No one seems to like the poor old IS-LM model except Paul Krugman. I thought I'd defend it as an effective macro theory. What does that mean? It means it is a theory that operates given a certain scope. For IS-LM, assuming the IT model is the "true" theory, that scope is low inflation. Here's how it works:
Start with the AD-AS model (with aggregate demand aka nominal output N and aggregate supply S), and introduce money (M)
N ⇄ M ⇄ S
The produces an information equilibrium relationship
P : N ⇄ M
The P just names the abstract price (the price level) and the above notation means:
We can solve this differential equation to obtain:
N ~ Mᵏ
P ~ k Mᵏ⁻¹
That should be a k - 1 in the exponent, not (M^k)^-1. If k ≈ 2, then we have P ~ M (a "quantity theory of money" where inflation is money growth). If k ≈ 1 (one way of stating our scope condition), then
P ~ constant
so that inflation π ≈ 0 (another way of stating our scope condition ). If inflation is "small", then
where Y is real output. Now let's introduce another set of information equilibrium relationships
i ⇄ p
where MB is the monetary base, p is the abstract price of money and i is the nominal (short term) interest rate. If π ≈ 0, then we can write this as:
r ⇄ p
using N ~ Y and i ~ r where r is the real interest rate. The differential equations associated with this pair of information equilibrium relationships has a similar solution in general equilibrium to the one above, but I'll write it differently:
Now which variables do you think will react the fastest to monetary expansion? I'd go with the monetary base and the interest rate, rather than real output. That puts us on the partial equilibrium solution where Y "moves last" (in economics parlance). In physics terms, this is analogous to an isothermal expansion of an ideal gas (an 'iso-output' expansion of an economy) that looks like this:
Those angle brackets mean ensemble averages (weighted average over random markets).
The interest rate falls and the monetary base expands from MB₁ to MB₂. Note at this level of effective theory, it is not important which one causes the other one -- increasing the base lowers interest rates or lowering interest rates expands the base. As Y rises, interest rates will come back up and equation (1) will hold -- general equilibrium restored.
What about fiscal policy? That's the second diagram above. In that case, the central bank "moves last". Fiscal expansion should take us from Y₁ to Y₂, but if the central bank didn't let rates rise or the base expand, you could end up back where you started. This is the basic monetary offset picture that happens when inflation is low and nominal (and real) interest rates are positive.
That captures most of the mechanics of the IS-LM model; you could rewrite this in terms of investment if you wanted (as I do here). And there is a bit more to the details of how it works, but the theories are roughly isomorphic to each other.
Paul Krugman points out in the liquidity trap argument everything changes at the zero lower bound (or basically, anytime interest rates are very low ... remember we're still in a limit where inflation is very low as well). This is the limit where r ≈ 0⁺. Here we have a different-looking pair of diagrams:
The general idea is the same, but the magnitudes are very different. A large expansion of the base drops your interest rate to zero without moving output very far to the right (monetary expansion is ineffective). A large fiscal expansion can happen without large movement in the monetary base (i.e. monetary offset doesn't work). In this case, we have a model that captures the essence of the liquidity trap.
Recall that we built this out of the AD-AS model, but we did it by restricting the scope to r ≈ 0⁺, π ≈ 0 (or just π ≈ 0 for the IS-LM model) . This doesn't mean this model is right, but it does mean if you are using the AD-AS model where inflation is low and interest rates are low, then you are effectively using the IS-LM model (if you are doing it correctly).
And since the above model does pretty well empirically (see the paper or the forecasts), it means that any theory that describes the data (when π ≈ 0) will also be approximated by the IS-LM model.
An analogy: since Newton's inverse square law describes the orbits of planets fairly well, any theory that describes the orbits of planets fairly well will be approximated by Newton's inverse square law. This is actually true of general relativity (which reduces to Newton's inverse square law) and will also be true of the fundamental quantum theory of gravity -- should it exist.
 Strictly speaking, these inflation and interest rates are small compared to some other scale in the theory, in this case we could use the average growth rate of the monetary base (minus reserves) r << μ, π << μ. In the US, this has been about 7% per year. Inflation is about 1-1.5% and (short term, nominal) interest rates are 0.3-0.5% -- both are less than 7% and that gives us an idea that the error in this model could be ~10% ... i.e. 0.005/0.07 ~ 0.01/0.07 ~ 0.1.
I learned some things here that had escaped me previously. Thanks. Just to be clear, nowhere above did you explicitly write out the partial equilibrium solutions that define each of the four curves, right? (I know where to find those)ReplyDelete
You can get the behavior from equation (1) by holding Y or MB constant.Delete
But to draw the blue curve, say, you have Delta Y, r, and MB all changing. I realize the Delta Y only describes a change in the surrounding "bath" (in the short term, since Y "in the system" is held constant)... But still there's no explicit Delta Y in equation (1).Delete
You'd tend to get something like
Delta Y ~ log /constant
Anyway, right? And a similar expression for Delta MB for the red curves, right?
