As I’ve been trying to point out – and as others, notably Ben Bernanke, have also tried to point out – such monetary wisdom as we possess starts with Knut Wicksell’s concept of the natural interest rate. Try to keep rates too low, and inflation accelerates; try to keep them too high, and inflation decelerates and heads toward deflation.So, I was thinking, what happens if we write that down and work it out?

To keep it simple, we'll just deal with a deterministic world. It's more or less New Keynesian, but a little different. To start, we have the standard Euler equation, which prices a one-period nominal bond - after taking logs and linearizing:

(1)

*R(t) = r* + ag(t+1) + i(t+1),*

where

*R(t)*is the nominal interest rate,

*r**is the subjective discount rate,

*a*is the coefficient of relative risk aversion (assumed constant),

*g(t+1)*is the growth rate in consumption between period

*t*and period

*t+1,*and

*i(t+1)*is the inflation rate, between period

*t*and period

*t+1.*Similarly, the real interest rate is given by

(2)

*r(t) = r* + ag(t+1).*

Assume there is no investment, and all output is consumed.

To capture Krugman's concept of Wicksellian inflation dynamics, first let

*r* + ag**denote the Wicksellian natural rate of interest, where

*g**is the economy's long-run growth rate. Krugman says that inflation goes up when the the real interest rate is low relative to the natural rate, and inflation goes down when the opposite holds. So, write this as a linear relationship,

(3)

*i(t+1) - i(t) = -b[r(t) - r* - ag*],*

where

*b > 0.*Then, from (2) and (3),

(4)

*i(t) = ba[g(t+1)-g*] + i(t+1),*

which is basically a Phillips curve - given anticipated inflation, inflation is high if the growth rate of output is high.

Then, substitute for

*g(t+1)*in equation (1), using (4), and write

(5)

*i(t+1) = -[b/(1-b)][R(t) - r* - ag*] + [1/1-b]i(t).*

So this is easy now, as to determine an equilibrium we just need to solve the difference equation (5) for the sequence of inflation rates, given some path for

*R(t),*or some policy rule for

*R(t),*determined by the central bank.

First, suppose that

*R(t) = R,*a constant. Then, from (5), the unique steady state is

(6)

*i = R - r* - ag*.*

That's just the long-run Fisher relation - the inflation rate is the nominal interest rate minus the natural real rate of interest. But what about other equilibria? If

*0 < b < 1,*or

*b > 2,*then in fact the steady state given by (6) is the only equilibrium. If

*1 < b < 2*then there are many equilibria which all converge to the steady state.

Next, suppose that

*R(t) = R1,*for

*t = 0, 1, 2, ..., T-1,*and

*R(t) = R2,*for

*t = T, T+1, T+2,...,*where

*R2 > R1.*This is an experiment in which the nominal interest rate goes up, once and for all, at time

*T,*and this change in monetary policy is perfectly anticipated. In the case where

*0 < b < 1,*there is a unique equilibrium that looks like this:

So, inflation increases prior to the nominal interest increase, and achieves the Fisherian steady state in period

*T,*and the growth rate in output and the real interest rate are low and falling before the nominal interest rate increase occurs.

We can look at the other cases, in which

*b > 1,*and the dynamics will be more complicated. Indeed, we get multiple equilibria in the case

*1 < b < 2.*But, in all of these cases, a higher nominal interest rate implies convergence to the Fisherian steady state with a higher inflation rate. Increasing the nominal interest rate serves to increase the inflation rate. Keeping the nominal interest rate at zero serves only to keep the inflation rate low, in spite of the fact that this model has Wicksellian dynamics and a Phillips curve.

I'm not endorsing this model - just showing you its implications. And those implications certainly don't conform to "try to keep rates too low, and inflation accelerates; try to keep them too high, and inflation decelerates and heads toward deflation," as Krugman says. The Wicksellian process is built into the model, just as Krugman describes it, but the model has neo-Fisherian properties.

You're assuming convergence to the steady state. If instead we treat the current inflation rate as a state variable, then given an initial condition i(0), and a path of the nominal interest rate R (our control variable), (5) determines the path of the inflation rate.

ReplyDeleteWe can rewrite (5) as a difference equation (suppressing time subscripts):

(5') di = -k * (R - i - r*)

Where di = i(t+1) - i(t) is the rate of change of the inflation rate, and k=b/(1-b)>0. Now if we set R so that r > r*, inflation decreases, and if we set r < r*, then inflation increases.

This produces the well-known result that a nominal interest rate peg is unstable: if we set R = r* - i, then it is stable, but otherwise it produces an explosive path of inflation or deflation.

