【发布时间】:2016-11-15 18:38:00
【问题描述】:
这是我尝试过的,使用 mvtnorm 包
样本数据集
library(mvtnorm)
set.seed(2357)
df <- data.frame(
x = rnorm(1000, mean=80, sd=20),
y = rnorm(1000, mean=0, sd=5),
z = rnorm(1000, mean=0, sd=5)
)
head(df)
x y z
1 70.38 1.307 0.2005
2 59.76 5.781 -3.5095
3 54.14 -1.313 -1.9022
4 79.91 7.754 -6.2076
5 87.07 1.389 1.1065
6 75.89 1.684 6.2979
拟合多元正态分布并检查 P(x
# Get the dimension means and correlation matrix
means <- c(x=mean(df$x), y=mean(df$y), z=mean(df$z))
corr <- cor(df)
# Check P(x <= 80)
sum(df$x <= 80)/nrow(df) # 0.498
pmvnorm(lower=-Inf, upper=c(80, Inf, Inf), mean=means, corr=corr) # 0.8232
# Get the dimension means and correlation matrix
means <- c(x=mean(df$x), y=mean(df$y), z=mean(df$z))
corr <- cor(df)
# Check P(x <= 80)
sum(df$x <= 80)/nrow(df) # 0.498
pmvnorm(lower=-Inf, upper=c(80, Inf, Inf), mean=means, corr=corr) # 0.8232
为什么拟合结果是 0.82?我哪里做错了?
【问题讨论】:
标签: r normal-distribution