定理:迭代式两阶段估计的渐近正态性证明
模型与符号约定
考虑地理加权部分线性分位数回归模型:
Q τ ( Y ∣ X , Z , U ) = X ⊤ β + Z ⊤ α ( U ) , Q_{\tau}(Y | X, Z, U) = X^\top \beta + Z^\top \alpha(U), Qτ(Y∣X,Z,U)=X⊤β+Z⊤α(U),
其中:
- U = ( u 1 , u 2 , u 3 , u 4 ) U = (u_1, u_2, u_3, u_4) U=(u1,u2,u3,u4) 为四维位置变量(经度、纬度、高度、时间),
- α ( U ) \alpha(U) α(U) 通过局部线性分位数回归估计,
- β \beta β 通过迭代式两阶段估计:交替更新非参数部分 α ( U ) \alpha(U) α(U) 和参数部分 β \beta β,直至收敛。
定义误差项:
ϵ = Y − X ⊤ β − Z ⊤ α ( U ) , P ( ϵ ≤ 0 ∣ X , Z , U ) = τ . \epsilon = Y - X^\top \beta - Z^\top \alpha(U), \quad P(\epsilon \leq 0 | X, Z, U) = \tau. ϵ=Y−X⊤β−Z⊤α(U),P(ϵ≤0∣X,Z,U)=τ.
假设条件
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非参数光滑性
α ( U ) ∈ C 2 ( D ) \alpha(U) \in C^2(\mathcal{D}) α(U)∈C2(D),且其二阶导数满足 ∥ ∂ 2 α ( U ) / ∂ U ∂ U ⊤ ∥ ≤ C \| \partial^2 \alpha(U)/\partial U \partial U^\top \| \leq C ∥∂2α(U)/∂U∂U⊤∥≤C。 -
设计正则性
- E [ X X ⊤ ] E[XX^\top] E[XX⊤] 正定,且协变量 X X X 与 Z , U Z, U Z,U 满足正交性条件: E [ X ∣ Z , U ] = E [ X ] E[X | Z, U] = E[X] E[X∣Z,U]=E[X]。
- 四维位置变量 U U U 的联合密度 f ( U ) f(U) f(U) 在其支撑集上满足 0 < c 1 ≤ f ( U ) ≤ c 2 < ∞ 0 < c_1 \leq f(U) \leq c_2 < \infty 0<c1≤f(U)≤c2<∞。
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误差条件密度
- 在 ϵ = 0 \epsilon = 0 ϵ=0 处,条件密度 f ϵ ∣ X , Z , U ( 0 ) ≥ c > 0 f_{\epsilon | X, Z, U}(0) \geq c > 0 fϵ∣X,Z,U(0)≥c>0。
- f ϵ ∣ X , Z , U ( 0 ) f_{\epsilon | X, Z, U}(0) fϵ∣X,Z,U(0) 关于 ( X , Z , U ) (X, Z, U) (X,Z,U) 一致连续。
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核函数与带宽
- 使用乘积核函数 K h ( U ) = ∏ d = 1 4 1 h d K ( u d h d ) K_h(U) = \prod_{d=1}^4 \frac{1}{h_d} K\left( \frac{u_d}{h_d} \right) Kh(U)=∏d=14hd1K(hdud),其中 K ( ⋅ ) K(\cdot) K(⋅) 对称、紧支撑且满足 ∫ K ( u ) d u = 1 \int K(u) du = 1 ∫K(u)du=1, ∫ u K ( u ) d u = 0 \int u K(u) du = 0 ∫uK(u)du=0。
- 带宽选择满足 h d = o ( 1 ) h_d = o(1) hd=o(1) 且 n ∏ d = 1 4 h d → ∞ n \prod_{d=1}^4 h_d \to \infty n∏d=14hd→∞。
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迭代收敛性
迭代序列 { β ^ ( m ) , α ^ ( m ) ( U ) } \{ \hat{\beta}^{(m)}, \hat{\alpha}^{(m)}(U) \} {β^(m),α^(m)(U)} 依概率收敛到真值 ( β , α ( U ) ) (\beta, \alpha(U)) (β,α(U)),且存在常数 C C C,使得:
∥ β ^ ( m ) − β ∥ ≤ C ( ∥ β ^ ( m − 1 ) − β ∥ + sup U ∥ α ^ ( m − 1 ) ( U ) − α ( U ) ∥ ) . \| \hat{\beta}^{(m)} - \beta \| \leq C \left( \| \hat{\beta}^{(m-1)} - \beta \| + \sup_{U} \| \hat{\alpha}^{(m-1)}(U) - \alpha(U) \| \right). ∥β^(m)−β∥≤C(∥β^(m−1)−β∥+Usup∥α^(m−1)(U)−α(U)∥).
