I hope that I am reinventing the wheel by writing this post. And that some game theorist has already written something much better on the same subject. (If not, then they should have done.) But I need a game theorist to tell me about it and explain it to me, in simple language (if possible).

[The macro punchline comes at the end of this post.]

This is my crude attempt to define "*fragility/robustness*" of Nash equilibria:

Let G be a game, let S be a set of strategies in that game (one for each player), and let S* be a Nash equilibrium in that game. Assume a large number of players, and a continuous strategy space, if it helps (because that's what I have in my mind).

Suppose that a small fraction n of the players deviate by a small amount e from S* (their hands tremble slightly), and that the remaining players know this. Let S*' (if it exists) be a Nash equilibrium in the modified game.

1. If S*' does not exist, then S* is *a fragile* Nash equilibrium.

2. If S*' does not approach S* in the limit as n approaches zero, then S* is *a fragile* Nash equilibrium.

3. If S*' does not approach S* in the limit as e approaches zero, then S* is *a fragile* Nash equilibrium.

4. But if S*' does exist, and S*' approaches S* in the limit as n or e approaches zero, then S* is a *robust* Nash equilibrium.

I do not like fragile Nash equilibria. Any model which uses a fragile Nash equilibrium to derive results is very very suspect. Because if the assumptions are false by even a very small amount, the results will be very different from the predictions of the model. It's a knife-edge.

Take an example, from my previous post on the collective speed limit game. [In that game, the cops care only about the average speed, and don't care about the speed of any individual driver, so they announce a system of fines where the (positive or negative) fine depends on the individual driver's speed relative to the average speed, and on the average speed relative to the target speed.]

Let's simplify it, by assuming that there is an infinite number of drivers, all drivers are identical, and each driver does not care about the speed he drives Si, except he prefers to drive at the average speed Sbar.

With no cops, any average speed, with all drivers driving the same speed, is a Nash Equilibrium.

If the cops announce a target average speed S*, to be enforced by a symmetric fine (that can be positive or negative) on each driver Fi=(Si-Sbar)(Sbar-S*) there is a unique Nash equilibrium Si=Sbar=S*. And it is a *robust* Nash equilibrium.

To see this, assume that a fraction n of the drivers are irrational, and drive at (S*+e). Notice that the (1-n) rational drivers will choose to drive at a speed Sr that is somewhere between Sbar and S*. (They like to drive at the average speed, but don't like paying fines like being paid money by the cops). And Sbar=(1-n)Sr + n(S*+e). So as n approaches 0, Sbar will approach S*.

If the cops **get the sign wrong**, and announce a fine Fi=(Si-Sbar)(**S*-Sbar**), then Si=Sbar=S* is also a Nash equilibrium. **But it is a fragile Nash equilibrium.**

To see this, notice that the (1-n) rational drivers will choose to drive at a speed Sr that is above Sbar if Sbar is above S* (and below Sbar if Sbar is below S*). (They like to drive at the average speed, but also like being given money by the cops). And Sbar=(1-n)Sr + n(S*+e).

If n=1, Sbar=S*+e. But as n falls below 1, Sbar gets larger if e > 0 (and Sbar gets smaller if e < 0). In the limit, as n approaches 0, the rational drivers drive as fast as possible if e > 0, and as slow as possible if e < 0.

[If you like math, assume each rational driver chooses Si to minimise a loss function Li = (1/2)(Si-Sbar)^{2} + Fi, and solve for the equilibrium Sbar as a function of S*, n, and e.] [edited to fix (I think) a problem spotted by Majromax. I really should not even try to do any math. Each driver sets Si=Sbar+(S*-Sbar)/2 if the cops get the sign right, and Si=Sbar+(Sbar-S*)/2 if the cops get the sign wrong. So there is strategic complementarity (positive feedback) in both cases, but too much strategic complementarity for stability if the cops get the sign wrong.]

**If the cops get the sign wrong, the Nash equilibrium Sbar=S* is not robust. If any drivers are irrational, the smaller the number of irrational drivers the worse the predictions of the model.**

To understand the metaphor, suppose the cops are the central bank, and want to target an inflation rate S*, by pegging a nominal interest rate equal to the natural rate plus S*. (The central bank forgets the Howitt/Taylor principle, and the cops get the sign wrong). And the representative rational firm increases its price by an amount equal to/less than/more than the average inflation rate Sbar if the real rate of interest is equal to/more than/less than the natural rate. But a fraction of firms are irrational and increase their prices by (S*+e) regardless of anything. And the real rate is the pegged nominal rate minus the average inflation rate Sbar.

