我试图计算期望值为 200 和标准差为 20 的正态分布密度的积分。从 -Inf 到 Inf 这应该是 1。

我得到以下信息:

> integrate(dnorm, mean=200, sd=20,-Inf, Inf)$value 
[1] 1.429508e-08 

对于低于 169 的预期值，我得到正确的值 1。 如何获得更大的期望值的正确值？

请您参考如下方法:

或者

integrate(dnorm, mean=200, sd=20, lower= -Inf, upper= Inf, abs.tol = 0)$value 
[1] 1 

见 here .

要查看发生了什么，请注意以下分割的数量:

js <- integrate(dnorm, mean=200, sd=20, lower = -Inf, upper = Inf) 
as <- integrate(dnorm, mean=200, sd=20, lower = -1e4, upper = 1e4) 
cj <- integrate(dnorm, mean=200, sd=20, lower = -Inf, upper = Inf, abs.tol = 0) 
 
str(js) 
List of 5 
 $ value       : num 1.43e-08 
 $ abs.error   : num 2.77e-08 
 $ subdivisions: int 2 
 $ message     : chr "OK" 
 $ call        : language integrate(f = dnorm, lower = -Inf, upper = Inf, mean = 200, sd = 20) 
 - attr(*, "class")= chr "integrate" 
 
str(as) 
List of 5 
 $ value       : num 1 
 $ abs.error   : num 2e-07 
 $ subdivisions: int 9 
 $ message     : chr "OK" 
 $ call        : language integrate(f = dnorm, lower = -10000, upper = 10000, mean = 200,      sd = 20) 
 - attr(*, "class")= chr "integrate" 
 
str(cj) 
List of 5 
 $ value       : num 1 
 $ abs.error   : num 9.37e-05 
 $ subdivisions: int 12 
 $ message     : chr "OK" 
 $ call        : language integrate(f = dnorm, lower = -Inf, upper = Inf, mean = 200, sd = 20,      abs.tol = 0) 
 - attr(*, "class")= chr "integrate"