假设我有三个时间序列:
y <- c(7, 8, 9, 2, 4, 5, 9, 4)
x <- c(9, 3, 5, 2, 7, 1, 6, 1) and
z <- c(NaN, NaN, NaN, 9, 10, 3, 5, 3)
现在我想用R:reg1 <- lm(y~x+z)
计算以下回归,然后summary(reg1)
给我以下输出:
Call:
lm(formula = y ~ x + z)
Residuals:
ALL 3 residuals are 0: no residual degrees of freedom!
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 6.2414 NA NA NA
x 0.5172 NA NA NA
z -0.5862 NA NA NA
Residual standard error: NaN on 0 degrees of freedom
(3 observations deleted due to missingness)
Multiple R-squared: 1, Adjusted R-squared: NaN
F-statistic: NaN on 2 and 0 DF, p-value: NA
看起来,系数是估计的,但没有t值等。我的问题是,1)为什么没有错误消息,2)我如何从回归中省略NaN,所以R给了我t值等?感谢
输出实际上回答了这个问题:
所有3个残差均为0:无剩余自由度!
因此不存在用于估计丢失输出的剩余自由度。根据这个公式,至少需要4行未遗漏的数据才能得到遗漏数量的估计值。