According to Bennett McCallum [1], the quantity theory of money (QTM) is the macroeconomic observation that the economy obeys long run neutrality of money (it's not just

*MV = PY*). This Implies supply and demand functions will be homogeneous of degree zero, i.e. ratios of $D$ to $S$*such that if $D \rightarrow \alpha D$ and $S \rightarrow \alpha S$ then $g(D,S) \rightarrow \alpha^{0} g(D,S) = g(D,S)$. The simplest differential equation consistent with this observation is*\text{(1) } \frac{dD}{dS} = \frac{1}{\kappa}\; \frac{D}{S}

$$

We can identify the RHS with the price level $P$ (~~ratio of NGDP to the money supply~~ the exchange rate for the marginal unit of AD for the marginal unit of AS should be proportional to the price level). The solution (for varying D and S) is $D \sim S^{1/\kappa}$, or

\text{(2) } P = \frac{1}{\kappa} S^{1/\kappa - 1}

$$

If we take

*D = NGDP*and*S = MB*, this equation does well over segments of the price level, with different values of*κ*(which I'll just call the IT index for now):
If we let $\kappa$ become $\log S / \log D$ [2] then with $D, S \gg 1$ and $\alpha$ small (i.e. small changes in NGDP or MB), there is still approximate short run homogeneity of degree zero. Additionally with

*S, D → ∞*with*S/D*finite, we have "long run" homogeneity of degree zero in a growing economy [3]. And it turns out the best fit to the local values of*κ*is approximately*log MB/log NGDP*(measured in billions of dollars):
But why should we be satisfied with (1)? In Part II, we'll motivate the equation via information theory. In this Part I, we'll resort to one of my favorite topics from physics:

*effective field theory*.
A general homogeneous differential equation (of first order) is given by

\frac{dD}{dS} = g(D/S)

$$

Where $g$ is an arbitrary function. This would capture any possible (first order) theory of supply and demand for money with long run neutrality. A Taylor expansion of $g$ around $D = S$ results in an equation of the form

\text{(3) } \frac{dD}{dS} = c_0 + c_1 \frac{D}{S} + c_2 \left( \frac{D}{S} \right)^{2} + c_3 \left( \frac{D}{S} \right)^{3} + \cdots

$$

Note that higher order derivatives by themselves are not consistent with homogeneity. If we take

*D→α D*,*S→α S*means that*d²D/dS² → (1/α) d²D/dS²*. Terms like*D d²D/dS²*would be necessary, which we'll subsume into "generalized" second order terms.\text{(4) }\mathcal{L} = \partial_{\mu} \phi \partial^{\mu} \phi + m^{2} \phi^2 + g_1 \phi^{4} + g_2 \phi^{6} + \cdots

$$

where all terms we consider must be consistent with Lorentz symmetry (i.e. special relativity, this theory is also symmetric under charge symmetry as well); the resulting theory is guaranteed to be consistent with Lorentz invariance. This way of coming up with particle theories is called an effective field theory. Generally, one writes down every possible term consistent the symmetries under consideration. Our process with equation (3) was analogous to writing down every consistent with long run neutrality of money (analogous to a symmetry).

The higher order terms in field theory (4) represent higher order interactions (2, 4-particle, etc interactions with coupling constants $g_1$, $g_2$). They tend to be "suppressed" (in physics) because the coefficient has dimensions of mass and that mass is considered "heavy". The higher terms (degree > 2) in the Lagrangian represent vertices (interactions) in Feynman diagrams.

It is possible that the analogous terms in our long run neutrality of money (money invariance) theory (3) represent three or more party transactions which would be heavily suppressed by the existence of money (you'd likely trade apples for money and then get oranges with some of that money rather than work out some complicated contract between the three parties allocating money, oranges and apples). I.e. the higher order coefficients $c_2$, $c_3$, ... might be suppressed by factors of

*1/MB*(!) where*MB*is the size of the monetary base (how much money is out there).
The previous paragraph is just reasoning to justify taking $c_2$, $c_3$,

*... = 0*[4]. But the best reason to do so is that the model fits the empirical data! If we use*κ = log S/log D*, then equation (2) does an excellent job of describing the price level:
In Part II, we'll motivate (1) with information theory.