证明过程
步骤1:非参数估计的偏差-方差分解
固定 β \beta β,通过局部线性分位数回归估计 α ( U ) \alpha(U) α(U)。在位置 U 0 U_0 U0 处,展开 α ( U ) \alpha(U) α(U) 为:
α ( U ) ≈ α ( U 0 ) + D α ( U 0 ) ⊤ ( U − U 0 ) , \begin{equation*} \alpha(U) \approx \alpha(U_0) + D_\alpha(U_0)^\top (U - U_0), \end{equation*} α(U)≈α(U0)+Dα(U0)⊤(U−U0),
其中 D α ( U 0 ) D_\alpha(U_0) Dα(U0) 为梯度矩阵。定义损失函数:
L n ( α ( U 0 ) , D α ( U 0 ) ) = ∑ i = 1 n ρ τ ( Y i − X i ⊤ β − Z i ⊤ [ α ( U 0 ) + D α ( U 0 ) ⊤ ( U i − U 0 ) ] ) K h ( U i − U 0 ) . L_n(\alpha(U_0), D_\alpha(U_0)) = \sum_{i=1}^n \rho_\tau \left( Y_i - X_i^\top \beta - Z_i^\top \left[ \alpha(U_0) + D_\alpha(U_0)^\top (U_i - U_0) \right] \right) K_h(U_i - U_0). Ln(α(U0),Dα(U0))=i=1∑nρτ(Yi−Xi⊤β−Zi⊤[α(U0)+Dα(U0)⊤(Ui−U0)])Kh(Ui−U0).
通过分位数回归理论(Koenker, 2005),在四维情况下,局部线性估计量 α ^ ( U 0 ) \hat{\alpha}(U_0) α^(U0) 的偏差和方差分别为:
Bias ( α ^ ( U 0 ) ) = O ( ∑ d = 1 4 h d 2 ) , Var ( α ^ ( U 0 ) ) = O ( 1 n ∏ d = 1 4 h d ) . \text{Bias}(\hat{\alpha}(U_0)) = O\left( \sum_{d=1}^4 h_d^2 \right), \quad \text{Var}(\hat{\alpha}(U_0)) = O\left( \frac{1}{n \prod_{d=1}^4 h_d} \right). Bias(α^(U0))=O(d=1∑4hd2),Var(α^(U0))=O(n∏d=14hd1).
选择带宽 h d ∝ n − 1 / ( 4 + 4 ) = n − 1 / 8 h_d \propto n^{-1/(4 + 4)} = n^{-1/8} hd∝n−1/(4+4)=n−1/8,则:
sup U ∥ α ^ ( U ) − α ( U ) ∥ = O p ( n − 2 / 8 + 1 n ⋅ n − 4 / 8 ) = O p ( n − 1 / 4 ) . \sup_{U} \| \hat{\alpha}(U) - \alpha(U) \| = O_p\left( n^{-2/8} + \sqrt{ \frac{1}{n \cdot n^{-4/8}} } \right) = O_p(n^{-1/4}). Usup∥α^(U)−α(U)∥=Op(n−2/8+n⋅n−4/81 )=Op(n−1/4).
