最小二乘回歸受數據中的離群點的影響較大,穩健回歸通過降低離群點的影響緩解此問題。M估計法是穩健回歸的重要方法之一,M 估計法的目標函數為:
m i n ∑ ρ ( ? i ) = m i n ∑ ρ ( y i ? β ^ ? X i ) min\sum\rho(\epsilon_i) = min\sum\rho(y_i - \hat{\beta} * X_i) min∑ρ(?i?)=min∑ρ(yi??β^??Xi?)
函數 ρ \rho ρ 具有特性:
- ρ ( ? ) ≥ 0 , ρ ( 0 ) = 0 \rho(\epsilon)\ge 0, \rho(0)=0 ρ(?)≥0,ρ(0)=0
- ρ ( ? ) = ρ ( ? ? ) \rho(\epsilon) = \rho(-\epsilon) ρ(?)=ρ(??)
目標函數關于帶估計參數 β ^ \hat{\beta} β^? 求導:
∑ ρ ( y i ? β ^ X i ) ? β ^ = ∑ ? ρ ( y i ? β ^ X i ) ? β ^ X i ? ∑ ψ ( ρ ( y i ? β ^ X i ) ) X i \frac{\sum\rho(y_i-\hat{\beta}X_i)}{\partial \hat{\beta}} = \sum -\frac{\rho(y_i - \hat{\beta}X_i)}{\partial \hat{\beta}}X_i\triangleq \sum \psi(\rho(y_i - \hat{\beta}X_i)) X_i ?β^?∑ρ(yi??β^?Xi?)?=∑??β^?ρ(yi??β^?Xi?)?Xi??∑ψ(ρ(yi??β^?Xi?))Xi?
其中 ψ ( ? ) = ? ρ ( ? ) ? β ^ \psi(\epsilon) = \frac{\partial \rho(\epsilon)}{\partial \hat{\beta}} ψ(?)=?β^??ρ(?)?
Andrews 估計
Andrews 1974年提出:
ψ ( z ) = { s i n ( z / c ) ∣ z ∣ ≤ c 0 ∣ z ∣ ≥ c \psi(z)=\left\{ \begin{aligned} sin(z/c) & & |z| \le c \\ 0 & & |z| \ge c \end{aligned} \right. ψ(z)={sin(z/c)0??∣z∣≤c∣z∣≥c?
ρ ( z ) = { 1 ? c o s ( z ) ∣ z ∣ < π 0 ∣ z ∣ ≥ π \rho(z)=\left\{ \begin{aligned} 1 - cos(z) & & |z| < \pi \\ 0 & & |z| \ge \pi \end{aligned} \right. ρ(z)={1?cos(z)0??∣z∣<π∣z∣≥π?
Huber 估計
ψ ( z ) = { z ∣ z ∣ < c c ? s i g n ( z ) ∣ z ∣ ≥ c \psi(z)=\left\{ \begin{aligned} z & & |z| < c \\ c\cdot sign(z) & & |z| \ge c \end{aligned} \right. ψ(z)={zc?sign(z)??∣z∣<c∣z∣≥c?
ρ ( z ) = { 1 2 z 2 ∣ z ∣ < c c ∣ z ∣ ? 1 2 c 2 ∣ z ∣ ≥ c \rho(z)=\left\{ \begin{aligned} \frac 12 z^2 & & |z| < c \\ c|z| - \frac12c^2 & & |z| \ge c \end{aligned} \right. ρ(z)=? ? ??21?z2c∣z∣?21?c2??∣z∣<c∣z∣≥c?
Tuerky 估計
ψ ( z ) = { z ( c 2 ? z 2 ) 2 ∣ z ∣ < c c ? s i g n ( z ) ∣ z ∣ ≥ c \psi(z)=\left\{ \begin{aligned} z(c^2 - z^2)^2 & & |z| < c \\ c\cdot sign(z) & & |z| \ge c \end{aligned} \right. ψ(z)={z(c2?z2)2c?sign(z)??∣z∣<c∣z∣≥c?
ρ ( z ) = { 1 6 ( c 6 ? ( c 2 ? z 2 ) 3 ) ∣ z ∣ < c 0 ∣ z ∣ ≥ c \rho(z)=\left\{ \begin{aligned} \frac 16(c^6 - (c^2-z^2)^3) & & |z| < c \\ 0 & & |z| \ge c \end{aligned} \right. ρ(z)=? ? ??61?(c6?(c2?z2)3)0??∣z∣<c∣z∣≥c?
L p L_p Lp? 估計
ρ ( z ) = ∣ z ∣ p \rho(z) = |z|^p ρ(z)=∣z∣p