[1]

*Long-Run Monetary Neutrality and Contemporary Policy Analysis*Bennett T. McCallum (2004)
[2] This prescription of

*κ = log S/log D*is reminiscent of the beta function in quantum field theory; it is motivated by empirical evidence and the information theory in Part II because*κ*is the ratio of the number of symbols used to describe*S*to the number of symbols used to describe*D*. In an older post, I refer to it as the unit of account effect when used to describe the price level: the size of the money supply defines the unit of money in which aggregate demand is measured.
[3] Series expansions around

*α*~ 0 have a small coefficient for the linear term (if*S, D >> 1*) and the limit as*S, D → ∞*is independent of*α.*
[4] We'll take

*c0*to be zero, too. Although we shoud be careful. Einstein famously took the equivalent of*c0*(the cosmological constant) in general relativity to be non-zero in order to allow a steady-state universe. He later regretted that action, but more recent results show that it is not actually zero, but very very close.
I did an update to convert the equations to mathjax ... hopefully that didn't mess too much up.

ReplyDelete1) "A Taylor expansion of g around D=S results in an equation of the form..."

ReplyDeleteSo we could consider g(u) with u=D/S, and thus a Taylor expansion around D=S would be an expansion around u=1:

g(u-1) = g(1) + g'(1)*(u-1) + [g''(1)*(u-1)^2]/2 + ...

Where g' = dg/du and g'' = d(dg/du)/du, etc.

Correct?

2) Why about D=S (equivalently u=1)?

3) Also, what do you mean by MB here? Base money or base money minus reserves (currency)? (i.e. which did you use to make the plots of your model results with? It looks pretty good, so I'd assume you meant currency).

4) You use M0 in the title to a plot above, but I assume that you don't mean currency by that, but instead a normalization parameter?

5) I read this before (I think), but it's a fairly mind blowing post. I'd forgotten about it I guess. Basically it's an alternative derivation (alternative to the information theory approach) of the price level formula based on long run neutrality of money, and the kinds of differential equations that implies.

6) "κ (which I'll just call the IT index for now)"

I know we've been over this before, but 1/κ = k you now call the IT index, right? (I just checked your draft paper, and yes that seems true)

7) I had all kinds of questions about the plots, but I'm pretty sure I resolved them all by staring at them long enough... my 1st attempt to comment (w/ those questions) was lost by me pressing the wrong key... it may have benefited other readers.

8) "Our process with equation (3) was analogous to writing down every consistent with long run neutrality of money (analogous to a symmetry)."

DeleteI think you forgot the word "term."

1. I'm not sure what your point is here. If it is that the form doesn't look exactly the same -- you can expand all the terms (u-1)^n and collect powers of u. In reality it's not a Taylor expansion and you need terms like (D/S) dD/dS ... but I mention those as generalized 2nd order terms immediately after.

Delete2. That is expanding around the equilibrium where demand = supply.

Delete3. Yes.

Delete4. Yes.

5. Yes, it's an alternative approach based on "effective field theory".

6. I call both "kappa" and k the IT index, but I now try to be consistent in using k for a general index that is in the numerator on the RHS of the information transfer equation.

DeleteBut it really doesn't matter as the map 1/x : R⁺ → R⁺ is a diffeomorphism :)

Thanks Jason.

ReplyDeleteYou write:

"In reality it's not a Taylor expansion and you need terms like (D/S) dD/dS ... but I mention those as generalized 2nd order terms immediately after."

This is the part immediately after, right?:

"Note that higher order derivatives by themselves are not consistent with homogeneity. If we take D→α D, S→α S means that d²D/dS² → (1/α) d²D/dS². Terms like D d²D/dS² would be necessary, which we'll subsume into "generalized" second order terms."

Sorry to be so slow on the uptake here. What if g(D/S) = (D/S-1)^3? (I'm trying to make a super simple example, but if you've got a better one...)

I don't see any partials, so it's not a 2D Taylor series:

http://www.math.ubc.ca/~feldman/m200/taylor2dSlides.pdf

So you're just expanding about either D or S then (rather than the ratio as a whole)? E.g., I need to replace my

g'(u) with g'(u)*du/dD

?

As I said, it's not really a Taylor expansion. A Taylor expansion around D = S is just a conceptual jumping-off point for people not familiar with effective field theory. It's an effective field theory expansion (the Lagrangian that goes with a Feynman diagram expansion) -- you write down all the terms consistent with the symmetry principle (invariant under D→αD, S→αS). But neither the coefficients nor the terms are necessarily "small" ... unless you have good physical reasons for it.

DeleteAh! Ok. That's what you meant by "it's not really a Taylor expansion" ... I thought perhaps that was a comment on my attempt with substituting u=D/S.

DeleteThanks!