\documentclass[11pt]{article}
\author{Jeremy Morris}
\title{Exam 2}

\usepackage{amssymb,amsmath,amsfonts}
\usepackage[dvips]{graphicx}
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\addtolength{\textwidth}{2.5cm}
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\newcommand{\bs}[1]{
\boldsymbol{#1}
}

\begin{document}
\maketitle

\begin{enumerate}
\item[4.2] Consider a bivariate normal population with $\mu_{1} = 0,\mu_{2} = 2,\sigma_{11}=2,\sigma_{22}=1$ and $\rho_{12} = .5$. \label{a}
	\begin{enumerate}
	\item Write out the bivariate normal density.

	The multivariate normal density is defined by the following equation.
	\begin{eqnarray*}
	f(\boldsymbol{x}) & = & \frac{1}{(2\pi)^{p/2}|\boldsymbol{\Sigma}|^{1/2}} e^{-\frac{1}{2}(\boldsymbol{x} - \boldsymbol{\mu})'\boldsymbol{\Sigma}^{-1}(\boldsymbol{x} - \boldsymbol{\mu})}
	\end{eqnarray*}
	With $\boldsymbol{\Sigma} = \left(\begin{array}{ll} \sigma_{11} & \sigma_{12} \\ \sigma_{12} & \sigma_{22} \end{array}\right)$ and $\sigma_{12} = \rho_{12}\sqrt{\sigma_{11}}\sqrt{\sigma_{22}}$.  For this example, we have the following definitions for $\boldsymbol{\Sigma}$, and $|\boldsymbol{\Sigma}|$.
	\begin{eqnarray*}
	\boldsymbol{\Sigma} & = & 
		\left(\begin{array}{cc} 
		2 & \frac{\sqrt{2}}{2} \\ 
		\frac{\sqrt{2}}{2} & 1 
		\end{array}\right) \\
	|\boldsymbol{\Sigma}| & = & (2)(1) - (\sqrt{2}/2)^{2} = \frac{7}{4} \\
	\boldsymbol{\Sigma}^{-1} & = & \frac{4}{7}
		\left(\begin{array}{cc} 
		1 & \frac{-\sqrt{2}}{2} \\ 
		\frac{-\sqrt{2}}{2} & 2 
		\end{array}\right)
	\end{eqnarray*}
	To complete the definition of the density function we will derive the squared generalized distance expression $(\boldsymbol{x} - \boldsymbol{\mu})'\boldsymbol{\Sigma}^{-1}(\boldsymbol{x} - \boldsymbol{\mu})$.
	\begin{equation*}
	(\boldsymbol{x} - \boldsymbol{\mu})'\boldsymbol{\Sigma}^{-1}(\boldsymbol{x} - \boldsymbol{\mu}) =
		\left( \begin{array}{ll} x_{1} & x_{2} - 2 \end{array} \right )
		\left(\begin{array}{cc} 
		\frac{4}{7} & -\frac{2\sqrt{2}}{7} \\ 
		-\frac{2\sqrt{2}}{7} & \frac{8}{7} 
		\end{array}\right)
		\left( \begin{array}{l} x_{1} \\ x_{2} - 2 \end{array} \right )
	\end{equation*}
	\begin{equation*}
		\left( \begin{array}{ll} x_{1} & x_{2} - 2 \end{array} \right )
	\left( \begin{array}{l}
	\frac{4}{7}x_{1} - (x_{2} - 2)\frac{2\sqrt{2}}{7} \\
	-\frac{2\sqrt{2}}{7}x_{1} + (x_{2} - 2)\frac{8}{7}
	\end{array} \right)
	\end{equation*}
	\begin{equation*}
		\frac{1}{7}\left( \begin{array}{ll} x_{1} & x_{2} - 2 \end{array} \right )
	\left( \begin{array}{l}
	4x_{1} - 2\sqrt{2}x_{2} + 4\sqrt{2} \\
	-2\sqrt{2}x_{1} + 8x_{2} - 16
	\end{array} \right)
	\end{equation*}
	\begin{equation*}
		\frac{1}{7}
	x_{1}(4x_{1} - 2\sqrt{2}x_{2} + 4\sqrt{2}) + 
	\frac{1}{7}(x_{2} - 2)(-2\sqrt{2}x_{1} + 8x_{2} - 16)
	\end{equation*}
	\begin{equation}
	\varphi(x_{1},x_{2}) = \frac{1}{7}(4x_{1}^{2} + 8\sqrt{2}x_{1} - 4\sqrt{2}x_{1}x_{2} - 32x_{2} + 8x_{2}^{2} + 32) \label{gde}
	\end{equation}
	Then the bivariate normal distribution function is defined as follows.
	\begin{eqnarray*}
	f(x_{1},x_{2}) & = & \frac{1}{\sqrt{7}\pi}e^{-\frac{1}{2}\varphi(x_{1},x_{2})}
	\end{eqnarray*}

