Pi/P = M/P - Qi

Pi is the ith producer's nominal price; P is the average price, Qi is the ith producer's quantity. M is the money supply. Each has no variable costs, only fixed, for simplicity. So we know that the optimal relative price - that which maximizes revenues- is simply half of M/Pi, as is the optimal quantity. In a world without menu costs, the price level adjusts so that M/P = 2. Suppose it were less than 2. Then each producer's optimal relative price is less than 1, so all would cut prices until M/P went back to 2 , at which point each would be happy charging what others are charging.

I want to add menu costs and examine two cases. We cut M in half first. Then we double it.

First case: Here we have two equilibria, one where none adjust and one where all adjust, if menu costs lie in the range .25-1. Suppose all others adjust optimally cutting their price in half. Then real money is unchanged at 2. If you also adjust, you make profits of 1- MC (MC for menu cost). If not, you charge a relative price of 2 and sell nothing, for profits of 0. All adjust is a symmetric Nash equilibrium if MC is less than 1.

If all others fail to adjust, each faces demand curves with intercepts of 1. The firm that doesn't adjust charges a relative price of 1 and sells nothing. The firm that adjusts charges a relative price of 1/2 and sells 1/2, earning profits of 1/4 -MC. So if menu costs are greater than 1/4, none adjusting is also a symmetric Nash equilibrium. This case obviously echoes Ball and Romer's much more sophisticated "Sticky prices as Coordination failure", which inspired the whole thing. Now, there's lots of experimental evidence for coordination failure, so let's say with a negative money shock, we get the sticky price equilibrium, with output falling to 0!

The second case, however, the positive money shock, is very different. So double the money supply. If all others adjust, I face a demand curve with an intercept of 2. If I adjust as well, I gets profits of 1- MC. If I don't adjust, I charge a relative price of 1/2 and sell 3/2, for profits of 3/4 . So I - and everyone else - will NOT want to adjust if all others adjust just in case MC is greater than 1/4. In other words, all adjust is not an SNE.

On the other hand, suppose all but you fail to adjust. You face a demand curve with an intercept of 4. If you adjust, your relative price is 2 and you sell 2, for profits of 4 - MC. If you fail to adjust, you charge a relative price of 1 and sell 3, for profits of 3. So for MC less than 1, you would want to adjust if all others fail to adjust, so stickiness is not an SNE either.

In the second case, for the positive money shock, we have partial adjustment. We have a chicken game, with a mixed-strategy SNE in which each adjusts with a certain probability ( or a proportion adjust equal to that same probability).(Is this a possible micro-foundation for a Calvo fairy?)

So: sticky prices for negative shocks; partial price adjustment - and so less of an output increase - with positive shocks.

It's almost like the Old Keynesian idea of a reversed-L- shaped aggregate supply, except here the vertical segment of the L is leaning to the right.

OK so it's silly. But think about the reason for the asymmetry: It comes about because the benefit of adjusting optimally

*gross of menu costs*is greater when demand is greater. For the negative shock, demand is greater when all adjust, so if the range is right we get two equilibria: MC greater than benefit of adjusting when when demand is low (because none adjust); MC smaller than benefit of adjusting when demand is high (because all are adjusting).

For the positive shock, demand is greater when

*none*adjust, so the benefits of adjusting gross of menu costs are greater when none adjusts, and smaller when all adjust - but you. That's why we get the chicken structure.

I know I need pictures, as Nick Rowe told me about my last post. I will work on it.!

Happy New Year!

## 8 comments:

Good post Kevin. But I am suspicious of your result. I *suspect* it comes from your particular functional form plus the increase in M being "bigger" (in absolute size) than the decrease in M.

This is how I think of it: My old post with pictures is here: http://worthwhile.typepad.com/worthwhile_canadian_initi/2010/01/macroeconomics-with-monopolistic-competition-in-pictures.html But I didn't have menu costs in that model.

Let y be our firm's output (same as your Qi), and Y be average output of all the other firms. Number of firms is very large. In Symmetric Nash Equilibrium, y=Y.

Real Total Revenue for our firm is some function of y and Y: T(y,Y).

Real Marginal Revenue for our firm is defined as dT/dy, holding Y constant. It is also a function of y and Y. Call it M(y,Y).

We can draw a "Marginal Revenue *Curve*" as M(y,Y) varying y and holding Y constant. Standard partial equilibrium micro curve. It must slope down, to satisfy the second order condition.

We can also draw the macro "Marginal Revenue *Locus*" as M(y,Y) varying both y and Y together, for y=Y. It must slope down, if we want a unique and stable macroeconomic equilibrium in a model with no menu costs.

If marginal costs are zero, and there are no menu costs, then SME is defined as Y* such that M(Y*,Y*)=0.

If the MR micro *curve* is steeper than the MR macro *locus*, then we get a positive feedback system. If all other firms increase Y, our firm will want to increase y too. And if all other firms decrease Y, our firm will want to decrease Y too.

In other words, an increase in Y by other firms will cause the MR micro curve for our firm to shift right. But if our menu costs are bigger than the area of the triangle defined by our MR curve and the zero axis, we will not want to change our firm's price.

With positive feedback, plus menu costs, we get a range of macroeconomic equilibria around Y*. The LRAS curve vertical is but thick. The left side is Y*min, and the right side is Y*max.

Start at Y*. A small increase (or decrease) in M will cause no change in P, and an increase (or decrease) in Y. A big increase (or decrease) in M will cause some percentage of firms to raise (or lower) their prices, until Y falls (or rises) back to the edge of the thick LRAS curve at Y*max (or Y*min).

I *think* that is right.

Damn! When I wrote "It must slope down, if we want a unique and stable macroeconomic equilibrium in a model with no menu costs." That is only true in a model (like yours) with zero marginal costs. More generally, for a unique and stable macro equilibrium, the Marginal Revenue *locus* must cut the Marginal Cost *locus* (similarly defined) from above.

The MR function in your model is (I think) M(y,Y)= 1+Y-2y

So the slope of the MR micro curve is dM/dy = -2.

And the slope of the MR macro locus is dM/dy+dM/dY (y=Y) = -1.

So your model has positive feedback, and has a unique stable macro equilibrium with no menu costs.

Thanks, Nick, for your comments. I did have a nagging worry about that difference in the absolute amounts of contraction and expansion of M, but I guess I put it out of my mind. And the functional form: well, I thought, as I say in the last part, that fact that the cost of suboptimal adjustment is greater when demand is high with this functional form might be more general, and that if so, we might expect asymmetry. I will study your different approach - I haven't quite got it yet!

Kevin: on second thoughts, don't worry about your functional form. It's OK. Nice and simple.

But it's a linear MR function, so you need to compare increases and decreases of M with the same absolute size, I think.

What I have learned from your post is to think about mixed strategy equilibria. I thought I understood this stuff (I actually published on this stuff in the late 1980's), but I hadn't thought about that. That was new to me. It was an interesting point. It could be developed.

We could change the assumptions slightly, so that different firms have different menu costs. A distribution. Then we solve for the percentage of firms that change their prices.

That percentage will be less than 100%, and (given positive feedback) money will still be non-neutral in aggregate.

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