Re: Tomsky's Theorem: P(A|A) = 1
- From: m00es <m00es@xxxxxxxxx>
- Date: Sat, 18 Aug 2007 17:52:03 -0700
On Aug 19, 12:57 am, Jack Tomsky <jtom...@xxxxxxxxxxxxx> wrote:
Since Afonso has given me credit for this profound theorem, I will now provide a detailed derivation.
If P(A) > 0, then P(A|A) = P(A,A)/P(A) = P(A)/P(A) = 1.
In particular, if A is the event that a team has won 62 of its first 120 games, then the conditional probability that it has won 62 of its first 120 games, given that it has won 62 of its first 120 games, is one.
Jack
Here is an interesting extension of your theorem. Suppose my prior
probability that the team has won 62 of its first 120 games is P
(which may not necessarily be equal to 1 -- in fact, I suspect there
may be people somewhere around here that may not assign a prior
probability of 1 to an event that has already happened). Then by
Bayes' theorem:
p(A|A) = p(A|A) * P / ( p(A|A) * P + p(A|not A)*(1-P) )
and since p(A|not A) is 0 (i.e., the probability that the team has won
62 of its first 120 games, given that is has NOT won 62 of its first
120 games is zero), we find that the posterior probability that the
team has won 62 of its first 120 games is 1. Therefore, regardless of
which prior probability I use (as long as P > 0), the posterior
probability must be 1.
Now suppose I consider it impossible that the team has won 62 of its
first 120 games (even though it has already won 62 of its first 120
games) prior to observing that the team has won 62 of its first 120
games. Therefore P = 0. Then:
p(A|A) = 0 / 0
which is undefined. I can conclude from this that there is not
sufficient evidence that the team has won 62 of its first 120 games.
m00es
.
- References:
- Tomsky's Theorem: P(A|A) = 1
- From: Jack Tomsky
- Tomsky's Theorem: P(A|A) = 1
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