{i_1\ne i_2}\textrm{E}\left[ h(X_{i_1},X_j)^2h(X_{i_2},X_j)^2\right] \\= & {} \frac{q(q-1)}{2}\textrm{E}[h(X_1,X_2)^4]\\{} & {} +6\sum _{j=2}^q(j-1)(j-2)\textrm{E}\\{} & {} \left[ h(X_1,X_3)^2h(X_2,X_3)^2\right] \\= & {} o(B^{-4}), \end{aligned}$$

which implies (23). Therefore we have

$$\begin{aligned} \frac{1}{\sqrt{n}} \sum _{k=1}^nh(Z_k)= & {} \sqrt{\frac{2}{q(q-1)}}\sum _{1\le i<j\le q}h(X_i,X_j)\nonumber \\{} & {} \overset{d}{\rightarrow }N(0,\sigma ^2_h). \end{aligned}$$

(25)

$\square$

Lemma 10

Consider $\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}$ defined in (13). Then

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^{\delta _n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) \overset{d}{=}\frac{3}{4q_m}\sqrt{\frac{m_n(m_n-1)}{2\delta _n}}{\mathcal {Z}}+o_P(1), \end{aligned}$$

where ${\mathcal {Z}}\sim N(0,\sigma _h^2).$

Proof

Write

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^{\delta _n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) =\frac{1}{\delta _n}\sum _{k=1}^{\delta _n}\frac{1}{m_n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) . \end{aligned}$$

Using (5) we have $\eta =m_n/4+O(1)$. Consider

$$\begin{aligned}{} & {} \frac{1}{m_n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) \\{} & {} \quad = \frac{1}{q_m\cdot m_n}\sum _{\ell =1}^{m_n}\left[ {m_n\atopwithdelims ()2}{\hat{\theta }}_{{\mathcal {B}}_k}-{r\atopwithdelims ()2}{\hat{\theta }}_{{\mathcal {S}}_\ell }^{(k)}-\theta _0\right] \\{} & {} \quad = \frac{1}{q_m\cdot m_n}\sum _{\ell =1}^{m_n}\left[ \sum _{(i,j)\in {\mathcal {V}}_{{\mathcal {B}}_k}}h(X_i,X_j)\right. \\{} & {} \qquad \left. - \sum _{(i.j)\in {\mathcal {V}}_{S_\ell }}h(X_i^{(k)},X_j^{(k)})\right] +R'\\{} & {} \quad =\frac{3m_n}{4q_m\cdot m_n}\sum _{(i,j)\in {\mathcal {V}}_{{\mathcal {B}}_k}}h(X_i,X_j)+R'\\{} & {} \quad =\frac{3}{4q_m}\sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})+R', \end{aligned}$$

where $R'=o_P(n^{-1})$. From Lemma 9, it is easy to see that

$$\begin{aligned} \frac{3}{4q_m}\sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})\overset{d}{=} \frac{3}{4q_m}\sqrt{\frac{m_n(m_n-1)}{2}}{\mathcal {Z}}_k, \end{aligned}$$

where ${\mathcal {Z}}_k\sim N(0,\sigma _h^2).$ Therefore, by the Central Limit Theorem,

$$\begin{aligned}{} & {} \frac{1}{n}\sum _{k=1}^{\delta _n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) \\{} & {} \quad = \frac{1}{\delta _n}\sum _{k=1}^{\delta _n}\left[ \frac{3}{4q_m}\sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})+R_k'\right] \\{} & {} \quad \overset{d}{=}\frac{3}{4q_m}\sqrt{\frac{m_n(m_n-1)}{2\delta _n}}\frac{1}{\sqrt{\delta _n}}\sum _{k=1}^{\delta _n}{\mathcal {Z}}_k\\{} & {} \qquad +R''\overset{d}{=}\frac{3}{4q_m}\sqrt{\frac{m_n(m_n-1)}{2\delta _n}}{\mathcal {Z}}. \end{aligned}$$

Thus the conclusion follows. $\square$

Lemma 11

For $\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}$ defined in (13), we have

