R中置信区间的仿真 [英] Simulation for Confidence interval in R

问题描述

我有一个R函数，可为t分布的 ncp (非中心参数)提供95％的置信区间.

I have an R function that provides the 95% confidence Interval for the ncp (non-centrality parameter) of a t distribution.

通过R中的仿真，有可能表明，从长远来看，该R函数的CI会捕获给定的 TRUE ncp (此处的"2"与输入 t )95％的时间?

Via simulation in R, is it possible to show that in the long-run the CIs from this R function capture a given TRUE ncp (here "2" same as input t) 95% of the time?

(感谢您对如何执行此操作的任何想法)

(I appreciate any ideas as to how to do this)

CI.ncp <- function(t, N){

  f <- function (ncp, alpha, q, df) {  
abs(suppressWarnings(pt(q = t, df = N - 1, ncp, lower.tail = FALSE)) - alpha) }

sapply(c(0.025, 0.975),
function(x) optim(1, f, alpha = x, q = t, df = N - 1, control = list(reltol = (.Machine$double.eps)))[[1]]) }

#Example of Use:
CI.ncp(t = 2, N = 20) # gives: -0.08293755  4.03548862 

#(in the long-run 95% of the time, "2" is contained within these
# two numbers, how to show this in R?)

这是我尝试的没有成功的事情:

Here is what I have tried with no success:

fun <- function(t = 2, N = 20){

  ncp = rt(1, N - 1, t)
  CI.ncp(t = 2, N = 20)
  mean(ncp <= 2 & 2 <= ncp )
   }

 R <- 1000
 sim <- t(replicate(R, fun()))
 coverage <- mean(sim[,1] <= 2 & 2 <= sim[,2])

推荐答案

问题是，我们需要在CI.ncp中提供从fun获得的随机ncp:

The problem is the that we need to feed the random ncp obtained from the fun in the CI.ncp:

 fun <- function(t = 2, N = 20){ ;

   ncp = rt(1, N - 1, t);
   CI.ncp(t = ncp, N = 20);
   }

  R <- 1e4 ;
  sim <- t(replicate(R, fun()));
  coverage <- mean(sim[,1] <= 2 & 2 <= sim[,2])

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