Exercise 4.19 - Decision boundary for LDA with semi tied covariances

We begin from the Bayes rule:

where we have omitted the terms independent of $y$. With a uniform prior on two classes:

The decision boundary, in which we are interested, is a curve depicted by $f(\mathbf{\text{𝐱}})=0$, where:

Therefore the decision boundary is:

where:

The decision boundary is a quardratic curve unless $k=1$, which is geometrically very intuitive.

Let us consider a 2d case, focusing on the transformed coordinates where ${\mathrm{\Sigma }}_0$ is the identity matrix. With out loss of generality, let ${\mu }_0=(0,0)$, ${\mu }_1=(z,0)$, denote the distance between a point $p$ in this place from ${\mu }_0$ and ${\mu }_1$ by $a(p)$ and $b(p)$. The decision boundary is exactly:

Plugging in the Cartesian representation $p(x,y)$, we ends up with a conical curve. The linear transform of the space would not change its conical nature.