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Math Help - Naive Bayes shortened?

  1. #1
    Junior Member
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    Naive Bayes shortened?

    Hi,

    Wikipedia states Bayes theorem as:
    Naive Bayes shortened?-3174021f44ba0d31f6ede772624c5523.png

    In my A-level Statistics book, it states the following:
    Naive Bayes shortened?-scans1.jpg

    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?

    Thanks in advance
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  2. #2
    Grand Panjandrum
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    Re: Naive Bayes shortened?

    Quote Originally Posted by webguy View Post
    Hi,

    Wikipedia states Bayes theorem as:
    Click image for larger version. 

Name:	3174021f44ba0d31f6ede772624c5523.png 
Views:	10 
Size:	1.6 KB 
ID:	22867

    In my A-level Statistics book, it states the following:
    Click image for larger version. 

Name:	scans1.jpg 
Views:	11 
Size:	582.0 KB 
ID:	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?

    Thanks in advance
    We usually start with the half Bayes' theorem:

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

    Which immediately gives the full Bayes' theorem:

    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
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