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Math Help - estimator of theta

  1. #1
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    estimator of theta

    If X1, X2.......,Xn constitute a random sample of size n from an exponential population, show that "X bar" is a sufficient estimator of the parameter theta

    Can somebody please, please help me with this? It is giving me a lot of trouble
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  2. #2
    Moo
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  3. #3
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    Quote Originally Posted by Moo View Post
    Thank you for the hint, but I already knew that factorization theorem was involved but I just can't seem to solve it. I was hoping someone could give me a more step by step answer because I am really trying but have no outside help for this course and the professor is hardly ever available to assist.
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    MHF Contributor matheagle's Avatar
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    Just write your likelihood function which is the joint density....

    f(x_1,\cdots,x_n)={1\over \theta^n}e^{-\sum_{i=1}^nx_i/\theta}

    where I left out the fact that all the x's are positive, that usually just confuses people.
    meaning........

    f(x_1,\cdots,x_n)={1\over \theta^n}e^{-\sum_{i=1}^nx_i/\theta}I(x_1>0)\cdots I(x_n>0)

    This shows that the SUM is suff for theta, but so is twice the sum....
    so the sum divided by n is suff....

    f(x_1,\cdots,x_n)={1\over \theta^n}e^{-n\bar X/\theta}I(x_1>0)\cdots I(x_n>0)

    shows that the sample mean is suff for theta.
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  5. #5
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    please verify my answer. Need help fast

    Not sure if I have this right but I want to make sure the way I am doing it is correct. Thanks

    If X1,.......,Xn make up a random sample of size n from an exponential population, show that 'X bar' is a sufficient estimator of the parameter theta

    Not sure how to write symbols, so I am going to let
    Theta = Z
    Summation = S
    e^ = exp(...)

    So,

    Z^-n * exp(-S(Xi)/Z)

    take LN

    = -n LN(Z) - S(Xi)/Z

    Take derivate w.r.t. Z

    = -n/Z + S(Xi)/Z^2

    Set = 0 and solve for Z

    0 = -nZ + S(Xi)
    nZ = S(Xi)
    Z = S(Xi)/n

    Is this right?
    Last edited by mr fantastic; March 17th 2010 at 03:26 AM.
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