	\item Write out the squared generalized distance expression $(\boldsymbol{x} - \boldsymbol{\mu})'\boldsymbol{\Sigma}^{-1}(\boldsymbol{x} - \boldsymbol{\mu})$ as a function of $x_{1}$ and $x_{2}$.

	Derived in part (a) equation \ref{gde}.

	\item Determine (and sketch) the constant-density contour that contains 50\% of the probability.

	To determine the constant-density contour, we need to calculate the eigenvalues and eigenvectors of $\boldsymbol{\Sigma}$.  This is done by solving for the roots of the equation $|\boldsymbol{\Sigma} - \lambda I|$.
	\begin{eqnarray*}
	(2-\lambda)(1-\lambda) - 1/2 & = & 0 \\
	2 - 2\lambda - \lambda + \lambda^{2} - 1/2 & = & 0 \\
	\lambda^{2} - 3\lambda + 3/2 & = & 0
	\end{eqnarray*}
	Using the quadratic equation, we get the eigenvalues
	\begin{eqnarray*}
	\lambda_{1} & = & \frac{3 + \sqrt{3}}{2}\\
	\lambda_{2} & = & \frac{3 - \sqrt{3}}{2}
	\end{eqnarray*}
	Then we find the eigenvector for $\lambda_{1}$.
	\begin{eqnarray*}
	\boldsymbol{\Sigma}\boldsymbol{e}_{1} - \lambda_{1}\boldsymbol{e}_{1} & = & 0
	\end{eqnarray*}
	\begin{equation*} \Rightarrow
	\begin{array}{lcl}
	(2 - \frac{3}{2} - \frac{\sqrt{3}}{2})x_{1} + \frac{\sqrt{2}}{2}x_{2} & = & 0 \\ 
	\frac{\sqrt{2}}{2}x_{1} + (2 - \frac{3}{2} - \frac{\sqrt{3}}{2})x_{1} & = & 0
	\end{array}
	\end{equation*}
	\begin{equation*}
	\Rightarrow
	\left( \begin{array}{cc} 1-\sqrt{3} & \sqrt{2} \\
		\sqrt{2} & -1-\sqrt{3} 
	\end{array} \right) \boldsymbol{e}_{1} = \boldsymbol{0}
	\end{equation*}
	\begin{equation*}
	\Rightarrow
	\boldsymbol{e}_{1} = \left( \begin{array}{c} \frac{-\sqrt{2}}{1-\sqrt{3}} \\ 1 \end{array} \right)
	\end{equation*}
	And the eigenvector for $\lambda_{2}$.
	\begin{eqnarray*}
	\boldsymbol{\Sigma}\boldsymbol{e}_{2} - \lambda_{2}\boldsymbol{e}_{2} & = & 0
	\end{eqnarray*}
	\begin{equation*} \Rightarrow
	\begin{array}{lcl}
	(2 - \frac{3}{2} + \frac{\sqrt{3}}{2})x_{1} + \frac{\sqrt{2}}{2}x_{2} & = & 0 \\ 
	\frac{\sqrt{2}}{2}x_{1} + (2 - \frac{3}{2} + \frac{\sqrt{3}}{2})x_{1} & = & 0
	\end{array}
	\end{equation*}
	\begin{equation*}
	\Rightarrow
	\left( \begin{array}{cc} 1+\sqrt{3} & \sqrt{2} \\
		\sqrt{2} & -1+\sqrt{3} 
	\end{array} \right) \boldsymbol{e}_{1} = \boldsymbol{0}
	\end{equation*}
	\begin{equation*}
	\Rightarrow
	\boldsymbol{e}_{2} = \left( \begin{array}{c} \frac{\sqrt{2}}{1+\sqrt{3}} \\ -1 \end{array} \right)
	\end{equation*}


