Statistics/Probability/Bayesian

Bayesian analysis is the branch of statistics based on the idea that we have some knowledge in advance about the probabilities that we are interested in, so called a priori probabilities. This might be your degree of belief in a particular event, the results from previous studies, or a general agreed-upon starting value for a probability. The terminology "Bayesian" comes from the Bayesian rule or law, a law about conditional probabilities. The opposite of "Bayesian" is sometimes referred to as "Frequentist Statistics."

Example
Consider a box with 3 coins, with probabilities of showing heads respectively 1/4, 1/2 and 3/4. We choose arbitrarily one of the coins. Hence we take 1/3 as the a priori probability $$P(C_1)$$ of having chosen coin number 1. After 5 throws, in which X=4 times heads came up, it seems less likely that the coin is coin number 1. We calculate the a posteriori probability that the coin is coin number 1, as:


 * $$\begin{align}

P(C_1|X=4)&=\frac{P(X=4|C_1)P(C_1)}{P(X=4)}\\ &=\frac{P(X=4|C_1)P(C_1)}{P(X=4|C_1)P(C_1)+P(X=4|C_2)P(C_2)+P(X=4|C_3)P(C_3)}\\ &=\frac{ {5 \choose 4} (\frac 14)^4 \frac 34 \frac 13} {{5 \choose 4} (\frac 14)^4 \frac 34 \frac 13+{5 \choose 4} (\frac 12)^4 \frac 12 \frac 13+{5 \choose 4} (\frac 34)^4 \frac 14 \frac 13} \end{align} $$ In words:
 * The probability that the Coin is the first Coin, given that we know heads came up 4 times... Is equal to the probability that heads came up 4 times given we know it's the first coin, times the probability that the coin is the first coin. All divided by the probability that heads comes up 4 times (ignoring which of the three Coins is chosen). The binomial coefficients cancel out as well as all denominators when expanding 1/2 to 2/4. This results in

\frac{3}{3 + 32 + 81}=\frac{3}{116} $$

In the same way we find:
 * $$P(C_2|X=4)=\frac{32}{3 + 32 + 81}=\frac{32}{116}$$

and
 * $$P(C_3|X=4)=\frac{81}{3 + 32 + 81}=\frac{81}{116}$$.

This shows us that after examining the outcome of the five throws, it is most likely we did choose coin number 3.

Actually for a given result the denominator does not matter, only the relative Probabilities $$p(C_i) = P(C_i|X=4)/P(X=4)$$ When the result is 3 times heads the Probabilities change in favor of Coin 2 and further as the following table shows: