
Our discriminat function (\ref{eq:discriminant}) for 2-D Gaussian pdf is
\begin{eqnarray}\label{eq:2-d-gaussian}
g_i(\mbox{\bf x}) 
 &=& - \frac{1}{2} (\mbox{\bf x}-\mu_i)^T
            \Sigma_i^{-1}(\mbox{\bf x}-\mu_i) + \ln p(\omega_i) + c_i
\end{eqnarray}
\noindent
where
\begin{equation}\label{eq:2-d-const}
c_i = - (1/2)\ln (2\pi) - (1/2) \ln \vert \Sigma_i \vert.
\end{equation}

\hspace*{1in}\\
\noindent
This is simplified\\

\begin{itemize}
\item[$\star$] if

\begin{eqnarray}
\nonumber
\Sigma_i =
\left(
\begin{array}{cc}
\sigma_1^2 & 0          \\
0          & \sigma_2^2 \\
\end{array}
\right)
\end{eqnarray}
\item[]then

\begin{equation}
g = \frac{(x_1-\mu_1)^2}{-2\sigma_1^2}
    + \frac{(x_2-\mu_2)^2}{-2\sigma_2^2}
    -\frac{1}{2}\ln(\sigma_1^2\sigma_2^2)
+ \ln p(\omega_i) - \frac{1}{2}\ln(2\pi) \\
\end{equation}
\end{itemize}

\hspace*{1in}\\
Note that the last two terms in the right hand side of 
               the above equation will be cancelled when obtaining 
               the border $g_i-g_j=0$.


\hspace*{1in}\\
\begin{itemize}
\item[-] This kind of two distributions construct a {\it quadratic}
               decision surface.
   \begin{itemize}
   \item[$\cdot$] {\it ellipsoid}
   \item[$\cdot$] {\it hyperbola}
   \item[$\cdot$] {\it parabola}
   \item[$\cdot$] {\it pair of lines}
   \end{itemize}
\item[-] If expanded to $l>2$ case, then it is called 
               {\it hyper-quadratic}. \\

\item[-] If $\sigma_1 = \sigma_2$ holds in both $\Sigma_i$ and 
         $\Sigma_j$ then the decision surface is a {\it circle}.
\item[-] Furthermore, if $\Sigma_i = \Sigma_j$ holds then 
         the decision surface is a {\it straight line.}
\end{itemize}



%%koko


\end{displaymath}
Then our two discriminant fanction $g_1$ and $g_2$ will be
\begin{eqnarray}
\nonumber
g_1 = - \frac{1}{2} (x \hspace{.1in} y)
\left(
\begin{array}{cc}
a_1 & 0             \\
0   & b_1
\end{array}
\right)
\left(
\begin{array}{c}
x             \\
y
\end{array}
\right)
- \frac{1}{2} \ln (a_1b_1)
\end{eqnarray}
and 
\begin{eqnarray}
\nonumber
g_2 = - \frac{1}{2} (x-1 \hspace{.1in} y)
\left(
\begin{array}{cc}
a_2 & 0             \\
0   & b_2
\end{array}
\right)
\left(
\begin{array}{c}
x-1             \\
y
\end{array}
\right)
- \frac{1}{2} \ln (a_2b_2)
\end{eqnarray}
Then, the equation of the border between $\omega_1$ and $\omega_2$ is obtained from 
$g_1(\mbox{\bf x} - g_2(\mbox{\bf x}=0$ which leads to
\begin{equation} \label{eq:2d-quadratic}
(a_1-a_2)x^2 + 2a_2 x + (b_1^2 - b_2^2)y^2 = \ln\frac{a_2b_2}{a_1b_1}
\end{equation}

\begin{equation} \label{eq:2d-quadratic}
(\frac{1}{a_1}-\frac{1}{a_2})x^2 + \frac{2}{a_2}x + (\frac{1}{b_1}-\frac{1}{b_2})y^2
= \frac{1}{a_2} + \ln \frac{a_2b_2}{a_1b_1}
\end{equation}

\noindent
Then question is what condition is needed for the border of 
two classes to be each of the above mentioned decision surfaces?\\

\noindent
First of all, the simplest case is \\

\noindent
$\star$ When $a_1 = b_1$ and $a_2 = b_2$.\\
In this case, Eq (\ref{eq:2d-quadratic}) is a (i) {\it straight line} 
perpendicular to y-axis.\\

\noindent
$\star$ When $\Sigma_1=\Sigma_2$\\
In this case, Eq (\ref{eq:2d-quadratic}) is a (ii) {\it circle.} \\

\noindent
$\star$ Otherwise\\
let's  use a formula from {\it Linear Algebra}:

\begin{formula}[Shape of 2-D quadratic form in general] \label{rule-2}
2-D quadratic form $ax^2+2bxy+cy^2=0$ will be (1) ellipse, (ii) parabola, and (iii) hyperabola
when $9b^2 - ac)$ is, respectively, positive, zero, and negative.
\end{formula}

\noindent
So, when we choose parameters so as to be $a_1b_1 = a_2b_2$, the right-hand side becomes zero,
and our conclusion is:

\begin{itemize}
\item When $(a_1-a_2)(b_1-b_2)$ is
   \begin{itemize}
   \item positive
      \begin{itemize}
      \item ellipsoid
      \end{itemize}
   \item zero
      \begin{itemize}
      \item parabola
      \end{itemize}
   \item positive
      \begin{itemize}
      \item hyperbola
      \end{itemize}
   \end{itemize}
\end{itemize}

\noindent
Finally, \\

\noindent
$\star$ When $a_2 = 0$, $a_1b_1=a_2b_2$ and $(a_1-a_2)(b_1-b_2)<0$.\\

\noindent
In this case, Eq (\ref{eq:2d-quadratic}) becomes like
$Ax^2-BY^2=0$, or  $(\sqrt{A}x+\sqrt{B}y)(\sqrt{A}x+\sqrt{B}y)=0$, that is 
(vi) {\it two straight line}.



\bigskip
\begin{excersize}\label{excersize:border-2}
When $p(\omega_1)=p(\omega_2)$ again, but $\mu_1=\mu_2 = (0, 0)$
this time, obtain decision surface.
\begin{small}
\begin{eqnarray}
\nonumber
(6) \hspace{.2in}\Sigma_1 =
\left(
\begin{array}{cc}
0.30 & 0             \\
0    & 0.30
\end{array}
\right),
\hspace{.5in}\Sigma_2 =
\left(
\begin{array}{cc}
0.10 & 0             \\
0    & 0.90
\end{array}
\right)
\end{eqnarray}
\end{small}
\end{excersize}

