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Suppose I am thinking about a classification problem and I have my features $X = (X_1,...,X_n)$ and my classification $C$ (taking values in some finite set of classes $c_1,...,c_k$). The naive Bayes classifier assumes that the different individual features $X_1,...,X_n$ are independent when conditioned on $C$. We then construct a classifier by taking the Baysian classifier $$ c(x) = \arg\max_{c_i} P(C=c_i | X = x) $$ and by Bayes rule together with our independence assumption, we have $$ P(C = c_i | X = x) = \frac{P(C = c_i) \prod_j{P(X_j=x_j | C = c_i)}}{P(X = x)} $$

To get a classifier out of this, we need to be able to compute these probabilities. One way to do this is to make distributional assumptions about the conditional distributions of the $X_i$ (for example assuming that they are Bernoulli or Gaussian and then using the data to estimate the parameters of these distributions as well as the marginal distribution of $C$).

On the wikipedia page for Naive Bayes it brings up what is called multinomial naive Bayes classification where we assume that the features $X$ follow multinomial distributions when conditioned on $C$. To get a classifier from this, the parameters of these multinomial distributions are estimated from the data (where we actually do not need the "number of rolls" parameter $n$ as it comes out in the wash when taking the argmax) and then used to compute $P(X = x | C = c_i)$ (without splitting it up).

One thing to note with a random vector that follows a multinomial distribution is that the different components are not independent (e.g., if you roll a 2 on a dice, you know you do not roll a 5). So the multinomial "naive Bayes" classifier does not satisfy the "naive" assumption that the components of $X$ are independent when conditioned on $C$.

Am I understanding this correctly? Why multinomial naive Bayes a naive Bayes classifier (my guess, assuming I am understanding everything else correctly: it is naive (making the distributional assumption) and it uses Bayes' rule).

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