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\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Problem Set 6}
\centerline{\it Due: In class March 26, 2004}
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\noindent
A)  In this problem you will complete the derivation of the 
maximum likelihood estimators for the parameters $\mu$ and 
$\sigma^2$ of a normal distribution that we started in class
on Wednesday, March 19.  Recall that we saw the likelihood function
is 
$$L = L(y_1,\ldots,y_n\ |\ \mu,\sigma^2) = \left({1\over \sqrt{2\pi\sigma^2}}\right)^n
\exp\left(-\sum_{i=1}^n(y_i - \mu)^2/(2\sigma^2)\right)$$
so 
$$\ln(L) = {-n\over 2}\ln(2\pi\sigma^2) - \sum_{i=1}^n (y_i - \mu)^2/(2\sigma^2)$$
We saw that
$${\partial \ln(L)\over \partial\mu} = \sum_{i=1}^n (y_i - \mu)/\sigma^2$$
and 
$${\partial \ln(L)\over \partial\sigma^2} = {-n\over 2\sigma^2} + 
{\sum_{i=1}(y_i - \mu)^2/2\over (\sigma^2)^2}$$
Setting ${\partial \ln(L)\over \partial\mu} = 0$, we got
$$\widehat{\mu}_{ML} = \overline{Y}$$
(the sample mean).
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\item{1)} Substituting $\widehat{\mu}_{ML} = \overline{Y}$ for
$\mu$, solve the equation ${\partial \ln(L)\over \partial\sigma^2} = 0$
for $\sigma^2$ to derive $\widehat{\sigma^2}_{ML}$.  Is this 
estimator biased or unbiased?
\item{2)} Check that the values of $\widehat{\mu}_{ML}$ and 
$\widehat{\sigma^2}_{ML}$ are actually give a {\it maximum}
(not just a critical point) of $\ln(L)$ using the second derivative
test for functions of two variables.
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\noindent
{\it From the text}
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9.78, 9.79, 9.92 (in part c, the efficiency is the ratio of the 
variances of the estimators -- see Definition 9.1), 10.4, 10.7, 10.9,
10.23.
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