Relative Weights: 计算每个解释变量对R方的贡献

2019-10-01 本文已影响0人别花春水

question:

还是不太懂为什么 $\beta^2$ （表示z对y的贡献）可以与 $\Lambda^2$ （x对z的贡献）相乘，乘了之后就是x对y的贡献。
x'z=z'x，转置一下lambda就真的没有correlation了？看上去很像操纵数据，但统计学意义上也uncorrelated了吗？
怎么证明R方等于β²各元素的和？

R语言程序示例

ref: R in Action, 2nd edition, page:209
ps:亲测该版本的R in Action有印刷错误，而且部分宏包更新后代码失效，但是它是freedownload，我懒得找其他版本了，慎重参阅。
relative weights :find contribution each predictor makes to R-square

relweights <- function(fit,...){
  R <- cor(fit$model) #相关系数矩阵Rxx
  nvar <- ncol(R)
  rxx <- R[2:nvar, 2:nvar] #自变量的相关系数矩阵
  rxy <- R[2:nvar, 1] 
  svd <- eigen(rxx) #计算矩阵特征值、特征向量
  evec <- svd$vectors  #特征向量
  ev <- svd$values #特征值
  delta <- diag(sqrt(ev))  #以特征值的平方根为对角线创建矩阵delta
  lambda <- evec %*% delta %*% t(evec)    # correlations between original predictors and new orthogonal variables。转化为对角化矩阵
  lambdasq <- lambda ^ 2  
  beta <- solve(lambda) %*% rxy   # regression coefficients of Y on orthogonal variables正交矩阵.$AA=I$则A是正交矩阵。求lambda的逆矩阵，再乘rxy
  rsquare <- colSums(beta ^ 2) #R^2是模型对总体的解释力度
  rawwgt <- lambdasq %*% beta ^ 2  #自变量单独对总体的解释力度。
  import <- (rawwgt / rsquare) * 100
  import <- as.data.frame(import)
  row.names(import) <- names(fit$model[2:nvar])
  names(import) <- "Weights" #设定列名
  import <- import[order(import),1, drop=FALSE]
  dotchart(import$Weights, labels=row.names(import),
    xlab="% of R-Square", pch=19,
    main="Relative Importance of Predictor Variables",
    sub=paste("Total R-Square=", round(rsquare, digits=3)),  #点线
    ...)
return(import)
}

states <- as.data.frame(state.x77[,c("Murder", "Population",
"Illiteracy", "Income", "Frost")])
fit <- lm(Murder ~ Population + Illiteracy + Income + Frost, data=states)
relweights(fit, col="blue")

基础知识：矩阵形式ols的求解

已知 $X\beta=Y$ ，求 $\beta$ :
$\because X^TX\beta=X^TY$
$\therefore \beta=(X^TX)^{-1}X^TY$

calculation of relative weight

ref:Jeff Johnson,2000. A Heuristic Method for Estimating the Relative Weight of Predictor Variables in Multiple Regression. Multivariate Behavioral Research, 35:1-19

每个变量的贡献包括单独贡献以及包含与其他变量的correlation的贡献。

主要动作是将原自变量矩阵转化为不互相关的正交矩阵，再obtaining the bestfitting (in the least squares sense) set of orthogonal variables（正交矩阵）

先求矩阵 $X{'}X$ 的eigenvectors和eigenvalues
再求 $X$ 的singular value decomposition （奇异值分解），求类似主成分分析（PCA）那样的退化矩阵
$X=P\Delta Q{'} \\ \Delta=\sqrt{eigenvalues}$
这里P和Q都是eigenvectors
ps：If no two predictor variables in X are perfectly correlated with each other, X is of full rank and no diagonal elements of $\Delta$ will be equal to zero.
找到与 $X$ 最接近的正交矩阵 $Z$ ，因为The columns of Z are the best-fitting approximations to the columns of X in that they minimize the residual sum of squares between the original variables and the orthogonal variables (Johnson, 1966)
$Z=PQ{'}$
让X在Z上回归， $X=\Lambda Z$ ，因此

$\Lambda=(Z'Z)^{-1}Z'X=Q\Delta Q' \tag{1}$

X is a linear transformation of Z
Because the Z variables are uncorrelated, the relative contribution of each z to each x is represented by the squared standardized regression coefficient (which is the same as the squared zero-order correlation) of each z for each x, represented by the squared column elements of $\Lambda (\lambda_{jk^2})$

由于 $Z'X=X'Z$ ，因此any particular $\lambda_{jk^2}$ represents the proportion of variance in $z_k$ accounted for by $x_j$ , just as it represents the proportion of variance in $x_j$ accounted for by $z_k$ .

找到Y被Z解释的部分 $\beta$ 。The vector of beta weights when regressing y on Z is obtained by

$\beta=(Z'Z)^{-1}Z'y=QP'y$

求relative weights
转化为方差的平方单位，再将其scaled by $R^2$
$\varepsilon=\frac{\Lambda^2*\beta^2}{\sum{\beta^2_i}} \tag{2}$
其中，Z是X的近似替代，所以模型解释力度R方不应变化，因而 $R^2=\sum{\beta^2_i}$
因此在计算relative weights的时候我们有X的correlation matrix等于
$X=P\Delta Q' \\ R_{XX}=X'X=Q\Delta P'P\Delta Q'=Q\Delta^2Q'$
Q正好是 $R_{XX}$ 的eigenvalues
由（1）知
$R^{1/2}_{XX}=Q\Delta Q'=\Lambda \tag{3}$
$\therefore R_{XZ}=X'Z=Q\Delta P'P\Delta Q'=Q\Delta Q'=R^{1/2}_{XX}$
由于 $R_{XZ}R_{YZ}=R_{XY}$ ，即 $R_{XZ}\beta=R_{XY}$ ，因此
$\beta=R^{-1}_{XZ}R_{XY}=\Lambda^{-1}R_{XY} \tag{4}$