**If the central bank pegs a nominal interest rate, the Nash equilibrium inflation rate gets further away from the inflation target S*, the smaller the percentage of firms that are irrational. The Nash equilibrium inflation rate in the limit does not equal the Nash equilibrium inflation rate at the limit. It's very far away from** it.

That punchline will be unsurprising to most macroeconomists. [They knew there was something fishy about the Neo-Fisherite equilibrium.]

[I think there is some relationship between "fragility" of equilibrium as I have defined it here and "instability" of equilibrium in the old-fashioned sense.]

> [If you like math, assume each rational driver chooses Si to minimise a loss function Li = (1/2)(Si-Sbar)2 + Fi, and solve for the equilibrium Sbar as a function of S*, n, and e.]

Okay, this doesn't quite give what you want.

With the proper fine, Li' = (Si-Sb) + (Sb-S*) = Si-S*, which is zero (finding the minimum loss) for Si=S*. The actual average speed will be Sb = (1-n)S* + n(S* + ε) = S* + nε, which is great, and everything's nicely determined.

With the improper fine, Li' = (Si-Sb) + (S*-Sb) = Si + S* - 2Sb. This is minimized for Si = 2Sb-S* (rational drivers want to overshoot the average by the same amount the average overshoots the target). The actual average speed will be Sb = (1-n)[2Sb - S*] + n(S* + ε). This is solved to give (2n - 1)Sb = n ε + (2n - 1)S*, giving Sb = S* + n/(2n-1) ε. In turn, with model-consistent expectations the rational drivers will drive at 2Sb - S* = S* + 2n/(2n-1)ε

This gives a breakpoint of undefined behaviour at n=1/2. If drivers expect fewer than half of their compatriots to behave irrationally, then the average speed is actually

oppositeS* from the preferences of irrational drivers (without loss of generality, ε > 0 and n < 1/2 means that the average speed is slower than S*). This happens because the irrational drivers don't have "much" effect on the average speed, so the rational ones can pick up some cash whilst driving only a little bit away from their preference.This changes above n=1/2, where now the irrational drivers have a greater weight in defining the average. At this point, rational drivers treat the eventual average as closer to (S*+ε) no matter what they do, so it's best to drive ever so slightly faster (positive ε) than that to pick up the cash. At n very near 1/2, those two factors almost cancel out allowing rational drivers to tolerate very large deviations from the average speed.

The issue is that your loss function is strictly decreasing as (Si-Sb) goes to positive or negative infinity, since the speed difference has quadratic weight whilst the fine is linear in (Si-Sb).

Posted by: Majromax | November 18, 2014 at 01:56 PM

Interestingly and possibly relevantly, this particular set of loss functions give qualitatively different results with rational expectations.

Without rational expectations, our enlightened driver thinks that they're the only one -- n ~= 1. The proper fine structure still has them drive at S*, but an improper fine structure will have them drive at (S* + 2ε).

With rational expectations, our enlightened driver thinks that everyone else is the same, so n ~= 0. Then, both models give Si = S*.

Posted by: Majromax | November 18, 2014 at 02:00 PM

Majro: Curses! So I need to assume the fine is quadratic in (Si-Sbar) too, to get a nice neat result? So when the cops get the sign wrong, Fi=(S*-Sbar)(Si-Sbar)^2 ?

Posted by: Nick Rowe | November 18, 2014 at 02:16 PM

Actually, wouldn't it be simpler to change the loss function to Li = (Si-Sbar)^2 + Fi ? (I only put the (1/2) in there to simplify the math!)

Posted by: Nick Rowe | November 18, 2014 at 02:30 PM

Yes, if we drop the (1/2), I think we get:

Si = Sb + (S*-Sb)/2 if the cops get the sign right.

Si = Sb + (Sb-S*)/2 if the cops get the sign wrong.

Is that right?

Posted by: Nick Rowe | November 18, 2014 at 02:41 PM

The economic intuition of this arithmetic is that I am assuming strategic complementarity between price-setting firms, so that Si is a positive function of Sbar, even if the central bank follows the Howitt/Taylor principle. But if the central raises the real interest rate by a very large amount when inflation rises above target, an individual firm would not just want to raise its price less than inflation, it would want to raise its price less than the target rate of inflation.

