Decision boundaries and Gaussian density functions This is for my hw, and if anyone can solve the first part of the question it will be great.
Here is the question:
Assume a two-class problem with equal a priori class probabilities and Gaussian class-conditional densities as follows:
$$p(x\mid w_1) = {\cal N}\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix},\begin{bmatrix} a & c \\ c & b \end{bmatrix}\right)\quad\text{and}\quad p(x\mid w_2) = 
{\cal N}\left(\begin{bmatrix} d \\ e \end{bmatrix},\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right)$$
where $ab-c^2=1$.


*

*Find the equation of the decision boundary between these two classes
in terms of the given parameters, after choosing a logarithmic
discriminant function.

*Determine the constraints on the values of a, b, c, d and e, such
that the resulting discriminant function results with a linear
decision boundary.

 A: I will use $(X,Y)$ for the observation.  Given $w = w_1$, we have that the variances of $X$ and $Y$ are $a$ and $b$ respectively, while the covariance is $c$.
Thus, the correlation coefficient $\rho = \frac{c}{\sqrt{ab}}$ and so 
$1-\rho^2  = 1 - \frac{c^2}{ab} = \frac{1}{ab}$. The two conditional (joint) densities are joint normal densities given by
$$\begin{align}
f_1(x,y) &= \frac{1}{2\pi \sqrt{ab(1-\rho^2)}}\exp\left[-\frac{1}{2(1-\rho^2)}\left(\frac{x^2}{a} - 2\rho\frac{xy}{\sqrt{ab}} + \frac{y^2}{b}\right)\right]\\
&= \frac{1}{2\pi}\exp\left[-\frac{ab}{2}\left(\frac{x^2}{a} - 2\frac{c}{\sqrt{ab}}\frac{xy}{\sqrt{ab}} + \frac{y^2}{b}\right)\right]\\
&= \frac{1}{2\pi}\exp\left[-\frac{1}{2}\left(bx^2 - 2cxy + ay^2\right)\right],\\
f_2(x,y) 
&= \frac{1}{2\pi}\exp\left[-\frac{1}{2}\left((x-d)^2+(y-e)^2\right)\right]\\
\end{align}$$
The decision boundary is the set of all $(x,y)$ for which $f_1(x,y) = f_2(x,y)$, and so the decision boundary is the conic section specified by
$$bx^2 -2cxy +ay^2 - \left((x-d)^2 + (y-e)^2\right) = 0.$$
Edit: As pointed out in the comment by whuber,
this conic section can be a hyperbola as well as an ellipse or
parabola (including as a special case a straight line).
The discriminant $c^2-(a-1)(b-1) = (a+b)-2$ (since we are
given that $ab-c^2=1$)
can be positive, negative, or zero depending on the variances
$a$ and $b$.
I suspect that this will be work out to be
either an ellipse or a parabola (but not
a hyperbola) depending on
the parameters $a,b,c,d,e$, including, as a special case of parabola,
a straight line.  A specific example of a straight-line decision
boundary is when $c = 0$ and $a = b = 1$ so that the
only difference between the two conditional distributions is the means:
$X$ and $Y$ are conditionally independent unit-variance
normal random variables under either hypothesis.  In this instance,
the decision boundary is the perpendicular bisector of the
straight line segment with end-points $(0,0)$ and $(d,e)$.
What puzzles me, though, is a hyperbola as a decision boundary
since a hyperbola partitions the plane into three regions
(two of which are congruent). The joint density surfaces are
a (possibly) flattened bell and a circular bell, and
so one of these bells subsumes the other in two non-contiguous regions
of the plane: I just can't visualize it.  Perhaps someone will
create a nice illustration....
