Saturday, September 27, 2014

The economic combinatorial problem


I mentioned in this post that an economy is a combinatorial problem (along with some hints at entropy); let me sketch out the mathematical side of the argument.

I connected $\log M$ (where M is the currency supply) with $\beta = 1 / T$ in the partition function awhile ago (setting $k_{B} = 1$). If we take the thermodynamic definition of temperature:

$$
\frac{1}{T} = \frac{dS}{dE}
$$

as an analogy (where $S$ is entropy and $E$ is energy), we can write (putting in a constant $c_{0}$ that corresponds to the constant $\gamma M_{0}$ in the information transfer model):

$$
\log M/c_{0} = \frac{dS}{dN}
$$

where we've used the correspondence of the demand (NGDP, or $N$ -- i.e. aggregate demand AD) with the energy of the system. We don't know what the economic entropy is at this point. However, if we take (using the solution shown here):

$$
N = M^{1/k}
$$

Then we can write down

$$
k \log N/c_{0} = \frac{dS}{dN}
$$

So that, integrating both sides,

$$
S = k \; (N/c_{0}) (\log N/c_{0} - 1) + C
$$

Which, via Stirling's approximation, gives us (dropping the integration constant $C$)

$$
S \simeq k \log \; (N/c_{0})!
$$

If we compare this equation with Boltzmann's grave:

$$
S = k \log W
$$

We can identify $(N/c_{0})!$ with the number of microstates in the economy. The factorial $N!$ counts the number of permutations of $N$ objects and we can see that $c_{0}$ adjusts for the distinguishability of  given permutations -- all the permutations where dollars are moved around in the same company or industry are likely indistinguishable. This could lend itself to an interpretation of the constant $\gamma$ mentioned above: large economies are diverse and likely have the same relative size manufacturing sectors and service sectors, for example -- once you set the scale of the money supply $M_{0}$, the relative industry sizes (approximately the same in advanced economies) are set by $\gamma$.

This picture provides the analogy that a larger economy ($N$) has larger entropy (economic growth produces entropy) and lower temperature ($1/\log M$).

5 comments:

  1. Jason, last paragraph:

    "(approximately the same advanced economies)"

    Do you need to insert an "in" in there?:

    "(approximately the same in advanced economies)"

    I'm going to stew on this post a bit.

    I'm a very occasional skier. I think I've been about 8 times in my life, over the course of nearly 20 years. I can work up to making it down an intermediate slope w/o too much panic. Last time I recall looking at the two unaccompanied toddler aged girls in the ski-lift in front of me and thinking to myself "Someday I hope to be as good as those two."

    I feel the same way about your intuition regarding this stuff. Someday I hope to have at least a respectable percentage of what you have myself.

    ReplyDelete
    Replies
    1. Cheers, Tom.

      And yes, there should be an "in" in there.

      Delete
  2. Given that you're slowly introducing S, T, W etc. maybe there's a way of recasting it all akin to the fundamental thermodynamics relation. I don't want this to come across as trying to shoehorn things into thermo. looking equations like every engineering economics crossover has tried to do for 50 years - but maybe the analogies are strong given the IT framework is fairly fundamental?

    ReplyDelete
    Replies
    1. I am aware of many attempts to recast economics as thermodynamics and in particular Paul Samuelson's acerbic quotes against it (you can search this blog for Samuelson and pull them up) ... And you are right: the IT framework could be used to build equilibrium thermodynamics (so is more fundamental, and that's why the analogies can work so well sometimes). However, the framework can also be used to build other theories that aren't thermodynamics ... so you have to be careful not to go too far.

      There might be a way to introduce the fundamental relation ... however that is equilibrium thermodynamics. Equilibrium thermodynamics would apply e.g. to the post here (where I make use of a "temperature" log M), but it wouldn't always apply in a very important sense.

      The information transfer framework is like a generalized thermodynamics where you don't really have/need a concept of temperature or equilibrium (it is basically more general than stat mech, but the trade off is you don't get as complicated dynamics). One of the interesting things that may come out of the framework is a rigorous way to discover those concepts -- but that is speculative.

      In a sense, the thermodynamic picture where I resort to the partition function, make note of that diagram with several countries on it and use log M as a temperature is "equilibrium" economics, but most of what I've been doing with "the" information transfer model doesn't need/have a temperature or equilibrium.

      Another way to say this is that sometimes using the partition function (log M as a temperature) will get you the wrong answer -- it will give you the equilibrium answer.

      I have a little more on this in this post:

      http://informationtransfereconomics.blogspot.com/2014/07/information-transfer-is-state-of-mind.html

      Delete
  3. As a fun side note: Using the best fit value of $c_{0}$ = 0.478 billion dollars and the most recent quarter of GDP data, we get 36192! which is approximately 10^149270 states ... roughly equivalent to a system with 496,000 (distinguishable) on/off states.

    That means the US economy is roughly as complex as an 2D Ising model with ~700 spin states on a side ...

    http://en.wikipedia.org/wiki/Ising_model

    ReplyDelete

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Also, try to avoid the use of dollar signs as they interfere with my setup of mathjax. I left it set up that way because I think this is funny for an economics blog. You can use € or £ instead.