D范数(x，mean=200，sd=20)的积分不是1

Question

我试着计算正态分布密度的积分，期望值为200，标准差为20，从-Inf到Inf应该是1。

我得到了以下信息：

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

对于低于169的期望值，我得到了正确的值，1。对于更大的期望值，我如何获得正确的值？

Answer 1

另一个选择

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

参见here。

要查看发生了什么，请注意以下内容中的细分数量：

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"

Answer 2

设置间隔有限似乎有帮助

integrate(dnorm, mean=200, sd=20, -1e4, 1e4)
# 1 with absolute error < 2e-07