# 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 $ab-c^2=1$.

1. Find the equation of the decision boundary between these two classes in terms of the given parameters, after choosing a logarithmic discriminant function.
2. 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.
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I see a study group forming... –  cardinal Nov 13 '11 at 15:41
i don't understand what you mean but, if this is the exact duplicate of a question asked before, i can't see the question or answer. Where can i find? –  ceylan Nov 13 '11 at 17:06
See the link to "Possible duplicate" at the top of this question? –  cardinal Nov 13 '11 at 17:10
@cardinal, i found the question... thanks. –  ceylan Nov 13 '11 at 17:16
Should this question be re-opened since the one that it duplicates has been removed voluntarily by the author, or should the author be asked to consider removing this one too? –  Dilip Sarwate Nov 14 '11 at 12:41
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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 $(d,e)$.
Good beginning! I fixed a subtraction error in the final formula. As a result, you might want to revisit your speculations in the last paragraph. To get started, note that the equation is linear if and only if all the quadratic coefficients are zero, which is equivalent to $a=1$, $b=1$, and $c=0$. (There is another special case where the equation represents a pair of lines.) Whether the locus is elliptical or hyperbolic depends on the sign of the determinant $(a-1)(b-1)-c^2$; both signs are possible. –  whuber Mar 17 at 16:06