[Hans Blom, 951003b]
(Bill Powers (951003.0500 MDT))
I appreciate your efforts to clarify my thinking about the
strategy of your model, but you're still giving me too much
credit for familiarity with your methods. What I'm trying to do
is eliminate all the statistical aspects of the strategy and
just see what is going on in the absence of noise.
You're loosing me here. Modelling is curve fitting. How can I
describe curve fitting without at least some statistics?
This is useful:
delta_K := delta_K + constant1 * estimate (dK/dt) * (x-y)
delta_D := delta_D + constant2 * estimate (dD/dt) * (x-y)
The constants, I presume, are to prevent computational
oscillations -- to keep the corrections from undershooting and
overshooting.
Yes, the equations implement a kind of hill-climbing toward the
"optimal" values of delta_K and delta_D. Not knowing where the
mountain top is and not knowing exactly how large your steps will
turn out to be (after all, you only have, maybe quite noisy,
_estimates_ of dK/dt and dD/dt), the most reliable strategy is to go
slow: pick small constants. But that may be too slow in practice,
especially if delta_K and delta_D keep varying -- the mountain top
keeps shifting around. So you'll have to find some practical compro-
mise.
In general, x will be some function f, with d being added to x
(because if d is an unknown waveform, it doesn't matter whether
it is passed through some function before having its effect on
x):
x = f(u) + d
In the second step of your method, it seems to me that it's
necessary to compute
-1 u = f (r - d)
What this step entails is that you want to pick that value of u that
brings the predicted x to r, i.e. solve for
r = x = f(u) + d
There are a variety of ways to do this. One way is by simulation:
vary u from a minimum to a maximum value and use the value of u that
brings f(u) + d as closely to r as you can discover. Binary search
over the range of u may be faster. Varying u, not from some min to
some max but in the vicinity of its previous value, is another way.
These methods work with any (known) function f.
If f is invertible, finding u is much simpler:
f (u) = r - d, i.e.
u = f^-1 (r - d)
which I presume your equation above ought to have been.
Hope this helps,
Hans