I put MB in angle braces, and now its gone. I meantDelete
Delta Y ~ log MB / constant
Y = ΔY + YrefDelete
This comment has been removed by the author.Delete
Sorry, I feel like a block head, but say I wanted to replicate your blue curves by using eq (1) and by holding Y constant. I can change either r or MB, but eq (1) doesn't inform me how to change both ΔY and Yref such that their sum (Y) stays constant as either r or MB change. And even if it did, then the result would still satisfy G.E. (eq (1)), so when you write that Y "moves last" andDelete
"As Y rises, interest rates will come back up and equation (1) will hold -- general equilibrium restored."
I'm wondering why it moves at all, since by satisfying eq. (1) the whole time we were in G.E. before it moved. It didn't need to be "restored" to G.E., it was already there.
Instead, as per your draft paper, I'd expect the actual formulas you used to plot the blue curves to look like the partial equilibrium solutions:
(2) r = constant1 * exp(-ΔY * constant2)
and the formulas you used to draw the red curves to look like the other partial equilibrium solution:
(3) r = constant3 * exp(ΔMB * constant4)
If not those exact expressions, then approximations to them. Those come from your partial equilibrium solutions, not the G.E. (eq. (1)), and they tell me exactly how to draw those curves.
In fact the bottom curves (especially) look just like what those exponential P.E. formulas describe. I can clearly see the curves crossing the vertical ΔY=0 and ΔMB=0 lines (respectively) at r=1 just like exp(-ΔY*constant2) and exp(ΔMB*constant4), respectively, would do. Are you sure you didn't use (2) and (3) to draw those?
I'm probably missing something obvious.
"doesn't inform me how to change both ΔY and Yref such that their sum (Y) stays constant as either r or MB change"Delete
"doesn't inform me how to change both ΔY and Yref such that their sum (Y) stays constant as both r and MB change"
If (2) and (3) do indeed describe those curves, then your statements about "moves last" and G.E. being "restored" make sense, since they imply a transition from satisfying (2) and (3) respectively to satisfying (1).Delete
It doesn't matter the order in which you apply the "approximately constant": either to the differential equation itself or its solution.Delete
As a *function*, the curves are just a function of ΔY (or ΔMB), i.e. r = f(ΔY). The ⟨MB⟩ or ⟨Y⟩ parameterizes length along the curve.
But the general equilibrium solution always applies, the partial equilibrium solutions are approximations to it for small values of ΔY.
"r = f(ΔY)"Delete
My eq. (2) has that form.
"But the general equilibrium solution always applies, the partial equilibrium solutions are approximations to it for small values of ΔY."
In the ideal gas law, it's not an approximation for the isothermal case, is it? ΔW (the energy transferred from the gas in the cylinder to the bath when the piston is moved (i.e. when the volume is changed), so as to keep the gas at a constant temperature) is described exactly by an equation like (2) for any sized ΔW, isn't it?
Does the function you plotted to create both blue curves, i.e. your f inDelete
r = f(ΔY)
Have the same form as my eq (2)? If not, what form?
I guess I don't understand your question.Delete
The curves I plotted in the graphs are notional curves. The result is as given in the paper. I probably should have drawn them free hand like here:
TL:DR version: I'm satisfied! (you'll be glad to know!)Delete
"The result is as given in the paper."
From the paper (pg 7, section 2.1), equations (2.14) and (2.15) produce a function for P in terms of ΔD, with the same form as my eq (2) expressing r in terms of ΔY. Likewise eqs (2.16) and (2.17) express P in terms of ΔS with the same form as my eq (3) expressing r in terms of ΔMB. Namely, exponentials.
All hail Krugtron the InvincibleReplyDelete
Also, this seems rather appropriate as well...Delete
Just one last thing bothers me. In the comments above you write:ReplyDelete
"But the general equilibrium solution always applies..."
But then why do you state in the post:
"As Y rises, interest rates will come back up and equation (1) will hold -- general equilibrium restored."
What is general equilibrium "restored" from if it already "always applies?"
If dx << 1, thenDelete
sin(dx) ≈ dx
But if dx ~ 1, then
dx → sin(dx)
It is "restored". The sin(dx) form always applied; for a brief period the form dx was valid.
"Restored" is not a technical mathematical term.Delete
To continue your analogy, I was imagining something different. Something like this:
f(x(t)) = x(t) + LPF(t)*[-x(t)^3/3! + x(t)^5/5! - ...]
Where LPF(t) is some unspecified low pass filter of its argument with unity gain at frequency=0.
So that if x(t) is a step function (say), then f(x) -> sin(x) as t -> infinity, but
f(x) ≈ x for t << 1
So then in the post, the unspecified LPF would be what causes Y to react more slowly (to say a step change in MB).