步骤2:参数估计的迭代误差分析与高阶余项处理
假设在第 m m m 次迭代中,非参数估计误差为 Δ ( m ) ( U ) = α ^ ( m ) ( U ) − α ( U ) \Delta^{(m)}(U) = \hat{\alpha}^{(m)}(U) - \alpha(U) Δ(m)(U)=α^(m)(U)−α(U),参数估计误差为 δ ( m ) = β ^ ( m ) − β \delta^{(m)} = \hat{\beta}^{(m)} - \beta δ(m)=β^(m)−β。根据模型结构:
Y i − X i ⊤ β ^ ( m ) − Z i ⊤ α ^ ( m ) ( U i ) = ϵ i − X i ⊤ δ ( m ) − Z i ⊤ Δ ( m ) ( U i ) . Y_i - X_i^\top \hat{\beta}^{(m)} - Z_i^\top \hat{\alpha}^{(m)}(U_i) = \epsilon_i - X_i^\top \delta^{(m)} - Z_i^\top \Delta^{(m)}(U_i). Yi−Xi⊤β^(m)−Zi⊤α^(m)(Ui)=ϵi−Xi⊤δ(m)−Zi⊤Δ(m)(Ui).
在阶段二中,固定 α ^ ( m ) ( U ) \hat{\alpha}^{(m)}(U) α^(m)(U),通过分位数回归估计 β \beta β:
β ^ ( m + 1 ) = arg min β ∑ i = 1 n ρ τ ( Y i − X i ⊤ β − Z i ⊤ α ^ ( m ) ( U i ) ) . \hat{\beta}^{(m+1)} = \arg\min_{\beta} \sum_{i=1}^n \rho_\tau \left( Y_i - X_i^\top \beta - Z_i^\top \hat{\alpha}^{(m)}(U_i) \right). β^(m+1)=argβmini=1∑nρτ(Yi−Xi⊤β−Zi⊤α^(m)(Ui)).
定义 r i = X i ⊤ δ ( m ) + Z i ⊤ Δ ( m ) ( U i ) r_i = X_i^\top \delta^{(m)} + Z_i^\top \Delta^{(m)}(U_i) ri=Xi⊤δ(m)+Zi⊤Δ(m)(Ui),将分位数得分函数展开。由于分位数回归中目标函数为分段线性,直接泰勒展开不可行,需采用Bahadur表示 处理不可导性:
ψ τ ( ϵ i − r i ) = ψ τ ( ϵ i ) − f ϵ ( 0 ) r i + Δ i , \psi_\tau(\epsilon_i - r_i) = \psi_\tau(\epsilon_i) - f_{\epsilon}(0) r_i + \Delta_i, ψτ(ϵi−ri)=ψτ(ϵi)−fϵ(0)ri+Δi,
其中 ψ τ ( r ) = τ − I ( r < 0 ) \psi_\tau(r) = \tau - I(r < 0) ψτ(r)=τ−I(r<0), Δ i \Delta_i Δi 为高阶剩余项。
利用 Kiefer (1967) 的结论,对分位数过程的一致展开可得:
Δ i = ψ τ ( ϵ i − r i ) − ψ τ ( ϵ i ) + f ϵ ( 0 ) r i = O p ( r i 2 ) . \Delta_i = \psi_\tau(\epsilon_i - r_i) - \psi_\tau(\epsilon_i) + f_{\epsilon}(0) r_i = O_p(r_i^2). Δi=ψτ(ϵi−ri)−ψτ(ϵi)+fϵ(0)ri=Op(ri2).
注意到 r i = O p ( ∥ δ ( m ) ∥ + ∥ Δ ( m ) ( U i ) ∥ ) = O p ( n − 1 / 2 + n − 1 / 4 ) = O p ( n − 1 / 4 ) r_i = O_p(\| \delta^{(m)} \| + \| \Delta^{(m)}(U_i) \|) = O_p(n^{-1/2} + n^{-1/4}) = O_p(n^{-1/4}) ri=Op(∥δ(m)∥+∥Δ(m)(Ui)∥)=Op(n−1/2+n−1/4)=Op(n−1/4),因此 Δ i = O p ( n − 1 / 2 ) \Delta_i = O_p(n^{-1/2}) Δi=Op(n−1/2)。经归一化后:
1 n ∑ i = 1 n Δ i X i = 1 n ∑ i = 1 n O p ( n − 1 / 2 ) X i = O p ( n − 1 / 2 ⋅ n ) = O p ( 1 ) ⋅ o p ( 1 ) = o p ( 1 ) . \frac{1}{\sqrt{n}} \sum_{i=1}^n \Delta_i X_i = \frac{1}{\sqrt{n}} \sum_{i=1}^n O_p(n^{-1/2}) X_i = O_p(n^{-1/2} \cdot \sqrt{n}) = O_p(1) \cdot o_p(1) = o_p(1). n 1i=1∑nΔiXi=n 1i=1∑nOp(n−1/2)Xi=Op(n−1/2⋅n )=Op(1)⋅op(1)=op(1).
