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Math Help - Find the best unbiased estimator of 1/b of gamma dist.

  1. #1
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    Exclamation Find the best unbiased estimator of 1/b of gamma dist.

    Hi guys. I have a question and I hope someone can help me out

    Let X1,.....Xn be a random sample from gamma(a,b) with a known.
    Find the best unbiased estimator of 1/b


    Waiting for your response as soon as you can

    Thanks in advance
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  2. #2
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    Quote Originally Posted by survivor1980 View Post
    Hi guys. I have a question and I hope someone can help me out

    Let X1,.....Xn be a random sample from gamma(a,b) with a known.
    Find the best unbiased estimator of 1/b


    Waiting for your response as soon as you can

    Thanks in advance
    Read this: Best Unbiased Estimators
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  3. #3
    MHF Contributor matheagle's Avatar
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    1 I need to know how you're writing your gamma density. Sometimes the b is in the numerator, sometimes it in the denomiator.

    2 is best=min variance? Hence UMVUE
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    Thumbs up

    Thank you so much guys for speedy response.

    Let X₁. X₂, X₃,, Xn be a random sample from gamma(α, Β) with α known. Find the best unbiased estimator of 1/ Β

    Note 1-α = alpha, Β=beta

    2- Yes the best unbiased estimatore is the same as UMVUE

    thanks
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  5. #5
    MHF Contributor matheagle's Avatar
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    I cannot help you until I see how you write your density.
    I said that yesterday, nor do I know what '1-α = alpha, Β=beta' means.
    Last edited by matheagle; May 6th 2009 at 05:48 PM.
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    Thumbs up

    Hi all

    This is the function


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    MHF Contributor matheagle's Avatar
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    OK, I'll work with that, BUT that is not how most people write the gamma density. I place beta in the denomiator.
    The likelihood function is L(\beta )={\beta^{n\alpha}\over \bigl(\Gamma (\alpha )\bigr)^n}\bigl(\Pi_{i=1}^nx_i\bigr)^{\alpha-1}e^{-\beta\sum_{i=1}^nx_i}.

    Thus with \alpha known we have \sum_{i=1}^nx_i suficient for \beta.

    And since E\biggl(\sum_{i=1}^nX_i\biggr)={n\alpha\over \beta} we have {\sum_{i=1}^nX_i\over n\alpha} UMVUE for {1\over \beta}.

    I could have done this yesterday, but the mean of my gamma's is \alpha\beta
    and getting an unbiased estimator of 1/\beta is a lot harder in that case.
    And I wasn't going to do this until I knew how you were writing your density.
    Last edited by matheagle; May 6th 2009 at 05:48 PM.
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  8. #8
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    Smile

    Quote Originally Posted by matheagle View Post
    OK, I'll work with that, BUT that is not how most people write the gamma density. I place beta in the denomiator.
    The likelihood function is L(\beta )={\beta^{n\alpha}\over \bigl(\Gamma (\alpha )\bigr)^n}\bigl(\Pi_{i=1}^nx_i\bigr)^{\alpha-1}e^{-\beta\sum_{i=1}^nx_i}.

    Thus with \alpha known we have \sum_{i=1}^nx_i suficient for \beta.

    And since E\biggl(\sum_{i=1}^nX_i\biggr)={n\alpha\over \beta} we have {\sum_{i=1}^nX_i\over n\alpha} UMVUE for {1\over \beta}.

    I could have done this yesterday, but the mean of my gamma's is \alpha\beta
    and getting an unbiased estimator of 1/\beta is a lot harder in that case.
    And I wasn't going to do this until I knew how you were writing your density.
    I met the same problem with this question.
    if the the mean of gamma's is \alpha\beta. How to get an unbiased estimator of 1/\beta? Anyone can help? THANKS~
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    It can be shown that Sum(Xi) is not only sufficient but complete for beta (as w(beta) of exponential function contains an open set), so try 1/sum(xi) to estimate 1/beta. Lehmann-Scheffe tells us that an unbiased estimator that is a function of a complete statistic is the best unbiased estimator.

    It can be shown that, assuming iid xi, Sum(xi) is distributed as Gamma(n*alpha,beta).

    Given the above, let Y=(1/sum(xi)). Y is distributed as an inverted gamma(n*alpha, 1/beta) with mean=(1/beta)/(n*alpha-1). Thus, E(1/sum(xi)) = E(Y) = (1/beta)/(n*alpha-1), which is obviously a biased estimator of 1/beta.

    Now let T = (n*alpha - 1)/sum(xi) be the unbiased estimator of 1/beta which is a function of a complete statistic. Thus, T is the best unbiased estimator of 1/beta, but it can be shown that it does not attain the lower Cramer-Rao bound.
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  11. #11
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by survivor1980 View Post
    Hi all

    This is the function


    THE mean here is {\alpha\over \beta} since you're writing the beta in the numerator instead of in the denominator.
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    Quote Originally Posted by survivor1980 View Post
    Hi all

    This is the function


    But if you write down your pdf like this, doesn't this mean you have a gamma(a,1/b) instead of gamma(a,b)?
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  13. #13
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    Quote Originally Posted by mnazam View Post
    But if you write down your pdf like this, doesn't this mean you have a gamma(a,1/b) instead of gamma(a,b)?
    There are 2 versions of the gamma distribution's pdf. You can have a look at the wikipedia page for the gamma distribution.
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  14. #14
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    Quote Originally Posted by Moo View Post
    There are 2 versions of the gamma distribution's pdf. You can have a look at the wikipedia page for the gamma distribution.
    I see. Thanks for the info. If the pdf can be written like this, it'd be so easy. Like we can show that it is an exponential family thus it is a complete sufficient estimator for beta. Then we can use MLE to get 1/beta(^). Then to show that 1/beta(^) is an unbiased estimator we calculate the mean of it and we get 1/beta. So 1/beta(^) in unbiased estimator.

    So we can conclude that 1/beta(^) is the best estimator. Is it right?
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