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Math Help - Sufficient statistic

  1. #1
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    Sufficient statistic

    Hi, I am preparing for my test. I have a question about sufficient statistics. More to follow later.

    X_1, \cdots, X_n$ \mbox{is a random sample with pdf:}

     f(x|\mu, \sigma) = \dfrac{1}{\sigma} e^{-(x-\mu)/\sigma} \;,\;\mu<x<\infty\;,\;0<\sigma<\infty. \mbox{Find a two dimesional sufficient statistic for}(\mu, \sigma)

    The problem has been worked this way:

    Let x_{(1)}=min_i\;x_i. \mbox{Then the joint pdf is}

    f(x_1,....., x_n)= \prod_{i=1}^n \dfrac{1}{\sigma} e^{-(x-\mu)/\sigma} \;I_{(\mu,\sigma)}(x_i)\;=\;\left(\dfrac{e^{\mu/\sigma}}{\sigma}\right)^n\cdot e^{-\sum_i x_i/\sigma}\;I_{(\mu,\sigma)}(x_1)

    and thus factorization theorem concludes that (X_{(1)}, \sum_i X_i) is a suff stat for (mu,sigma).

    I am conused in why we are taking the indicator function and why was the minmum order statistic used. Could anybody explain? Thank You.
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  2. #2
    Moo
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    Hello,

    We are taking the indicator function because \forall i, x_i\in(\mu,\infty) (this is a consequence of the pdf of X)

    For the product (in the joint pdf) to be different from 0, we need \forall i,I_{(\mu,\infty)}(x_i)\neq 0 \Leftrightarrow I_{(\mu,\infty)}(x_i)=1

    So \forall i,x_i\geq \mu, which means that the minimum among the x_i's should be \geq \mu. Hence the result.
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  3. #3
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    Thanks, and one more thing. If i wanted to know whether the sufficient stat above is also a minimum sufficient what shall I do?

    I looked at few examples in the book where two sample points x and y are considered and the ratio of their pdfs are taken, but I am confused on how to use that here. Could you tell?

    Thank you.
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  4. #4
    MHF Contributor harish21's Avatar
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    just take the ratio of \dfrac{f(x|\mu, \sigma)}{f(y|\mu, \sigma)}, and find the conditions that would make this ratio a constant as a function of \mu \;and\; \sigma^2, which will give you the minimum sufficient statistic.
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