Posted by: Nick Rowe | November 18, 2014 at 02:57 PM

> So I need to assume the fine is quadratic in (Si-Sbar) too, to get a nice neat result?

That doesn't work. It will result in the speed preference strictly dominating the fine if (S*-Sb) > -1, or be capable of strictly positive and unbounded gains otherwise. Also, a quadratic fine is always either strictly positive or strictly negative, which breaks the intuitive understanding.

I'm at a bit of a loss for a loss function. What we need is one where the sign of the second derivative at the "good" equilibrium (Si=Sbar) depends on whether the fine is correct or incorrect.

> Yes, if we drop the (1/2), I think we get:

Yes, that's right. If the cops get the sign right, the rational driver will drive at the mean of the target and presumed-average speed. If the cops get the sign wrong, the rational driver will drive half again as far off-target as the presumed average [(Si-S*) = 3/2 (Sb-S*)]. It's qualitatively the same as with the (1/2) present.

> But if the central raises the real interest rate by a very large amount when inflation rises above target, an individual firm would not just want to raise its price less than inflation, it would want to raise its price less than the target rate of inflation.

So in Chuck Norris terms, if inflation rises above target the central bank promises to mug my customers. Therefore, to remain profitable I will want to increase my prices by less than the inflation rate, or else my freshly-mugged customers will not be able to afford my goods.

In the counterfactual with the wrong sign, the CB pays interest on money. By raising interest rates, the CB is actually giving my customers more base money. Therefore, I want to raise my prices more than I would otherwise, to capture some fraction of those gains.

So the salient question is: if the CB raises interest rates, is it a reward or punishment? Equivalently, does the LM curve slope upwards or downwards?

Posted by: Majromax | November 18, 2014 at 03:36 PM

Majro: "So in Chuck Norris terms, if inflation rises above target the central bank promises to mug my customers. Therefore, to remain profitable I will want to increase my prices by less than the inflation rate, or else my freshly-mugged customers will not be able to afford my goods."

The bit about Chuck Norris and promises is right.

But it's not a question of being able to *afford*. Interest rates in (simple) NK models have a substitution effect but no income effect. If r goes up, they choose to postpone consumption till later.

If the cops get the sign wrong, then if inflation rises above target, the real interest rate falls, customers want to buy more, so each firm wants to raise its price even more than inflation, and so much more than target inflation.

If the cops get the sign right, then if inflation rises above target, the real interest rate r rises, and customers want to buy less. There are two cases:

a) If r rises a small amount, demand falls a little, and each firm will want to raise its price by somewhere between actual inflation and target inflation. (This is the case I want.)

b) If r rises a large amount, demand falls a lot, and each firm will want to raise its price by less than target inflation.

Posted by: Nick Rowe | November 18, 2014 at 04:20 PM

And thanks for checking my math!

Posted by: Nick Rowe | November 18, 2014 at 04:21 PM

> There are two cases:

I don't think those are two exclusive cases, but rather strength-based effects. For the obligatory car metaphor, it's the difference between tapping on the brakes and stomping on them, not the difference between pressing the break pedal or gas pedal to slow down.

Posted by: Majromax | November 18, 2014 at 04:29 PM

Majro: yes. But if you stomp too hard on the brakes if you are above the speed limit, or too hard on the gas if you are below, weird things can happen. And you were seeing some of those weird things in your first comment. With my original loss function, the cops were reacting too strongly relative to the desire of drivers to drive the average speed. So I changed the loss function to give drivers a stronger preference for driving at the average speed.

Posted by: Nick Rowe | November 18, 2014 at 04:34 PM

Short version, Neo-Fisherite ideology is the equivalent of saying gravity causes people to float up into the sky.

It's too bad central bankers will never take them seriously. Waiting out a depression for inflation to rise up and meet interest rates would make the same mess of the economy the liquidationists made in the 1930s. It would be brutal enough for the people to rise up against the "robber barons." But in a Japan-style liquidity trap, the misery can go on for decades without demand for real change.

The real motivation for reviving this flaky hypothesis, however, is to put pressure on the US Fed to raise interest rates sooner rather than later. They will get their way to some degree, and that will likely cause the economy to fall back into recession. Who knows, maybe there will be another derivatives meltdown as a result.

The only way to turn the tide on disaster capitalism is more disaster. The neo-cons think the disaster will work in their favor: more spending cuts, more tax cuts, more cuts to wages and benefits. I think the next crisis will provide the opportunity to kill neoclassical ideology, which is what happened during the Great Depression. But this time the people need to drive a steak in its heart to make sure its economy-destroying days are over.

