User:Thepigdog/Bayes' theorem

Bayesian Inference
Bayes' theorem is about conditional probabilities. Probability is about sets of outcomes. We start by assuming that these outcomes are equally likely. Suppose we have a bag full of balls, each ball is either red or blue. Each ball is also either Small or Big. Taking a ball from the bag is an outcome.

The conditional probability of a ball taken from the bag being Red if we already know it is Big is 10/40. This is written,


 * $$P(Red|Big) = \frac{Count(Red \cap Big)}{Count(Big)} = \frac{10}{40}.$$


 * $$P(Big|Red) = \frac{Count(Big \cap Red)}{Count(Red)} = \frac{10}{30}.$$

These are conditional probabilities. $$P(Red|Big)$$ means,

First I found that the ball was Big. What then is the probability of it being red.

The probabilities for a a ball being red $$P(Red)$$ is,
 * $$P(Red) = \frac{Count(Red)}{Count(All)}.$$

Note that $$P(Red|Big)$$ has no meaning by itself. Instead probability has two sets,
 * The set of events that register success.
 * The domain from which those events are taken.

Note that,
 * $$P(Red) = P(Red|All).$$

The probabilities for a a ball being Big $$P(Big)$$ is,
 * $$P(Big) = \frac{Count(Big)}{Count(All)}.$$

Now the probability of a ball being Red and Big $$P(Red \cap Big)$$ is,
 * $$P(Red\cap Big) = \frac{Count(Red\cap Big)}{Count(All)}.$$

or,
 * $$P(Red\cap Big) = \frac{Count(Red\cap Big) * Count(Red)}{Count(Red)) * Count(All)}.$$

so,
 * $$P(Red\cap Big) = P(Big|Red) * P(Red).$$

similarly,
 * $$P(Red\cap Big) = P(Red|Big) * P(Big).$$

so the result is,
 * $$P(Big|Red) * P(Red) = P(Red|Big) * P(Big). \!$$

This is Bayes' theorm usually written as,


 * $$P(Big|Red) = \frac{P(Red | Big)\, P(Big)}{P(Red)}.$$

it is also true that,
 * $$p(Red) = P(Red|Big) * P(Big) + P(Red|Small) * P(Small). \!$$

so,
 * $$P(Big|Red) = \frac{P(Red | Big)\, P(Big)}{P(Red|Big) * P(Big) + P(Red|Small) * P(Small)}.$$