为什么负二项式回归的Pearson残差比Poisson回归的残差小?
我有这些数据: set.seed(1) predictor <- rnorm(20) set.seed(1) counts <- c(sample(1:1000, 20)) df <- data.frame(counts, predictor) 我进行了泊松回归 poisson_counts <- glm(counts ~ predictor, data = df, family = "poisson") 负二项式回归: require(MASS) nb_counts <- glm.nb(counts ~ predictor, data = df) 然后我为泊松回归计算色散统计量: sum(residuals(poisson_counts, type="pearson")^2)/df.residual(poisson_counts) # [1] 145.4905 负二项式回归: sum(residuals(nb_counts, type="pearson")^2)/df.residual(nb_counts) # [1] 0.7650289 在不使用方程式的情况下,谁能解释为什么负二项式回归的色散统计量远小于泊松回归的色散统计量?