How can a multinomial naive Bayes classifier be used here?

Question

I'm doing a problem from a textbook where the data (found here) are students and the features include sex, family size, health, age, etc. The problem asks to make a naive Bayes classifier from scratch in order to predict whether or not a student's final grade will be above or below a threshold, and it says to use a multinomial likelihood for features that take on a range of values (e.g. age).

To my understanding, a multinomial distribution is used when each of the features are of the same type and a feature's value represents a count. That is, a feature vector is a histogram. But this data doesn't look like a histogram at all, and so I can't wrap my head around using a multinomial likelihood.

Wouldn't it make more sense to compute $$P(x^{(i)}=v | c) = \frac{\text{# of times that }x^{(i)}=v \text{ when class is } c}{\text{total # of entries when class is } c}$$ ?

Answer 1

In multinomial Naive Bayes a feature is multinomial rather than binary in Berlluli Naive Bayes.

Take for example the first four features in your dataset:

1 school - student's school (binary: 'GP' - Gabriel Pereira or 'MS' - Mousinho da Silveira)
2 sex - student's sex (binary: 'F' - female or 'M' - male)
3 age - student's age (numeric: from 15 to 22)
4 address - student's home address type (binary: 'U' - urban or 'R' - rural)

All except the third are binary.

I think you are right that the parameters can be learned this way:

$$P(x^{(i)}=v | c) = \frac{\text{# of times that }x^{(i)}=v \text{ when class is } c}{\text{total # of entries when class is } c}$$

For each feature $\sum_{v\in V} P(x^{(i)}=v|c)=1$ holds, and in Berlluli Naive Bayes $|V| =2$ for all features while in multinomial Naive Bayes $|V|>2$ for at least one feature.