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%\date{January 23, 2003}

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\title{Stat 312: Lecture 4\\
Maximum Likelihood Estimation}

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

\begin{enumerate}
\item For a random sample $X_1,\cdots,X_n$  the {\em likelihood function} is given as the product of probability or density functions, i.e. $L(\theta)= f(x_1;\theta)f(x_2;\theta)\cdots f(x_n;\theta)$.
\item The {\em maximum likelihood estimatate} of $\theta$ 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 $\hat \theta = \hat \theta(X_1,\cdots,X_n)$. {\bf Note:} $x_i$ are numbers while $X_i$ are random variables.

\item When the sample size is large, the maximum likelihood estimator of $\theta$ is approximately unbiased. The MLE of $\theta$ is approximately the MVUE of $\theta$. This is why it is the most widely used parameter estimation technique.

\item If explicit density function is not available, you can not apply MLE. In this case apply the method of moment matching. 

\item (Invariance Principle) If $\hat \theta_1, \hat \theta_2$ are the MLE's of $\theta_1,\theta_2$, the MLE of $h(\theta_1,\theta_2)$ is $h(\hat \theta_1, \hat \theta_2)$.
\end{enumerate}
\section*{In-class problems}
Example 6.15.
Example 6.16.
Example 6.21.\\
Exercise 6.25. Assuming normal distribution, find $c$ such that $P(\mbox{strength} \leq c) = 0.95$. 

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\caption{Density of $N(0,1)$}
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\begin{verbatim}
> x<-strength
> length(x)
[1] 10
> sd(x)
[1] 19.87852
> sigma<-sqrt((length(x)-1)/length(x))*sd(x)
> sigma
[1] 18.85842
> qnorm(0.95,mu,sigma)
[1] 415.4193
> qnorm(0.95)
[1] 1.644854
> pnorm(415,mu,sigma)
[1] 0.9476644
> y<- -30:30/10
> plot(y,dnorm(y),'l')
\end{verbatim}
\section*{Self-study problems}
Example 6.17., Example 6.18., Exercise 6.23., Exercise 6.29.


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