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Topic: Conditional Probability + Partition Proof (Read 1302 times) |
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william wu
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Conditional Probability + Partition Proof
« on: Jan 28th, 2003, 3:44am » |
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This is probably not difficult, but I can't seem to get it. The question is to determine whether this statement is true or false, and justify (if true provide proof, if false provide counterexample):
where the Ei's form a partition on the sample space.
Intuitively I think it is true. We say, given that B has happened, let's look at a particular region of the universe Ei, and then measure the probability of A happening in that region. Then we look at a different region, and another, and another -- with all these regions being disjoint. In the end we add up the contributions from each region toward the probability that A happens, given that B happened. My problem is how to prove the statement formally. I tried doing algebraic manipulation on the right hand side, hoping everything would cancel out to make the left hand side. I applied the identity P(X | Y) = P(X AND Y) / P(Y) a few times. Stuck.
Note 1: what is a partition you ask? if you imagine the sample space as some kind of shape like an oval, carve up the shape into pieces. the set of pieces comprise a partition. it's simply a set of jigsaw puzzle pieces that can be fit together to make the whole sample space. formally, the set { Ej } partitions a space S if all the Ej are pairwise disjoint and Union(Ej) = S.
Note 2: thanks again to [towr] for super handy latex2png generator
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| « Last Edit: Jan 28th, 2003, 3:57am by william wu » |
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towr
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Re: Conditional Probability + Partition Proof
« Reply #1 on: Jan 28th, 2003, 8:46am » |
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we have
P(A|B) = sum(P(A|Ei)*P(Ei|B), i)
which expands to
P(A AND B)/P(B) = sum(P(A AND Ei)/ P(Ei) * P(Ei AND B)/P(B), i)
Suppose there is only one partition, then Ei is always true, and we're left with
P(A AND B)/P(B) = P(A) * P(B)/P(B)
P(A AND B) = P(A) * P(B)
which isn't true in general.. but only when A and B are independant stochastic variables..
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Wikipedia, Google, Mathworld, Integer sequence DB
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william wu
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Re: Conditional Probability + Partition Proof
« Reply #2 on: Jan 28th, 2003, 3:53pm » |
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I don't think that constitutes a proof because it only works for the case where there is only one partition ... ?
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| « Last Edit: Jan 28th, 2003, 3:53pm by william wu » |
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william wu
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Re: Conditional Probability + Partition Proof
« Reply #3 on: Jan 28th, 2003, 4:47pm » |
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Ah I think I may know how to prove it. Use the following fact:
P(A) = sumi [ P(Ei) P(A|Ei) ]
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| « Last Edit: Jan 28th, 2003, 4:48pm by william wu » |
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Icarus
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Re: Conditional Probability + Partition Proof
« Reply #4 on: Jan 28th, 2003, 5:35pm » |
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Towr was not proving the formula - he disproved it: If A and B are not independent, then the formula is false, at least for a single partition.
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william wu
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Re: Conditional Probability + Partition Proof
« Reply #5 on: Jan 28th, 2003, 5:39pm » |
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Oh, whoops. Hmm, I guess my intuition isn't too hot. Thanks guys.
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| « Last Edit: Jan 28th, 2003, 5:40pm by william wu » |
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