Re: Naive Bayes shortened?

Quote:

Originally Posted by

**webguy** Hi,

Wikipedia states Bayes theorem as:

Attachment 22867
In my A-level Statistics book, it states the following:

Attachment 22868
Using a single feature (F) in Bayes theorem, would it be logical to state the following?

$\displaystyle P(C|F) = \frac{P(C)P(F|C)}{P(F)}$

The book states that $\displaystyle P(F|C) = \frac{P(C \cap F)}{P(C)}$

Hence, in Bayes theorem one may say:

$\displaystyle P(C|F) = \frac{\frac{P(C)P(C \cap F)}{P(C)}}{P(F)}$

The P(C)s cancel out, hence one is left with:

$\displaystyle P(C|F) = \frac{P(C \cap F)}{P(F)}$

Is this correct, or have I made a crucial mistake?

Thanks in advance :)

We usually start with the half Bayes' theorem:

$\displaystyle P(C \wedge F)=P(C|F)P(F)$

Which immediately gives the full Bayes' theorem:

$\displaystyle P(C \wedge F)=P(C|F)P(F)=P(F|C)P(C)$

So what you have is OK, its just a rearrangement of the half Bayes' theorem.

CB