# Thread: min order stats and unbiased estimator.

1. ## min order stats and unbiased estimator.

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

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.

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