Re: My claim on Omega's defn
From: Daryl McCullough (stevendaryl3016_at_yahoo.com)
Date: 02/02/05
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Date: 2 Feb 2005 07:34:52 -0800
r.e.s. says...
>
>"Daryl McCullough" <stevendaryl3016@yahoo.com> wrote ...
>
>> Finally, define Omega to be the limit as n-->infinity of S(n).
>> This is the same as
>> Omega = sum over all valid bit strings p of 2^{-length(p)}
You're right that interpreting Omega as a probability requires
the consideration of infinite bit strings. However, the definition
of Omega can be made in terms limits of finite bit strings.
>Letting A be the prefix-free set {s: s is a valid bitstring},
>with some measure theory I can show pretty directly that
>
>Omega(A) := SUM[s in A] 1/2^|s| = Pr(X has a prefix in A), [1]
>
>where X is a random bitstring of *infinite* length.
>OTOH, the type of approach you've used here shows that
>
>Omega(A) = LIM[n -> oo] Pr(X_n has a prefix in A), [2]
>
>where X_n is a random bitstring of *finite* length n.
>(Your S(n) = Pr(X_n has a prefix in A).)
>
>The second approach has avoided a measure on an uncountable
>set, but *hasn't* established convergence of the infinite
>sequence [2] -- and I'm only sure the limit exists because
>of its connection to [1].
What we know about S(n) is:
1. S(n) <= 1
2. S(n+1) >= S(n)
Don't these two facts together imply that the sequence S(n) has
a limit?
Let s = the least upper bound of the set { S(n) | n >= 0 }.
To prove that lim n --> infinity S(n) = s, we need to show
forall e > 0, exists n, all m > n
s - S(m) < e
That's easy enough: for each e > 0, define E = { S(n) | S(n) + e > s }.
Then let n_e = the smallest n such that S(n) is in E.
-- Daryl McCullough Ithaca, NY
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