Elementary Probability
- A sample-space (e.g. S={a,c,g,t}) is the set of possible outcomes of some experiment.
- Events A, B, C, ..., H, ... An event is a subset (possibly a singleton) of the sample space, e.g. Purine={a,g}.
- Events have probabilities P(A), P(B), etc.
- Random variables X, Y, Z, ... A random variable X takes values, with certain probabilities, from the sample space.
- We may write P(X=a), P(a) or P({a}) for the probability that X=a.
Thomas Bayes (1702-1761)
Thomas Bayes made an early study of probability and games of chance.
Bayes's Theorem
If B1, B2, ..., Bk is a partition of a set B (of causes) then
-
P(Bi|A) =
P(A|Bi) P(Bi)
/ ∑j=1..k P(A|Bj) P(Bj)
- i = 1, 2, ..., k
Inference
Bayes's theorem is relevant to inference because we may be entertaining a number of exclusive and exhaustive hypotheses H1, H2, ..., Hk, and wish to know which is the best explanation of some observed data D. In that case P(Hi|D) is called the posterior probability of Hi, "posterior" because it is the probability after the data has been observed.
- ∑j=1..k P(D|Hj) P(Hj) = P(D)
- P(Hi|D) = P(D|Hi) P(Hi) / P(D) --posterior
P(Hi) is called the prior probability of Hi, "prior" because it is the probability before D is known.
Notes
- T. Bayes.
An essay towards solving a problem in the doctrine of chance.
Phil. Trans. of the Royal Soc. of London, 53, pp.370-418, 1763.
Reprinted in Biometrika, 45, pp.296-315, 1958.
Conditional Probability
The probability of B given A is written P(B|A). It is the probability of B provided that A is true; we do not care, either way, if A is false. Conditional probability is defined by:
- P(A&B) = P(A).P(B|A) = P(B).P(A|B)
- P(A|B) = P(A&B) / P(B)
- P(B|A) = P(A&B) / P(A)
There are four combinations for two Boolean variables:
-
A not A margin B A & B not A & B (A or not A)& B = B not B A & not B not A & not B (A or not A)& not B = not B margin A = A&(B or not B) not A = not A &(B or not B) LA 1999
- P(A) = P(A & B) + P(A & not B)
- P(B) = P(A & B) + P(not A & B)
Independence
A and B are said to be independent if the probability of A does not depend on B and v.v.. In that case P(A|B)=P(A) and P(B|A)=P(B) so
- P(A&B) = P(A).P(B)
- P(A & not B) = P(A).P(not B)
- P(not A & B) = P(not A).P(B)
- P(not A & not B) = P(not A).P(not B)
- P(A & not B) = P(A).P(not B)
A Puzzle
I have a dice (made it myself, so it might be "tricky") which has 1, 2, 3, 4, 5 & 6 on different faces. Opposite faces sum to 7. The results of rolling the dice 100 times (good vigorous rolls on carpet) were:
1- 20:
3 1 1 3 3 5 1 4 4 2 3 4 3 1 2 4 6 6 6 621- 40:
3 3 5 1 3 1 5 3 6 5 1 6 2 4 1 2 2 4 5 541- 60:
1 1 1 1 6 6 5 5 3 5 4 3 3 3 4 3 2 2 2 361- 80:
5 1 3 3 2 2 2 2 1 2 4 4 1 4 1 5 4 1 4 281-100:
5 5 6 4 4 6 6 4 6 6 6 3 1 1 1 6 6 2 4 5
— LA 1999