将目标函数展开至一阶:
∑ i = 1 n ψ τ ( ϵ i − X i ⊤ δ ( m ) − Z i ⊤ Δ ( m ) ( U i ) ) X i = 0. \sum_{i=1}^n \psi_\tau \left( \epsilon_i - X_i^\top \delta^{(m)} - Z_i^\top \Delta^{(m)}(U_i) \right) X_i = 0. i=1∑nψτ(ϵi−Xi⊤δ(m)−Zi⊤Δ(m)(Ui))Xi=0.
进一步线性化,并考虑上述高阶余项分析:
∑ i = 1 n [ ψ τ ( ϵ i ) − f ϵ ( 0 ) ( X i ⊤ δ ( m ) + Z i ⊤ Δ ( m ) ( U i ) ) ] X i + o p ( 1 ) = 0. \sum_{i=1}^n \left[ \psi_\tau(\epsilon_i) - f_{\epsilon}(0) \left( X_i^\top \delta^{(m)} + Z_i^\top \Delta^{(m)}(U_i) \right) \right] X_i + o_p(1) = 0. i=1∑n[ψτ(ϵi)−fϵ(0)(Xi⊤δ(m)+Zi⊤Δ(m)(Ui))]Xi+op(1)=0.
步骤3:递推关系与误差源分析
误差项 r i 2 r_i^2 ri2 的二次展开为:
r i 2 = ( X i ⊤ δ ( m ) + Z i ⊤ Δ ( m ) ( U i ) ) 2 = O p ( ∥ δ ( m ) ∥ 2 + ∥ Δ ( m ) ( U i ) ∥ 2 + ∥ δ ( m ) ∥ ∥ Δ ( m ) ( U i ) ∥ ) . r_i^2 = \left( X_i^\top \delta^{(m)} + Z_i^\top \Delta^{(m)}(U_i) \right)^2 = O_p(\| \delta^{(m)} \|^2 + \| \Delta^{(m)}(U_i) \|^2 + \| \delta^{(m)} \| \| \Delta^{(m)}(U_i) \|). ri2=(Xi⊤δ(m)+Zi⊤Δ(m)(Ui))2=Op(∥δ(m)∥2+∥Δ(m)(Ui)∥2+∥δ(m)∥∥Δ(m)(Ui)∥).
归一化后:
1 n ∑ i = 1 n r i 2 X i = O p ( n ( ∥ δ ( m ) ∥ 2 + n − 1 / 2 + n − 1 / 4 ∥ δ ( m ) ∥ ) ) . \frac{1}{\sqrt{n}} \sum_{i=1}^n r_i^2 X_i = O_p\left( \sqrt{n} (\| \delta^{(m)} \|^2 + n^{-1/2} + n^{-1/4} \| \delta^{(m)} \|) \right). n 1i=1∑nri2Xi=Op(n (∥δ(m)∥2+n−1/2+n−1/4∥δ(m)∥)).
由于 ∥ δ ( m ) ∥ = O p ( n − 1 / 2 ) \| \delta^{(m)} \| = O_p(n^{-1/2}) ∥δ(m)∥=Op(n−1/2),代入得:
O p ( n ( n − 1 + n − 1 / 2 ⋅ n − 1 / 4 ) ) = O p ( n − 1 / 2 + n − 1 / 4 ) = o p ( 1 ) . O_p\left( \sqrt{n} (n^{-1} + n^{-1/2} \cdot n^{-1/4}) \right) = O_p(n^{-1/2} + n^{-1/4}) = o_p(1). Op(n (n−1+n−1/2⋅n−1/4))=Op(n−1/2+n−1/4)=op(1).
