dnorm(x,均值=200,标准差=20)的积分不是1
我试图计算正态分布密度的积分,期望值为200,标准偏差为20。从-Inf到Inf这应该是1 我得到以下信息:dnorm(x,均值=200,标准差=20)的积分不是1,r,integral,probability-density,R,Integral,Probability Density,我试图计算正态分布密度的积分,期望值为200,标准偏差为20。从-Inf到Inf这应该是1 我得到以下信息: > integrate(dnorm, mean=200, sd=20,-Inf, Inf)$value [1] 1.429508e-08 对于低于169的预期值,我得到正确的值1。 如何为更大的期望值获取正确的值?设置区间有限似乎有帮助 integrate(dnorm, mean=200, sd=20, -1e4, 1e4) # 1 with absolute error <
> integrate(dnorm, mean=200, sd=20,-Inf, Inf)$value
[1] 1.429508e-08
对于低于169的预期值,我得到正确的值1。
如何为更大的期望值获取正确的值?设置区间有限似乎有帮助
integrate(dnorm, mean=200, sd=20, -1e4, 1e4)
# 1 with absolute error < 2e-07
integrate(dnorm,平均值=200,标准差=20,-1e4,1e4)
#1绝对误差<2e-07
或者
integrate(dnorm, mean=200, sd=20, lower= -Inf, upper= Inf, abs.tol = 0)$value
[1] 1
看
要查看发生了什么,请注意以下细分的数量:
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"
js当R提供CDF时,为什么要积分正态密度函数,即pnorm
?