## Are all probabilities conditional probabilities?

We know that $P(A|B) = \frac{P(A \cap B)}{P(B)}$. So $P(B) = P(A|B)P(A \cap B)$. Thus are all probabilities conditional probabilities? Can one make a probability more accurate by introducing a conditional component? For example, the probability of rolling a six on a fair die is $1/6$. But can we make this more accurate by assuming prior events?

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 The definition of the words "fair die" is that the probability that each face comes up is $\frac{1}{6}$. What would it mean to make this more accurate? (Of course, in the physical world it is likely that there is no such thing as a fair die.) – JBL Jun 13 2010 at 2:53 @JBL, that's the kind of accuracy the OP is referring to: introducing a bias based on prior knowledge about the die etc. – Suresh Venkat Jun 13 2010 at 3:27

There's an obvious algebra error in Tony's question.

In one trivial sense all probabilities are conditional: just condition on the whole "sample space". In (another?) sense, probabilities are necessarily conditional on what is known.

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In a sense this is precisely the Bayesian persective on what probabilities mean. Rather than being the limit fraction of a number of times an event occurs, it is a strength of belief conditioned on a prior. Bayesian approaches to statistical reasoning always start with a prior belief and condition based on that.

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It is possible to develop probability theory taking conditional probability as one of the basic definitions; see section 3.2 in this book and the references mention there. Renyi was one of the first mathematicians to favor this approach, which is described in his book Probability Theory. Part of the motivation for this approach is to directly build in the ability to condition on measure-zero events without having to make a limiting argument. Another key word related to this idea is disintegration.

So, as Kjetil mentions, it all depends on what one takes as axioms. But certainly it is possible to develop theories that take conditional probability as the centerpiece.

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I think that's the main difference between the Bayesian approach vs. the frequentist approach, having prior knowledge is supposed to make your predictions more accurate. See [LessWrong][1]

Bailey was trained in statistics, and when he joined an insurance company he was horrified to see them using Bayesian techniques developed in 1918. They asked not "What should the new rates be?" but instead "How much should the present rates be changed?" But after a year of trying different things, he realized that the Bayesian actuarial methods worked better than frequentist methods. Bailey "realized that the hard-shelled underwriters were recognizing certain facts of life neglected by the statistical theorists." For example, Fisher's method of maximum likelihood assigned a zero probability to nonevents. But since many businesses don't file insurance claims, Fisher's method produced premiums that were too low to cover future costs.cover future costs.

[1]: http://lesswrong.com/lw/774/a_history_of_bayes_theorem/cover future costs.

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I think this is a tricky question! and the answer depends on the exact meaning of the question. In the usual standard probability theory, based on Kolmogoro'vs axioms, in the axioms itself there are no conditional probabilities, and only later conditional probability are separately axiomatized. This seems to me a most strange procedure, and I am not sure why it is done that way. But in the lighjt of this theory, there are in fact nn-conditional probabilities. From an applied point of view, that of course do not make sense.

Others can maybe comment on alternative axiomatizations.

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