全文链接：http://tecdat.cn/?p=27695 

原文出处：拓端数据部落公众号

 

参数引导：估计 MSE

统计学问题：级别(k\)修剪后的平均值的MSE是多少？

我们如何回答它：估计从标准柯西分布（t 分布 w/df = 1）生成的大小为 20 的随机样本的水平 k" role="presentation" style="position: relative;">k$k$ 修剪均值的 MSE。目标参数 θ" role="presentation" style="position: relative;">θ$\theta$ 是中心或中位数。柯西分布不存在均值。在表中总结 MSE 的估计值 k=1,2,...9" role="presentation" style="position: relative;">k=1,2,...9$k=1,2,...9$。

 
result=rep(0,9)
for(j in 1:9){
  n<-20
 
  for(i in 1:m){
    x<-sort(rcauchy(n))

参数自抽样法：经验功效计算

统计问题：随着零假设与现实之间的差异发生变化，功效如何变化？

我们如何回答：绘制 t 检验的经验功效曲线。

t 检验的原假设是 。另一种选择是。

您将从具有 的正态分布总体中抽取大小为 20 的样本。您将使用 0.05 的显着性水平。

显示当总体的实际平均值从 350 变为 650（增量为 10）时，功效如何变化。

y 轴是经验功效（通过 bootstrap 估计），x 轴是 μ" role="presentation" style="position: relative;">μ$\mu$ 的不同值（350、360、370 … 650）。

 
 
    x <- rnorm(n, mean = muA, sd = sigma) #抽取平均值=450的样本
    ts <- t.test(x, mu = mu0) #对无效的mu=500进行t检验
    ts$p.value
 

 

参数自抽样法：经验功效计算

统计问题：样本量如何影响功效？

我们如何回答：创建更多的功效曲线，因为实际均值在 350 到 650 之间变化，但使用大小为 n = 10、n = 20、n = 30、n = 40 和 n = 50 的样本生成它们。同一图上的所有 5 条功效曲线。

 
pvals <- replicate(m, pvalue())
power <- mean(pvals <= 0.05)
 
 
points(sequence,final2[2,],col="red",pch=1)
 
points(sequence,final2[3,],col="blue",pch=2)

 

参数自抽样法：经验置信水平

统计问题：在制作 95% CI 时，如果我们的样本很小并且不是来自正态分布，我们是否仍有 95% 的置信度？

我们如何回答它：根据样本为总体的平均值创建一堆置信区间 (95%)。

您的样本大小应为 16，取自具有 2 个自由度的卡方分布。

找出未能捕捉总体真实均值的置信区间的比例。（提醒：自由度为 k" role="presentation" style="position: relative;">k$k$ 的卡方分布的平均值为 k" role="presentation" style="position: relative;">k$k$。）

 
for(i in 1:m){
  samp=rchisq(n,df=2)
  mean=mean(samp)
  sd=sd(samp)
  upper=mean+qt(0.975,df=15)*sd/4

 

非参数自抽样法置信区间

统计问题：基于一个样本，我们可以为总体相关性创建一个置信区间吗？

我们如何回答：为相关统计量创建一个 bootstrap t 置信区间估计。

 
boot.ti <-
  function(x, B = 500, R = 100, level = .95, stattic){
    
    x <- as.matrix(x)
 
library(boot)       #for boot and boot.ci
 
data(law, package = "bootstrap")
 
dat <- law
 
ci <- boot.t.ci(dat, statistic = stat, B=2000, R=200)
ci

 

自抽样法后的Jackknife

统计问题：R 的标准误差的 bootstrap 估计的标准误差是多少？

我们如何回答它： data(law) 像上一个问题一样使用。在 bootstrap 后执行 Jackknife 以获得标准误差估计的标准误差估计。（bootstrap 用于获得总体中 R 的 SE 的估计值。然后使用折刀法获得该 SE 估计值的 SE。）

 
indices <- matrix(0, nrow = B, ncol = n)
 
# 进行自举
for(b in 1:B){
    i <- sample(1:n, size = n, replace = TRUE)
    LSAT <- law$LSAT[i]
 
#  jackknife
 
for(i in 1:n){
    keepers <- function(k){
         !any(k == i)   
    }

 

自测题

Submit the rendered HTML file. Make sure all requested output (tables, graphs, etc.) appear in your document when you submit.

Parametric Bootstrap: Estimate MSE

Statistical question: What is the MSE of a level k" role="presentation" style="position: relative;">k$k$ trimmed mean?

How we can answer it: Estimate the MSE of the level k" role="presentation" style="position: relative;">k$k$ trimmed mean for random samples of size 20 generated from a standard Cauchy distribution (t-distribution w/df = 1). The target parameter θ" role="presentation" style="position: relative;">θ$\theta$ is the center or median. The mean does not exist for a Cauchy distribution. Summarize the estimates of MSE in a table for k=1,2,...9" role="presentation" style="position: relative;">k=1,2,...9$k=1,2,...9$.

Parametric Bootstrap: Empirical Power Calculations

Statistical question: How does power change as the difference between the null hypothes and the reality changes?

How we can answer it: Plot an empirical power curve for a t-test.

The null hypothesis of the t-test is μ=500" role="presentation" style="position: relative;">μ=500$\mu =500$. The alternative is μ≠500" role="presentation" style="position: relative;">μ≠500$\mu \ne 500$.

You will draw samples of size 20, from a normally distributed population with σ=100" role="presentation" style="position: relative;">σ=100$\sigma =100$. You will use a significance level of 0.05.

Show how the power changes as the actual mean of the population changes from 350 to 650 (increments of 10).

On the y-axis will be the empirical power (estimated via bootstrap) and the x-axis will be the different values of μ" role="presentation" style="position: relative;">μ$\mu$ (350, 360, 370 … 650).

Parametric Bootstrap: Empirical Power Calculations

Statistical question: How does sample size affect power?

How we can answer it: Create more power curves as the actual mean varies from 350 to 650, but produce them for using samples of size n = 10, n = 20, n = 30, n = 40, and n = 50. Put all 5 power curves on the same plot.

Parametric Bootstrap: Empirical Confidence Level

Statistical question: When making a 95% CI, are we still 95% confident if our samples are small and do not come from a normal distribution?

How we can answer it: Create a bunch of Confidence Intervals (95%) for the mean of a population based on a sample.

Your samples should be of size 16, drawn from a chi-squared distribution with 2 degrees of freedom.

Find the proportion of Confidence Intervals that fail to capture the true mean of the population. (Reminder: a chi-squared distribution with k" role="presentation" style="position: relative;">k$k$ degrees of freedom has a mean of k" role="presentation" style="position: relative;">k$k$.)

Non Parametric Bootstrap Confidence Interval

Statistical question: Based on one sample, can we create a confidence interval for the correlation of the population?

How we can answer it: Create a bootstrap t confidence interval estimate for the correlation statistic.

Jackknife after bootstrap

Statistical question: What is the standard error of the bootstrap estimate of the standard error of R?

How we can answer it: Use data(law) like the previous problem. Perform Jackknife after bootstrap to get a standard error estimate of the standard error estimate. (The bootstrap is used to get an estimate of the SE of R in the population. The jackknife is then used to get an SE of that SE estimate.)

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