DFFITS的计算对杠杆的影响以及对回归的影响 [英] Calculation of DFFITS as diagnostic for Leverage and Influence in regression

问题描述

我正在尝试手动计算 DFFITS.获得的值应等于通过 dffits 函数获得的第一个值.但是我自己的计算肯定有问题.

I am trying to calculate DFFITS by hand. The value obtained should be equal to the first value obtained by dffits function. However there must be something wrong with my own calculation.

attach(cars)

x1 <- lm(speed ~ dist, data = cars) # all observations

x2 <-  lm(speed ~ dist, data = cars[-1,]) # without first obs

x <- model.matrix(speed ~ dist) # x matrix

h <- diag(x%*%solve(crossprod(x))%*%t(x)) # hat values

num_dffits <- x1$fitted.values[1] - x2$fitted.values[1] #Numerator

denom_dffits <- sqrt(anova(x2)$`Mean Sq`[2]*h[1]) #Denominator

df_fits <- num_dffits/denom_dffits #DFFITS

dffits(x1)[1] # DFFITS function

推荐答案

您的分子是错误的.从第二个模型中删除了第一个基准后，相应的预测值不在 fitted(x2)中.我们需要使用 predict(x2，cars [1，])代替 fitted(x2)[1] .

Your numerator is wrong. As you have removed first datum from the second model, corresponding predicted value is not in fitted(x2). We need to use predict(x2, cars[1, ]) in place of fitted(x2)[1].

帽子值可以通过以下方式有效地计算

Hat values can be efficiently computed by

h <- rowSums(qr.Q(x1$qr) ^ 2)

或使用其R包装函数

h <- hat(x1$qr, FALSE)

R还具有用于获取帽子值的通用功能:

R also has a generic function for getting hat values, too:

h <- lm.influence(x1, FALSE)$hat

或其包装函数

h <- hatvalues(x1)

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您也不必致电 anova 即可获得MSE:

c(crossprod(x2$residuals)) / x2$df.residual

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