2
$\begingroup$

In the Naive Bayes classifier, why do we have to normalize the probabilities after calculating the probabilities of each hypothesis?

$\endgroup$
3
  • 2
    $\begingroup$ If you want to have the responsibility of each class for a single datapoint, you must normalize to 1, otherwise it can not be interpreted as a probability. $\endgroup$ Dec 5, 2016 at 15:42
  • $\begingroup$ This question appears to be squarely on-topic to me - it's about probability and statistics. $\endgroup$
    – Sycorax
    Dec 5, 2016 at 16:48
  • $\begingroup$ Maybe duplicate: stats.stackexchange.com/q/129666/103153 $\endgroup$ Jan 20, 2019 at 16:24

1 Answer 1

6
$\begingroup$

You do not have to normalize the probabilities if you only care about knowing which class ($\hat{y}$) your input ($\mathbf{x}=x_1, \dots, x_n$) most likely belongs to, since the maximum a posteriori (MAP) decision rule is as follows:

$\hat{y} = \underset{k \in \{1, \dots, K\}}{\operatorname{argmax}}p(C_k \vert x_1, \dots, x_n) = \underset{k \in \{1, \dots, K\}}{\operatorname{argmax}} \ p(C_k) \displaystyle\prod_{i=1}^n p(x_i \vert C_k)$

Since

$$\begin{align} p(C_k \vert x_1, \dots, x_n) & \varpropto p(C_k, x_1, \dots, x_n) \\ & \varpropto p(C_k) \ p(x_1 \vert C_k) \ p(x_2\vert C_k) \ p(x_3\vert C_k) \ \cdots \\ & \varpropto p(C_k) \prod_{i=1}^n p(x_i \vert C_k)\,. \end{align}$$

If you do want the class probabilities, then you indeed need to normalize:

$p(C_k \vert x_1, \dots, x_n) = ~\frac{p(C_k) \prod_{i=1}^n p(x_i \vert C_k)\,}{ \sum_{1 \leq j \leq |C|}{ \left( p(C_j) \prod_{i=1}^n p(x_i \vert C_j\,) \right) }}$

But keep in mind that, from {1}

the winning class in NB classification usually has a much larger probability than the other classes and the estimates diverge very significantly from the true probabilities. NB classifiers estimate badly, but often classify well.


References:

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.