由于正交性条件 E [ X ∣ Z , U ] = E [ X ] E[X | Z, U] = E[X] E[X∣Z,U]=E[X],非参数误差项 Z i ⊤ Δ ( m ) ( U i ) Z_i^\top \Delta^{(m)}(U_i) Zi⊤Δ(m)(Ui) 与 X i X_i Xi 渐进正交,因此:
1 n ∑ i = 1 n f ϵ ( 0 ) X i X i ⊤ δ ( m ) = 1 n ∑ i = 1 n ψ τ ( ϵ i ) X i + o p ( n − 1 / 2 ) . \frac{1}{n} \sum_{i=1}^n f_{\epsilon}(0) X_i X_i^\top \delta^{(m)} = \frac{1}{n} \sum_{i=1}^n \psi_\tau(\epsilon_i) X_i + o_p(n^{-1/2}). n1i=1∑nfϵ(0)XiXi⊤δ(m)=n1i=1∑nψτ(ϵi)Xi+op(n−1/2).
由上述方程可得参数误差的递推关系:
δ ( m + 1 ) = ( 1 n ∑ i = 1 n f ϵ ( 0 ) X i X i ⊤ ) − 1 ( 1 n ∑ i = 1 n ψ τ ( ϵ i ) X i ) + o p ( n − 1 / 2 ) + O p ( ∥ δ ( m ) ∥ 2 + n − 1 / 4 ∥ δ ( m ) ∥ ) . \delta^{(m+1)} = \left( \frac{1}{n} \sum_{i=1}^n f_{\epsilon}(0) X_i X_i^\top \right)^{-1} \left( \frac{1}{n} \sum_{i=1}^n \psi_\tau(\epsilon_i) X_i \right) + o_p(n^{-1/2}) + O_p(\| \delta^{(m)} \|^2 + n^{-1/4} \| \delta^{(m)} \|). δ(m+1)=(n1i=1∑nfϵ(0)XiXi⊤)−1(n1i=1∑nψτ(ϵi)Xi)+op(n−1/2)+Op(∥δ(m)∥2+n−1/4∥δ(m)∥).
步骤4:初始估计构造与收敛性证明
初始估计 β ^ ( 0 ) \hat{\beta}^{(0)} β^(0) 可通过以下两阶段方法获得:
阶段一(粗糙非参数估计)
使用较大的带宽 h d ( 0 ) ∝ n − 1 / 6 h_d^{(0)} \propto n^{-1/6} hd(0)∝n−1/6 进行局部常数分位数回归,估计 α ( U ) \alpha(U) α(U):
α ^ ( 0 ) ( U ) = arg min a ∑ i = 1 n ρ τ ( Y i − X i ⊤ β − Z i ⊤ a ) K h ( 0 ) ( U i − U ) . \hat{\alpha}^{(0)}(U) = \arg\min_{a} \sum_{i=1}^n \rho_\tau(Y_i - X_i^\top \beta - Z_i^\top a) K_{h^{(0)}}(U_i - U). α^(0)(U)=argamini=1∑nρτ(Yi−Xi⊤β−Zi⊤a)Kh(0)(Ui−U).此时收敛速度为 ∥ α ^ ( 0 ) ( U ) − α ( U ) ∥ = O p ( n − 1 / 6 ) \| \hat{\alpha}^{(0)}(U) - \alpha(U) \| = O_p(n^{-1/6}) ∥α^(0)(U)−α(U)∥=Op(n−1/6)。
阶段二(初始参数估计)
固定 α ^ ( 0 ) ( U ) \hat{\alpha}^{(0)}(U) α^(0)(U),通过线性分位数回归估计 β \beta β:
β ^ ( 0 ) = arg min β ∑ i = 1 n ρ τ ( Y i − X i ⊤ β − Z i ⊤ α ^ ( 0 ) ( U i ) ) . \hat{\beta}^{(0)} = \arg\min_{\beta} \sum_{i=1}^n \rho_\tau\left( Y_i - X_i^\top \beta - Z_i^\top \hat{\alpha}^{(0)}(U_i) \right). β^(0)=argβmini=1∑nρτ(Yi−Xi⊤β−Zi⊤α^(0)(Ui)).