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^{\delta _n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) ^2\overset{p}{\rightarrow }\frac{3m_n(m_n-1)}{8q_m^2}(\sigma _h^2+o_P(1)). \end{aligned}$$

Proof

We first consider

$$\begin{aligned} \sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) ^2= & {} \frac{1}{q_m^2}\sum _{\ell =1}^{m_n}\left[ \sum _{1\le i<j\le m_n}h\left( X_i^{(k)},X_j^{(k)}\right) \right. \\{} & {} \left. - \sum _{(i.j)\in {\mathcal {V}}_{S_\ell }}h\left( X_i^{(k)},X_j^{(k)}\right) \right] ^2+R, \end{aligned}$$

where R is a small order term compared to the leading term. By simple calculations, it is easy to find that the term R plays a negligible role in the proof. For simplicity, in the sequel, we omit the term R. We first look at

$$\begin{aligned}{} & {} \left[ \sum _{1\le i<j\le m_n}h\left( X_i^{(k)},X_j^{(k)}\right) \right] ^2\\{} & {} \quad {=} \sum _{ i<j}h^2(X_i^{(k)},X_j^{(k)}){+}2\!\!\sum _{ i<j<\ell }h(X_i^{(k)},X_j^{(k)})h(X_i^{(k)},X_\ell ^{(k)})\\{} & {} \qquad +2\sum _{ i<j<\ell }h(X_i^{(k)},X_j^{(k)})h(X_j^{(k)},X_\ell ^{(k)})\\{} & {} \qquad +2\sum _{ i<j<\ell }h(X_i^{(k)},X_\ell ^{(k)})h(X_j^{(k)},X_\ell ^{(k)})\\{} & {} \qquad +2\sum _{i<j<\ell <t}\Big [h(X_i^{(k)},X_j^{(k)})h(X_\ell ^{(k)},X_t^{(k)})\\{} & {} \qquad +h(X_i^{(k)},X_\ell ^{(k)})h(X_j^{(k)},X_t^{(k)})\\{} & {} \qquad +h(X_i^{(k)},X_t^{(k)})h(X_j^{(k)},X_\ell ^{(k)})\Big ] \end{aligned}$$

We respectively consider the terms above. For the first term, by Hoeffding (1992), it is easy to obtain that

$$\begin{aligned} {m_n\atopwithdelims ()2}^{-1}\sum _{ i<j}h^2(X_i^{(k)},X_j^{(k)})\overset{p}{\rightarrow }\sigma _h^2. \end{aligned}$$

(26)

For the second to the fourth terms, we take $\sum _{i<j}^{\ell -1}h(X_i^{(k)},X_\ell ^{(k)})h(X_j^{(k)},X_\ell ^{(k)})$ as an example for illustration. The others are handled similarly. So we omit them. Let

$$\begin{aligned} U_{\ell }^{(k)}=\sum _{i<j}^{\ell -1}h(X_i^{(k)},X_\ell ^{(k)})h(X_j^{(k)},X_\ell ^{(k)}), \end{aligned}$$

and hence

$$\begin{aligned} \sum _{ i<j<\ell }h(X_i^{(k)},X_\ell ^{(k)})h(X_j^{(k)},X_\ell ^{(k)})=\sum _{\ell =3}^{m_n}U_{\ell }^{(k)}. \end{aligned}$$

For simplicity, we remove the superscript “(k)” when there is no confusion. Note that from the condition that $\textrm{Var}(\textrm{E}[h(X_i,X_j)|X_j])=0$, we have $\textrm{E}[h(X_i,X_j)|X_j]=0$ with probability one. Thus, we have $\textrm{E}U_\ell =0$. Now we consider $\textrm{E}(\sum _{\ell =3}^{m_n}U_{\ell })^2$, i.e.,

$$\begin{aligned} \textrm{E}(\sum _{\ell =3}^{m_n}U_{\ell })^2=\sum _{\ell =3}^{m_n}\textrm{E}U_{\ell }^2+\sum _{\ell _1\ne \ell _2\ge 3}\textrm{E}U_{\ell _1}U_{\ell _2}. \end{aligned}$$

