min order stats and unbiased estimator.

**lllll** · November 15th 2008, 10:55 PM

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

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.

**mr fantastic** · November 16th 2008, 01:13 AM

Originally Posted by lllll

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

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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