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% Make title bold and 14 pt font (Latex default is non-bold, 16pt)
\title{Stat 312: Lecture 04\\
Moment Matching}

% For single author (just remove % characters)
\author{Moo K. Chung\\
mchung@stat.wisc.edu}
\date{September 14, 2004}
\maketitle \thispagestyle{empty}
\begin{enumerate}

\item Given a random sample $X_1,\cdots, X_n$, a linear estimator of
parameter $\theta$ is an estimator of form $$\hat \theta =
\sum_{i=1}^n c_i X_i.$$ Then it can be shown that $\bar X$ is the
MVUE for population mean among all linear unbiased estimators.\\
{\em Proof.} Case $n=2$ will be proved. The general statement
follows inductively. Consider linear estimators $$\hat \mu = c_1 X_1
+ c_2 X_2.$$ To be unbiased, $c_1 + c_2 =1$. To be most efficient
among all unbiased linear estimators, the variance has to be
minimized. The variance is
$$\mathbb{V} \hat \mu = c_1^2 \mathbb{V} X_1 + c_2^2\mathbb{V} X_2= \big[c_1^2 + (1-c_1)^2\big]\sigma^2$$
The quadratic term in the bracket $2c_1^2 -2c_1 + 1$  is minimized
when $c_1=1/2$.


\item Given a random sample $X_1,\cdots, X_n$, the {\em $k$-th sample moment} is $M_k=\sum_{j=1}^n X_j^k/n$.
The {\em moment estimators} of population parameters are obtained by
matching the sample moments to correspond population moments and
solving the resulting equations simultaneously.

\item {\em Exponential distribution.} $X$ is an
exponential distribution with parameter $\lambda$ if the density
function is $f(x) = \lambda e^{-\lambda x}$ for $ x \geq 0$. It can
be shown that $\mathbb{X}= \frac{1}{\lambda}, \mathbb{V}X =
\frac{1}{\lambda^2}$.


\item Given random sample $X_1,\cdots,X_n$,
the {\em likelihood function} is given as the product of
probability or density functions, i.e.
$$L(\theta)= f(x_1;\theta)f(x_2;\theta)\cdots f(x_n;\theta).$$
The {\em maximum likelihood estimatate} of $\theta$ is an estimate
that
 maximizes $L(\theta)$. If we denote $\hat \theta=\theta (x_1,\cdots,x_n)$ to
be the maximum likelihood estimate, The {\em maximum likelihood
estimator} (MLE) of $\theta$ is denoted by $\hat \theta = \hat
\theta(X_1,\cdots,X_n)$.
\end{enumerate}


{\em Review Problems.}

Example 6.12. Example 6.16. Example 6.17.


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