Tuesday, August 9, 2016

The Economy at the End of the Universe, part II


Nick Rowe asked in a comment on this post if I could look at the second model (a Cagan monetary demand model) he describes in his post and that took me on an adventure that might be enlightening. I also thought about calling this post Life, the Universe and Economics (or something like that) as the sequel to the previous post.

Anyway, a Cagan model has $M/P$ (real money supply) as a negative function of expected inflation, so using rational expectations (model-consistent expectations) and taking $\log M \equiv m$ and $\log P \equiv p$ we have

$$
\log M/P = m - p = - \tau \pi^{E} = - \tau \frac{d}{dt} \log P = - \tau \dot{p}
$$

The general (forward-looking) solution to this equation is an integral of the money supply (or monetary base or what have you)

$$
p(t) = \frac{1}{\tau} \int_{t}^{\infty} dt' e^{\frac{t-t'}{\tau}} m(t')
$$

As an aside, this is an example of using a Laplace transform to solve a differential equation. We can see that $\tau$ is essentially a time horizon. It weights the future (expected) money supply by a decreasing exponential factor (similar to a discount factor).

But that infinity is also a time horizon -- it tells us where to stop considering the future at all. Note that times $T \gg \tau > t$ don't contribute much to the integral. If we look at the limiting behavior of this integral (at times $t$ in the distant future after any impact of a shock to monetary policy such that $m(t) = \mu t$ -- constant money growth) we have:

$$
P(t) = \exp \; p(t) = \exp \; \left( \mu (t + \tau) - e^{(t-T)/\tau} (T + \tau) \right)
$$

This has two different limits depending on whether you take $\tau$ to infinity first (it's 0) or $T$ to infinity first (it's infinite). Again, we have the same issue as we had with Nick Rowe's reductio ad absurdum and the conclusion we should draw is that only the limits $T \gg \tau$ and $\tau \gg T$ could make sense. There's either rapid discounting or slow discounting (those limits just mentioned, respectively) at the horizon. And in those two cases we have either $P \sim \exp \mu t$ (steady growth in the price level) or $P \sim 1$ (i.e. constant, but with a growing money supply).

If we take the first limit we have something that looks basically like the previous post (except the steady state has constant growth, which we could have chosen for the previous model but didn't for simplicity):


In any case, we have the same problem that as the time horizon $t_{0}$ at which $m(t_{0})$ returns to "normal" moves out, the result differs by a larger and larger amount from $P \sim \exp \mu t$. Here's the graph of inflation ($\dot{p}(t)$):


Note that the Cagan model has different results [pdf] for adaptive expectations. Depending on parameters, inflation is either basically monetary (it depends on $m(t)$ Nick Rowe's concrete steppes) or basically expectations (i.e. is independent of $m(t)$). The former limit is either rapid discounting, rapid adaptation, or both. The latter is either slow discounting, slow adaptation. or both (but results in exploding inflation).

...

Addendum

It is interesting to look at this model as an information transfer model $X : M \rightleftarrows P$ where "expectations" $X$ are the detector of information equilibrium between the money supply and the price level

$$
k \frac{M}{P} = \frac{dM}{dP} \equiv X = e^{-\tau \pi^{E}}
$$

Or in the form above, using $x = \log X$

$$
m - p + \log k = x
$$

In general equilibrium we have

$$
X \sim P^{k -1}
$$

so if we use rational expectations, we must equate the price $X$ with the value of $X$ above such that

$$
X \sim e^{(k-1) \log P} \sim e^{- \tau \pi^{E}} \equiv e^{-\tau \frac{d}{dt} \log P}
$$

and we discover the operator formula

$$
-\tau \frac{d}{dt} = k - 1
$$

when acting on $p = \log P$. That means

$$
P \sim \exp \left( \exp \left( \frac{1-k}{\tau} t \right) \right)
$$

That's not a typo -- it's a double exponential. In order to have $P$ not explode (given rational expectations), we need $k = 1$, which implies that the price level is constant. This is exactly the same finding as a more traditional approach to the Cagan model [pdf].  The resolution here depends on what model you trust more. If you trust rational expectations, then you question the equilibrium. If you trust information equilibrium, you question rational expectations.

...

PS Gosh, I had to think about this for nearly two weeks and over three flights. I may still have gotten it wrong. Weird and subtle things can happen when you have to deal with infinity.

PPS You can see the limits pretty directly just looking at the equation

$$
m - p = - \tau \dot{p}
$$

If $\tau \gg T$, then you have $- \tau \dot{p} \approx 0$, and thus $P \sim 1$. If $T \gg \tau$, then you have $m - p \approx 0$ and so $P \sim M(t) \sim \exp \; \mu t$ in my example.

2 comments:

  1. Thanks Jason.

    I'm (as usual) not following all of this. Two points:

    1. "Note that times T≫τ>t don't contribute much to the integral."

    Yes (if I understand it correctly). That was the point I was making in my post. What people expect to happen in the very distant future doesn't matter much for what happens today, in the Cagan model. And in the limit, as "very distant" goes to infinity, it doesn't matter at all. That is very different to the Neo-Fisherian model.

    2. What you are graphing is an *anticipated* change to money growth at time 0. What I was talking about in the post was an *unanticipated* change in money growth at time 0. (people think that money growth will always be 0%, then at time 0 the central bank makes a surprise announcement, and says that from now on it will grow at 1% instead).

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    Replies
    1. 1. I agree that the neo-Fisher model critically depends on what happens "at infinity". As you correctly state, the big difference between the neo-Fisher view and the more mainstream view comes from future times T where T ≫ τ > t (where t is now and τ is the discounting time).

      In a sense, the neo-Fisher view is that M(T) for τ < T < ∞ matters and the more traditional view is that it doesn't. However, I'd argue that the neo-Fisher view is a kind of "hyperrational expectations" or "hypobolic discounting" and that the traditional view essentially says that people could expect any value for M(T) for τ < T < ∞. Monetary policy could be anything beyond the horizon.

      But if you have rational expectations up to some horizon and then ignore what's beyond it, as you move that horizon out the traditional view deforms into the neo-Fisher view (depending on the rate of discounting).

      In any case, the question is what effect the boundary condition at infinity has on the present. Neo-Fisher says it is important, the contra argument (which I am more inclined to believe) says it isn't. But it's hard to argue the contra argument is as devoted to rational expectations as the neo-Fisher argument -- in fact, I think the neo-Fisher argument is the correct application of rational expectations for all times T, even those where τ < T < ∞.

      2. It's funny you should mention that because that was what was taking me so long to get right (in my head at least). The unanticipated version can be effectively modeled by a perfect foresight with near-zero horizon up until the monetary policy change (monetary policy e.g. one second ahead is anticipated) at which point the anticipated solution kicks in.

      This turns out to have limited effect on the solution shown above, just making the initial transition a bit sharper and deeper -- but not changing the main conclusion that the path gets farther away. So I left it out. You can think of it as a weighted average of the rational expectations solution and an adaptive expectations solution. Agents would jump immediately to the perfect foresight solution in the way Robinson Crusoe immediately jumps to the optimal saddle path. This jump might take some time, which leads to a bit of dynamics at the leading edge, but the trailing edge looks the same.

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