Now usually we impose stability by assuming a Taylor rule with coefficient > 1 on current inflation. So for instance, suppose that we follow the rule:

R = r* + (1+m)*(i-i*) - n*x

where x is the output gap. Substituting this into (5'), we obtain:

(7) di = -k * (m*(i-i*) - n*x)

This produces convergence to the steady state (inflation i*, output gap x=0). We would have deviations when there was an output gap, but they would be temporary.

Now what happens if we hit the ZLB? Then we can't use the Taylor rule, or rather we use a constrained (and therefore asymmetric) Taylor rule, with R>=0. If we set R = 0, and this results in r > r*, then this produces a deflationary spiral: we get stuck at R=0, and i goes off to -inf.

Consider this a (belated) explanation of my comment on your last post. This is the deflationary spiral that Krugman (and Summers, and others) are worried about, and why they consider raising rates to pose asymmetric risks for the economy.

"You're assuming convergence to the steady state."

DeleteNot assuming that. It's a result.

"Now if we set R so that r > r*..."

That is a specific policy rule. You have to specify the rule, and then work out the dynamics.

On Taylor rules you're not making any sense. I worked out some of that, but didn't report it. That's pretty standard. With a target inflation rate, there's a steady state where the central bank hits the target. But then, given the Taylor principle, there are many equilibria that converge to the zero lower bound, with i = -r* - ag*.

Let me help you out a bit. There need to be upper and lower bounds on g(t). Clearly g(t) > -1, as we can't have negative output. Also g(t) has to bounded from above, as there is a technological limit on how fast output can grow. Then if R is fixed, from (1), any path along which i goes off to minus infinity or plus infinity is not an equilibrium, as this will violate the upper or the lower bound on g(t), respectively. When b < 1, there are candidate equilibria that have these characteristics, so we can rule them out, and the only one left is the steady state.

DeleteNext, suppose that the central bank tries to keep the real interest rate permanently high or permanently low, relative to the natural rate. We can specify that policy as a rule:

R(t) = r* + ag* + i(t+1) + e,

where e is any real number. This will imply a difference equation for inflation:

i(t+1) = i(t) + be.

So, if e > 0, then the central bank tries to keep the real interest rate permanently high by increasing the inflation rate, and thus the nominal interest rate, without bound. That seems to be a valid hyperinflationary equilibrium. If e < 0, then the policy is not feasible, as the nominal interest rate must fall forever. Thus the central bank will hit the ZLB, and cannot keep the real rate low permanently.

"Clearly g(t) > -1, as we can't have negative output."

DeleteI thought g(t) was the log difference in consumption? That is not bounded below. You could argue that an equilibrium path that approaches 0 output asymptotically is silly, but it's an equilibrium admitted by the model.

Here's another way to put it. Suppose that the economy is initially at a steady state with a nominal interest rate R, so that i = R - r* (letting g* = 0 for simplicity). Everyone expects to stay there forever.

Now suppose that at time T>0, the Fed raises R to a higher level, R' > R, and keeps it there. This change is unannounced and unanticipated. What happens?

I think in this case, the only equilibrium is the divergent one.

log(1+x) is approximately x. g(t) is the growth rate of output.

Delete"You could argue that an equilibrium path that approaches 0 output asymptotically is silly..."

No, growth rate in output goes to minus infinity, which is not possible.

"...the only equilibrium is the divergent one."

No, you're talking about paths that are not equilibria.

The problem is that inflation is not a state variable, so one cannot simply "take it as one". Why not just say "let's assume inflation is equal to 2%' and discovering that, lo and behold, your model says inflation will be 2%.

DeleteThis is a deterministic model. The inflation rate and the growth rate of output (also the growth rate of consumption) are endogenous, and the nominal interest rate is exogenous. We could also specify monetary policy as a policy rule (a Taylor rule for example) in which case the nominal interest rate would be endogenous. The problem is that the inflation is not a state variable? What problem? I have no idea what you're getting at.

DeleteUm, Stephen, jonathan above is right, although perhaps not expressing himself clearly. If R(t)=R and 0 < b < 1 then the difference equation for inflation is

Deletei(t+1)=-q+(1+r)*i(t), where q is a constant which could be positive or negative, and r is 1/(1-b) > 1. So you got an explosive process there. Put i(t+1) on y-axis, put i(t) on x-axis, draw a 45 degree line and map the dynamics.

Then suppose we do *start* (not converge to) i=R-r*-ag* > 0 and then R goes up. That means q goes up. The line i(t+1)=-q+(1+r)*i(t) shifts down. Next period economy experiences deflation. Then even greater deflation. Then even greater... It's an explosive dynamic. Those time diagrams you drew have nothing to do whatsoever with the actual model as written down.

