Example: Let \(X_1,...,X_n\) be iid \(uniform(0,\theta)\).
\[ f(x)\begin{cases}{1\over \theta}~~~0\leq x\leq \theta,\\0,~~~elsewhere\end{cases} \]
k=1, only need the first moment
\[ \mu_1=E(X_1)={\theta\over 2}\\ \hat\mu_1={1\over n}\sum_{i=1}^nX_i \]
Setting \(\mu_1=\hat \mu_1\), we have \({\theta\over 2}={1\over n}\sum_{i=1}^nX_i\Rightarrow\hat\theta=2\bar X\)
\[ lik(\theta)=\prod^n_{i=1}f(x_i|\theta)=\begin{cases}{1\over\theta^n},~~~0\leq X_1,...,X_n\leq \theta\\0,~~~ elsewhere\end{cases} \]
We can use the sample mean \(\bar X\) as a point estimate for the population mean \(\mu\).
LLN: \(\bar X\) approximates \(\mu\) more accurately as the sample size increases.
CLT: allows us to access the probability that \(\bar X\) will be within a certain distance from \(\mu\).
use only \(\bar X\)to estimate \(\mu\), it is not very informative since it does not quantify accuracy.
A confidence interval is an interval for which we can asserts, with a given degree of confidence/certainty, that it includes the true value of the parameter being estimated.
\[ z_\alpha=100(1-\alpha)-th~percentile \]
\(z_\alpha\) is also called as upper percentile sometime.
For example:
\[ \alpha =0.025\\ z_\alpha=qnorm(1-\alpha)=1.96 \]
Let \(X_1,...,X_n\) be iid random variables with parameter \(\theta\).
\[ \hat\theta\sim N(\theta,\sigma_{\hat\theta}^2), approximately\\ {\hat\theta-\theta\over \sigma_{\hat\theta}}\sim N(0,1), approximately \]
you can also replace \(\sigma_{\hat\theta}\) by \(S_{\hat\theta}\) (estimated standard error) if \(\sigma_{\hat\theta}\) is unknown
\[ P(-1.96<{\hat\theta-\theta\over \sigma_{\hat\theta}}<1.96)=0.95 \]
\[ \hat \theta-1.96S_{\hat \theta}\leq \theta\leq \hat\theta+1.96S_{\hat \theta} \]
\[ {\hat\theta-\theta\over \sigma_{\hat\theta}}\leq1.96 \Leftrightarrow \hat\theta-\theta\leq1.96\sigma_{\hat\theta} \Leftrightarrow \theta\geq\hat \theta-1.96\sigma_{\hat\theta} \]
\[ {\hat\theta-\theta\over \sigma_{\hat\theta}}\geq-1.96 \Leftrightarrow \hat\theta-\theta\geq -1.96\sigma_{\hat\theta} \Leftrightarrow \theta\leq\hat \theta+1.96\sigma_{\hat\theta} \]
you can replace \(\sigma_{\hat\theta}\) by \(S_{\hat \theta}\)
\[ \hat\theta-z_{\alpha/2}S_{\hat\theta}\leq\theta\leq\hat\theta+z_{\alpha/2}S_{\hat\theta} \]
gives us a range of possible values of the true parameter \(\theta\), such that we are \((1-\alpha)100%\) confident that it contains the true \(\theta\)
Common choice of \(\alpha\): 0.01 (99%), 0.05 (95%), 0.01 (90%).
only works well if \(\sigma\) is known in mean estimation or sample size is very large.
Confidence intervals that use percentiles from the T distribution.
\(T_v\) is a random variable from the T distribution with v degrees of freedom. (or df)
degree of freedom often depends on the sample size.
\[ \lim_{v\to+\infty}t_v\sim N(0,1) \]
In R:
dt(x, v) gives the PDF of \(T_v\) for x
pt(x, v) gives the CDF of \(T_v\) for x
qt(s, v) gives the s100 percentile of \(T_v\)
rt(n, v) generates a random sample of n \(T_v\) random variables
Many estimator \(\hat\theta\) of \(\theta\) satisfy
\[ {\hat\theta-\theta\over S_{\hat\theta}}\sim T_v \]
when you replace the estimated by \(S_{\hat \theta}\), the result shall have greater variability.
\({\hat\theta-\theta\over S_{\hat\theta}}\) has a greater variability than \({\hat\theta-\theta\over \sigma_{\hat\theta}}\)
\[ (\hat\theta-t_{v,\alpha/2}S_{\hat\theta},\hat\theta+t_{v,\alpha/2}S_{\hat\theta}) \]
Notes:
T confidence intervals will generally be used for means.
\(t_{v,\alpha/2}\to z_{\alpha/2}\) as \(n\to \infty\)
Common values of \(\alpha\) are 0.1,0.05,and 0.01