This is question 1 of chapter 2 from Pattern Classification by Richard O.Duda (W

Question

This is question 1 of chapter 2 from Pattern Classification by Richard O.Duda (Wiley)

Eq.5 : P (error) = Integral-+ (P (error, x)) dx = Integral-+ (P (error|x)p(x)) dx.

Eq.7 : [P (1|x), P (2|x)]

In the two-category case, under the Bayes’ decision rule the conditional error is given by Eq.7. Even if the posterior densities are continuous, this form of the conditional error virtually always leads to a discontinuous integrand when calculating the full error by Eq.5.

(a) Show that for arbitrary densities, we can replace Eq.7 by P(error|x)=2P(1|x)P(2|x) in the integral and get an upper bound on the full error.

(b) Show that if we use P (error|x) = P (1|x)P (2|x) for < 2, then we are not guaranteed that the integral gives an upper bound on the error.

(c) Analogously, show that we can use instead P(error|x) = P(1|x)P(2|x) and get a lower bound on the full error.

(d) Show that if we use P (error|x) = P (1|x)P (2|x) for > 1, then we are not guaranteed that the integral gives an lower bound on the error.

Explanation / Answer

Show that if we use P (error|x) = ?P (?1|x)P (?2|x) for ? < 2, then we are not guaranteed that the integral gives an upper bound on the error.

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