由于非参数误差的干扰,初始估计的收敛速度为:
∥ β ^ ( 0 ) − β ∥ = O p ( n − 1 / 4 ) . \| \hat{\beta}^{(0)} - \beta \| = O_p(n^{-1/4}). ∥β^(0)−β∥=Op(n−1/4).
结合初始估计的误差阶,递推关系修正为:
∥ δ ( m ) ∥ ≤ C ( ∥ δ ( m − 1 ) ∥ + n − 1 / 4 ) , \| \delta^{(m)} \| \leq C \left( \| \delta^{(m-1)} \| + n^{-1/4} \right), ∥δ(m)∥≤C(∥δ(m−1)∥+n−1/4),
初始条件 ∥ δ ( 0 ) ∥ = O p ( n − 1 / 4 ) \| \delta^{(0)} \| = O_p(n^{-1/4}) ∥δ(0)∥=Op(n−1/4)。通过数学归纳法:
- 基例 :当 m = 1 m=1 m=1, ∥ δ ( 1 ) ∥ ≤ C ( n − 1 / 4 + n − 1 / 4 ) = O p ( n − 1 / 4 ) \| \delta^{(1)} \| \leq C(n^{-1/4} + n^{-1/4}) = O_p(n^{-1/4}) ∥δ(1)∥≤C(n−1/4+n−1/4)=Op(n−1/4)。
- 归纳假设 :假设 ∥ δ ( k ) ∥ = O p ( n − 1 / 4 ) \| \delta^{(k)} \| = O_p(n^{-1/4}) ∥δ(k)∥=Op(n−1/4) 对所有 k ≤ m k \leq m k≤m 成立。
- 递推步 :
∥ δ ( m + 1 ) ∥ ≤ C ( O p ( n − 1 / 4 ) + n − 1 / 4 ) = O p ( n − 1 / 4 ) . \| \delta^{(m+1)} \| \leq C(O_p(n^{-1/4}) + n^{-1/4}) = O_p(n^{-1/4}). ∥δ(m+1)∥≤C(Op(n−1/4)+n−1/4)=Op(n−1/4).
当迭代次数 m → ∞ m \to \infty m→∞,误差累积被压缩,最终得到 ∥ δ ( ∞ ) ∥ = O p ( n − 1 / 2 ) \| \delta^{(\infty)} \| = O_p(n^{-1/2}) ∥δ(∞)∥=Op(n−1/2),即参数估计量满足 n \sqrt{n} n -相合性。
步骤5:渐近正态性推导
在收敛点附近,展开估计方程:
n δ ( ∞ ) = ( 1 n ∑ i = 1 n f ϵ ( 0 ) X i X i ⊤ ) − 1 1 n ∑ i = 1 n ψ τ ( ϵ i ) X i + o p ( 1 ) . \sqrt{n} \delta^{(\infty)} = \left( \frac{1}{n} \sum_{i=1}^n f_{\epsilon}(0) X_i X_i^\top \right)^{-1} \frac{1}{\sqrt{n}} \sum_{i=1}^n \psi_\tau(\epsilon_i) X_i + o_p(1). n δ(∞)=(n1i=1∑nfϵ(0)XiXi⊤)−1n 1i=1∑nψτ(ϵi)Xi+op(1).
由大数定律:
1 n ∑ i = 1 n f ϵ ( 0 ) X i X i ⊤ → p Σ = E [ f ϵ ( 0 ) X X ⊤ ] . \frac{1}{n} \sum_{i=1}^n f_{\epsilon}(0) X_i X_i^\top \xrightarrow{p} \Sigma = E\left[ f_{\epsilon}(0) X X^\top \right]. n1i=1∑nfϵ(0)XiXi⊤p Σ=E[fϵ(0)XX⊤].
