I have two features which are both continuous. How to perform a classification task based on them? I've read the Wikipedia entry on Naive Bayes, but this is only for discrete outcome and one feature.

  • $\begingroup$ What is the outcome you want to classify? $\endgroup$
    – chl
    Commented Mar 24, 2012 at 9:36
  • $\begingroup$ The outcome is discrete class (nominal). $\endgroup$
    – fikr4n
    Commented Mar 24, 2012 at 13:33
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    $\begingroup$ Err, actually, not. The wikipedia page labels the features $F_1\ldots F_n$. $\endgroup$ Commented Mar 24, 2012 at 13:48
  • $\begingroup$ @fkr Does this question/answer help ?: stats.stackexchange.com/questions/4298/… $\endgroup$
    – steffen
    Commented Mar 24, 2012 at 19:58
  • $\begingroup$ @ConjugatePrior Does it mean P(C).P(F1|C).P(F2|C) rather than P(C).P(F|C)? $\endgroup$
    – fikr4n
    Commented Mar 25, 2012 at 2:36

1 Answer 1


I think I've found the solution in the same page. It might because I was dumb or being stressed :).


$$ \text{posterior}(\text{male})=\frac{P(\text{male})P(\text{height}\mid\text{male})P(\text{weight}\mid\text{male})P(\text{footsize}\mid\text{male})}{\text{evidence}} $$

Thanks @ConjugatePrior.

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    $\begingroup$ Glad to hear it now makes sense. btw you don't have to post the image - you can write it in latex math notation, e.g. Bayes theorem is written p(C|F) = \frac{p(F|C) p(C)}{p(F)} which when you surround it with $ signs renders automatically as $p(C|F) = \frac{p(F|C) p(C)}{p(F)}$. $\endgroup$ Commented Mar 25, 2012 at 8:04
  • $\begingroup$ @fkr so, concluding from your answer your actual question was how to perform NB with more than one feature ? I ask because your answer does not aim at the special case of continuous features ;). Some have problems calculating the conditional probabilities for cont. features, so I thought your question aims in the same direction. $\endgroup$
    – steffen
    Commented Mar 25, 2012 at 10:15
  • $\begingroup$ @steffen Actually, I need simply two numerical features and a class of nominal :). $\endgroup$
    – fikr4n
    Commented Mar 25, 2012 at 11:46
  • $\begingroup$ To point out a nuance about BornToCode's answer, the expression he wrote for the probability of a class is not technically correct. For continuous features such as height, the conditional probability of a single value is always 0. Fortunately, we can use the height of the conditional probability density function corresponding to the value of height as a proxy for the conditional probability of that value occurring. See this question for more details. $\endgroup$ Commented Apr 21, 2015 at 15:31

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