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        riddles >> medium >> Conditional Probability + Partition Proof
        
(Message started by: william wu on Jan 28^th, 2003, 3:44am)

Title: Conditional Probability + Partition Proof
Post by william wu on Jan 28^th, 2003, 3:44am

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):

http://www.ocf.berkeley.edu/~wwu/images/riddles/condProbPartition.png

where the E_i'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 E_i, 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 { E_j } partitions a space S if all the E_j are pairwise disjoint and Union(E_j) = S.

Note 2: thanks again to [towr] for super handy latex2png generator

Title: Re: Conditional Probability + Partition Proof
Post by towr on Jan 28^th, 2003, 8:46am

we have [hide]
P(A|B) = sum(P(A|Ei)*P(Ei|B), i)
[/hide]
which expands to[hide]
P(A AND B)/P(B) = sum(P(A AND Ei)/ P(Ei) * P(Ei AND B)/P(B), i)[/hide]

Suppose there is only one partition, then Ei is always true, and we're left with[hide]
P(A AND B)/P(B) = P(A) * P(B)/P(B)
P(A AND B) = P(A) * P(B)[/hide]

which isn't true in general.. but only when A and B are independant stochastic variables..

Title: Re: Conditional Probability + Partition Proof
Post by william wu on Jan 28^th, 2003, 3:53pm

I don't think that constitutes a proof because it only works for the case where there is only one partition ... ?

Title: Re: Conditional Probability + Partition Proof
Post by william wu on Jan 28^th, 2003, 4:47pm

Ah I think I may know how to prove it. Use the following fact:

[hide]
P(A) = sum_i [ P(E_i) P(A|E_i) ]
[/hide]

Title: Re: Conditional Probability + Partition Proof
Post by Icarus on Jan 28^th, 2003, 5:35pm

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.

Title: Re: Conditional Probability + Partition Proof
Post by william wu on Jan 28^th, 2003, 5:39pm

Oh, whoops. Hmm, I guess my intuition isn't too hot. Thanks guys.