Which brings up an interesting question: what if MB was a large step function in time? Your sin(dx) analogy seems to indicate Y should move quickly (be a step itself actually), simply because MB is large, and thus thus the higher order terms immediately make a big contribution. Or do you think it would rise slowly?Delete
There is no time dimension in these, so the only time is implicit: Y(t) and MB(t). MB moves "quickly" only relative to Y. It depends on the size of the MB move, not its "speed".Delete
... nobody cares, but my LPF() above should take all the higher order terms as an argument, not t.Delete
Your scope conditions say this works for π ≈ 0. So why not π = 0? In that case:ReplyDelete
r = i
k = 1
Y = N = (Nref/MBref)*MB (from the opening section of the post and assuming M = MB)
In which case Y reacts just as fast as MB always. Unless there's some hidden time dynamics on Y which temporarily retards its reactions to changes in MB so it's temporarily out of G.E. for the expression resulting from
p : Y ⇄ MB
But Y ~ MB seems to contradict your assumption that either Y or MB could "move last" in relation to the other. Is π = 0 outside the scope conditions? Where did I go wrong?
I must be driving you nuts by now. I won't pester you about this anymore today! (ever?). I was just surprised to see that my assumption of unstated time dynamics (the "LPF" in the comment above) was apparently not what you were thinking of, which got me thinking about simplifying by setting k=1 precisely.
Last I heard 0 ≈ 0Delete
But in the two markets (inflation and interest rate models) don't have the same money supply -- one is M0 and one is MB.
In general, these are all time averages or equilibrium solutions. Don't treat them as Newtonian orbits.
If you do an isothermal expansion on a gas, it will induce some fluctuations in the density and temperature. After some time (quickly in physics with 10^23 particles, but slowly in economics with 10^9) it will approach the thermodynamic equilibrium isothermal expansion result.
"Last I heard 0 ≈ 0"Delete
"M0 and one is MB"
Ah, excellent point! When you used "M" instead of "M0" it kind of threw me, but that makes sense, and I should have figured that out since your inflation models all work best when M = M0. (The casual reader might have even less chance than I do of catching onto that, however). Also, doesn't classic IS-LM work just fine in an all cash society away from a liquidity trap (i.e. away from r ≈ 0)? Don't you just require π ≈ 0?
Well, maybe it doesn't if we're setting MB = M0. I don't know.
"In general, these are all time averages or equilibrium solutions. Don't treat them as Newtonian orbits."
I guess I usually have been assuming that F&B's "ideal point particles" condition holds (condition C8 in their 1st draft, which justifies assuming E[V] = V)
"it will approach the thermodynamic equilibrium isothermal expansion result"
Ah, I almost used that myself. I was thinking of an imperfect thermal coupling between the bath and the gas cylinder, so a short time delay would be required to bring the gas back to the bath's temperature after an exogenous change in volume, but you're thinking of something much much faster I think.
But it sounds like there is a little bit of time dynamics assumed then? (especially with 10^9 "particles")? After all "approach" implies it's not truly at "thermodynamic equilibrium" during each and every 5.4e-44 second interval when reacting to an exogenous change in volume (or to a change in MB).
OK, thanks. I'll stew on all that a bit.
Shoot, I went and pestered you again. Oh well. You can ignore me.Delete
E[V] = V is not quite the same thing. Ideal point particles have zero volume, so the expectation value of the volume operator is an interesting concept ... especially since real atoms do have volume (or at least a thermodynamic size from the thermal wavelength). That is essentially a statement that our macro concept of volume is "emergent".Delete
"But it sounds like there is a little bit of time dynamics assumed then?"
No, there is no specific dynamics assumed. It just doesn't happen instantly. How "not instantly"?
And in economics it would probably be impossible to separate the "equilibrium fluctuations" (around eq values) from the non-equilibrium transition fluctuations (transition from one eq to another). The former are perhaps even larger than the latter ...
Y ~ MB^k1
r ~ p^k2
It seems like you're assuming 0 < k1 < 1 and 0 < k2, which seems reasonable.
Then another way to say what you do in the blue curve case is to say dY/dMB goes to 0 as MB goes to infinity, or alternatively as r goes to 0 (since k3 < 0, where r ~ MB^k3, and k3 = (k1 - 1)*k2).
I understand you're relating the argument to IS-LM, so you break it down accordingly.
...the nice thing about the above argument for me is it doesn't involve having to consider fluctuations: i.e. anything not "happening instantly." Fluctuations could cause Y to either "move first" (r up, then down) or "move last" (r down, then up, as you describe) in relation to MB and r, and you arrive at the same place: liquidity trap.Delete
The fiscal stimulus argument is different: there you propose that the CB exogenously fix r and MB in a liquidity trap (i.e. when dMB/dY >> 1), so it's plain to see that's not going to be very effective monetary offset.Delete