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Um, no?
...Oops, yes, said that without thinking. But this
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?
is correct.
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.
and P(A|C)/P(B|C) = P(A)/P(B) when A and B are subsets of C?
When A is a subset of C, P(A|C) = P(A).
In response to
Looking for proof of conditional probability
I agree with the OP: simply defining a probability concept doesn't by itself map it to our intuitions about it. For example, if we defined P(A|B) = P(AB) / 2P(B), it wouldn't correspond to our intuitions, and here's why.
Intuitively, P(A|B) is the probability of A happening if we know that B already happened. In other words, the entirety of the elementary outcome space we're taking into consideration now are those that correspond to B. Of those remaining elementary outcomes, the only ones that can lead to A are those that lie in AB. Their measure in absolute terms is equal to P(AB); however, their measure in relation to the elementary outcomes in B is equal to P(AB)/P(B).
Thus, P(A|B) is P(A) as it would be if the only elementary outcomes in existence were those yielding B. P(B) here is a normalizing coefficient: if we were evaluating the conditional probability of A in relation to a set of exhaustive and mutually exclusive experimental outcomes, as it is done in Bayesian reasoning, dividing by P(B) means renormalizing the elementary outcome space after B is fixed.
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I would love to see a post on hacking yourself poly from a woman's perspective. To be honest, I'm a little frustrated with the extent to which all dating advice on LessWrong is aimed at those attracted to females. (Luke's post is great and well considered, though).
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I'm actually staggered by the amount of so-called "dating advice" on LW in the first place.
May defame my image with women (who would date an immortalist after all .....)
...
The thing is, we support his trolling.
Speak for yourself.
In response to
Help me transition to human society!
I think you're taking this roleplaying thing too far.
I think the problem is the definitely not the language. From the original post:
- Presuming the US by default when it is assumed that no country name needs to be given.
- Expecting reader familiarity with US-specific cultural concepts.
- A tendency to focus on the US first and foremost when talking about worldwide problems and scenarios.
Expecting people to self-modify to "correct" these is wrong, although I don't think you were suggesting this.
(EDIT: read "wrong" as "unreasonable")
How exactly is it wrong?
Seems to me you could replace "fringe technophiles" with 'white guys not named Harold,' and have just as valid a statement.
I assume you mean that your description gets at what a random member of the public would likely notice first. This does seem close to the truth. (Although a more literal account of what they'd perceive first would involve some word like 'rationality' or 'reason', perhaps in connection with the term 'worship'.) But the term "fringe technophiles," by itself, would not lead anyone to expect a community that asks you to justify your belief in detail if you say technology will not destroy the world.
Except the agenda of the fraction I'm speaking about is not "technology will not destroy the world". It's "friendly AI and uploads will lead us to a bright perfect techno-utopia". And as I said multiple times before, I don't buy it.
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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.