Components of variance

[Martin Taylor 950504 16:30]

I'm not going to quote, but this is in further response to Bill Powers
(950503.0530 MDT)

you keep
returning to the idea that control is somehow made possible by the
_component_ of the perceptual signal that contains information (or
passes information, if you like) about the effect of the disturbance.

Here is an attempt to induce a different viewpoint, that may help in
understanding why I object to being characterised in this way.

···

------------------
Let us forget about control systems for the moment, and think only about
an arbitrary signal S carried by an observable wire.

S varies over time, and an outside observer measures it (S can't measure
itself, can it?). The outside observer determines that S has a variance
Vs.

The outside observer notes that the wire that carries S is connected into
a black box, and there are two other wires entering that black box. Call
them X and Y. The observer measures the variances of the signals on these
wires and finds them to be Vx and Vy. Is there any connection among
these three wires in the black box? How can we tell? We can't, with
the measurements available so far.

So our observer starts to look at the relationships among the three wires,
and finds that always, possibly apart from small time-shifts that he ignores,
S = X-Y. (Figure 1)
           ______
X----->----| |
           > - |---->----S
Y----->----|_____|

Interesting. But then our observer notices that S < X and S < Y.
More interesting, because if X and Y are independent and S = X-Y, then
on average S^2 should = X^2 + Y^2, and it doesn't. Vs < Vx + Vy. In
fact Vs < Vx < Vy (at least most of the time).

So X and Y cannot be independent. Not only does our black box look as if
it is a subtractor, but also X and Y look as if they are influenced by
some common factor, even though there is no connection visible to the
observer between where X seems to come from and where Y seems to come from.

There's nothing that can account for X and Y being related if the sources
that are driving them are at the ends of their wires away from the black box.
So, perhaps the black box is not a subtractor after all, but is an adder
instead. The finding S = X-Y could equally well be represented X = S+Y,
with the interpretation that the output of the black box is X, not S
(Figure 2).
           _____
S----->---| |
          > + |---->---X
Y----->---|_____|

But even though S is much smaller than X or Y, our observer still finds
that Vx != Vs + Vy. Now S and Y are apparently related, even though the
observer can find nowhere that the other ends of the S and Y wires have
any interconnections. Mystery! X and Y are closely related, S and Y are
slightly related, but there is no connection of the other end of either
the X wire or the S wire to the other end of the Y wire. How can this be?

But wait--our observer sees that there is some kind of complicated path
through which the other end of the S wire DOES connect to the other end
of the _X_ wire. Maybe the equation should be Y = X-S, and there is some
kind of a common influence (Z) affecting both X and S (Figure 3).
                       ____
(other)--->---X--->---| |
      / | |
Z-->-- | - |---->---Y
      \ | |
(other)-------S--->---|____|

This seems fine, especially since the careful observer can see a wire that
does seem to go to both S and X, and that wire could readily carry the Z
signal.

But now our observer tries cutting the Y wire. Then what happens? Vy remains
unchanged, but Vx and Vs go essentially to zero, as does the variance of the
signal on the black box end of the cut Y wire.

Oh boy! Y was an input after all, and what happened to signal Y affected
what happened at X and S. So figure 1 or 2 could be right, but 3 could
not. And yet 1 or 2 both seems wrong, in that either S or X has to be
an output from the black box and yet their ends are interconnected, albeit
by some complicated path.

So let's look at the configurations of Figures 1 or 2 if S and X are connected
together.

1a

-----<---|stuff|--<-------

          _____ |

X----->----| | |
           > - |---->----S
Y----->----|_____|

2a
-----<---|stuff|--<-----

         _____ |

S----->---| | |
          > + |---->---X
Y----->---|_____|

Both of these configurations are consistent with what the observer now knows.
Y is independently driven, and somehow X and S are related. But why is it
that X and Y are related, or X and S? It must have to do with "stuff" that
is in the link between X and S. If "stuff" is an attenuator, then Y must
drive X, which has a small effect on S and the result added to Y. But since
X is less than Y as a rule, "stuff" must also invert the sign of its signal.
And since S is only slightly related to X, S must also have input from a
more powerful other place that provides mostly noise. That must be the Z
line from figure 3. This looks fine.

Until our observer cuts the wire between X and the black box. Now what
happens? All of a sudden, S = Y regardless of anything that might be
happening on the Z wire, and X starts going wild, but in a way that
can clearly be related to variations in S even though it doesn't replicate
those variations. Figure 2a doesn't work.

But Figure 1a works, if "stuff" has some of the functions of an amplifier.
X must be an input to the black box, and S an output that feeds X through
"stuff" that amplifies it. So if Y changes, that change is amplified in X
and largely nullified through the subtraction of X from Y.

Now, if our observer cuts the wire connecting X and S, what happens? Again
S takes on the value of Y, but now X goes to zero or varies in some way
unrelated to S or Y. Figure 1a works again, if any variation in X is
assumed to have some connection with the Z wire. Our black box must be a
subtractor, as the observer first thought, and the information that
determines the variations in the values of X comes from Y through S.

But there is no "component" of S that separately observes Y and passes
the result to "stuff" for passage to X. And our observer doesn't postulate
one.

----------------

I don't know if this way of talking about it helps at all. I'm just casting
around for different ways of saying the same thing.

Martin