As here a mean of sample $X \sim \chi^2_2$ is estimated by a 95% symmetric t-interval, then we can get $\frac{\bar{X}-\mu}{\sqrt{\frac{\sum{(X_i-\bar{X})^2}}{n-1}}} \sim t_{n-1}$, so the interval estimated would be $P(\frac{\bar{X}-\mu}{\sqrt{\frac{\sum{(X_i-\bar{X})^2}}{n-1}}} > t_{n-1}(\alpha))$, which could be transformed as $P(\bar(X)>\mu+\sqrt{\frac{\sum{(X_i-\bar{X})^2}}{n-1}}t_{n-1}(\alpha))$. And we use Monte-Carlo method to get empirical confidence level.
set.seed(19036) n <- 20 m=10000 alpha <- .05 UCL <- replicate(1000, expr = { x <- rnorm(n, mean = 0, sd = 2) (n-1) * var(x) / qchisq(alpha, df = n-1) }) y=rchisq(n,df=2) ybar=mean(y)/qt(df=n-1,1-alpha) esti_int=replicate(10000,expr={ y=rchisq(n,df=2) (mean(y)-sd(y)*qt(df=n-1,alpha)) #(mean(y)-2)/sd(y) }) ##the empirical estimate of intervals for variance in example6.4 (mean_x=mean(UCL>4)) (sd_x=sd(UCL>4)) ##the empirical estimate of intervals for mean (mean_t=mean(esti_int>2)) (sd_t=sd(esti_int>2))
From the results, we can see that the t-interval estimate is much larger than the $\alpha$ which means that this method is much more robust.
n=300 p=c(0.025,0.05,0.95,0.975) ##true quantile of the skewness distribution cp=qnorm(p,sd=sqrt(6/n)) m=10000 sb1=replicate(m,expr = { x=rnorm(n) 1/n*sum((x-mean(x))^3)/(((n-1)/n*var(x))^(3/2)) }) esti_b1=rep(0,length(cp)) for(l in 1:length(cp)){ esti_b1[l]=quantile(sb1,p[l]) } ##estimate of the quantile of the skewness distribution esti_b1 knitr::kable(rbind(cp,esti_b1),col.names = p)
The table above is the comparison of the estimated quantiles with the true quantiles of $\sqrt{b_1}$.
sd1=rep(0,length(cp)) for(l in 1:length(cp)){ sd1[l]=p[l]*(1-p[l])/n/(dnorm(cp[l],sd=sqrt(6/n)))^2 } sd1
I compute the $f(x_q)$ by plug the true quantiles into the density function of $\sqrt{b_1}$.
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