# 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 "Classical 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 ${\displaystyle P(C_{1})}$ 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:

{\displaystyle {\begin{aligned}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 {1}{4}})^{4}{\frac {3}{4}}{\frac {1}{3}}}{{5 \choose 4}({\frac {1}{4}})^{4}{\frac {3}{4}}{\frac {1}{3}}+{5 \choose 4}({\frac {1}{2}})^{4}{\frac {1}{2}}{\frac {1}{3}}+{5 \choose 4}({\frac {3}{4}})^{4}{\frac {1}{4}}{\frac {1}{3}}}}\end{aligned}}}

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

In the same way we find:

${\displaystyle P(C_{2}|X=4)={\frac {32}{3+32+81}}={\frac {32}{116}}}$

and

${\displaystyle P(C_{3}|X=4)={\frac {81}{3+32+81}}={\frac {81}{116}}}$.

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 ${\displaystyle p(C_{i})=P(C_{i}|X=4)/P(X=4)}$ When the result is 3 times heads the Probabilities change in favor of Coin 2 and further as the following table shows:

Heads ${\displaystyle p(C_{1})}$ ${\displaystyle p(C_{2})}$ ${\displaystyle p(C_{3})}$
5 1 32 243
4 3 32 81
3 9 32 27
2 27 32 9
1 81 32 3
0 243 32 1