	\end{enumerate}

\item[4.6] Let $\boldsymbol{X}$ be distributed as $N_{3}(\boldsymbol{\mu},\boldsymbol{\Sigma})$, where $\boldsymbol{\mu} = [1,-1,2]$ and
\begin{equation*}
\boldsymbol{\Sigma} = 
\left[\begin{array}{rrr}
	4 & 0 & -1 \\
	0 & 5 & 0 \\
	-1 & 0 & 2
	\end{array}\right]
\end{equation*}
Which of the following random variables are independent?  Explain.
	\begin{enumerate}
	\item \textit{$X_{1}$ and $X_{2}$.}

	$X_{1}$ and $X_{2}$ are independent because $cov(X_{1},X_{2}) = cov(X_{2},X_{1}) = 0$.

	\item \textit{$X_{1}$ and $X_{3}$.}

	$X_{1}$ and $X_{3}$ are not independent because $cov(X_{1},X_{3}) = cov(X_{3},X_{1}) = -1$.

	\item \textit{$X_{2}$ and $X_{3}$.}

	$X_{2}$ and $X_{3}$ are independent because $cov(X_{2},X_{3}) = cov(X_{3},X_{2}) = 0$.

	\item \textit{$(X_{1},X_{3})$ and $X_{2}$.}

	To determine the answer, we need to rearrange the covariance matrix and partition it.  The new covariance matrix follows.
	\begin{equation*}
	\boldsymbol{\Sigma} = \left[\begin{array}{rr|r}
		4 & -1 & 0 \\
		-1 & 2 & 0 \\ \hline
		0 & 0 & 5
		\end{array}\right]
	\end{equation*}
	We conclude that $(X_{1},X_{3})$ and $X_{2}$ are independent.

	\item \textit{$X_{1}$ and $X_{1} + 3X_{2} - 2X_{3}$.}

	We have the multivariate normal distribution 
	\begin{equation*}
	\boldsymbol{AX} = 
	\left[\begin{array}{c}
		X_{1} \\
		X_{1} + 3X_{2} - 2X_{3}
	\end{array} \right]
	\end{equation*}
	Where
	\begin{eqnarray*}
	\boldsymbol{A} & = & 
	\left[\begin{array}{rrr}
		1 & 0 & 0 \\
		1 & 3 & -2
	\end{array}\right] \\
	\boldsymbol{X} & = & 
	\left[\begin{array}{c}
		X_{1} \\
		X_{2} \\ 
		X_{3}
	\end{array} \right]
	\end{eqnarray*}
	And $\boldsymbol{AX}$ has the distribution $N_{2}(\boldsymbol{A\mu},\boldsymbol{A\Sigma A'})$.  Here we show the matrix $\boldsymbol{A\Sigma A'}$ to determine independence.
	\begin{eqnarray*}
	\boldsymbol{A\Sigma A'} & = & 
	\left[\begin{array}{rrr}
		1 & 0 & 0 \\
		1 & 3 & -2
	\end{array}\right]
	\left[\begin{array}{rrr}
		4 & 0 & -1 \\
		0 & 5 & 0 \\
		-1 & 0 & 2
	\end{array}\right]
	\left[\begin{array}{rr}
		1 & 1 \\
		0 & 3 \\
		0 & -2 
	\end{array}\right]\\
	& = & 
	\left[\begin{array}{rrr}
		4 & 0 & -1 \\
		6 & 15 & -5
	\end{array}\right]
	\left[\begin{array}{rr}
		1 & 1 \\
		0 & 3 \\
		0 & -2 
	\end{array}\right]\\
	& = & 
		\left[\begin{array}{rr}
		4 & 6 \\
		6 & 61
		\end{array}\right]
	\end{eqnarray*}
	And we conclude that $(X_{1},X_{3})$ and $X_{2}$ are not independent.

