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附录 A.1 引理
引理A.1[18]引理5] 令α,β∈Rp, 有‖α‖2≤M,‖β‖2≤M. 令Xk∈Rp并且Xk~N(0,Σ*). 则对于m≥2, 有下式成立: E|αTXkXTkβ-EαTXkXTkβ|m/(2M2K2)m≤m!2.
引理A.2[9,引理1] 设X∈Rp且X~N(0,Σ*), 令δτ(n,r)=8(1+12K2)maxi Σ*ii 2log(4r)/n, 则对于所有的K>0,γ>2,δτ(n,pγ) <8(1+12K2)maxi Σ*ii有下式成立:
P(|Σ︿-Σ*|∞≥ δτ(n,pγ))≤1pγ-2.
引理A.3[18,引理1] 设X∈Rp且X~N(0,Σ*). 假设Θ*=(Σ*)-1存在并且满足条件2.1, 则有下式成立: |Θ* W |∞=maxi=1,…,p|Θ*iWej| =OP(logp/n).
引理A.4[10, 定理2-3] 设X∈Rp且X~N(0,Σ*).假设Θ*=(Σ*)-1存在并且满足条件2.1和条件2.2. 设Θ︿为SCIO估计量, 调节参数λni logp/n. 则以大于1-O(p-1+n-δ/8)的概率有下式成立:
(a) Θ︿S∈Θ*S. (b) |Θ︿-Θ*|∞≤CMplogpn. A.2 证明定理2.1
证明 将式(3)简单变化得出以下分解式: n(T︿ij-Θ*ij)σij=-n(Θ*WΘ*)ijσij+n(rem)ijσij(A1)
随后将分别证明 n(Θ*WΘ*)ij/σijdN(0,1)和 n|rem|∞/σij=oP(1).
A.2.1 证明n(Θ*WΘ*)ij/σijdN(0,1)
n(Θ*WΘ*)ijσij=
nσij∑nk=11n(Θ*TiXkXTkΘ*j-Θ*ij)=
1σijn∑nk=1 Zijk,
这里Zijk=Θ*TiXkXkTΘ*j-Θ*ij.
由引理A.1,可得对于m≥2,σ0>0(条件2.1得出‖Θ*i‖2≤L,‖Θ*j‖2≤L)有下式成立: E|Zij,k|m/(2σ0L2K)m≤m!2(K/σ0)m-2.
即对于每个固定的m有:supΘ*E|Zij,k|m=O(1). 则存在常数w>0, 有E|Zij,k|2=σ2ij≥w2.
最后由文献[25]中的Berry-Esseen定理, 可得 |Fn(z)-Φ(z)|≤3E|Z1|3(E|Z1|2)3/2n≤Cn,
其中,Fn(z)是1σijn∑nk=1Zijk的分布函数,Φ(z)是N(0,1)的分布函数, 并且常数C独立于Θ*和n.
由此可以证明: n(Θ*WΘ*)ijσijdN(0,1).
A.2.2 证明n|rem|∞/σij=oP(1) 由矩阵范数的运算可得 |rem|∞ ≤|(Θ︿-Θ*)WΘ*|∞ +|(Θ︿Σ︿-I)(Θ︿-Θ*)|∞≤‖Θ︿-Θ*‖∞|WΘ*|∞ +|Θ︿Σ︿-I|∞‖Θ︿-Θ*‖∞, 其中,
|Θ︿Σ︿-I|∞=|(Θ︿-Θ*)(Σ︿-Σ*)+(Θ︿-Θ*)Σ*+Θ*(Σ︿-Σ*) |∞≤‖Θ︿-Θ*‖∞|Σ︿-Σ*|∞+ |Θ︿-Θ*|∞‖Σ*‖∞+|Θ* W|∞.
因此可得
|rem|∞ ≤2‖Θ︿-Θ*‖∞|WΘ*|∞+|Σ︿-Σ*|∞‖Θ︿-Θ*‖∞2+|Θ︿-Θ*|∞‖Σ*‖∞‖Θ︿-Θ*‖∞.
令
|rem1|∞=‖Θ︿-Θ*‖∞|WΘ* |∞,|rem2|∞= |Σ︿-Σ*|∞‖Θ︿-Θ*‖∞2,
|rem3|∞=|Θ︿-Θ*|∞‖Σ*‖∞‖Θ︿-Θ*‖∞.
又有:
①由引理A.2, K=O(1),maxi Σ*ii=O(1), 可得|Σ︿-Σ*|∞=OP (logp/n).
②由引理A.4得‖Θ︿-Θ*‖∞≤d|Θ︿-Θ*|∞(Θ︿S∈Θ*S)及|Θ︿-Θ*|∞=OP(logp/n)(Mp=O(1)).
③在条件2.1成立下, 由引理A.3得|Θ*W|∞=OP(logp/n).
则由①,②,③可得
|rem|∞ =OPmax dlogpn, d2logpn3/2, d‖Σ*‖∞logpn =OPdlogpn.
再由定理2.1中的稀疏性假设d=o(n/logp)以及1/σij=O(1),可得 n|rem|∞/σij=oP(1).
综合A.2.1和A.2.2可以证明 n(T︿ij-Θ*ij)σijdN(0,1).
证毕. A.3 证明引理2.1
证明 因为X~N(0,Σ*), 所以Θ* X~N(0,Θ*). 令Z=Θ*TX, 则Z~N(0,Θ*).
则有 σ2ij=Var(Θ*TiXΘ*Tj X) =Var(eTiΘ*TXXTΘ*ej)=Var(eTiZZTej).
从而, σ2ij =Var(ZiZj)=E((Zi)2(Zj)2)-(E(ZiZj))2 =Θ*iiΘ*jj+2Θ*ij2-Θ*ij2=Θ*iiΘ*jj+Θ*ij2.
再由条件2.1中1/L≤Λmin(Θ*)≤Λmax(Θ*)≤L, 可得Θ*iiΘ*jj+Θ*ij2≥Λ2min(Θ*)≥1/L2>0, 这里L1, 因此1/σij=O(1).
通过引理A.4以及Mp=O(1)可得|Θ︿-Θ*|∞=OP(logp/n), 因此有下式成立:
|σ︿2ij-σ2ij| ≤|Θ︿iiΘ︿jj-Θ*iiΘ*jj|+|Θ︿2ij-Θ*ij2| ≤|ΔiiΔjj+Θ*iiΔjj+Θ*jjΔii|+|Δij(Δij+2Θ*ij)|=OP(logp/n),
此处定义Δij=Θ︿ij-Θ*ij.
证毕.
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