Posted by: Ron Waller | November 18, 2014 at 09:16 PM

Ron: "It's too bad central bankers will never take them seriously."

Except, some of them are, or were, central bankers.

"The real motivation for reviving this flaky hypothesis, however, is to put pressure on the US Fed to raise interest rates sooner rather than later."

Except, one of them (Kocherlakota) changed his mind, and is now one of the strongest advocates of raising interest rates later rather than sooner. He changed his economics, in the light of theory and evidence. I see no reason to believe he changed his politics.

Not everything is politics. (And I am certainly more right-wing than some of the people I am arguing against.)

This post is about game theory, and money/macro theory, and a bit of math too. It is not a good place for random rants about capitalism, neo-conservatism, and "neoclassical ideology". I am 'playing with my Lego', as you would describe it. It is not a good topic for practicing high school debating tactics.

If you want to contribute, start by reading (e.g.) Wiki on Nash equilibria, and Trembling hand perfection. Because that Lego stuff matters for the real world.

Posted by: Nick Rowe | November 18, 2014 at 09:44 PM

Ron: "Short version, Neo-Fisherite ideology is the equivalent of saying gravity causes people to float up into the sky."

No. It's not that at all. Read the post again. A better analogy, would be balancing a broomstick upright in the palm of your hand (an inverted pendulum). Which way do you move your hand if you want to broomstick to move North? Except the broomstick has expectations.

Posted by: Nick Rowe | November 18, 2014 at 09:58 PM

Nick,

With the fine structure:

Fi = (Sbar - S*) * (Si - Sbar)

Can the cops change S* without introducing a whole other level of instability? It would seem that as soon as they change S*, negative fine maximization could send the average speed oscillating wildly.

Posted by: Frank Restly | November 19, 2014 at 02:11 AM

Frank: there are no dynamics in this model, so we can't say. If there were dynamics, it would depend on how expectations are formed. Do the cops announce the new S*, or do they let drivers learn it?

Posted by: Nick Rowe | November 19, 2014 at 04:55 AM

Broomstick - it may have expectations, but they depend on what it "wants." Does it want to be balanced? Back in the realm of just assuming long-run equilibrium.

Posted by: Robert | November 19, 2014 at 11:22 AM

Robert: the broomstick is a bundle of little sticks. None of the individual sticks tries to be balanced; it just wants to do a mixture of: stay with the other sticks and go where it expects they are going to go; fall over if it's leaning. So you have to move your hand if it leans, to keep the whole mess balanced. (That's if the central bank sets a nominal interest rate.)

So no. The whole point of the broomstick metaphor is that we are NOT back in the realm of just assuming we get to long run equilibrium.

Posted by: Nick Rowe | November 19, 2014 at 11:38 AM

Nick,

"Frank: there are no dynamics in this model, so we can't say."

Okay,

So we start with the fine structure:

Fi = (Sbar - S*) * (Si - Sbar)

And then pick some strategies that drivers may adopt:

1. Maximize the negative value of Fi without regard to risk

2. Drive the speed limit always

3. Try to drive the average speed

Assume that the game starts with n drivers driving the speed limit and a fixed endowment, that everyone is made aware of the speed limit simultaneously at the beginning of every turn, that everyone is made aware of the average speed from the previous turn, that everyone decides on their next speed simultaneously, and everyone picks a strategy and sticks with it until they are broke.

Finally, assume that all drivers do not initially know what strategies all the other drivers are going to adopt.

Strategy #1: After the speed limit is changed from S*(t) to S*(t+1), the driver trying to maximize his negative fine may choose a speed

Si(t+1) = n/2 x [ S*(t+1) - S*(t) ] - (n-1) x Sbar (t)

*** Note that if this driver had instead chose speed Si(t+1) = n x S*(t+1) - (n-1) x Sbar(t), then he will have single handedly brought the average speed up / down to equal the speed limit if no one else changes speed. But he will also have reduced his negative fine to 0. The really greedy driver does not maximize his negative fine by doing all of the lifting.

Strategy #2: After the speed limit is changed from S*(t) to S*(t+1), the driver trying to drive the speed limit may choose a speed

Si(t+1) = S*(t+1)

Strategy #3: After the speed limit is changed from S*(t) to S*(t+1), the driver trying to drive the average may choose a speed

Si(t+1) = Sbar(t)

With a fixed endowment, and a random distribution of strategies in driver population n, does any one strategy become dominant over time?