	\end{enumerate}

\item[4.9] Refer to Exercise 4.8, but modify the construction by replacing the break point 1 by $c$ so that
\begin{equation*}
X_{2} = \left\{ \begin{array}{rl}
	-X_{1} & \textrm{if } -c \leq X_{1} \leq c \\
	X_{1} & \textrm{elsewhere}
	\end{array} \right.
\end{equation*}
Show that $c$ can be chosen so that Cov($X_{1},X_{2}$) = 0, but that the two random variables are not independent.

First, we note (from exercise 4.8) that $X_{1} \sim N(0,1)$.  For $c = 0$, we have
\begin{eqnarray*}
\textrm{Cov}(X_{1},X_{2}) & = & E[X_{1}(X_{1})] - E[X_{1}]E[X_{1}] \\
 & = & E[(X_{1})^{2}] = \textrm{var}(X_{1}) + E[X_{1}]^{2} \\
 & = & 1
\end{eqnarray*}

For $c$ very large we have
\begin{eqnarray*}
\textrm{Cov}(X_{1},X_{2}) & = & E[X_{1}(-X_{1})] + E[X_{1}]E[X_{1}] \\
 & = & -E[(X_{1})^{2}] = -(\textrm{var}(X_{1}) + E[X_{1}]^{2}) \\
 & = & -1
\end{eqnarray*}

Since the covariance is a smooth function of $c$, then by the mean value theorem, Cov($X_{1},X_{2}$) = 0 at some point, but that the two random variables are not independent.

\item[4.10] Show each of the following.
	\begin{enumerate}
	\item $\begin{array}{|ll|} \bs{A} & \bs{0} \\ \bs{0} & \bs{B} \end{array} = |\bs{A}||\bs{B}|$
	
	Factor $\begin{array}{|ll|} \bs{A} & \bs{0} \\ \bs{0} & \bs{B} \end{array}$ so that we have the following.
	\begin{eqnarray*}
	\begin{array}{|ll|} \bs{A} & \bs{0} \\ \bs{0} & \bs{B} \end{array} & = & 
	\begin{array}{|ll|} \bs{A} & \bs{0} \\ \bs{0}' & \bs{I} \end{array} 
	\; \begin{array}{|ll|} \bs{I} & \bs{0} \\ \bs{0}' & \bs{B} \end{array}
	\end{eqnarray*}
	Then take the determinants to get
	\begin{eqnarray*}
	\begin{array}{|ll|} \bs{A} & \bs{0} \\ \bs{0}' & \bs{I} \end{array} 
	\; \begin{array}{|ll|} \bs{I} & \bs{0} \\ \bs{0}' & \bs{B} \end{array} & = & (|\bs{AI}| - |\bs{00}'|) (|\bs{IB}| - |\bs{00}'|) \\
	 & = & |\bs{A}||\bs{B}|
	\end{eqnarray*}

	\item $\begin{array}{|ll|} \bs{A} & \bs{C} \\ \bs{0}' & \bs{B} \end{array} = |\bs{A}||\bs{B}|$ for $|\bs{A}| \neq 0$.

