Statistics/Probability/Bayesian

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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(C1) 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:

P(C1|X=4)=P(X=4|C1)P(C1)P(X=4)=P(X=4|C1)P(C1)P(X=4|C1)P(C1)+P(X=4|C2)P(C2)+P(X=4|C3)P(C3)=(54)(14)43413(54)(14)43413+(54)(12)41213+(54)(34)41413

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
33+32+81=3116

In the same way we find:

P(C2|X=4)=323+32+81=32116

and

P(C3|X=4)=813+32+81=81116.

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(Ci)=P(Ci|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 p(C1) p(C2) p(C3)
5 1 32 243
4 3 32 81
3 9 32 27
2 27 32 9
1 81 32 3
0 243 32 1

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