如何为 r 中的指数分布生成估计参数（标准差和均值）？

Question

例如：

#Using QQ-plots to decide which of the Normal, Gumbel and Exponential distribution best fits Qc
qqplot(x=qexp(ppoints(length(data$Qc))), y=data$Qc, main="Exponential Q-Q Plot",
       xlab="Theoretical Quantiles", ylab= "Data Quantiles")
qqplot(x=qnorm(ppoints(length(data$Qc))), y=data$Qc, main="Normal Q-Q Plot",
       xlab="Theoretical Quantiles", ylab= "Data Quantiles")
qqplot(x=qgumbel(ppoints(length(data$Qc))), y=data$Qc, main="Gumbel Q-Q Plot",
       xlab="Theoretical Quantiles", ylab= "Data Quantiles"

在假设我认为正态/指数分布是适合我的变量的好模型之后，如何生成正态分布或指数分布的估计参数（即标准偏差和平均值）？

Answer 1

指数的

MASS::fitdistr(Qc, dexp, start = list(rate = 0.1))
## 0.0338
## however, the maximum likelihood estimate of the exponential rate parameter
##  is just 1/mean(x):
(r <- 1/mean(Qc))
## 0.0338
pexp(30, rate = r, lower.tail = FALSE)  ## 0.362

普通的

样本均值和样本标准差可以很好地估计 Normal 的均值和 SD 参数（尽管如果我们真的想使用fitdistr可以使用）：

MASS::fitdistr(Qc, dnorm, start = list(mean = 25, sd = 7))
pnorm(30, mean = mean(Qc), sd = sd(Qc), lower.tail = FALSE)  ## 0.473

贡贝尔

library(fitdistrplus)
library(VGAM)
fitdist(Qc, "gumbel", start = list(location = 25, scale = 1)) 
pgumbel(30, location = 27.03, scale = 7.56, lower.tail = FALSE) ## 0.491

经验

我们可以计算观察到的值 > 30 的概率：

mean(Qc>30)  ## 0.469