I think we (and the central bank) would hope that strategy #2 becomes dominant. But I don't have a program to run multivariable iterative solutions and plot probability results to determine that strategy #2 has a better than 50/50 chance of becoming dominant.

Posted by: Frank Restly | November 19, 2014 at 08:53 PM

And then there's the part about assuming any of this has something to do with actual human beings living on Earth. Whatevs, his was a beautiful mind.

Posted by: Thornton Hall | November 19, 2014 at 09:36 PM

Thornton, of course it has *something* to do with actual earth-beings. As with any model, it only captures *part* of what is going on, and, as with any model, the question is *how much*?

Posted by: Gene Callahan | November 20, 2014 at 12:27 AM

Gene: exactly. And the *how much* question is central to this post. God, the anti-intellectualism of some of these sneering artsies!

Posted by: Nick Rowe | November 20, 2014 at 07:06 AM

@Frank Restly:

> 2. Drive the speed limit always

I think that part of the premise of this model is that individual drivers do not care about the speed target in the abstract. They only care insofar as it affects their income (payments/fines).

In non-metaphor land, this reflects that the absolute price level is arbitrary.

> I think we (and the central bank) would hope that strategy #2 becomes dominant.

I don't think that's quite it. I think that us/the CB hope that drivers on average *act* as if they're following strategy #2, even if none of them is deliberately doing so. If we're also turning this into a dynamical model, then note that the cops should only infrequently change the speed target.

Also, a "low-information" dynamical model may also be interesting. At each period, drivers know the average speed (Sb) and the fine that they individually pay/receive, but *not* the speed target. I think this would converge to the target speed S* for the proper fine structure, converge to a randomly-walking Sb for an always-pay-it fine structure, and diverge for the wrong-sign fine structure.

Intuition: drivers know the payment is proportional to the square of the difference between their speed and the average speed. Drivers who pay a fine therefore minimize their expected loss by driving at the average speed; drivers who receive a payment minimize their expected loss by driving modestly away from the average speed. If drivers don't know the piecewise sign of the fine structure, they won't intentionally switch sides of the average.

For proper fines and an average speed faster than target, every driver faster than average will, next period, drive at the expected average; every driver slower will drive somewhat slower than average. This brings the average down, iteratively, until it reaches target.

For wrong-sign fines, the iteration works the other way, and the average diverges away from target.

For always-pay-it fines, drivers on each side of the average will wish to drive the average speed next period, but that average is not itself pinned down in that perturbations accumulate.

The economic intuition seems to be that the equilibrium is most robust if individuals have a modest incentive to "be wrong" on the target side of the average.

Posted by: Majromax | November 20, 2014 at 10:04 AM

@Majormax

2. Drive the speed limit always

The reason that I included this strategy is that drivers who always drive the speed limit never pay a positive fine. If driver always sets Si = S*, then their fine is Fi = (Sbar - Si)*(Si - Sbar), which is always either 0 or a negative number. On an infinite time scale, a driver who follows strategy #2 never loses his/her endowment. The question I asked was can other strategies become dominant when strategy #2 is a never lose strategy.

Obviously if strategy #2 is the dominant strategy from the start of the game, it will remain so. But if we start with a random mix of strategies #1, #2, and #3 players, what does that mix look like after several turns and lost endowments? Can we end up with more strategy #1/#3 players than strategy #2 players by the time the cops change the speed limit again?

"If we're also turning this into a dynamical model, then note that the cops should only infrequently change the speed target."

Agreed. But what measure of frequency should the cops use? Should they change the speed limit only when the number of players (n) has stabilized, when the average speed has stabilized, or some other measure?

Posted by: Frank Restly | November 20, 2014 at 04:03 PM

Frank: if the cops get the sign right, the driver's loss function is:

Li = (Si-Sbar)^2 + (Si-Sbar)(Sbar-S*)

The fine (the second term) is not the only thing that matters.

And n is not the number of drivers. It is the fraction of drivers who are irrational.

Posted by: Nick Rowe | November 20, 2014 at 05:19 PM

Nick,

I am a little confused on the signs. When the second term is negative the driver realizes a negative loss (a positive gain). But shouldn't driving faster than average also generate a negative loss (positive gain)? I get where I am going more quickly than most other people driving faster than average.

