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DanielLC comments on Looking for proof of conditional probability - Less Wrong
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Comments (33)
Basically, P(A|B) = 0 when A and B are disjoint, and P(A|C)/P(B|C) = P(A)/P(B) when A and B are subsets of C?
It's better, but it's still not that good. I have a sneaking suspicion that that's the best I can do, though.
Now, a hopefully intuitive explanation of independent events.
By definition, A is independent from B if P(A|B) = P(A), or equivalently P(AB) = P(A)P(B). What does it mean in terms of measures?
It is easy to prove that if A is independent from B, then A is also independent from ~B: P(A|~B) = P(A ~B) / P(~B) = (P(A) - P(AB)) / (1 - P(B)) = (P(A) - P(A)P(B)) / (1 - P(B)) = P(A).
Therefore, A is independent from B iff P(A) = P(AB) / P(B) = P(A ~B) / P(~B), which implies that P(AB) / P(A ~B) = P(B) / P(~B).
Geometrically, it means that A intersects B and ~B with subsets of measures proportionate to the measures of B and ~B. So if P(B) = 1/4, then 1/4 of A lies in B, and the remaining 3/4 in ~B. And if B and ~B are equally likely, then A lies in equal shares of both.
And from an information-theoretic perspective, this geometric interpetation means that knowing whether B or ~B happened gives us no information about the relative likelihood of A, since it will be equally likely to occur in the renormalized outcome space either way.
I feel like independence really is just a definition, or at least something close to it. I guess P(A|B) = P(A|~B) might be better. Independence is just another way of saying that A is just as likely regardless of B.
P(A|B) = P(A|~B) is equivalent to the classic definition of independence, and intuitively it means that "whether B happens or not, it doesn't affect the likelihood of A happening".
I guess that since other basic probability concepts are defined in terms of set operations (union and intersection), and independence lacks a similar obvious explanation in terms of sets and measure, I wanted to find one.
When A is a subset of C, P(A|C) = P(A).
Um, no?
...Oops, yes, said that without thinking. But this
is correct.