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\begin{document}

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%\date{January 23, 2003}

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\title{Stat 312: Lecture 3}

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\author{Moo K. Chung\\
mchung@stat.wisc.edu}
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\section*{Concepts}
\begin{enumerate}
\item There may be many unbiased estimators of $\theta$. The {\em Minimum variance unbiased estimator} (MVUE) is an unbiased estimator with the smallest variance. Given a random sample $X_1,\cdots,X_n$ with population mean $\mu$
and population variance $\sigma^2$, $\bar X$ is MVUE of $\mu$.
\item $\hat \theta_1$ and $\hat \theta_2$ are two unbiased estimator of $\theta$.
If ${\bf V(\hat \theta_1) \leq {\bf V(\hat \theta_2)$, $\hat
\theta_1$ is more efficient than $\hat \theta_2$. MVUE is the
most efficient estimator. An efficient estimator $\hat \theta$
will produce an estimate closer to the true parameter $\theta$.
\item The {\em $k$-th sample moment} is $M_k=\sum_{j=1}^n X_j^k/n$.
\item The {\em moment estimators} of population parameters
are obtained by solving $\mathbb{E}(X)= M_1$,
$\mathbb{E}(X^2)=M_2, $\mathbb{E}(X^3)=M_3$, etc.
\end{enumerate}

\section*{In-class problems}
Let $X_1,X_2$ be a random sample. Among all estimator of the form
$\hat \mu = aX_1 + bX_2$, find MVUE.\\
Exercise 6.25. Assume that shear strength is normally
distributed. Find the moment estimators of $\mu$ and $\sigma^2$.
\begin{verbatim}
> str(ex06.25)
`data.frame':   10 obs. of  1 variable:
 $ strength: int  392 376 401 367 389 362 409 415 358 375
> X<-strength
> M1<-mean(X)
> M2<-sum(X^2)/10
> M1
[1] 384.4
> M2
[1] 148119
> sigma<-sqrt(M2-M1^2)
> sigma
[1] 18.85842

\end{verbatim}
 Example 6.11.

\section*{Self-study problems}
Example 6.5, Example 6.7. Example 6.12.
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