In looking for inequality data for the previous post, I came across a presentation at Emmanuel Saez's homepage with the slide that looked very much like the information transfer model turned on its side. It was on matching theory; here is slide from Pascal Michaillat and Saez alongside a slide describing the information transfer model:Editor's note, 6 Feb 2017:This is the beginning of a post that approached the topic incorrectly. I have subsequently wrote up a new approach a couple of years later; it is available here.

I made some suggestive edits to the slide from Michaillat and Saez. Can we use the information transfer model to arrive at the same conclusions as matching theory does in the labor market? Yes, we can. First we'll start with some matching theory. Essentially, I'd like to derive a matching function in Cobb-Douglas form

$$

\text{(1) } h = M(U, V) \sim U^{\alpha} V^{1-\alpha}

$$

where $h$ is the number of hires, $U$ is the number of unemployed and $V$ is the number of job vacancies (assuming constant returns to scale, i.e. the exponents are $\alpha$ and $1-\alpha$). The diagram from Michaillat and Saez above shows what we mean by "matching". There are so many job applicants and so many job vacancies and the matching function describes how these match up and turn into new hires. Where do we start in the information transfer model?

...

Please refer to this post.

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Please refer to this post.

The equation after (1) is the solution to the differential equation

ReplyDelete$$

\frac{dJ}{dL} = \frac{1}{\kappa} \; \frac{J}{L}

$$

I forgot to explicitly say what the matching function I used is. We take the equilibrium situation to be $E = J_{ref} = L_{ref}$ such that the unemployed and job vacancies are perturbations from a reference level of employment. Therefore

ReplyDelete$$

M(U, V) = (E+U)^{1-1/\kappa}(E+V)^{\kappa - 1} - \lambda_{ref}

$$

This has the effect of subtracting off the large overall constant (the existing matches, i.e. the currently employed) that is ignored in the Cobb-Douglas form. I subsequently normalized to the civilian labor force $C$, exchanging $\lambda_{ref}$ for $\lambda$ the fit parameter.

Jason, starting from:

ReplyDelete(1) J/Jref = (L/Lref)^(1/kappa)

raise both sides to the power of kappa:

(2) (J/Jref)^kappa = L/Lref

Now divide the left side of (2) by the left side of (1) and divide the right side of (2) by the right side of (1) gives us:

(3) (J/Jref)^(kappa-1) = (L/Lref)^(1-1/kappa)

Now divide the left side of (3) by the right side of (3):

(4) [(J/Jref)^(kappa-1)]*[(L/Lref)^(1/kappa-1)] = 1

My equation (4) looks like your 3rd equation from the top, except the exponent I have on (L/Lref) is the negative of the exponent you have on (L/Lref).

I should have written:

Delete"Now divide both sides of (3) by the right side of (3):"

You need to use the price equation

DeleteP = f(L)

P = f(J)

Well since your 2nd equation in the post is the solution to the "floating information source" (i.e. general equilibrium) problem, the associated price equation should be:

DeleteP = (1/kappa)*J/L

It seems P is a function of both L and J together.

Do I know ahead of time what the numerical value P is or something?

You can solve the general equilibrium solution and write P as a function of either J or L alone.

DeleteJason, I must be trying your patience at this point (so perhaps don't bother looking at the rest of this comment just yet... I'll come back and double check this later to see if I screwed up). But I'll leave a comment anyway so I have a record here of what I did:

DeleteLooking at the general equilibrium solution for P in terms of J, and then solving for J we have:

J = Jref*(kappa*Lref*P/Jref)^(1/(1-kappa))

Substituting that in for my (4) in my comment above, (repeated here for convenience):

(4) [(J/Jref)^(kappa-1)]*[(L/Lref)^(1/kappa-1)] = 1

we get:

(5) [(kappa*Lref*P/Jref)^(-1)]*[(L/Lref)^(1/kappa-1)] = 1

Which gives the usual general equilibrium expression for P in terms of L:

(6) P = (Jref/(kappa*Lref))*[(L/Lref)^(1/kappa-1)]

So it looks as if my (4) checks out. Essentially I have:

X/Y = 1

and you have

X*Y = 1

I don't know how you get yours and if they are both true then X = Y = 1 (assuming X and Y are both non-negative real numbers)

I will check it out more closely. The key is

DeleteP' * P = (dL/dJ)*(dJ/dL) = 1

but it should be consistent with inverting J = f(L).

I will have a look at the code as well.

This comment has been removed by the author.

DeleteYou write:

DeleteP' * P = (dL/dJ)*(dJ/dL) = 1By P' you mean 1/P, right?

You can write P = f(L) like this:

P = (Jref/(k*Lref)) * (L/Lref)^(1/k - 1)

You can write P = f(J) like this:

P = (Jref/(k*Lref)) * (Jref/J)^(k - 1)

Now if you write:

(1/P)*P = (1/f(J))*f(L) = [(J/Jref)^(k-1)]*[(L/Lref)^(1/k - 1)] = (dL/dJ)*(dJ/dL) = 1

Which agrees with what I came up with, but disagrees with what you have.

But as far as I can tell, you don't need to invoke the price equation since you can start with your first equation:

J/Jref = (L/Lref)^(1/k)

Raise both sides to the (k-1):

(J/Jref)^(k-1) = (L/Lref)^(1 - 1/k)

And then divide the left side by the right side:

[(J/Jref)^(k-1)]*[(L/Lref)^(1/k - 1)] = 1

Again, the same result. I'll be amazed if I've done it wrong now for the 4th time in a row.

But I've done that kind of thing before!

So I'd say go ahead and give it another check to see. I think you got the sign of the exponent wrong on L/Lref.

Here's a more human readable format

Delete