So yes, R-r*-ag* is an equilibrium, and yes, it is unique, but it is unstable and the economy will never converge to it. jonathan says "You're assuming convergence to the steady state." You say "It's not an assumption, it's a result". No, no it's not. It's not even an assumption because that would create a contradiction.

I've been seeing this done by Macro folks repeatedly in the recent past (always in the context of them "proving" some "interesting result", which here just means "wrong but I don't realize it's wrong"), even at the ASSA meetings so forgive me if I put it in all caps:

YOU CAN'T DO COMPARATIVE STATICS ON AN UNSTABLE EQUILIBRIUM!

"So you got an explosive process there."

DeleteExactly. There are some paths that solve the difference equation, but not all paths that solve the difference equation are equilibria. The price level cannot be negative, output cannot be negative, there is an upper bound on the quantity of output that can be produced. All of those things have to be satisfied on any equilibrium path. If you have a candidate equilibrium path that solves the difference equation, but it doesn't satisfy all of those properties for all time, then that's not an equilibrium.

Some advice: Sometimes when you see a large number of people who are supposed to be experts in something, and they're all doing the same thing, and you think they're doing the wrong thing, there's a high probability that they are right and you are wrong.

But there's no path that converges to the steady state. The only paths that solve the difference equation are explosive ones. Check the eigenvalues of the characteristic matrix (you don't have to actually, just graph i(t+1) versus i(t))

DeleteI believe you're thinking of a situation like in John Cochrane's model on his "Wither Inflation" post, where there is a stable path and unstable path and we can rule out the unstable path because it's not an equilibrium. But that's a different model (you have an extra equation). Here there are only unstable paths.

You *could* argue that because the steady state is unstable, no matter what the shock, the economy will always immediately jump right back to it, with no transitional dynamics. That's a bit of a stretch and it sort of depends on what underlies the Phillips Curve and which one is the control variable (as jonathan also notes). But even in that case you don't get the impulse response functions you have above. You'd just have immediate move from one (unstable) steady state to another (unstable) steady state.

In your version of the model the Taylor Principle has to hold for it to be stable. Central bank has to respond more than 1 for 1 to inflation. A constant interest rate is a response of 0 to inflation. So unstable.

"The only paths that solve the difference equation are explosive ones."

DeleteAnd the steady state. And there's nothing funny going on in the steady state - it's an equilibrium.

"You *could* argue that because the steady state is unstable, no matter what the shock..."

But there are no shocks in the model. We could introduce shocks, but that would be another model. In this model, we can only talk about equilibria. And there is only one, which is the steady state.

"But even in that case you don't get the impulse response functions you have above. You'd just have immediate move from one (unstable) steady state to another (unstable) steady state."

No. What the pictures show is a unique dynamic equilibrium given the path for the policy rate.

"In your version of the model the Taylor Principle has to hold for it to be stable. Central bank has to respond more than 1 for 1 to inflation. A constant interest rate is a response of 0 to inflation. So unstable."

Wrong. You can consider Taylor rules in here alright, and get equilibria with and without the Taylor principle. Typically, what New Keynesians are worried about is determinacy. Here, with R constant and 0 < b < 1, you get determinacy - unique equilibrium.

The "instability" you are talking about is not instability the way people usually think about it. In this case it doesn't make any sense to think about a perturbation that would take you away from the unique equilibrium.

Stephen, the steady state is unstable, that's obvious just from the math. The only way you get there is if you start there. Otherwise, no go. If the Central Bank increases the interest rate then either a) the economy somehow immediately jumps to the new (unstable) steady state and there's no dynamics or "paths" or impulse response or anything like that or b) the economy moves AWAY from the new steady state. That's what unstable means.

DeleteIf you still think otherwise then please write down the path for inflation that the economy will follow. The one that's suppose to converge to the steady state. The one which is supposedly illustrated by your graphs above. I'm saying that such a path does not exist in your model.

(and determinacy and instability are two different things. And I'm using "instability" in the sense that everyone uses it when talking about dynamic systems, nothing special about that)

Well, we're not really getting anywhere. Repeating the same incorrect things does not make them right.

Delete"...the steady state is unstable, that's obvious just from the math."

Unstable, under the usual interpretation, in this context, would be a situation where, if we had an initial condition i(0) close to the steady state, but not the steady state, i(t) would follow an equilibrium path that would take us away from the steady state. But, there is no such equilibrium path, so we cannot say that the steady state is unstable. Sorry.

