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.
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Acknowledgements
The authors would like to express their gratitude to the editor, Prof. Jeremie Houssineau and two anonymous referees for their careful reading and useful comments which led to an improved presentation of the paper. Yiming Liu’s research was partially supported by the China Postdoctoral Science Foundations under Grant 2022M711342. Shaochen Wang’s research was partially supported by Guangdong Basic and Applied Basic Research Foundation (No. 2023A1515012125). Wang Zhou’s research was partially supported by a Grant A-8000440-00-00 at the National University of Singapore.
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Appendix
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
Lemma 2
Under conditions of Theorem 1, we have
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
Proof
Recalling (1), write
where \(R=o(n^{-1})\). Consider the above first sum,
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
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
Proof
Let \(\psi (x)=\max (0,x)\). Without loss of generality, we assume that \(\theta _0=0\), and hence
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
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
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
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
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
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
By using Lemmas 6, 8 and the fact that \(\lambda =O_P(m_n/n^{1/2})\), we have
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
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
Proof
Write
Let
and define the filtration \({\mathcal {F}}_r=\sigma \{Z_1,\ldots ,Z_r\}\). Note that from the condition that
we have \(\textrm{E}[h(X_i,X_j)|X_j]=0\) with probability one. So \(\textrm{E}[M_r]=0\) and
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
for B such that
Note that
and
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
Notice that
Hence, under Condition C1, combining Chebyshev’s inequality, one can obtain (24). To show (23), it suffices to prove that
Note that
which implies (23). Therefore we have
\(\square \)
Lemma 10
Consider \(\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}\) defined in (13). Then
where \({\mathcal {Z}}\sim N(0,\sigma _h^2).\)
Proof
Write
Using (5) we have \(\eta =m_n/4+O(1)\). Consider
where \(R'=o_P(n^{-1})\). From Lemma 9, it is easy to see that
where \({\mathcal {Z}}_k\sim N(0,\sigma _h^2).\) Therefore, by the Central Limit Theorem,
Thus the conclusion follows. \(\square \)
Lemma 11
For \(\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}\) defined in (13), we have
Proof
We first consider
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
We respectively consider the terms above. For the first term, by Hoeffding (1992), it is easy to obtain that
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
and hence
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.,
Here
If \(\ell _1<\ell _2\), when \(j_1\ne j_2\), we have
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,
Consequently, one has
Combining Condition C1, we thus have
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),
Hence, by the law of large numbers for U-statistics, one has
Note that
Now, write
By (26), (27) (28) and the law of large numbers for U-statistics, we have
Similarly, one can also prove that
Therefore, we have
and the conclusion follows. \(\square \)
Lemma 12
For the \(\tilde{\theta }_{{\mathcal {S}}_j}^{(i)}\) as defined in (13), we have
Proof
From the proof of Lemma 11, by tedious computations, it is easy to verify that
Write
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
By Lemma 12, we have
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
Hence, \(-8\log R(\theta _0)/(3m_n)\overset{d}{\rightarrow }\chi ^2_1\), and the conclusion follows. \(\square \)
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Liu, Y., Wang, S. & Zhou, W. General Jackknife empirical likelihood and its applications. Stat Comput 33, 74 (2023). https://doi.org/10.1007/s11222-023-10245-z
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DOI: https://doi.org/10.1007/s11222-023-10245-z