Here

$$\begin{aligned} \textrm{E}U_{\ell }^2= & {} \sum _{i_1<j_1}^{\ell -1}\sum _{i_2<j_2}^{\ell -1}\textrm{E}\left[ h(X_{i_1},X_{\ell })h(X_{j_1},X_\ell )h(X_{i_2},X_{\ell })\right. \\{} & {} \left. h(X_{j_2},X_\ell )\right] \\= & {} \sum _{i<j}\textrm{E}\left[ h^2(X_{i},X_{\ell })h^2(X_{j},X_\ell )\right] . \end{aligned}$$

If $\ell _1<\ell _2$, when $j_1\ne j_2$, we have

$$\begin{aligned}{} & {} \textrm{E}\left[ h(X_{i_1},X_{\ell _1})h(X_{j_1},X_{\ell _1})h(X_{i_1},X_{\ell _2})h(X_{j_2},X_{\ell _2})\right] \\{} & {} \quad =\textrm{E}\left( \textrm{E}\left[ h(X_{i_1},X_{\ell _1})h(X_{j_1},X_{\ell _1})h(X_{i_1},X_{\ell _2})\right. \right. \\{} & {} \quad \quad \left. \left. h(X_{j_2},X_{\ell _2}) \Big |X_{i_1},X_{\ell _1},X_{\ell _2}\right] \right) \\{} & {} \quad =\textrm{E}\left( h(X_{i_1},X_{\ell _1})h(X_{i_1},X_{\ell _2})\textrm{E}\left[ h(X_{j_1},X_{\ell _1})\Big |X_{\ell _1}\right] \right. \\{} & {} \quad \quad \left. \quad \textrm{E}\left[ h(X_{j_2},X_{\ell _2})\Big |X_{\ell _2}\right] \right) =0, \end{aligned}$$

and similarly $\textrm{E}\left[ h(X_{i_1},X_{\ell _1})h(X_{j_1},X_{\ell _1})h(X_{i_2},X_{\ell _2})\right. \left. h(X_{j_2},X_{\ell _2})\right] =0$ for $i_1\ne j_1\ne i_2\ne j_2$. Thus,

$$\begin{aligned} \textrm{E}U_{\ell _1}U_{\ell _2}= & {} \sum _{i_1<j_1}^{\ell _1-1}\sum _{i_2<j_2}^{\ell _2-1}\textrm{E}\left[ h(X_{i_1},X_{\ell _1})h(X_{j_1},X_{\ell _1})\right. \\{} & {} \left. h(X_{i_2},X_{\ell _2})h(X_{j_2},X_{\ell _2})\right] \\= & {} \sum _{i_1<j_1}^{\ell _1-1}\textrm{E}\left[ h(X_{i_1},X_{\ell _1})h(X_{j_1},X_{\ell _1})h(X_{i_1},X_{\ell _2})\right. \\{} & {} \left. h(X_{j_1},X_{\ell _2})\right] \\= & {} \sum _{i_1<j_1}^{\ell _1-1}\textrm{E}H^2(X_{i_1},X_{j_1}). \end{aligned}$$

Consequently, one has

$$\begin{aligned} \textrm{E}\left( \sum _{\ell =3}^{m_n}U_{\ell }\right) ^2= & {} C_1m_n^3\textrm{E}\left[ h^2(X_{i},X_{\ell })h^2(X_{j},X_\ell )\right] \\{} & {} +C_2m_n^4\textrm{E}H^2(X_{i_1},X_{j_1}). \end{aligned}$$

Combining Condition C1, we thus have

$$\begin{aligned} \frac{\sum _{ i<j<\ell }h(X_i^{(k)},X_\ell ^{(k)})h(X_j^{(k)},X_\ell ^{(k)})}{\frac{m_n(m_n-1)}{2}\sigma _h^2}=o_P(1). \end{aligned}$$

(27)

