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Let ri be components of a residual vector
and consider perturbation components
.We study the properties of the new residual
| ![\begin{displaymath}
\tilde {\bold r} \quad =\quad\bold r +\alpha \Delta \bold r\end{displaymath}](img7.gif) |
(1) |
where
is a scaling factor.
Consider the implications of
| ![\begin{eqnarray}
0 &=& \mbox{median}_i \left( { \tilde r_i \over \Delta r_i} \ri...
...t)
= \mbox{median}_i \left( { r_i\over \Delta r_i}\right) +\alpha\end{eqnarray}](img9.gif) |
(2) |
| (3) |
Thus choosing
causes
the median of
to vanish
which means
has as many positive terms
as negative ones.
Consequently, the number of polarity agreements
in the component pairs
of the new residual
and the regressor
equals their polarity disagreements.
In other words,
| ![\begin{displaymath}
0 \quad =\quad\sum_i \mbox{sgn}(\tilde r_i)\;\mbox{sgn}(\Delta r_i)
\quad =\quad\mbox{agreements} -
\mbox{disagreements}\end{displaymath}](img15.gif) |
(4) |
In what sense have we created the smallest new residual
?
It is smallest in the sense that if we change
by adding or subtracting some
of the perturbation
,the count of growing components less the decreasing components
of
is
| ![\begin{displaymath}
\sum_i \mbox{sgn}(\tilde r_i+\epsilon \Delta r_i)\;\mbox{sgn}(\epsilon\Delta r_i)
\quad \ge \quad 0\end{displaymath}](img19.gif) |
(5) |
The best residual is one in which any perturbation
serves only to increase the polarity agreements with the regressor.
We chose
to reduce as many components
of
as we could.
Next: MEDIANS IN AUTOREGRESSION
Up: Claerbout: Medians in regression
Previous: INTRODUCTION
Stanford Exploration Project
11/12/1997