---

Let $X_1$, $X_2$, $X_3$, $...$, $X_n$ be a random sample from a distribution with a parameter $\theta$ that is to be estimated. Our goal is to find two estimators for $\theta$:

1. the low estimator, $\hat{\Theta}_l=\hat{\Theta}_l(X_1, X_2, \cdots, X_n)$, and
2. the high estimator, $\hat{\Theta}_h=\hat{\Theta}_h(X_1, X_2, \cdots, X_n)$.

The interval estimator is given by the interval $[\hat{\Theta}_l, \hat{\Theta}_h]$. The estimators $\hat{\Theta}_l$ and $\hat{\Theta}_h$ are chosen such that the probability that the interval $[\hat{\Theta}_l, \hat{\Theta}_h]$ includes $\theta$ is larger than $1-\alpha$. Here, $1-\alpha$ is said to be confidence level. We would like $\alpha$ to be small. Common values for $\alpha$ are $0.1$, $.05$, and $.01$ which correspond to confidence levels $90\%$, $95\%$, and $99\%$ respectively. Thus, when we are asked to find a $95 \%$ confidence interval for a parameter $\theta$, we need to find $\hat{\Theta}_l$ and $\hat{\Theta}_h$ such that \begin{align}%\label{} P\bigg(\hat{\Theta}_l < \theta \, \textrm{and} \, \hat{\Theta}_h > \theta \bigg) \geq 0.95 \end{align} The above discussion will become clearer as we go through examples. Before doing that let's formally define interval estimation. Note that the condition \begin{align}%\label{} P\bigg(\hat{\Theta}_l \leq \theta \, \textrm{and} \, \hat{\Theta}_h \geq \theta \bigg) \geq 1-\alpha \end{align} can be equivalently written as \begin{align}%\label{} P\bigg(\hat{\Theta}_l \leq \theta \leq \hat{\Theta}_h \bigg) \geq 1-\alpha, \quad \textrm{or} \quad P\bigg(\theta \in \big[\hat{\Theta}_l,\hat{\Theta}_h\big] \bigg) \geq 1-\alpha. \end{align} The randomness in these terms is due to $\hat{\Theta}_l$ and $\hat{\Theta}_h$, not $\theta$. Here, $\theta$ is the unknown quantity which is assumed to be non-random (frequentist inference). On the other hand, $\hat{\Theta}_l$ and $\hat{\Theta}_h$ are random variables because they are functions of the observed random variables $X_1$, $X_2$, $X_3$, $...$, $X_n$.

---