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Math Help - Bayesian Statistics - Factorization Theorem

  1. #1
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    Bayesian Statistics - Factorization Theorem

    Hi all, having real problems doing this question. Can anyone explain the steps of how its done (leaving the process of actually carrying out the steps to me! :-))

    If X1,X2, . . . ,Xn are independent random variables from the Binomial distribution
    Bin(k, \theta), show that S = \displaystyle\sum\limits_{i=1}^n Xi is sufficient for \theta
    (i) by using the Factorization Theorem;
    (ii) by finding the distribution of X1,X2, . . . ,Xn conditional on S = s.
    Last edited by sirellwood; February 24th 2011 at 11:07 AM.
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  2. #2
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    Quote Originally Posted by sirellwood View Post
    Hi all, having real problems doing this question. Can anyone explain the steps of how its done (leaving the process of actually carrying out the steps to me! :-))

    If X1,X2, . . . ,Xn are independent random variables from the Binomial distribution
    Bin(k, \theta), show that S = \displaystyle\sum\limits_{i=0}^n i^3 Xi is sufficient for \theta
    (i) by using the Factorization Theorem;
    (ii) by finding the distribution of X1,X2, . . . ,Xn conditional on S = s.
    Did you write S correctly? T = \sum X_i is minimal sufficient for theta in this problem, and it doesn't look to me like the S you gave is a function of T.
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    Argh! My apologies.... original post is now edited. Thank you theodds
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    MHF Contributor harish21's Avatar
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    Note that X_is are independent , and for binomial RVs, the sum \displaystyle S=\sum_{i=1}^n X_i \sim Binomial(kn,\theta)
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    ok so im guessing that i begin like that for part (ii) of this question?
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    Ok i can do part (ii), the direct method now, but still having problems with the factorization method. I know that f \theta(x) = g(T(x), \theta)h(x)

    but im not able to express this for the binomial distribution :-(
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  7. #7
    MHF Contributor harish21's Avatar
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    the RVs X_1, \cdots, X_n form Binomial (k, \theta), and the pmf is

    f(x|\theta)= \dbinom{k}{x} {\theta}^x (1-\theta)^{k-x}\;I_{0,\cdots,k}(x)

    Now,

    \displaystyle f(x_1,\cdots,x_n|\theta)= \prod_{i=1}^n f(x_i|\theta)=\prod_{i=1}^n \bigg[\dbinom{k}{x_i}\theta^{x_i} (1-\theta)^{k-x_i}\;I_{0,\cdots,k}(x_i)\bigg]

    \displaystyle = {\theta}^{\sum_{i=1}^n x_i} (1-\theta)^{\sum_{i=1}^n 1-x_i}\;\prod_{i=1}^n\bigg[\dbinom{k}{x_i} I_{0,\cdots,k}(x_i)\bigg]

    simplify further and remember the factorization theorem to come to your conclusion.
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