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Math Help - min order stats and unbiased estimator.

  1. #1
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    min order stats and unbiased estimator.

    f(y) =<br />
\begin{cases}<br />
3\beta^3y^{-4} & \mbox{for}~\beta \leq y \\<br />
0 & \mbox{otherwise}<br />
\end{cases}

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

    derive the bias estimator for \hat{\beta}

    the minimum density function is

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

    so if I integrate f(y)

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

    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 \beta and \infty. I'm thinking there has to be some change of variable performed with respect to \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
    f(y) =<br />
\begin{cases}<br />
3\beta^3y^{-4} & \mbox{for}~\beta \leq y \\<br />
0 & \mbox{otherwise}<br />
\end{cases}

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

    derive the bias estimator for \hat{\beta}

    the minimum density function is

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

    so if I integrate f(y)

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

    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 \beta and \infty. I'm thinking there has to be some change of variable performed with respect to \left(\frac{\beta}{y}\right)^3, but I can't seem to figure it out.
    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 g(u) = n [1-F(u)]^{n-1} f(u) = n u^{-3n - 1} \beta^{-3n}.


    Therefore 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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