[From Bill Powers (951205.1245 MST)]
Bruce Abbott (951205.1320 EST) --
I think I do get the picture, but I need to communicate it better.
You do get the picture, but you need to get it better.
I am maintaining that if the reinforcer did not tend to reduce
error, it would not serve as a reinforcer. Within the control
system, a constant disturbance would tend to increase the error;
the system will reach an equilibrium in which the reduction in
error supplied by the reinforcer will just balance the increase in
error supplied by the disturbance, so one will not _observe_ any
further reduction in error once this state is reached;
nevertheless, the reinforcer continues to supply an error-reducing
effect on the controlled variable each time it occurs as a result
of control-system action. If it ceased to provide this service, it
would cease to function as a reinforcer.
Your premise that "a constant disturbance would tend to increase the
error" is false. A disturbance may act either to increase or decrease
the error; in either case it is a disturbance, and in either case it is
resisted by an opposing change in the action of the control system. Let
me illustrate using an algebraic model, which gives a correct steady-
state picture of the behavior of a stable system. I'll set up the
equations and then show how an increasing positive disturbance reduces
the error and the output.
qc = controlled quantity (in units of rate)
p = perceptual signal, assumed numerical equal to qc
r = reference signal
e = error signal
qo = output quantity (rate of action rate in environment)
d = disturbing quantity, in units of effect on qc
1/m = environmental function (ratio = m)
g = output gain
Basic equations:
(a) qc = qo/m + d
(b) p = qc
(c) e = r - p
(d) qo = g*e
Solving for controlled quantity by substition:
qc = g*(r - qc)/m + d, or
(1) qc = [(g/m)*r + d]/(1+g/m)
Solving for output quantity by sustitution:
qo = g*(r - qo/m - d), or
(2) qo = g*(r-d)/(1 + g/m)
Solving for error signal by substitution:
e = r - g*e/m - d, or
(3) e = (r - d)/(1 + g/m)
First, look at equation 1. We can extract a common factor of (g/m)/(1 +
g/m) to get
g/m
(1a) qc = --------[r + d/(1 + g/m)]
1 + g/m
The factor g/m is the negative feedback loop gain, which is the negative
of
(the output sensitivity g times the environmental function 1/m times the
input function 1 times the comparison function -1 -- because the
perceptual signal is subtracted).
The output quantity can be expressed in the same form by extracting
(g/m)/(1+g/m):
g/m
(2) qo = ------- m*(r-d)
1 + g/m
Let's make a table now. Assume the output sensitivity g = 50 and the
ratio m = 5, with the reference signal r = 100. We can vary the
disturbance and see the effect on the controlled quantity and output
quantity (that is, the reinforcement rate and the behavior rate). The
loop gain g/m is 10.
The formulas now are
qc = 90.9 + d/11 and
qo = 4.545*(100 - d)
d qc qo error, r-p
0 90.9 454.5 9.1
3 91.2 440.9 8.8
5 91.4 431.8 8.6
10 91.8 409.1 8.2
30 93.6 318.2 6.4
50 95.4 227.3 4.6
100 100.0 0.0 0.0
···
------------------------------------------------------
300 118.2 -791.0 -18.2
Notice that the output just goes to zero when the disturbance alone is
making qc exactly equal to the reference level.
-------------------------
Compare these results with your reasoning. A disturbance of +10 units
causes the controlled quantity to increase from 90.9 to 91.8 units, an
increase of +0.9 units. The error is reduced from 9.1 units to 8.2
units. This reduces the behavior rate from 545.5 units to 409.1 units.
Such a 25% drop should be easily observable.
So _increasing_ the reinforcement rate causes a _decrease_ in the
behavior rate.
Note 1: these equations apply only when it has been shown that a control
system does exist. This demonstration would have to include showing that
there is an actual change in behavior rate, not an apparent one as you
demonstrated by taking into account the collection time.
Note 2: If you plug in higher ratios, you will see the loop gain
becoming smaller and the error becoming larger. The same relationships
among disturbance, qc, and qo, however, will still be seen. If the
output sensitivity is 50, the loop gain will drop to 1 when the ratio is
50, and only half of the effect of the error will be resisted. So for
high ratios we are not looking at good control.
--------------------------------------------------------------------
Martin Taylor 951205 14:00 --
There is no indication that it has been displaced until
you reach out for something.
I take it this means you've never tried it. I'd suggest you get a
prism and try it. There's no way you'd mistake the prism world for
a normal world.
If you used a high-quality achromatic prism, there should be no
difference at all (except for the position of your nose, which isn't
normally seen anyway). The effect will be just as though the room has
actually rotated. If it's not, then your prisms are introducing effects
other than image displacement, and all conclusions are null and void.
You should ask yourself WHY you'd never mistake the prism world for the
normal one. All the answers have to do with effect of movement that are
abnormal -- not with the appearance of the world.
Anyway, the same comment applies to left-right or up-down inverting
spectacles, at least when worn by humans.
Inverting spectacles produce a scene that could not, in a gravity field,
be mistaken for normal. They do not displace the field, they rotate it
about the optic axis. An astronaut suspended in the center of a
spherical room, however, would not notice anything abnormal unless he
moved.
You said,
Chicks don't seem to learn, which suggests that the related
control system hierarchy is not easily reorganized, as compared to
ours.
and I said
That's a pretty wild generalization from the limited evidence.
And you now object,
The fact
that chickens (adult) can be operantly conditioned is a pretty good
indication that grown birds can learn.
Hey--just a minute. How do you reconcile this statement with the
one quoted above:
>>The only interesting point about the chicks is that they seem
>>unable to realign the kinesthetic and visual spatial maps.
Either they can or they can't. You can't have it both ways in the
same message.
I don't seen anything to reconcile. Adult chickens can certainly learn
many things. The fact that a chick can't learn to realign the
kinesthetic and visual maps doesn't show that chicks can't learn. It
just shows that they can't learn to do that particular thing. Chicks
can't learn algebra, either, but that doesn't prove that they can't
learn. You said that because they can't learn, their hierarchies must
not be easily reorganized, which is just what I called it: a wild
generalization.
----------------------------------------------------------------------
Best,
Bill P.