# [OPE-L:5304] Some Simple Dynamics

andrew kliman (Andrew_Kliman@msn.com)
Mon, 23 Jun 1997 12:27:32 -0700 (PDT)

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A reply to part of Ajit's ope-l 5301.

In ope-l 5289, he wrote: "As far as your 'difference equation' is concerned.
I must admit, I don't understand what is going on there."

So I explained it to him in ope-l 5290.

His response (ope-l 5301)?: "I have never understood your mathematics. By my
standard[s] they are simply far out!"

Ignorance is bliss. Ignorance is strength. To encourage Ajit to raise his
standards a bit, I will concentrate on the following points I made in ope-l
5290:

"Your mistake was to assume wrongly that, in the long run, stationary (growth)
equilibria and breakdown are the only possibilities. You've made that mistake
before. For instance, you confuse average prices with stationary equilibrium
prices, though the two will be equal in general only if the system in question
is a static equilibrium one. You also seem to argue at times that the mere
persistence of a system in the long-run means that its behavior can be
approximated by means of static equilibrium equations."

Assume that the price of some commodity at time t is

P(t) = 0.85 + 0.2*X(t) (1)

where X is a variable that behaves as follows:

X(t+1) = 4*X(t)*[1 - X(t)]. (2)

Assume that the initial value of X, X(0), is greater than zero and less than
1, and not equal to 0.5. Then X(t) will also be greater than zero and less
than 1 forever after (I won't prove this, but it is well known). Under this
condition, there is no "breakdown" of P -- as time proceeds, it does not
"explode" to infinity, nor does it become negative. Indeed its movement is
strictly bounded, with its minimum being Pmin = 0.85 and its maximum being
Pmax = 1.05. This is verified by plugging the minimum and maximum values of X
into equation (1).

What is the equilibrium value of P, Pe? "Equilibrium" doesn't mean market
clearing here. Rather Pe is that value such that, if P(t) = Pe, then also
P(t+1) = Pe, so that P(t) = P(t+1). Clearly, (1) implies that P(t) = P(t+1)
only if X(t) = X(t+1). So we can write

Pe = 0.85 + 0.2*Xe (1')

where Xe is that value of X(t) such that X(t) = X(t+1). To find Xe, we set
X(t) = X(t+1) in equation (2), and find that Xe = 0.75. (If X(t) = 0.75,
then X(t+1) = 0.75, and so X(t+2) = 0.75, etc.) Hence, substituting 0.75 for
Xe in (1'), we find that Pe = 1.

Were it the case that X(0) = 0.75, then all X(t) would = 0.75, and all P(t)
would equal 1. So one possible "realization" of this system is a stationary
equilibrium. The only other possibility Ajit seems to recognize is
"breakdown." But we already know that, given 0 < X(0) < 1, "breakdown" of P
is impossible. Indeed, since we now know that Pe = 1, we know that the P
never falls below its equilibrium value by as much as 15% and it never exceeds
its equilibrium value by a much as 5%.

What happens if X(0) is not equal to 0.75? Answer: The movements in X and P
are chaotic. This means that (a) the system is strictly deterministic; the
whole time series of values for depends on X(0) and only on X(0). Thus, if
X(0) and equations (1) and (2) are *known*, then the movement of P can be
predicted *exactly*. However, (b), the system is *unpredictable*. Even if
one knows the equations determining the movement of the system are (1) and
(2), and even if one knows the value of X at time t to any *possible* degree
of accuracy, then the values of X and P at later dates cannot be predicted
accurately. The longer the elapse of time, the less is the ability to
predict. This is because (c) X and P do NOT converge to their stationary
equilbrium values, nor do they move in a periodic cyclical fashion.

So it is simply a mistake to think that, in the long run, stationary
equilibrium (or stable growth equilibrium) and breakdown are the only
possibilities. Among the other possibilities in deterministic systems are
limit cycles and chaos. And then there are (perhaps) nondeterministic
systems, which can, like chaotic systems, lead to bounded instability (the
system neither converges to anything, nor repeats itself, nor explodes).

It is also a mistake to confuse average prices with stationary equilibrium
prices. In the system (1)-(2), the equilibrium price is Pe = 1. What is the
*average* price? This depends on the time period for which one takes the
average, and especially X(0). We know that Xe = 0.75. And we know that if
X(0) = Xe, then the price will always equal 1. So what if X(0) is very close
to Xe, say 0.7499? It may seem "reasonable" that the average price will be
very close to 1. But it isn't TRUE. The average price over the first 1000
periods is 0.9462. Are, however, most of the values for P close to 1, say
between 0.97 and 1.03? No, fewer than 20 0.000000all in that range. How about
between 0.95 and 1.05? No, nearly 53 0.000000all outside than range. The prices
don't even cluster around the average price. More than 430f the time, P is
either less than 0.87 or more than 1.03.

So, this simple example shows conclusively that the mere persistence of a
system in the long-run, i.e., non-explosive behavior, does not mean that its
behavior can be approximated by means of static equilibrium equations.
Therefore, if one wants to claim that the system P = PA(1+r) + wl or something
like it gives something close to the average relative prices and even the
average simultaneist "rate of profit" when in fact things happen even if all
profit rates aren't equal, one needs something a good deal better than the
persistence of capitalism and the boundedness of relative price and profit
rate variations to justify that claim. One pretty much needs to prove that
the "out-of-equilibrium" behaviors actually lead the system to converge to
this static system. Investigations of this question to date indicate that
there is no a priori reason to believe that that they do.

The foregoing is totally independent of the challenge I have put before Ajit
to show that his price theory isn't self-contradictory. That challenge
concerns the inconsistent way in which he defines "input prices."

Andrew Kliman