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Thread: 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:
    $\displaystyle \begin{array}{c|ccccccc}x&1&2&3&...\\prob&p&(1-p)p&(1-p)^2 ...\\\hline\end{array}$
    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:
    $\displaystyle E(c(x)) = p$

    $\displaystyle \sum c(x) \times p(1-p)^{x-1} =p$
    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

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

    So your estimator is $\displaystyle c(x)$ 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

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

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

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

    Isn't $\displaystyle I(X=1)$ 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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