Distribution of $\hat{d}({\bf X}_{opt})$

Special case: $Y = -X^2 + \epsilon, \epsilon \sim N(0, \sigma^2)$
d(Y) is of type (-1, 0, 1, 1, 1), therefore ${\bf X}_{opt} = 0$ is known.

Fitting quadratic model
$\hat{Y} = \hat{\beta_0} + \hat{\beta_1} X + \hat{\beta_3} X^2$ results in $d(\hat{Y}({\bf X}_{opt}) = \hat{d}(0) = d(\hat{\beta_0}).$

It is known, that $\hat{\beta_i} \sim \beta_i + t_{n - B} \sqrt{\hat{var}(\hat{\beta_i})}$

with $\hat{var}(\hat{\beta_i}) = \hat{\sigma}^2 ((X'X)^{-1})_{ii}$ and $\hat{\sigma}^2 = \frac{1}{n - B}\sum_{i=1}^n \hat{\epsilon_i}$ . Now:

$\begin{eqnarray*}E( \hat{d}(0)) &= & P(\hat{\beta_0} \in [-1, 0)) E(\hat{\beta... ...\vert \hat{\beta_0} > 0)\\ &= & 1- E(\vert\hat{\beta_0}\vert) \end{eqnarray*}$

Written down for the density function: $f_{\hat{d}(0)} = 1 - \vert t_{n - B}\vert$
In ${\bf X}_{opt}$ the desirability is systematically underestimated.

Dipl.-Stat. Detlef Steuer
1999-10-04