Exercise 4.21 - Gaussian decision boundaries

Answers

We have:

p (x | μ_{i}, σ_{i}^{2}) = \frac{1}{\sqrt{2 π σ_{i}^{2}}} \cdot \exp (- \frac{(x - μ_{i}^{2})}{2 σ_{i}^{2}}),

so:

p (x | μ_{1}, σ_{1}^{2}) = \frac{1}{\sqrt{2 π}} \cdot \exp (- \frac{x^{2}}{2}),

p (x | μ_{2}, σ_{2}^{2}) = \frac{1}{1 0^{3} \cdot \sqrt{2 π}} \cdot \exp (- \frac{{(x - 1)}^{2}}{2 \times 1 0^{6}}) .

The decision region satisfies:

\frac{p (x | μ_{1}, σ_{1}^{2})}{p (x | μ_{1}, σ_{1}^{2})} \geq 1,

this is tantamount to:

\frac{{(x - 1)}^{2}}{1 0^{6}} - x^{2} \geq - 6 \cdot \ln 10 .

Denote $x_{1}$ and $x_{2}$ as the zeros of this quardratic form, then

R_{1} = [x_{1}, x_{2}] .

When $σ^{2} = σ^{1}$ , $R_{1}$ is exactly $(\infty, \frac{1}{2})$ .

One can solve for (a) and (b) by plugging in (4.289).

solour_lfq

2021-03-24 13:42