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Math Help - statistics help needed

  1. #1
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    statistics help needed

    Let X1 X2,...Xn be a random sample from exponential distribution with parameter q that is f(x,q)=(1/q)e^{(-x/q)} for x>0 and q>0.



    Find the Uniformly Minimum Variance Unbiased Estimator for P[X<c] where c is a known positive constant.



    please help me on this. thank you.
    Last edited by Kat-M; May 1st 2009 at 10:36 AM. Reason: hard to read without typing it in tex
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  2. #2
    MHF Contributor matheagle's Avatar
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    The suff stat for q is the sum, S_n=\sum_{i=1}^nX_i.
    Next we need to find a function of this sum that's unbiased for P(X<c)=1-e^{-c/q}.

    Since each X_i\sim\Gamma(1,q), we have S_n\sim\Gamma(n,q).
    Last edited by matheagle; May 1st 2009 at 09:47 PM.
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  3. #3
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    Would you help a little more?
    How does knowing that S~  Gam(n,q)help me figure out the Uniformly Minimum Variance Estimator?
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  4. #4
    MHF Contributor matheagle's Avatar
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    You will need to do the Rao-Blackwell Theorem directly.

    Compute E(I(X_1<c)|S_n=s)
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    i am sorry to ask you this but what is I(X1<c)? Is Ia funtion?
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Kat-M View Post
    i am sorry to ask you this but what is I(X1<c)? Is Ia funtion?
    I stands for indicator.
    It's 1 when that event happens and zero when it doesn't.
    EI(X\in A)=P(X\in A).
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  7. #7
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    i thought about this for a long time but still i am not sure how to compute E(I(X1<c)|S=s).
    X1 is part of S. and i dont know how it affects in computing \int I(X1<c)f(x1|s).
    i got f(x1|s)=\Gamma(n)/s^{n-1} So when i integrate it from 0 to c, i got c\Gamma(n)/s^{n-1}. did i do it right?
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  8. #8
    MHF Contributor matheagle's Avatar
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    Back to E(I(X_1<c)|S_n=s) where s>c.

    We need the density  f(x_1|s_n) ={f(x_1) f(s_{n-1}) \over f(s_n) }, where s_{n-1}=x_2+\cdots +x_n.

    The density of  X_1 is  f(x_1)=e^{-x_1/q}/q.

    While  f(s_{n-1})= {s_{n-1}^{n-2} e^{-s_{n-1}/q}\over \Gamma(n-1)q^{n-1}} and  f(s_n) ={s_n^{n-1} e^{-s_n/q}\over \Gamma(n)q^n}

    Noting that x_1+s_{n-1}=s_n we have

     f(x_1|s_n) ={(n-1) (x_2+\cdots +x_n)^{n-2} \over (x_1+\cdots +x_n)^{n-1}}

    Thus P(X_1<c|S_n=s)=\int_0^cf(x_1|S_n=s)dx_1=(n-1)\int_0^c {(s-x_1)^{n-2}dx_1\over s^{n-1}}

    which gives me 1-\biggl(1-{c\over s}\biggr)^{n-1}, which doesn't make sense when n=1.
    Last edited by matheagle; May 2nd 2009 at 10:39 PM.
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  9. #9
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by matheagle View Post
    Back to E(I(X_1<c)|S_n=s) where s>c.
    We need the density  f(x_1|s_n) ={f(x_1) f(s_{n-1}) \over f(s_n) }, where s_{n-1}=x_2+\cdots +x_n.

    The density of  X_1 is  f(x_1)=e^{-x_1/q}/q.

    While  f(s_{n-1})= {s_{n-1}^{n-2} e^{-s_{n-1}/q}\over \Gamma(n-1)q^{n-1}} and  f(s_n) ={s_n^{n-1} e^{-s_n/q}\over \Gamma(n)q^n}

    Noting that x_1+s_{n-1}=s_n we have

     f(x_1|s_n) ={(n-1) (x_2+\cdots +x_n)^{n-2} \over (x_1+\cdots +x_n)^{n-1}}

    Thus P(X_1<c|S_n=s)=\int_0^cf(x_1|S_n=s)dx_1=(n-1)\int_0^c {(s-x_1)^{n-2}dx_1\over s^{n-1}}

    which gives me 1-\biggl(1-{c\over s}\biggr)^{n-1}, which doesn't make sense when n=1.
    I don't think this is right.
    The ideas are good, but the answer seems lame.

    Don't mess with guard duck, he's one mean sob.
    Last edited by matheagle; May 4th 2009 at 05:57 PM. Reason: Added quote. (Given your current avatar, don't you mean lame duck?)
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