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Theoretical Basis

To estimate the smoothing variances, ${\sigma_x}^2, {\sigma_y}^2, {\sigma_z}^2$, the expectation of the partial derivative is used. That is,

\begin{displaymath}
E\left\{ \left( \frac{\partial S}{\partial x} \right)^2 \right\} =
\frac{1}{2 {\sigma_x}^2}
\end{displaymath} (30)

where the quantity $S$ is a scaled version of $F_S$ such that $E\{ S^2 \} = 1$.

This follows easily from equations 28 and 14 by setting $\ensuremath{\mathbf{x_1}}= \ensuremath{\mathbf{x_2}}= \ensuremath{\mathbf{x}}$. That is:

$\displaystyle E \{ ( F_{Dx}(\ensuremath{\mathbf{x}}) \, )^2 \}$ $\textstyle =$ $\displaystyle E \{ ( F_{S}(\ensuremath{\mathbf{x}}) \, )^2 \} \, \left( \frac{1}{2{\sigma_x}^2} \right)$ (31)
$\displaystyle E \{ ( F_{S}(\ensuremath{\mathbf{x}}) \, )^2\}$ $\textstyle =$ $\displaystyle \frac{1}{(4 \pi)^\frac{3}{2} \sigma_x \sigma_y \sigma_z}.$ (32)

Therefore, by setting $S(\ensuremath{\mathbf{x}}) = k F_S(\ensuremath{\mathbf{x}})$ with $k = (4 \pi)^\frac{3}{4} (\sigma_x \sigma_y \sigma_z)^\frac{1}{2}$ gives $E\{ S^2(\ensuremath{\mathbf{x}}) \} = 1$ and $\frac{\partial S(\ensuremath{\mathbf{x}})}{\partial x} = k F_{Dx}(\ensuremath{\mathbf{x}})$, which leads to equation 30.



Mark Jenkinson 2001-11-07