全文链接：http://tecdat.cn/?p=27695
参数疏导：预计 MSE
统计学问题：级别 (k\) 修剪后的平均值的 MSE 是多少？
咱们如何答复它：预计从规范柯西散布（t 散布 w/df = 1）生成的大小为 20 的随机样本的程度 \(k\) 修剪均值的 MSE。指标参数 \(\theta\) 是核心或中位数。柯西散布不存在均值。在表中总结 MSE 的估计值 \(k = 1, 2, … 9\)。
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：教训效用计算
统计问题：随着零假如与事实之间的差别发生变化，效用如何变动？
咱们如何答复：绘制 t 测验的教训效用曲线。
t 测验的原假如是 。另一种抉择是。
您将从具备 的正态分布总体中抽取大小为 20 的样本。您将应用 0.05 的显着性程度。
显示当总体的理论平均值从 350 变为 650（增量为 10）时，效用如何变动。
y 轴是教训效用（通过 bootstrap 预计），x 轴是 \(\mu\) 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) #对有效的 mu=500 进行 t 测验
ts$p.value
参数自抽样法：教训效用计算
统计问题：样本量如何影响效用？
咱们如何答复：创立更多的效用曲线，因为理论均值在 350 到 650 之间变动，但应用大小为 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 条效用曲线。
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：教训置信水平
统计问题：在制作 95% CI 时，如果咱们的样本很小并且不是来自正态分布，咱们是否仍有 95% 的置信度？
咱们如何答复它：依据样本为总体的平均值创立一堆置信区间 (95%)。
您的样本大小应为 16，取自具备 2 个自由度的卡方散布。
找出未能捕获总体实在均值的置信区间的比例。（揭示：自由度为 \(k\) 的卡方散布的平均值为 \(k\)。）
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，咱们能够为总体相关性创立一个置信区间吗？
咱们如何答复：为相干统计量创立一个 bootstrap t 置信区间预计。
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 预计的标准误差是多少？
咱们如何答复它：data(law) 像上一个问题一样应用。在 bootstrap 后执行 Jackknife 以取得标准误差预计的标准误差预计。（bootstrap 用于取得总体中 R 的 SE 的估计值。而后应用折刀法取得该 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 \(k\) trimmed mean?
How we can answer it: Estimate the MSE of the level \(k\) trimmed mean for random samples of size 20 generated from a standard Cauchy distribution (t-distribution w/df = 1). The target parameter \(\theta\) is the center or median. The mean does not exist for a Cauchy distribution. Summarize the estimates of MSE in a table for \(k = 1, 2, … 9\).
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 \(\mu = 500\). The alternative is \(\mu \ne 500\).
You will draw samples of size 20, from a normally distributed population with \(\sigma = 100\). 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 \(\mu\) (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.
\[\bar{x} \pm t^{*} \times \frac{s}{\sqrt{n}}\]
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 \(k\) degrees of freedom has a mean of \(k\).)
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.)
最受欢迎的见解
1. 应用 R 语言进行 METROPLIS-IN-GIBBS 采样和 MCMC 运行
2.R 语言中的 Stan 概率编程 MCMC 采样的贝叶斯模型
3.R 语言实现 MCMC 中的 Metropolis–Hastings 算法与吉布斯采样
4.R 语言 BUGS JAGS 贝叶斯剖析 马尔科夫链蒙特卡洛办法（MCMC）采样
5.R 语言中的 block Gibbs 吉布斯采样贝叶斯多元线性回归
6.R 语言 Gibbs 抽样的贝叶斯简略线性回归仿真剖析
7.R 语言用 Rcpp 减速 Metropolis-Hastings 抽样预计贝叶斯逻辑回归模型的参数
8.R 语言应用 Metropolis- Hasting 抽样算法进行逻辑回归
9. R 语言中基于混合数据抽样 (MIDAS) 回归的 HAR-RV 模型预测 GDP 增长回归的 HAR-RV 模型预测 GDP 增长 ”)