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Math Help - Sufficiency question

  1. #1
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    Sufficiency question

    Let Y_1, \ Y_2, \ \dotso,\ Y_n denote a random sample from a Weibull dist. with known m and unknown \alpha. Show that \sum^n_{i=1} Y^m_{i} is sufficient for \alpha.

    Attempt:

    \frac{1}{\alpha} m y^{m-1} \exp \left( -\frac{y^m}{\alpha} \right) is my function.

    so the likelihood is:

    L(y_1, \ y_2, \ \dotso, \ y_n|\alpha) = f(y_1, \ y_2, \ \dotso, \ y_n|\alpha) = f(y_1|\alpha) \cdot f(y_2|\alpha) \cdot \dotso \cdot f(y_n|\alpha)

    =\prod^n_{i=1} \frac{1}{\alpha} m y^{m-1} \exp \left( -\frac{y^m}{\alpha} \right)

    =\bigg{[}\frac{1}{\alpha} m y_1^{m-1} \exp \left( -\frac{y_1^m}{\alpha} \right)\bigg{]} \times \bigg{[}\frac{1}{\alpha} m y_2^{m-1} \exp \left( -\frac{y_2^m}{\alpha} \right)\bigg{]} \times \dotso \times \bigg{[}\frac{1}{\alpha} m y_n^{m-1} \exp \left( -\frac{y_n^m}{\alpha} \right)\bigg{]}

    =\bigg{(}\frac{1}{\alpha} m\bigg{)}^n \prod^n_{i=1} y_i^{m-1} \exp \left( -\sum^n_{i=1}\frac{y_i^m}{\alpha} \right)

    at which point I get stuck, I know I'm supposed to factor it out and have a second function that isn't dependent on \alpha, but I'm not quite sure on how to separate it out.
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  2. #2
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    Quote Originally Posted by lllll View Post
    Let Y_1, \ Y_2, \ \dotso,\ Y_n denote a random sample from a Weibull dist. with known m and unknown \alpha. Show that \sum^n_{i=1} Y^m_{i} is sufficient for \alpha.

    Attempt:

    \frac{1}{\alpha} m y^{m-1} \exp \left( -\frac{y^m}{\alpha} \right) is my function.

    so the likelihood is:

    L(y_1, \ y_2, \ \dotso, \ y_n|\alpha) = f(y_1, \ y_2, \ \dotso, \ y_n|\alpha) = f(y_1|\alpha) \cdot f(y_2|\alpha) \cdot \dotso \cdot f(y_n|\alpha)

    =\prod^n_{i=1} \frac{1}{\alpha} m y_{\color{red}i}^{m-1} \exp \left( -\frac{y_{\color{red}i}^m}{\alpha} \right) Mr F says: Note the small errata.

    =\bigg{[}\frac{1}{\alpha} m y_1^{m-1} \exp \left( -\frac{y_1^m}{\alpha} \right)\bigg{]} \times \bigg{[}\frac{1}{\alpha} m y_2^{m-1} \exp \left( -\frac{y_2^m}{\alpha} \right)\bigg{]} \times \dotso \times \bigg{[}\frac{1}{\alpha} m y_n^{m-1} \exp \left( -\frac{y_n^m}{\alpha} \right)\bigg{]} Mr F says: I'll pick it up from here because the line below is wrong and not needed.

    =\bigg{(}\frac{1}{\alpha} m\bigg{)}^n \prod^n_{i=1} y_i^{m-1} \exp \left( -\sum^n_{i=1}\frac{y_i^m}{\alpha} \right)

    at which point I get stuck, I know I'm supposed to factor it out and have a second function that isn't dependent on \alpha, but I'm not quite sure on how to separate it out.
    = \frac{m^n}{\alpha^n} \, (y_1 \, y_2 \, .... \, y_n)^{m-1} \exp \left(-\frac{1}{\alpha} \sum_{i=1}^n y_i^m\right)


    = m^n \, (y_1 \, y_2 \, .... \, y_n)^{m-1} \, \frac{\exp \left(-\frac{1}{\alpha} u\right)}{\alpha^n}

    where U = \sum_{i=1}^n Y_i^m


    = \left(\frac{\exp \left(-\frac{u}{\alpha} \right)}{\alpha^n}\right) \cdot \left(m^n \, (y_1 \, y_2 \, .... \, y_n)^{m-1} \right)


    = g\left(\alpha, \, u \right) \cdot h(y_1 \, y_2 \, .... \, y_n).
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