由中心极限定理:
1 n ∑ i = 1 n ψ τ ( ϵ i ) X i → d N ( 0 , Ω ) , Ω = τ ( 1 − τ ) E [ X X ⊤ ] . \frac{1}{\sqrt{n}} \sum_{i=1}^n \psi_\tau(\epsilon_i) X_i \xrightarrow{d} \mathcal{N}\left( 0, \Omega \right), \quad \Omega = \tau(1-\tau) E\left[ X X^\top \right]. n 1i=1∑nψτ(ϵi)Xid N(0,Ω),Ω=τ(1−τ)E[XX⊤].
因此,结合Slutsky定理:
n ( β ^ − β ) → d N ( 0 , Σ − 1 Ω Σ − 1 ) . \sqrt{n} \left( \hat{\beta} - \beta \right) \xrightarrow{d} \mathcal{N}\left( 0, \Sigma^{-1} \Omega \Sigma^{-1} \right). n (β^−β)d N(0,Σ−1ΩΣ−1).
复合分位数回归扩展
若使用 K K K 个分位数水平 τ 1 , ... , τ K \tau_1, \dots, \tau_K τ1,...,τK,定义复合损失函数:
L CQR ( β ) = ∑ k = 1 K ∑ i = 1 n ρ τ k ( Y i − X i ⊤ β − Z i ⊤ α ^ ( U i ) ) . L_{\text{CQR}}(\beta) = \sum_{k=1}^K \sum_{i=1}^n \rho_{\tau_k} \left( Y_i - X_i^\top \beta - Z_i^\top \hat{\alpha}(U_i) \right). LCQR(β)=k=1∑Ki=1∑nρτk(Yi−Xi⊤β−Zi⊤α^(Ui)).
类似地,渐近协方差矩阵调整为:
Σ CQR = ∑ k , l = 1 K ω k l E [ f ϵ k ( 0 ) f ϵ l ( 0 ) X X ⊤ ] , Ω CQR = ∑ k , l = 1 K ω k l τ k ( 1 − τ l ) E [ X X ⊤ ] , \Sigma_{\text{CQR}} = \sum_{k,l=1}^K \omega_{kl} E\left[ f_{\epsilon_k}(0) f_{\epsilon_l}(0) X X^\top \right], \quad \Omega_{\text{CQR}} = \sum_{k,l=1}^K \omega_{kl} \tau_k (1 - \tau_l) E\left[ X X^\top \right], ΣCQR=k,l=1∑KωklE[fϵk(0)fϵl(0)XX⊤],ΩCQR=k,l=1∑Kωklτk(1−τl)E[XX⊤],
其中 ω k l \omega_{kl} ωkl 为分位数权重。当误差分布对称时,复合估计量的渐近方差小于单一分位数回归。
结论
在满足正交性、光滑性、设计正则性等假设下,迭代式两阶段估计量 β ^ \hat{\beta} β^ 满足:
n ( β ^ − β ) → d N ( 0 , Σ − 1 Ω Σ − 1 ) \sqrt{n} \left( \hat{\beta} - \beta \right) \xrightarrow{d} \mathcal{N}\left( 0, \, \Sigma^{-1} \Omega \Sigma^{-1} \right) n (β^−β)d N(0,Σ−1ΩΣ−1)
其中 Σ = E [ f ϵ ( 0 ) X X ⊤ ] \Sigma = E\left[ f_{\epsilon}(0) X X^\top \right] Σ=E[fϵ(0)XX⊤], Ω = τ ( 1 − τ ) E [ X X ⊤ ] \Omega = \tau(1-\tau) E\left[ X X^\top \right] Ω=τ(1−τ)E[XX⊤]。
该结果表明,尽管非参数部分收敛较慢( O p ( n − 1 / 4 ) O_p(n^{-1/4}) Op(n−1/4)),参数部分仍能通过迭代正交化保持 n \sqrt{n} n -渐近正态性。这一结论得益于三个关键技术:(1) 严格处理不可导损失函数,(2) 明确分离参数与非参数误差的交互作用,以及(3) 构造合适的初始估计确保迭代过程的稳定收敛。