R語言參數(shù)自抽樣法Bootstrap：估計(jì)MSE、經(jīng)驗(yàn)功效、杰克刀Jackknife、非參數(shù)自抽樣法可
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原文出處：拓端數(shù)據(jù)部落公眾號(hào)
?相關(guān)視頻：什么是Bootstrap自抽樣及應(yīng)用R語言線性回歸預(yù)測(cè)置信區(qū)間實(shí)例
什么是Bootstrap自抽樣及R語言Bootstrap線性回歸預(yù)測(cè)置信區(qū)間
，時(shí)長(zhǎng)05:38
參數(shù)引導(dǎo)：估計(jì) MSE
統(tǒng)計(jì)學(xué)問題：級(jí)別(k\)修剪后的平均值的MSE是多少？
我們?nèi)绾位卮鹚汗烙?jì)從標(biāo)準(zhǔn)柯西分布（t 分布 w/df = 1）生成的大小為 20 的隨機(jī)樣本的水平 \(k\) 修剪均值的 MSE。目標(biāo)參數(shù) \(\theta\) 是中心或中位數(shù)?？挛鞣植疾淮嬖诰?。在表中總結(jié) MSE 的估計(jì)值 \(k = 1, 2, ... 9\)。
result=rep(0,9)for(j in 1:9){n<-20for(i in 1:m){x<-sort(rcauchy(n))
參數(shù)自抽樣法：經(jīng)驗(yàn)功效計(jì)算
統(tǒng)計(jì)問題：隨著零假設(shè)與現(xiàn)實(shí)之間的差異發(fā)生變化，功效如何變化？
我們?nèi)绾位卮穑豪L制 t 檢驗(yàn)的經(jīng)驗(yàn)功效曲線。
t 檢驗(yàn)的原假設(shè)是?
。另一種選擇是
。
您將從具有
?的正態(tài)分布總體中抽取大小為 20 的樣本。您將使用 0.05 的顯著性水平。
顯示當(dāng)總體的實(shí)際平均值從 350 變?yōu)?650（增量為 10）時(shí)，功效如何變化。
y 軸是經(jīng)驗(yàn)功效（通過 bootstrap 估計(jì)），x 軸是 \(\mu\) 的不同值（350、360、370 … 650）。
x <- rnorm(n, mean = muA, sd = sigma) #抽取平均值=450的樣本ts <- t.test(x, mu = mu0) #對(duì)無效的mu=500進(jìn)行t檢驗(yàn)ts$p.value
?
參數(shù)自抽樣法：經(jīng)驗(yàn)功效計(jì)算
統(tǒng)計(jì)問題：樣本量如何影響功效？
我們?nèi)绾位卮穑簞?chuàng)建更多的功效曲線，因?yàn)閷?shí)際均值在 350 到 650 之間變化，但使用大小為 n = 10、n = 20、n = 30、n = 40 和 n = 50 的樣本生成它們。同一圖上的所有 5 條功效曲線。
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)
?
參數(shù)自抽樣法：經(jīng)驗(yàn)置信水平
統(tǒng)計(jì)問題：在制作 95% CI 時(shí)，如果我們的樣本很小并且不是來自正態(tài)分布，我們是否仍有 95% 的置信度？
我們?nèi)绾位卮鹚焊鶕?jù)樣本為總體的平均值創(chuàng)建一堆置信區(qū)間 (95%)。
您的樣本大小應(yīng)為 16，取自具有 2 個(gè)自由度的卡方分布。
找出未能捕捉總體真實(shí)均值的置信區(qū)間的比例。（提醒：自由度為 \(k\) 的卡方分布的平均值為 \(k\)。）
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
?
非參數(shù)自抽樣法置信區(qū)間
統(tǒng)計(jì)問題：基于一個(gè)樣本，我們可以為總體相關(guān)性創(chuàng)建一個(gè)置信區(qū)間嗎？
我們?nèi)绾位卮穑簽橄嚓P(guān)統(tǒng)計(jì)量創(chuàng)建一個(gè) bootstrap t 置信區(qū)間估計(jì)。
boot.ti <-function(x, B = 500, R = 100, level = .95, stattic){x <- as.matrix(x)library(boot) ? ? ? #for boot and boot.cidata(law, package = "bootstrap")dat <- lawci <- boot.t.ci(dat, statistic = stat, B=2000, R=200)ci
?
自抽樣法后的Jackknife
統(tǒng)計(jì)問題：R 的標(biāo)準(zhǔn)誤差的 bootstrap 估計(jì)的標(biāo)準(zhǔn)誤差是多少？
我們?nèi)绾位卮鹚?data(law)?像上一個(gè)問題一樣使用。在 bootstrap 后執(zhí)行 Jackknife 以獲得標(biāo)準(zhǔn)誤差估計(jì)的標(biāo)準(zhǔn)誤差估計(jì)。（bootstrap 用于獲得總體中 R 的 SE 的估計(jì)值。然后使用折刀法獲得該 SE 估計(jì)值的 SE。）
indices <- matrix(0, nrow = B, ncol = n)# 進(jìn)行自舉for(b in 1:B){i <- sample(1:n, size = n, replace = TRUE)LSAT <- law$LSAT[i]# ?jackknifefor(i in 1:n){keepers <- function(k){!any(k == i)}
?
自測(cè)題
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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.)
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