Next we consider the remaining terms, and also take $\sum _{i<j<\ell <t}h(X_i^{(k)},X_j^{(k)})h(X_\ell ^{(k)},X_t^{(k)})$ as an exmple, which can be viewed as a U statistic of degree 4 with mean zero. Following Hoeffding (1992), it is easy to verify that such a U statistic is stationary of order 3. Hence, by (5.13) in Hoeffding (1992),

$$\begin{aligned}{} & {} \textrm{Var}\Big (\sum _{i<j<\ell <t}h(X_i^{(k)},X_j^{(k)})h(X_\ell ^{(k)},X_t^{(k)})\Big )\\{} & {} \quad ={m_n\atopwithdelims ()4}\{\textrm{E}[h(X_1,X_2)^2]\}^2. \end{aligned}$$

Hence, by the law of large numbers for U-statistics, one has

$$\begin{aligned} \frac{1}{\delta _n}\sum _{k=1}^{\delta _n}\sum _{i<j<\ell <t}{m_n\atopwithdelims ()2}^{-1}h(X_i^{(k)},X_j^{(k)})h(X_\ell ^{(k)},X_t^{(k)})\overset{p}{\rightarrow }0.\nonumber \\ \end{aligned}$$

(28)

Note that

$$\begin{aligned}{} & {} \sum _{\ell =1}^{m_n}\left[ \sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})\right] \cdot \left[ \sum _{(i.j)\in {\mathcal {V}}_{S_\ell }}h(X_i^{(k)},X_j^{(k)})\right] \\{} & {} \quad =\eta \left[ \sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})\right] ^2 \end{aligned}$$

Now, write

$$\begin{aligned}{} & {} \frac{1}{n}\sum _{k=1}^{\delta _n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) ^2=\frac{1}{\delta _n}\sum _{k=1}^{\delta _n}\frac{1}{m_n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) ^2\\{} & {} \quad =\frac{1}{\delta _n}\sum _{k=1}^{\delta _n}\frac{m_n-2\eta }{m_nq_m^2}\left[ \sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})\right] ^2\\{} & {} \qquad + \frac{1}{\delta _n}\sum _{k=1}^{\delta _n}\frac{1}{m_nq_m^2}\sum _{\ell =1}^{m_n} \left[ \sum _{(i.j)\in {\mathcal {V}}_{S_\ell }}h(X_i^{(k)},X_j^{(k)})\right] ^2. \end{aligned}$$

By (26), (27) (28) and the law of large numbers for U-statistics, we have

$$\begin{aligned} \frac{1}{\delta _n}\sum _{k=1}^{\delta _n}\frac{m_n-2\eta }{m_nq_m^2}\left[ \sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})\right] ^2\overset{p}{\rightarrow }\frac{{m_n\atopwithdelims ()2}}{2q_m^2}\sigma _h^2.\end{aligned}$$

Similarly, one can also prove that

$$\begin{aligned} \frac{1}{\delta _n}\sum _{k=1}^{\delta _n}\frac{1}{m_nq_m^2}\sum _{\ell =1}^{m_n}\left[ \sum _{(i.j)\in {\mathcal {V}}_{S_\ell }}h(X_i^{(k)},X_j^{(k)})\right] ^2\overset{p}{\rightarrow }\frac{{m_n\atopwithdelims ()2}}{4q_m^2}\sigma _h^2. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \frac{1}{n}\sum _{k=1}^{\delta _n}\sum _{\ell =1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }-\theta _0\right) ^2 \overset{p}{\rightarrow }\frac{3m_n(m_n-1)}{8q_m^2}\sigma _h^2, \end{aligned}$$

and the conclusion follows. $\square$

Lemma 12

For the $\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}$ as defined in (13), we have

$$\begin{aligned} \max \limits _{1\le i\le \delta _n,~1\le j\le m}\left| (\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)\right| =O_P((\log n\cdot \delta _n/m^3_n)^{1/4}). \end{aligned}$$