	As in part (a), factor the determinant so that we have the following.
	\begin{eqnarray*}
	\begin{array}{|ll|} \bs{A} & \bs{C} \\ \bs{0}' & \bs{B} \end{array} & = & 
	\begin{array}{|cc|} \bs{A} & \bs{0} \\ \bs{0}' & \bs{B} \end{array} 
	\; \begin{array}{|cc|} \bs{I} & \bs{A}^{-1}\bs{C} \\ \bs{0}' & \bs{I} \end{array}
	\end{eqnarray*}
	Then, by part (a), we get $\begin{array}{|ll|} \bs{A} & \bs{0} \\ \bs{0} & \bs{B} \end{array} = |\bs{A}||\bs{B}|$, and $\begin{array}{|cc|} \bs{I} & \bs{A}^{-1}\bs{C} \\ \bs{0}' & \bs{I} \end{array} = 1$.
	\end{enumerate}

\item[4.10] Show that if $\bs{A}$ is square,
\begin{eqnarray*}
|\bs{A}| & = & |A_{22}||A_{11} - A_{12}A_{22}^{-1}A_{21}| \textrm{\hspace*{1cm}for } |A_{22}| \neq 0\\
 & = & |A_{11}||A_{22} - A_{21}A_{11}^{-1}A_{12}| \textrm{\hspace*{1cm}for } |A_{11}| \neq 0
\end{eqnarray*}

Let
\begin{equation*}
\bs{A} = \left( \begin{array}{ll} A_{11} & A_{12} \\
	A_{21} & A_{22} \end{array} \right)
\end{equation*}
and factor so that we have the following.
\begin{equation*}
\left( \begin{array}{cc}
	I & -A_{12}A_{22}^{-1} \\
	0' & I \end{array} \right)
\left( \begin{array}{ll} 
	A_{11} & A_{12} \\
	A_{21} & A_{22} \end{array} \right)
\left( \begin{array}{cc}
	I & 0 \\
	-A_{22}^{-1}A_{21} & I \end{array} \right)
\end{equation*}
\begin{equation*}
=
\left( \begin{array}{cc}
	A_{11} - A_{12}A_{22}^{-1}A_{21} & 0 \\
	A_{21} & A_{22} \end{array} \right)
\left( \begin{array}{cc}
	I & 0 \\
	-A_{22}^{-1}A_{21} & I \end{array} \right)
\end{equation*}
\begin{equation*}
=
\left( \begin{array}{cc}
	A_{11} - A_{12}A_{22}^{-1}A_{21} & 0 \\
	0 & A_{22} \end{array} \right)
\end{equation*}
Finally, we have
\begin{equation*}
\left( \begin{array}{cc}
	I & -A_{12}A_{22}^{-1} \\
	0' & I \end{array} \right)
\left( \begin{array}{ll} 
	A_{11} & A_{12} \\
	A_{21} & A_{22} \end{array} \right)
\left( \begin{array}{cc}
	I & 0 \\
	-A_{22}^{-1}A_{21} & I \end{array} \right)
=
\left( \begin{array}{cc}
	A_{11} - A_{12}A_{22}^{-1}A_{21} & 0 \\
	0 & A_{22} \end{array} \right)
\end{equation*}
Taking the determinant of both sides, we get the following.
\begin{equation*}
|I||A||I| = |A_{22}||A_{11} - A_{12}A_{22}^{-1}A_{21}|
\end{equation*}
\begin{equation*}
|A| = |A_{22}||A_{11} - A_{12}A_{22}^{-1}A_{21}|
\end{equation*}

For the second relation, we do the same by factoring $\bs{A}$ so that we have the following.
\begin{equation*}
\left( \begin{array}{cc}
	I & 0 \\
	-A_{21}A_{11}^{-1} & I \end{array} \right)
\left( \begin{array}{ll} 
	A_{11} & A_{12} \\
	A_{21} & A_{22} \end{array} \right)
\left( \begin{array}{cc}
	I & -A_{11}^{-1}A_{21} \\
	0' & I \end{array} \right)
\end{equation*}
\begin{equation*}
=
\left( \begin{array}{cc}
	A_{11} & A_{12} \\
	0 & A_{22} - A_{21}A_{11}^{-1}A_{12} \end{array} \right)
\left( \begin{array}{cc}
	I & -A_{11}^{-1}A_{21} \\
	0' & I \end{array} \right)
\end{equation*}
\begin{equation*}
\left( \begin{array}{cc}
	I & 0 \\
	-A_{21}A_{11}^{-1} & I \end{array} \right)
\left( \begin{array}{ll} 
	A_{11} & A_{12} \\
	A_{21} & A_{22} \end{array} \right)
\left( \begin{array}{cc}
	I & -A_{11}^{-1}A_{21} \\
	0' & I \end{array} \right)
=
\left( \begin{array}{cc}
	A_{11} & 0 \\
	0 & A_{22} - A_{21}A_{11}^{-1}A_{12} \end{array} \right)
\end{equation*}
Again, taking the determinant of both sides we get the following.
\begin{equation*}
|I||A||I| = |A_{11}||A_{22} - A_{21}A_{11}^{-1}A_{12}|
\end{equation*}
\begin{equation*}
|A| = |A_{11}||A_{22} - A_{21}A_{11}^{-1}A_{12}|
\end{equation*}

