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Thread: Finding the Sufficient Statistic of Weibull Distribution.

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    Finding the Sufficient Statistic of Weibull Distribution.




    So I have a Weibull Distribution with two parameters with the following pdf:

    f(x/\alpha,\beta) = \frac{\beta}{\alpha^\beta} x^{\beta-1} exp [ - (\frac{x}{\alpha})^\beta ]

    If I know \beta, how can I find the sufficient statistic for \alpha? I believe (correct me if I am wrong, I can use either the Neyman Factorization theorem or express the pdf in an exponential family?)

    Have no idea how to add spaces in math code..
    Last edited by MegaMan; 12-02-2011 at 01:30 PM.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    if you want to use fractions the latex code is something like this [math]\frac{\alpha}{\beta}[/math] which gives \frac{\alpha}{\beta}. Notice the curly braces.

    I think you also forgot to include x somewhere in the actually pdf itself.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    Quote Originally Posted by Dason View Post
    if you want to use fractions the latex code is something like this [math]\frac{\alpha}{\beta}[/math] which gives \frac{\alpha}{\beta}. Notice the curly braces.

    I think you also forgot to include x somewhere in the actually pdf itself.
    Was still editing when I posted. Hope I would be able to do it before anybody noticed :PP

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    I have a hunch that expressing it in exponential family form would probably be the easiest way to do it in this case. But that's just a hunch. I know if both parameters are unknown then I don't think we can do better than the set of order statistics. But if we know \beta it looks like things would be a little bit nicer.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    When I am simplifying it, do I need to group up certain parts? i.e. do I need to seperate alpha from beta
    Last edited by MegaMan; 12-02-2011 at 03:17 PM.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    Not sure what hint can be given because that is "straight forward" in the sense that it just require you to group the terms appropriately.

    What have you try? via Exponential family/factorization theorem?

    Beta is treated as a constant like 1, 2, 3. So do not worry about that.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    Ok, this is what I have so far:

    exp [ (-\frac{\sum{X_i}}{\alpha})^\beta + (\beta-1)\log\sum{X_i} + n\log(\frac{\beta}{\alpha^\beta}) ]

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    The first and the second term has mistake. Please try again.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    Hmm are you sure? This is by the way, f(X_1...X_n/\alpha,\beta)
    Last edited by MegaMan; 12-02-2011 at 03:37 PM.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    Yes he is sure. So am I. Note that x^2 + y^2 is not the same as (x+y)^2. And also be a little more careful with your log rules.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    I'm sorry but how can the first term be wrong. It is an EXACT copy of the definition in the pdf............

    I have the joint density likelihood function as:

    f(X_1..X_n/\alpha,\beta) = (\frac{\beta}{\alpha^\beta})^n \sum{X_i}^{\beta-1} exp[(-\frac{\sum{X_i}}{\alpha})^\beta)]

    Is this correct?
    Last edited by MegaMan; 12-02-2011 at 04:08 PM.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    I guess more directly...

    (\sum x_i)^\beta \neq \sum (x_i^\beta)

    for \beta \neq 1

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    And in the middle part, outside the exponential, how can you have a summation when you are actually having a product of marginal pdfs?

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    Re: Finding the Sufficient Statistic of Weibull Distribution.

    ok so in the middle part, I should have x_1^{\beta-1} * x_2^{\beta-1} .. * x_n^{\beta-1} ? which i can only put as (\prod{x_i})^{\beta-1} ?
    Last edited by MegaMan; 12-02-2011 at 05:16 PM.

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    Re: Finding the Sufficient Statistic of Weibull Distribution.


    Hi, if any of you guys are still around to help. I have worked out that

    f(x_i..x_n/\alpha,\beta) = exp [-\frac{1}{\alpha^\beta}\sum{x_i}^\beta + (\beta - 1)(\sum{\log x_i}) + nlog(\frac{\beta}{\alpha^\beta})] .

    Can I just confirm if this is now correct?.. and how I need to order this now, if I know beta and need to find the sufficient statistic for alpha
    Last edited by MegaMan; 12-02-2011 at 06:30 PM.

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