Proof

From the proof of Lemma 11, by tedious computations, it is easy to verify that

$$\begin{aligned} \textrm{E}\left| \sum _{1\le i< j\le m_n}h(X_i,X_j)\right| ^4=O(m_n^4). \end{aligned}$$

(29)

Write

$$\begin{aligned}{} & {} \max \limits _{1\le i\le \delta _n,~1\le j\le m_n}\left| (\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)\right| \\{} & {} \quad =\max \limits _{1\le k\le \delta _n,~1\le \ell \le m_n}\frac{1}{q_m}\left| \sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})\right. \\{} & {} \qquad \left. - \sum _{(i.j)\in {\mathcal {V}}_{S_\ell }}h(X_i^{(k)},X_j^{(k)})\right| \\{} & {} \quad \le \max \limits _{1\le k\le \delta _n}\frac{1}{q_m}\left| \sum _{1\le i<j\le m_n}h(X_i^{(k)},X_j^{(k)})\right| \\{} & {} \qquad +\max \limits _{1\le k\le \delta _n,~1\le \ell \le m_n}\frac{1}{q_m}\left| \sum _{(i.j)\in {\mathcal {V}}_{S_\ell }}h(X_i^{(k)},X_j^{(k)})\right| . \end{aligned}$$

Then combining Chebyshev’s inequality and (29), one can obtain the conclusion. $\square$

Proof of Theorem 3.

Proof

The proof of Theorem 3 is similar to that of Theorem 2. We only provide different parts. Note that Lemma 5 guarantees that the solution to (12) exists and is unique. From Lemma 11, we have ${\tilde{S}}=[\sigma _h^2+o_P(1)]/q_m$. Moreover, by Lemma 10, $\frac{1}{n}\sum _{k{=}1}^{\delta _n}\sum _{\ell {=}1}^{m_n}\left( \tilde{\theta }^{(k)}_{{\mathcal {S}}_\ell }{-}\theta _0\right) \overset{d}{=}\frac{3}{4q_m}\sqrt{\frac{m_n(m_n{-}1)}{2\delta _n}}{\mathcal {Z}}{+}o_P(1)$, where ${\mathcal {Z}}\sim N(0,\sigma _h^2).$ We thus have $\lambda =O_P(\sqrt{m_n^3/n})$. Let $\gamma _{ij}=\lambda (\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}-\theta _0)$. Applying $\frac{1}{1+x}=1-x+\frac{x^2}{1+x}$ to (12), one can obtain that

$$\begin{aligned} 0=\frac{1}{n}\sum _{i=1}^{\delta _n}\sum _{j=1}^m({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0)\left[ 1-\gamma _{ij}+\frac{\gamma _{ij}^2}{1+\gamma _{ij}}\right] . \end{aligned}$$

By Lemma 12, we have

$$\begin{aligned} \frac{1}{N} \sum _{i=1}^N\lambda ^2({\tilde{\theta }}_{{\mathcal {S}}_i}-\theta _0)^3=\left( \frac{\log n}{n^3}\right) ^{1/4}, \end{aligned}$$

and thus $\lambda ={\tilde{S}}^{-1}(\frac{1}{n}\sum _{i=1}^{\delta _n}\sum _{j=1}^m({\tilde{\theta }}_{{\mathcal {S}}_j}^{(i)}-\theta _0))+o_P(\sqrt{m_n^3/n})$. Applying Taylor’s expansion and plugging the expression of $\lambda$ into $-\log R(\theta _0)$, we have

$$\begin{aligned} -\log R(\theta _0)/m_n= & {} \left[ \sum _{i=1}^{\delta _n}\sum _{j=1}^m\gamma _{ij}-\frac{1}{2} \sum _{i=1}^{\delta _n}\sum _{j=1}^m\gamma _{ij}^2\right] /m_n\\{} & {} +o_P(1) \overset{d}{\rightarrow }\frac{3}{8}{\mathcal {Z}}^2/\sigma _h^2. \end{aligned}$$

Hence, $-8\log R(\theta _0)/(3m_n)\overset{d}{\rightarrow }\chi ^2_1$, and the conclusion follows. $\square$