\item[4.16] Let $\bs{X}_{i}$, $i = 1,2,\ldots,4$, be independent $N_{p}(\bs{\mu},\bs{\Sigma})$ random vectors.
	\begin{enumerate}
	\item Find the marginal distributions for each of the random vectors
	\begin{eqnarray*}
	\bs{V}_{1} & = & \frac{1}{4}\bs{X}_{1} - \frac{1}{4}\bs{X}_{2} + \frac{1}{4}\bs{X}_{3} - \frac{1}{4}\bs{X}_{4} \\
	\end{eqnarray*}
	and
	\begin{eqnarray*}
	\bs{V}_{2} & = & \frac{1}{4}\bs{X}_{1} + \frac{1}{4}\bs{X}_{2} - \frac{1}{4}\bs{X}_{3} - \frac{1}{4}\bs{X}_{4}
	\end{eqnarray*}

	By result 4.8 in the text, $\bs{V}_{1}$ and $\bs{V}_{2}$ have the following distribution.
	\begin{equation*}
	N_{p}\left(\sum_{j=1}^{n}c_{j}\bs{\mu}, \left(\sum_{j=1}^{n}c_{j}^{2}\right) \bs{\Sigma} \right)
	\end{equation*}
	Then we have $\bs{V}_{1} \sim N_{p}(\bs{0},\frac{1}{4}\bs{\Sigma})$ and $\bs{V}_{2} \sim N_{p}(\bs{0},\frac{1}{4}\bs{\Sigma})$.

	\item Find the joint distribution of the random vectors $\bs{V}_{1}$ and $\bs{V}_{2}$ defined in (a).

	Also by result 4.8, $\bs{V}_{1}$ and $\bs{V}_{2}$ are jointly multivariate normal with covariance matrix
	\begin{equation*}
	\left[ \begin{array}{cc}
		\displaystyle \left(\sum_{j=1}^{n}c_{j}^{2}\right) \bs{\Sigma} & (\bs{b}'\bs{c})\bs{\Sigma} \\
		(\bs{b}'\bs{c})\bs{\Sigma} &\displaystyle  \left(\sum_{j=1}^{n}b_{j}^{2}\right) \bs{\Sigma}
		\end{array} \right]
	\end{equation*}
	With $\bs{c} = (1/4,-1/4,1/4,-1/4)'$ and $\bs{b} = (1/4,1/4,-1/4,-1/4)'$.  So that we have the covariance matrix
	\begin{equation*}
	\left[ \begin{array}{rr}
		\frac{1}{4} \bs{\Sigma} & \bs{0} \\
		\bs{0} & \frac{1}{4} \bs{\Sigma}
		\end{array} \right]
	\end{equation*}
	\end{enumerate}

\item[6.11] A likelihood argument provides aditional support for pooling the two independent sample covariance matrices to estimate a common covariance matrix in the case of two normal populations.  Give the likelihood function $L(\bs{\mu}_{1},\bs{\mu}_{2},\bs{\Sigma})$, for the two independent samples of sizes $n_{1}$ and $n_{2}$ from $N_{p}(\bs{\mu}_{1},\bs{\Sigma})$ and $N_{p}(\bs{\mu}_{2},\bs{\Sigma})$ respectively.  Show that this likelihood is maximized by the choices $\hat{\bs{\mu}}_{1} = \bar{\bs{x}}_{1}, \hat{\bs{\mu}}_{2} = \bar{\bs{x}}_{2}$ and
\begin{equation*}
\hat{\bs{\Sigma}} = \frac{1}{n_{1} + n_{2}}\left[ (n_{1} - 1)\bs{S}_{1} + (n_{2} - 1)\bs{S}_{2} \right] = \frac{n_{1} + n_{2} - 1}{n_{1} + n_{2}}\bs{S}_{\textrm{pooled}}
\end{equation*}

