# Bayes Decision Theory With 3 Classes

I'm trying to create a Bayes classificator in 1 dimension with 3 classes. I have created the following graph, where you can see that from zero to $$x_{bnd1}$$ is the first area $$R1$$, then from $$x_{bnd1}$$ to $$x_{bnd2}$$ is $$R2$$ and finally $$R3$$, from $$x_{bnd2}$$ to one.

The different classes

My problem is, how do I integrate in order to find the error?

Take for example the first area $$R1$$ where the accepted class is $$ω_{1}$$. The error here, I think, will be the following :

$$P\left(\omega_2 \right) \int \limits_{R_1}p\left( x |\omega_2 \right) dx + P\left(\omega_3 \right) \int \limits_{R_1}p\left( x |\omega_3 \right)dx\\$$

However, I'm not sure that this is correct. By doing this, I add the area below the red line two times, aren't I? Or is it ok to add it 2 times because it's 2 different errors?

(Of course the error will have more terms related to areas $$R2, R3$$, where the same problem occurs.)

In case someone wonders, the conditional pdfs I have are $$p(x|\omega_1)=1$$ $$p(x|\omega_2)=6x(1-x)$$ $$p(x|\omega_3)=2x$$ and the prior probabilities of the classes are $$P(\omega_1)=1/6, P(\omega_2)=1/3$$ and $$P(\omega_3)=1/2$$.

Sorry if my thoughts are a bit jumbled.. Thanks in advance.

You formula is correct. The error when $$\omega_1$$ is accepted can be written as follows; $$P(\text{making error}|\omega_1 \text{is accepted})=P(\text{making error}|x\in R_1)=P(e|x\in R_1)$$ In order to make the error, $$x$$ is either from class 2 or 3. Using total probability theorem, we have $$P(e|x\in R_1)=P(\text{x belongs } \omega_2|x\in R_!)+P(\text{w belongs }\omega_3|x\in R_1)$$ Then, each summand can be calculated as below: $$P(\text{x belongs }\omega_i|x\in R_1)=P(\omega_i)\int_{R_1}p(x|w_i)dx$$
• It can happen. For example, what would you calculate if I ask $P(x<0.5)$? – gunes Nov 9 '19 at 21:03
• Without determining $x$'s pdf? Hm..I guess it would be the sum of the integrals from $0$ to $0.5$ for all 3 pdfs? – Thomas Nov 9 '19 at 21:07
• yes, but I wouldn't call them as pdfs since they're multiplied with priors, i.e. $p(\omega_1)f(x|\omega_1)$ is not a PDF. – gunes Nov 10 '19 at 9:26