# Naive Bayes shortened?

• Nov 23rd 2011, 05:07 PM
webguy
Naive Bayes shortened?
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?

$P(C|F) = \frac{P(C)P(F|C)}{P(F)}$
The book states that $P(F|C) = \frac{P(C \cap F)}{P(C)}$
Hence, in Bayes theorem one may say:
$P(C|F) = \frac{\frac{P(C)P(C \cap F)}{P(C)}}{P(F)}$
The P(C)s cancel out, hence one is left with:
$P(C|F) = \frac{P(C \cap F)}{P(F)}$

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

• Nov 23rd 2011, 09:39 PM
CaptainBlack
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?

$P(C|F) = \frac{P(C)P(F|C)}{P(F)}$
The book states that $P(F|C) = \frac{P(C \cap F)}{P(C)}$
Hence, in Bayes theorem one may say:
$P(C|F) = \frac{\frac{P(C)P(C \cap F)}{P(C)}}{P(F)}$
The P(C)s cancel out, hence one is left with:
$P(C|F) = \frac{P(C \cap F)}{P(F)}$

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

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