Here we note the density function for the multivariate normal distribution for reference.
\begin{eqnarray*}
f(\bs{x};\bs{\mu},\bs{\Sigma}) & = & \frac{1}{(2\pi)^{p/2}|\boldsymbol{\Sigma}|^{1/2}} e^{-\frac{1}{2}(\boldsymbol{x} - \boldsymbol{\mu})'\boldsymbol{\Sigma}^{-1}(\boldsymbol{x} - \boldsymbol{\mu})}
\end{eqnarray*}
Then we have the likelihood function for the two independent samples as defined below.
\begin{eqnarray*}
L(\bs{\mu}_{1},\bs{\mu}_{2},\bs{\Sigma}) & = & \prod_{i=1}^{n_{1}} f(\bs{X}_{i};\bs{\mu}_{1},\bs{\Sigma}) \prod_{j=1}^{n_{2}} f(\bs{X}_{j};\bs{\mu}_{2},\bs{\Sigma}) \\
 & = & L(\bs{\mu}_{1},\bs{\Sigma})L(\bs{\mu}_{2},\bs{\Sigma})
\end{eqnarray*}
This statement gives us directly that $\hat{\bs{\mu}}_{1} = \bar{\bs{x}}_{1}, \hat{\bs{\mu}}_{2} = \bar{\bs{x}}_{2}$.  Using equation (4-13) in the text, the likelihood function can defined as
\begin{eqnarray*}
L(\bs{\mu}_{1},\bs{\mu}_{2},\bs{\Sigma}) & = & \frac{1}{(2\pi)^{Np/2}|\boldsymbol{\Sigma}|^{N/2}} 
%\exp{\left\{ -\frac{1}{2} \; \textrm{ tr}\left[\bs{\Sigma}^{-1}\left(\sum_{i=1}^{n_{1}}\Phi_{1}(\bs{X}_{i}) + \sum_{j=1}^{n_{2}}\Phi_{2}(\bs{X}_{j})\right) \right] \right\} }
\exp{\left( \displaystyle -\frac{1}{2} \; \textrm{ tr}\left[\bs{\Sigma}^{-1}\left(\sum_{i=1}^{n_{1}}\Phi_{1}(\bs{X}_{i}) + \sum_{j=1}^{n_{2}}\Phi_{2}(\bs{X}_{j})\right) \right] \right) }
\end{eqnarray*}
Where
\begin{eqnarray*}
\Phi_{i}(\bs{x}) & = & (\bs{x} - \bs{\mu}_{i})(\bs{x} - \bs{\mu}_{i})'
\end{eqnarray*}
By result 4.10 in the text, we let $\displaystyle \bs{B} = \sum_{i=1}^{n_{1}}\Phi_{1}(\bs{x}_{i}) + \sum_{j=1}^{n_{2}}\Phi_{2}(\bs{x}_{j})$, $b = N = n_{1} + n_{2}$ and substitute the MLE's for each mean, then the maximum is reached at
\begin{eqnarray*}
\hat{\bs{\Sigma}} & = & \frac{1}{2(n_{1} + n_{2})}\bs{B}
\end{eqnarray*}
And we get the result
\begin{equation*}
\hat{\bs{\Sigma}} = \frac{1}{n_{1} + n_{2}}\left[ (n_{1} - 1)\bs{S}_{1} + (n_{2} - 1)\bs{S}_{2} \right] = \frac{n_{1} + n_{2} - 1}{n_{1} + n_{2}}\bs{S}_{\mathrm{pooled}}
\end{equation*}

\end{enumerate}

\end{document}
