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\date{September 2, 2004}

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\title{Stat 312: Lecture 02\\
Point Estimation}

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\author{Moo K. Chung\\
mchung@stat.wisc.edu}

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\begin{enumerate}
\item If $X$ and $Y$ are independent, $\mathbb{E} (X+Y) =
\mathbb{E}X + \mathbb{E}Y$ and $\mathbb{V} (X+Y) = \mathbb{V}X +
\mathbb{V}Y$. See pp. 244.
\item If a random sample $X_i \sim N(\mu,\sigma^2)$, $\bar X \sim
N(\mu,\sigma^2/n)$.\\
 {\em Proof.}
\bq \mathbb{V} \bar X &=&\mathbb{E} \Big(\frac{\sum X_i}{n}\Big)^2
-\Big[\mathbb{E} \Big(\frac{\sum X_i}{n}\Big)\Big]^2\\
&=& \frac{1}{n^2}\mathbb{E} (\sum X_i)^2 - \mu^2.\eq Now note that
\bq \mathbb{E} (\sum X_i)^2 &=&\mathbb{V} (\sum X_i) + (\mathbb{E}
\sum X_i)^2\\
&=& \sum \mathbb{V} X_i + (\sum \mathbb{E}X_i)^2\\
&=& n\sigma^2 + (n\mu)^2
 \eq
Hence $\mathbb{V}\bar X = \frac{\sigma^2}{n}.$

\item A {\em point estimate} of population parameter $\theta$ is a single
number that reasonably approximate $\theta$. A {\em point estimator}
of $\theta$ is a statistic (rule or formula) for getting the point
estimate given sample data.

\item
Let $X_i$ be a Bernoulli random variable with $P(X_i=1)=p$. Then
$\sum_{i=1}^n X_i$ is a Binomial random variable with $n$ and $p$
parameters. Note that a Binomial random variable with $n=1$ is a
Bernoulli random variable. The following example simulates 1000 coin
tosses.

\begin{verbatim}
> a<-rbinom(1000,1,0.5)
> a
   [1] 0 0 1 0 0 1 0 1 1 0 0 ...
> sum(a)/1000
[1] 0.517
> b<-rbinom(1000000,1,0.5)
> sum(b)/1000000
[1] 0.500545
\end{verbatim}

\item We can let
$$\hat \theta = \theta + \mbox{error of estimation.}$$
Taking expectation on both sides, we define the {\em bias} of
estimator $\hat \theta$ to be $$Bias(\hat \theta) = \mathbb{E}(\hat
\theta) - \theta.$$ It measures the biasness of estimator $\hat
\theta$ in average sense.


\end{enumerate}
{\em Reviw Problems.} example 6.1.,6.3.\\
{\bf Assignment I.} Due Sept 16 9:30am. Exercise. 6.2. 6.8. 6.10.
6.16. 6.20. 6.22. 


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