Re: A simple but confusing question
From: Henry (se16_at_btinternet.com)
Date: 10/03/04
- Previous message: Edward L: "brownian motion question"
- In reply to: Ian Jermyn: "Re: A simple but confusing question"
- Next in thread: Ian Jermyn: "Re: A simple but confusing question"
- Reply: Ian Jermyn: "Re: A simple but confusing question"
- Messages sorted by: [ date ] [ thread ]
Date: Sun, 3 Oct 2004 23:38:39 +0000 (UTC)
On Sun, 3 Oct 2004 18:32:37 +0200, "Ian Jermyn" <ianjermyn@wanadoo.fr>
wrote:
>This is only approximately true, for N >> n, where N is the number of balls
>in the bucket and n the number of draws.
Lewis Carroll thought N>n was enough (you want to take n+1 balls out)
Try to point out the error in the following, taking account of his
prior.
>Let w_{n} be the proposition 'the
>nth ball is white'. Let W_{n} be the proposition w_{n} & w_{n - 1} & ... &
>w_{2} &w_{1}. Assume we know how many balls (N) are in the bucket. Let the
>number of white balls in the bucket be M.
Lewis Carroll's prior was P(M|N) = C(N,M)/2^N
[C(M,N) means N choose M if N>=M>=0 and means 0 otherwise]
>The probability of the next ball
>being white after n white balls is given by
>Pr(w_{n + 1} | W_{n}, N) = Pr(W_{n + 1} | N) / Pr(W_{n} | N) .
and similarly
Pr(w_{n + 1} | W_{n}, M, N) = Pr(W_{n + 1} |M, N) / Pr(W_{n} | M,N) .
but Pr(W_{n} | M,N) = C(M,n)/C(N,n)
so Pr(w_{n + 1} | W_{n}, M, N) = (M-n)/(N-n)
and Pr(M | W_{n},N) = C(N-n,M-n)/2^(N-n) using Bayes theorem and
Carroll's prior
so Pr(w_{n + 1} | W_{n}, N) = Sum_M C(N-n,M-n)(M-n)/((N-n)*2^(N-n))
=1/2 which is what Carroll said
- Previous message: Edward L: "brownian motion question"
- In reply to: Ian Jermyn: "Re: A simple but confusing question"
- Next in thread: Ian Jermyn: "Re: A simple but confusing question"
- Reply: Ian Jermyn: "Re: A simple but confusing question"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
