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Thread: min order stats and unbiased estimator.

  1. #1
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    min order stats and unbiased estimator.

    $\displaystyle f(y) =
    \begin{cases}
    3\beta^3y^{-4} & \mbox{for}~\beta \leq y \\
    0 & \mbox{otherwise}
    \end{cases}$

    consider the estimator $\displaystyle \hat{\beta}=\min(Y_1,Y_2,\dotso , Y_n)$

    derive the bias estimator for $\displaystyle \hat{\beta}$

    the minimum density function is

    $\displaystyle g_{(1)}(y) = n[1-F(y)]^{n-1}f(y)$

    so if I integrate $\displaystyle f(y)$

    $\displaystyle F(y)= \int 3\beta^3y^{-4} dy = -\beta^3y^{-3} = -\left(\frac{\beta}{y}\right)^3$

    $\displaystyle E[\hat{\beta}]=\int_\beta^\infty n\bigg{[}1+\left(\frac{\beta}{y}\right)^3\bigg{]}^{n-1}3\beta^3y^{-4} \cdot y \ \ dy = 3n\int_\beta^\infty \bigg{[}1+\left(\frac{\beta}{y}\right)^3\bigg{]}^{n-1}\cdot\left(\frac{\beta}{y}\right)^3\ \ dy$

    at this point I don't know how to proceed, although the function is looking a lot like a beta function but my bounds aren't 0 and 1, but $\displaystyle \beta$ and $\displaystyle \infty$. I'm thinking there has to be some change of variable performed with respect to $\displaystyle \left(\frac{\beta}{y}\right)^3$, but I can't seem to figure it out.
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  2. #2
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    Quote Originally Posted by lllll View Post
    $\displaystyle f(y) =
    \begin{cases}
    3\beta^3y^{-4} & \mbox{for}~\beta \leq y \\
    0 & \mbox{otherwise}
    \end{cases}$

    consider the estimator $\displaystyle \hat{\beta}=\min(Y_1,Y_2,\dotso , Y_n)$

    derive the bias estimator for $\displaystyle \hat{\beta}$

    the minimum density function is

    $\displaystyle g_{(1)}(y) = n[1-F(y)]^{n-1}f(y)$

    so if I integrate $\displaystyle f(y)$

    $\displaystyle F(y)= \int 3\beta^3y^{-4} dy = -\beta^3y^{-3} = -\left(\frac{\beta}{y}\right)^3$

    $\displaystyle E[\hat{\beta}]=\int_\beta^\infty n\bigg{[}1+\left(\frac{\beta}{y}\right)^3\bigg{]}^{n-1}3\beta^3y^{-4} \cdot y \ \ dy = 3n\int_\beta^\infty \bigg{[}1+\left(\frac{\beta}{y}\right)^3\bigg{]}^{n-1}\cdot\left(\frac{\beta}{y}\right)^3\ \ dy$

    at this point I don't know how to proceed, although the function is looking a lot like a beta function but my bounds aren't 0 and 1, but $\displaystyle \beta$ and $\displaystyle \infty$. I'm thinking there has to be some change of variable performed with respect to $\displaystyle \left(\frac{\beta}{y}\right)^3$, but I can't seem to figure it out.
    $\displaystyle F(u) = \int_{\beta}^{u} 3 \beta^3 y^{-4} \, dy = \beta^3 \, \left[ -y^{-3} \right]_{\beta}^{u} = 1 - \frac{\beta^3}{u^3}$.


    Therefore $\displaystyle g(u) = n [1-F(u)]^{n-1} f(u) = n u^{-3n - 1} \beta^{-3n}$.


    Therefore $\displaystyle E(\hat{\beta}) = \int_{\beta}^{+\infty} u \, g(u) \, du = 3n \, \beta^{3n} \int_{\beta}^{+\infty} u^{-3n} \, du = \frac{3n}{3n-1} \, \beta$.
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