Tuesday, December 14, 2010

Deterministic and probabilistic causation

Suppose that some random process C selects uniformly a random number in the interval [0,1]. Let C* be a process just like C, except that it can't select the number 1/2. Basically, C* works like this: a random number x in [0,1] is randomly picked with uniform distribution; if x is not 1/2, then the result of C* is x; if x = 1/2, then the result of C* is 1/4.

Then, for any (measurable) subset S of [0,1], under normal unfinked circumstances, the probability that C selects a number in S is equal to the probability that C* selects a number in S. (Proof: C and C* assign the same probability to any subset that does not contain either 1/2 or 1/4. But {1/2, 1/4} has probability zero, so C and C* assign the same probability to any (measurable) subset.)

This shows that numeric probability values fail to characterize all there is to be said about probabilistic causation. In addition to the probability distribution, we need something else that we might call the "range" of the probabilistic causation. C has a larger range than C*: C can pick out 1/2, but C* can't. And just as we shouldn't try to define the probabilistic strength of causation in terms of conditional probabilities, as that can be finked, we also shouldn't try to define the range of the probabilistic causation in modal terms (tempting suggestion: in C's range we put all the events that it's logically possible for C to cause given the laws; this fails as God can miraculously make C cause something outside its range).

The range of probabilistic causation is relative to a partition of logically possible relevant states. Thus, relative to the partition { [0,1/2], (1/2,1], everything else }, C and C* have the same range, namely { [0,1/2], (1/2,1] }. On the other hand, relative to the partition { {1/2}, [0,1/2), (1/2,1], everything else }, the range of C is { {1/2}, [0,1/2), (1/2,1] } while the range of C* is { [0,1/2), (1/2,1] }.

This also solves the following little puzzle: Here's a game. Suppose you pick a number, and then the process C (just as above) picks a number, and you get hurt iff C picks the same number as you did, and otherwise you get a dollar. Sam picks 0.14159. Jane picks 2. Obviously Jane did the safer thing, even though her probability of getting hurt is zero, which is the same as Sam's. Puzzle: Why? Answer: Because {0.14159} is in the range of C relative to the partition given by all the real numbers, while 2 isn't in the range.

We can now define deterministic causation: C deterministically causes E iff C probabilistically causes E and the range of C relative to the partition { E, ~E } is { E }.

Interesting fact: if C is a probabilistic cause, and R is the range of C, then the union (or disjunction, if you will) of all the events in R is something that C deterministically causes. Therefore, in any normal situation where there is probabilistic causation, there is also deterministic causation.

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