"The one which is supposedly illustrated by your graphs above. I'm saying that such a path does not exist in your model."

Again, you're wrong. From equation (5), we already know from the analysis of a constant nominal interest rate that we have to be at the steady state given by (6), with R = R2, for t = T+1, T+2, ... . But then note that i(T) = i(T+1) = i(T+2) ... . Now, work backward from period T using equation (5). That is, given R(t) = R1 < R2, solve for i(T-1) given i(T), and then work backward to period 0. Now you have the whole path for i(t), t = 0, 1, 2, ..., and you can use equation (1) to solve for the growth rates of output. It's a unique dynamic equilibrium. QED

So, go work on that for a bit, and convince yourself.

We're done now.

That could mean one of three things:

Delete1. Wow, you are so smart.

2. Wow, you are so smart (sarcastic version).

3. Wow, you are so stupid (non-sarcastic version).

Given the intransigence revealed in the previous discussion, I suspect (3).

No, sorry. It was actually "Wow that's a stupid model", which is not directed at you personally but at the model. I do understand what you're doing now - given that we have to get to steady state, at time zero inflation has to be "just right" so that it "explodes" just enough so we get there by time T - but that's got to be the most ridiculous notion of equilibrium I've ever seen. I'm also having trouble understanding how this initial - and critical - level of inflation would be achieved in period zero but I guess that's buried in the microfoundations somewhere in the background.

DeleteOk. Putting that aside, this model probably is still not a good candidate for a Fisherite argument. If the change in the interest rate is unexpected, it all goes to hell, In fact, if anything unexpected happens, no matter how trivial, it all goes screwy.

But thanks for the illumination.

It is a stupid model. As I said, I'm not endorsing it. This is just my representation of what Krugman is saying, in a model he should like.

Delete"but that's got to be the most ridiculous notion of equilibrium I've ever seen."

I'm not sure what you're used to seeing. It's a rational expectations equilibrium in which everyone knows that the interest rate increase is coming in the future. Notice that, the further in the future the interest rate increase is, the closer the initial period's inflation rate (and everything else) is to the steady state for the case in which R(t) = R1 forever.

"I'm also having trouble understanding how this initial - and critical - level of inflation would be achieved in period zero..."

Everyone's forward looking. They see what's coming, and that's the only possible initial period inflation rate that gives you an equilibrium path into the future.

"this model probably is still not a good candidate for a Fisherite argument."

I certainly would not take the model seriously. The idea is that, when I've discussed these issues before, some people have been dismissive. They seem to think I've got some weird model that allows me to cook this up. It then seems useful to take the models that other people think are useful, and show they have neo-Fisherian properties too. We know that standard New Keynesian models have these properties. We now know that this Wicksellian-Krugmanian thing has neo-Fisherian properties too.

Steve, my understanding is that you define equilibrium as a sequence of nominal interest rates and inflation rates such that 1) the Euler equation holds and 2) the policy rule of the central bank holds, given that g>=0. Is this correct?

ReplyDeleteAn equilibrium is an infinite sequence {i(t),g(t)} solving the difference equations (1) and (4) given the policy rule for {R(t)}. This is leaving aside how we determine the price level and the level of output. As I said, I'm not endorsing this model. Just trying to write something down that represents what Krugman seems to be saying, and that people will recognize. It looks very reduced-form NK.

DeleteGreat, that helps, thanks!

Delete

ReplyDeleteMy friend Tom Brown said I should run this idea by you as it sounded a bit Neo-Fisherite to him. The idea is that the velocity of money depends on the interest rate and inflation rate. John Hussman has posted several times over the years showin how velocity depends on the interest rate. I wrote a post and appreciate any feedback you can give: http://www.howfiatdies.blogspot.com/2015/09/punchbowl-removal-difficulties.html

To be clear, it's this sentence fragment in your comment on your own blog that made me think "Neo-Fisherite":

Delete"if the Fed starts to raise at all the velocity of money starts to go up"

That sounded vaguely Neo-Fisherite to me. However don't be surprised if Stephen doesn't agree (and he'd know, not me). Worth a shot though.

Well, almost all economists tends to agree that the battle between Keynesians and Monetarists was all about Wicksell vs Fischer. Are you saying all of them are wrong?

ReplyDelete1. Be more precise. Would "almost all" be 80%, 90%, or maybe .05%?

Delete2. Does "tend to agree" mean agree somewhat, agree, strongly agree, or agree kinda sorta?

3. Is this "almost all" economists in the world, in Europe, in the U.S., or in Antarctica?

4. Wicksell vs. Fischer: Do you mean Stan Fischer or Bobby Fischer?