Neo-Fisherite fun. Plus concrete steppes fun.
1. Suppose you don't care what speed you drive. Anything between 0 and 200 is all the same to you. But the cops do care what speed you drive. They want you to drive at exactly the speed limit S*, neither faster nor slower. (They have a symmetric target.) So the cops announce that if your speed is Si, you must pay a fine Fi, where Fi=(Si-S*)2, for driving either faster or slower than the speed limit. In equilibrium you drive at exactly the speed limit, and never pay the fine.
In equilibrium, the cops do nothing, and you expect them to do nothing. It is your counterfactual conditional expectations that matter — what you expect the cops would do if you did do something you won't in fact do.
And if the cops announce they have increased the speed limit S*, you would increase your speed too, and yet the cops actually do nothing.
1a. But if the cops got the sign wrong in the formula, and announced that the fine would be Fi=minus[(Si-S*)2], it wouldn't work. It's a negative fine. The cops pay you money if they catch you driving above or below the speed limit. There is no interior equilibrium. You would either drive at 0, or else at 200.
2. Now let's change the game slightly.
Suppose you don't care what speed you drive. But you do care about your speed relative to the average speed of everyone else. And everyone else is just like you. Except that some drivers like driving faster than average, and some drivers like driving slower than average. But the average driver likes driving at exactly the average speed, regardless of what that average speed is.
And suppose the cops don't care what speed you drive either. But they do care about the average speed of all drivers, Sbar. They want the average speed to equal the speed limit S* exactly, neither faster nor slower. If some want to drive faster than average, and some want to drive slower than average, that's fine by the cops.
So the cops think about it for a bit, then announce that each driver must pay a fine Fi=(Sbar-S*)(Si-Sbar). [Update: note that this is fiscal-neutral, both in and out of equilibrium. Total net revenue from fines is always zero.]
If the average speed is greater than the speed limit, those driving faster than average pay a positive fine, and those driving slower than average pay a negative fine. So each individual driver has an incentive to slow down relative to other drivers.
If the average speed is less than the speed limit, those driving faster than average pay a negative fine, and those driving slower than average pay a positive fine. So each individual driver has an incentive to speed up relative to other drivers.
And if the average speed is exactly equal to the speed limit, nobody pays any fines, and so the average driver chooses to drive at the average speed (by assumption), with some driving slower and some driving faster.
The only equilibrium to this game is where the average speed is exactly equal to the speed limit.
In equilibrium, the cops do nothing, and everyone expects them to do nothing. And if the number of cars is very large, so that each individual driver has no effect on the average speed, each individual driver expects the cops would do nothing even if he did do something that he won't in fact do. The counterfactual conditional expectations that create this equilibrium are expectations about what the cops would do if everyone did something that they won't in fact do.
And if the cops announce they have increased the speed limit S*, all drivers increase their speed too, and yet the cops actually do nothing.
2a. But what would happen if the cops get the sign wrong in the formula, and announce that each driver must pay a fine Fi=(S*-Sbar)(Si-Sbar)? And would all the drivers still speed up if the cops raised S*?
Neo-Fisherites say it makes no difference if the cops get the sign wrong. There's still an equilibrium where Sbar=S*, and nobody pays any fines, so nobody has any incentive to drive either faster or slower than they are actually driving. And Neo-Fisherites say that if the cops raise S*, all the drivers will speed up.
I say there are two other equilibria: a second where everyone drives as slowly as possible; and a third where everyone drives as fast as possible. And we are much more likely to observe the second or third equilibria than the first. Because there is no way that all the individual drivers, alone in their cars, could coordinate on that first equilibrium. It's unstable. (If the average driver is driving faster than the speed limit, those driving slower than average pay a positive fine, and those driving faster than average pay a negative fine, so every driver has an incentive to speed up relative to other drivers, so the average speed rises even further above the speed limit.) [Update: is it a trembling right foot equilibrium?? I think it maybe isn't.] And if by sheer fluke they were initially at that first equilibrium, and the cops raised S*, we would see all the drivers slow down, and all drive as slowly as possible.
[S* is the inflation target, Sbar is average inflation, the cops are the central bank, and the average firm wants to increase its price by more/less than average inflation if the real interest rate is less/more than the natural rate. No firm cares about its dollar price; it only cares about its price relative to other prices. And the central bank cares only about the average inflation rate. And the cops getting the sign right means the central bank follows the Howitt/Taylor principle.
And to make it even more fun for the people from the concrete steppes, we could change the game slightly, so that the average driver wants to drive at what he expects the average speed will be next period. So even if there are no cops at all on this stretch of road, the possibility that there might be cops in the very far distance will be enough to make the average driver drive now at the distant future speed limit. By backward induction. Forward guidance works.]
