Re: A simple but confusing question
From: Ian Jermyn (Ian.Jermyn_at_sophia.inria.fr)
Date: 10/05/04
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Date: Tue, 5 Oct 2004 08:53:45 +0200
Very nice. I was not considering the 'sampling without replacement' case,
but it is simpler in fact. The result can be seen even more easily from the
fact that
Pr(w_{n + 1} | W_{n}, N) = Pr(W_{n + 1} | N) / Pr(W_{n} | N)
and, somewhat surprisingly,
Pr(W_{n} | N) = \sum_{M = n}^{N} [C(M, n)/C(N, n)] C(N, M) 2^{-N} = 2^{-n}
This also means that unknown N does not change things, since Pr(W_{n} | N)
does not depend on N.
Ian.
--
--------------------------------------------------
Ian Jermyn
ianjermyn@wanadoo.fr
"Henry" <se16@btinternet.com> a écrit dans le message de
news:btdul09se6s2t80kobmcbmjuop2bra4tae@4ax.com...
> 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
>
>
>
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