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Math Help - unbiased estimate for p in geometric(p)

  1. #1
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    unbiased estimate for p in geometric(p)

    let X be a geometric(p)
    find unbiased estimate c for p.
    please help.

    can und an unbiased estimate for 1/p but don't know how to find for p.
    please help. thanks.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Since E(X)=p.
    Obtain E(1/X), which won't be p, but it's a start.
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  3. #3
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    Lets start with the probability table for x. This is a geometric(p) distribution so the probabilities are:
    \begin{array}{c|ccccccc}x&1&2&3&...\\prob&p&(1-p)p&(1-p)^2 ...\\\hline\end{array}
    Notice that we have a value of p occuring in the probability table at x=1. We will exploit this to get an unbiased estimate later.



    suppose you have 1 realisation of x and your estimator is c(x)

    For your estimator to be unbiased you require:
    E(c(x)) = p

    \sum c(x) \times p(1-p)^{x-1} =p
    where the summation is taken over all possible values of x (1,2,3,4,5,6...)

    Now, consider the following function:

    c(x)=1 if x=1
    c(x)=0 otherwise

    The expected value of our function is then

    \sum c(x) \times p(1-p)^x =p
    =(f(1)\times p)  + (f(2)\times p(1-p)) + (f(3)\times p(1-p)^2)  +...
    =(1 \times p)  + (0 \times p(1-p)) + (0 \times p(1-p)^2)  +...
    =p
    as required

    So your estimator is c(x) where c(x) was defined above.

    In reality, this estimator is no practical use. But it is an unbiased which is all the question asked for. I did assume that your sample is a single realisation from the distribution. if you have a sample with multiple data points, just discard all but one of them and the proceedure still works.


    PS: not bad for an economist, eh?
    Last edited by SpringFan25; May 29th 2010 at 07:24 PM.
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  4. #4
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    Thank you MathEagle and SpringFan25 !

    MathEagle: Computing the expectation of E(1/x) seems quite hard since
    E(1/x)= sum[[p(1-p)^(x-1)]/x]. Not sure how to solve this.

    SpringFan25: I didn't think of that, very smart . However, I don't think this is going to be very applicable in my problem.

    I am trying to get the UMVUE for p when X~geometric(p).
    X is a complete sufficient statistic.
    if use I(c(x)=1) as my unbiased estimate,
    then T*(x)=E( I(c(x))|x) is UMVUE for p

    but E(I(c(x)=1)|x)=P(c(x)=1)... which will give me some function of p?
    but p is what we are trying to estimate, so this is probably not a good solution
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  5. #5
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    yes, my estimate is definately not the UMVUE, but you didn't ask for that :P
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  6. #6
    MHF Contributor matheagle's Avatar
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    yea, I was thinking about some way to take the derivative to obtain
    that sum. But it doesn't look good.

    Maybe try to integrate wrt p

    \int  \sum_{x=1}^{\infty} (1-p)^{x-1} dp = -\sum_{x=1}^{\infty} {(1-p)^x\over x}

    So \sum_{x=1}^{\infty} {(1-p)^x\over x}=-\ln p

    and p\sum_{x=1}^{\infty} {(1-p)^x\over x}=-p\ln p which doesn't help.

    Isn't I(X=1) the UMVUE then?
    Last edited by matheagle; May 29th 2010 at 08:22 PM.
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  7. #7
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    I think you are right!

    I(x=1) would be umvue because it is unbiased and a function of the complete sufficient statististic. (and therefore I don't need to take E(I(x=1)|x).
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  8. #8
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    horay then
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  9. #9
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    neg binomial

    Do you have any idea of how to find an unbiased estimate for the negative binomial?

    wikipedia says a UMVUE for p for Y~negative binomial is


    im trying to prove this result.

    since the geometric is a special case of the negative binomial, do you think I should get a similar result for the geometric?
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