Worthwhile Canadian Initiative 
Nick Rowe: “If inflation were below target, the central bank would cut the nominal interest rate more than the drop in inflation, which would cut the real rate, which would increase demand, which would give each firm an incentive to raise its price more than inflation.”
In your model #1, where Fi = (Sbar – S)(Si – Sbar), the incentive for the greedy firm is not just to raise its price above inflation, but to raise its price above the target.
Toy example: 10 firms, Sbar = 90, S = 100. If a greedy firm raises its price to 150 while every other firm keeps its the same, Sbar will rise to 96, and the firm will make 216 through the negative fine. If all the others raise their price to 100, Sbar will rise to 105, and the firm will lose 225. A risky strategy for the greedy firm.
Now suppose that the greedy firm raises its price to 150, but the other firms raise prices so that their new average is only 95. That makes Sbar 100.5, and the greedy firm loses only 24.75. Not so risky.
Now suppose that the greedy firm guesses that the other firms will raise their prices to 95 on average, which they do. The greedy firm raises its price to 102, making Sbar equal 95.7. The greedy firm gains 27.09. In the worst case, from the greedy firm’s point of view, the other firms raise their prices to average 100 8/9. Then the greedy firm loses 1. Them’s pretty good odds, as Maverick’s pappy used to say. 😉
Anyway, in real life it seems unlikely that an announcement by the CB of an inflation target will lead greedy firms to overshoot the target. 🙂
What’s the difference (for the private sector) between paying $100 in interest to the central bank vs $100 in taxes?
Miami: incentives. It’s not how much you pay; it’s what actions you would choose to take to pay less (or get paid more).
Are you talking about the incentives to lend or borrow?
I thought you were the quantity demanded, quantity supplied guy;)
Junior miners vs government?
Are the returns, adjusted for risk, different?
Didn’t make it through all the comments so this might have come up but regarding this:
“[Update: is it a trembling right foot equilibrium?? I think it maybe isn’t.]”
I think it is, basically, although getting technical about trembling hand with an infinite number of drivers is a little dicey since one trembling hand/foot technically still has no effect (and with a finite number, I believe it’s not an equilibrium at all). Incidentally, when I came across this concept in my graduate game theory book and asked my professor (a very good economist) something like “so is _______ not trembling-hand perfect?” his response was basically “beats me, I don’t really know what that means, you’ll have to just look it up.” And yet economics is full of such equilibria. It’s a concept that we should probably take more notice of.
Mike: It didn’t come up in comments, IIRC. I started doing a post on it, but abandoned it. Thanks for your thoughts on this. I’m still puzzling it over.
I think you are right, that with a finite number of players, Sbar=S* is not a Nash equilibrium if the cops get the sign wrong. Good point.
With an infinite number of players, if a mass of players trembled in the same direction, and if the other players observed that tremble, it is not trembling hand perfect. But I don’t know if that counts for “not trembling hand perfect”.
Mike: I don’t know if there is any formal definition of “fragility/robustness” of equilibria in game theory. But there should be:
Suppose a small random fraction of the drivers make a mistake, and all the other drivers observe this mistake and respond rationally. We get a new equilibrium.
If the cops get the sign right, that new equilibrium will be close to the S* Nash equilibrium. (And will approach it in the limit, as small goes to zero.) It is robust.
If the cops get the sign wrong, that new equilibrium will be very distant from the S* Nash equilibrium. (And will not approach it in the limit, as small goes to zero.) It is fragile.
Nick,
Yes I think you essentially have it right. I don’t have my textbook handy but I looked it up on Wikipedia to refresh my memory about the technical definition.
“First we define a perturbed game. A perturbed game is a copy of a base game, with the restriction that only totally mixed strategies are allowed to be played. A totally mixed strategy is a mixed strategy where every pure strategy is played with non-zero probability. This is the “trembling hands” of the players; they sometimes play a different strategy than the one they intended to play. Then we define a strategy set S (in a base game) as being trembling hand perfect if there is a sequence of perturbed games that converge to the base game in which there is a series of Nash equilibria that converge to S.”
The reason I don’t think this technically works as “not trembling hand perfect” is actually more subtle than I originally thought. It’s not that the effect of any individual driver is infinitesimal since we would be assuming that every driver had a small chance of hitting the “wrong” speed. However, I think the average of their speeds will still converge to the equilibrium speed as the number goes to infinity. There should be some refinement that rules that equilibrium out though. One of us should invent one haha.
Mike: please check out my new post, where I attempt to define a “fragile/robust” Nash equilibrium, based on the trembling hand idea. Tell me what you think.