Li = Abs(Sbar - Si) * (Sbar - Si) + (Si - Sbar)(Sbar - S*)

Abs() is the absolute value. Driving faster than Sbar generates a negative loss (positive gain) in the first term.

Also, we are mixing units here. In the second term we are paying / receiving fines (presumably in some currency). In the first term, no monetary transaction takes place and so the net monetary loss is partially a function of how each individual driver values an extra bit of speed in monetary terms.

For instance, if I place a value X on each extra mile an hour of speed I can drive above the average, my loss function would be:

Li = X^2 * Abs(Sbar - Si) * (Sbar - Si) + (Si - Sbar)(Sbar - S*)

Posted by: Frank Restly | November 20, 2014 at 06:44 PM

Just realized that I made a slight misstatement. It may be more important for me to drive faster than the median rather than the average.

Imagine 9 drivers on the road and one person is driving a Ferrari at 200 MPH. The other eight (including me) are driving Volvos at 50 MPH. Average is (200 + 50*8)/9 = 66.6 MPH. Median is 50 MPH. Driving 51 MPH puts me driving faster than 7 other people.

Posted by: Frank Restly | November 20, 2014 at 06:58 PM

Frank: you do not understand this at all. Please stop. You are just depressing and frustrating me.

Posted by: Nick Rowe | November 20, 2014 at 10:38 PM

Sorry Nick if you are depressed. From reading through all of the prior comments it appears that you

"...changed the loss function to give drivers a stronger preference for driving at the average speed."

Okay. I didn't realize that when you replied. I thought you were adding the second term to represent some added monetary gain from driving faster / getting to a destination more quickly. Though I still don't understand how the loss ( (Si-Sbar)^2 ) manifests itself if it is not part of the fine levied by the cops.

Posted by: Frank Restly | November 20, 2014 at 11:37 PM

Nick weeps in frustration.

Frank:

There have ALWAYS been TWO (and ONLY two) things drivers dislike: driving at a speed different from the average speed; paying fines. When I changed the loss function all I did was delete a "(1/2)", so that their dislike of driving at a speed different from the average speed became stronger.

Why do they dislike driving at a speed different from the average speed? Because they do. It's MY model. Plus, it's a METAPHOR. Or, if you really need an invented reason: it's because it increases the risk and severity of an accident, plus the stress of driving.

Driving at a speed different from the average speed creates a PSYCHIC loss. Just like paying a fine creates a psychic loss, because it means you can't afford to buy as many neat things that would create a psychic benefit.

So:

ASSUME Li = (Si-Sbar)^2 + (Si-Sbar)(Sbar-S*).

Minimise Li with respect to Si. (You know some calculus, right? Take the derivative, and set it = 0). So you solve for Si as a function of Sbar and S*.

Now assume that Sbar = (1-n)Si + n(S*+e)

Solve for Sbar as a function of S*, n, and e. (You know how to do simultaneous equations, right?)

Take the limit of your solution for Sbar as n approaches 0.

Now change the loss function to:

Li = (Si-Sbar)^2 + (Si-Sbar)(S*-Sbar)

and repeat the exercise. (Remember the distinction between "in the limit" and "at the limit", because it's a discontinuous function.)

Posted by: Nick Rowe | November 21, 2014 at 06:29 AM

And your fine for misunderstanding my last comment, or getting the math wrong, and failing your driving test, will be a permanent ban from driving. So read it very very carefully. And show your work.

And if I see an "ABS" operator anywhere in your work, you automatically fail the test. Because these cars do not have ABS.

Posted by: Nick Rowe | November 21, 2014 at 06:37 AM

> Take the limit of your solution for Sbar as n approaches 0.

If I'm not missing something, this is very similar to my worked example above; nothing explosive happens at n=0.

Posted by: Majromax | November 21, 2014 at 09:52 AM

Majro: if the cops get the sign wrong, I get: Sbar=2S* in the limit (for e < 0 as n goes to 0, and (mis?)using L'Hopital's rule), but Sbar=S* at the limit for n=0. But I never trust my math.

