Why should we not worry about the denominator in naive Bayes?

Why should we not worry about the denominator in naive Bayes?

Question. In this exercise about computing the denominator for the naive Bayes classifier, it is noted that we can ignore the denominator since we’re comparing P(positive | review) and P(negative | review) and so can cancel out their denominators to simplify our work.

Why is the probability of a continuous random variable 0?

The probability of a specific value of a continuous random variable will be zero because the area under a point is zero. Probability is area. The curve is called the probability density function (abbreviated as pdf). We use the symbol f(x) to represent the curve.

Is naive Bayes classifier optimal?

Because the Bayes classifier is optimal, the Bayes error is the minimum possible error that can be made.

Can the probability of a random variable be zero?

The probability of any particular value, e.g., P(X = 1) is zero because there is no area under a single point. 3. NOTE: We always use capital letters for random variables.

How is the denominator broken down in bayes’theorem?

I’ve noticed in many places that people tend to break down the denominator in the equation from Bayes’ Theorem. So instead of this: You can see that this convention is used in this Wikipedia article and in this insightful post by Tim Peters. I am baffled by this. Why is the denominator broken down like this? How does that help things at all?

Which is a nice equivalent form for bayes’rule?

Another reason is recognizing equivalent forms of Bayes’ Rule by manipulating that expression. For example: Which is a nice equivalent form for Bayes’ Rule, made even handier by subtracting this from the original expression to obtain:

How is the Bayes rule stated in terms of odds?

This is Bayes’ Rule stated in terms of Odds, i.e. posterior odds against B = Bayes factor against B times the prior odds against B. (Or you could invert it to get an expression in terms of odds for B.) The Bayes factor is the ratio of the likelihoods of your models.

Is the denominator an unsightly integral in Monte Carlo?

The trouble is that with more involved on-the-battlefield Bayesian problems the denominator is an unsightly integral, which may or may not have a closed-form solution. In fact, sometimes we need sophisticated Monte Carlo methods just to approximate the integral and churning the numbers can be a real pain in the rear.