simultaneous estimation of the mean and variance

A $100(1-alpha)%$ confidence interval for a parameter $theta$ is a random interval $[L_1, L_2]$ :

$$P[L_1 leq theta leq L_2] = 1- alpha$$

Confidence interval on $mu$ when $sigma^2 $ is known:

$$overline{X} pm z_{alpha/2} frac{sigma}{sqrt{n}}$$

Joint Sampling of Mean and Variance:

Let $X_{1}, ldots, X_{n}, n geq 2​$, be a random sample of size n from a normal distribution with mean $mu​$ and variance $sigma^{2}​$ . Then:

$$begin{array}{l}{text { 1. The sample mean } overline{X} text { is independent of the sample variance } S^{2} text { , }} \ {text { 2. } overline{X} text { is normally distributed with mean } mu text { and variance } sigma^{2} / n,} \ {text { 3. }(n-1) S^{2} / sigma^{2} text { is chi-squared distributed with } n-1 text { degrees of freedom. }}end{array}$$

Furthermore, if in a given situation we assume that X and S2 are independently distributed we are essentially assuming that the population is normally distributed.

Interval Estimation of Variability:

A $100(1-alpha)%$ confidence interval on $sigma^2$ is given by:

$$left[(n-1) S^{2} / chi_{alpha / 2, n-1}^{2},(n-1) S^{2} / chi_{1-alpha / 2, n-1}^{2}right]$$

Let Z be a standard normal variable and let $chi^2$ be an independent chi-squared random variable with $gamma$ degrees of freedom. The random variable

$$T_{gamma}=frac{Z}{sqrt{chi_{gamma}^{2} / gamma}}$$

is said to follow a T distribution with $gamma$ degrees of freedom.

The density:

$$f_{T_{gamma}}(t)=frac{Gamma((gamma+1) / 2)}{Gamma(gamma / 2) sqrt{pi gamma}}left(1+frac{t^{2}}{gamma}right)^{-frac{gamma+1}{2}}$$

Theorem:

Let $X_{1}, X_{2},$ … , $X_{n}$ be a random sample from a normal distribution with $mu$ and $sigma$. Then the random variable:

$$T_{n-1}=frac{overline{X}-mu}{S / sqrt{n}}$$

follows a T distribution with n-1 degrees of freedom.

A $100(1-alpha)%$ confidence interval on $sigma^2$ is given by:

$$overline{X} pm t_{alpha / 2, n-1} S / sqrt{n}$$

Let $X$ be a normally distributed random variable and $overline{X}$ and $S^2$ be the mean and variance obtained from a sample of size n. Then there exist $K=K(n, alpha, delta)$ such that the interval

$$overline{X} pm K(n, alpha, delta) S$$

covers $($ at least ) $ delta cdot 100 %$ of the population with $(1-alpha) cdot 100 %$ confidence.

there exist numbers $K=K(n, alpha, delta)$ such that the intervals

$$(-infty, overline{X}+K S) quad and quad(overline{X}-K S, infty)$$

covers $($ at least ) $ delta cdot 100 %$ of the population with $(1-alpha) cdot 100 %$ confidence.

$$begin{aligned} 1-alpha &=Pleft[left(X_{min }, X_{max }right) text { covers at least } delta cdot 100 % text { of the population }right] \ &=1-n delta^{n-1}+(n-1) delta^{n} end{aligned}$$

where $n approx dfrac{1}{2}+dfrac{1+delta}{1-delta} cdot dfrac{chi_{alpha, 4}^{2}}{4}$