Posted by: Nick Rowe | November 21, 2014 at 12:43 PM

Nick,

Li = (Si - Sbar)^2 + (Si - Sbar)(Sbar - S*) = (Si - Sbar)(Si - Sbar + Sbar - S*)

Li = (Si - Sbar)(Si - S*) = Si^2 - Si x (Sbar + S*) + S* x Sbar

dLi / dSi = 2Si - Sbar - S* = 0

Si = (Sbar + S*) / 2

Sbar = (1 - n) x Si + n x (S* + e)

Sbar = (1 - n) x (Sbar + S*)/2 + n x (S* + e)

Sbar/2 x (n + 1) = S*/2 x (1 - n) + n x (S* + e)

Sbar/2 x (n + 1) = S*/2 x (n + 1) + n x e

Sbar = S* + 2e * ( n / n + 1 )

As n approaches 0, Sbar approaches S*. Got it.

Also, since Si = (Sbar + S*) / 2

Then, Li = -[(S* - Sbar)^2]. With the loss function you prescribe, the total loss / gain of the group's "psychic well being" is no longer preserved like it was with just the monetary fine. In this case, all drivers gain.

We can achieve the same effect by loading everyone on a commie bus with one driver and sing "Kumbaya" :-)

Posted by: Frank Restly | November 21, 2014 at 01:04 PM

Frank: well done! your math is better than mine! (I made an arithmetic mistake (as usual) in my 12.43)

Posted by: Nick Rowe | November 21, 2014 at 01:16 PM

Nick,

I realize that this is just a metaphor, but have you considered that with a fine / reward system that is comprised of a resource that is not limited in supply (psychic well being), the incentive to increase the amount of award you receive is diminished?

Also, we are treating Sbar and Si as independent variables when they are not.

Posted by: Frank Restly | November 21, 2014 at 01:51 PM

Nevermind, I see it now, Sbar = (1-n) x Sr + n x (S* + e)

Sr = the rational speed we calculated = (Sbar + S*)/2

Posted by: Frank Restly | November 21, 2014 at 02:07 PM

Assuming symmetrical fines is exactly the same thing as saying that the discounted future value equals the present value.

Isn't it?

Posted by: Miami Vice | November 21, 2014 at 04:29 PM

Also, when you say "get the sign wrong" that's because it's not relative to the rate of inflation. It's not the real rate.

If the rate of inflation goes from 2% to 1% and the nominal rate falls but by less than the rate of inflation the sign is wrong. Even though the nominal rate moved in the right direction.

I hope I'm not making a fool of myself.

Posted by: Miami Vice | November 21, 2014 at 06:17 PM

The wrong sign is attributable to the fact that the probability the fine is asymmetrical is zero in one case and non zero in the other. That difference is what determines the right strategy to minimize the loss function and whether you have a robust or fragile equilibrium.

The cops should be a player of the game in both or neither or have I misunderstood?

Posted by: Miami Vice | November 21, 2014 at 07:07 PM

A somewhat lengthy reply: [Link here thanks Mike NR]

Posted by: Mike Freimuth | November 21, 2014 at 07:34 PM

Miami: the cops are not really players in this game. The cops decide what the game's payoff matrix looks like, before play begins. They are like the cop in prisoners' dilemma.

"If the rate of inflation goes from 2% to 1% and the nominal rate falls but by less than the rate of inflation the sign is wrong. Even though the nominal rate moved in the right direction."

Correct. The Howitt/Taylor principle says it must move by *more* than 1-for-1.

"Assuming symmetrical fines is exactly the same thing as saying that the discounted future value equals the present value. Isn't it?"

No. Just saying the central bank responds symmetrically to inflation rising above target and falling below target. It dislikes both equally. It's the Bank of Canada, not the ECB.

Posted by: Nick Rowe | November 22, 2014 at 07:19 AM

Nick, here is paper from Matt Jackson and co-authors that essentially argues that *all* (Bayesian) Nash equilibria are stable! http://web.stanford.edu/~jacksonm/emailgame.pdf

Instead of considering small trembles in actions, they consider small trembles in payoffs. For any game where you have the payoffs slightly wrong, there is a sequence of epsilon-Nash equilibrium that approaches the Nash equilibrium as the payoff mistake approaches zero.

My (very quick, and possibly wrong) attempt at reconciling this paper with your post is as follows: Your introduction of a fraction of non-rational types introduces a payoff discontinuity that violates the setup of the Jackson paper.

Posted by: Evan | November 24, 2014 at 08:12 PM

Reality is that in stage one of the game... forward price targets

Present Value(t) * (1+r)^{(T-t)}

whereby r is the risk free rate (nominal rate) and therefore... an increase in r would lead to an increase in future price and a lower r would lead to a lower price.

Posted by: Derivs | December 13, 2014 at 06:03 AM