tag:blogger.com,1999:blog-6837159629100463303.post3008847540373879509..comments2023-06-18T01:25:08.748-07:00Comments on Information Transfer Economics: Information transfer and Cobb-Douglas matching functionsJason Smithhttp://www.blogger.com/profile/12680061127040420047noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-6837159629100463303.post-30188331859432005782016-02-07T17:41:21.863-08:002016-02-07T17:41:21.863-08:00Here's a more human readable format<a href="http://banking-discussion.blogspot.com/2016/02/algebra-part-1.html" rel="nofollow">Here's a more human readable format</a>Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-509196899960735652016-02-06T22:24:00.746-08:002016-02-06T22:24:00.746-08:00You write:
P' * P = (dL/dJ)*(dJ/dL) = 1
By P...You write:<br /><br /><i>P' * P = (dL/dJ)*(dJ/dL) = 1</i><br /><br />By P' you mean 1/P, right?<br /><br />You can write P = f(L) like this:<br /><br />P = (Jref/(k*Lref)) * (L/Lref)^(1/k - 1)<br /><br />You can write P = f(J) like this:<br /><br />P = (Jref/(k*Lref)) * (Jref/J)^(k - 1)<br /><br />Now if you write:<br /><br />(1/P)*P = (1/f(J))*f(L) = [(J/Jref)^(k-1)]*[(L/Lref)^(1/k - 1)] = (dL/dJ)*(dJ/dL) = 1<br /><br />Which agrees with what I came up with, but disagrees with what you have.<br /><br />But as far as I can tell, you don't need to invoke the price equation since you can start with your first equation:<br /><br />J/Jref = (L/Lref)^(1/k)<br /><br />Raise both sides to the (k-1):<br /><br />(J/Jref)^(k-1) = (L/Lref)^(1 - 1/k)<br /><br />And then divide the left side by the right side:<br /><br />[(J/Jref)^(k-1)]*[(L/Lref)^(1/k - 1)] = 1<br /><br />Again, the same result. I'll be amazed if I've done it wrong now for the 4th time in a row.<br /><br />But I've done that kind of thing before!<br /><br />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.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-46468657473601166592016-01-27T12:39:54.098-08:002016-01-27T12:39:54.098-08:00This comment has been removed by the author.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-11142399879392809922016-01-26T15:50:36.849-08:002016-01-26T15:50:36.849-08:00I will check it out more closely. The key is
P&#...I will check it out more closely. The key is <br /><br />P' * P = (dL/dJ)*(dJ/dL) = 1<br /><br />but it should be consistent with inverting J = f(L).<br /><br />I will have a look at the code as well.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-17936063327903294492016-01-26T14:42:17.220-08:002016-01-26T14:42:17.220-08:00Jason, I must be trying your patience at this poin...Jason, 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:<br /><br />Looking at the general equilibrium solution for P in terms of J, and then solving for J we have:<br /><br />J = Jref*(kappa*Lref*P/Jref)^(1/(1-kappa))<br /><br />Substituting that in for my (4) in my comment above, (repeated here for convenience):<br /><br />(4) [(J/Jref)^(kappa-1)]*[(L/Lref)^(1/kappa-1)] = 1<br /><br />we get:<br /><br />(5) [(kappa*Lref*P/Jref)^(-1)]*[(L/Lref)^(1/kappa-1)] = 1<br /><br />Which gives the usual general equilibrium expression for P in terms of L:<br /><br />(6) P = (Jref/(kappa*Lref))*[(L/Lref)^(1/kappa-1)]<br /><br />So it looks as if my (4) checks out. Essentially I have:<br /><br />X/Y = 1<br /><br />and you have<br /><br />X*Y = 1<br /><br />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)Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-67402605678920334342016-01-26T10:49:54.576-08:002016-01-26T10:49:54.576-08:00You can solve the general equilibrium solution and...You can solve the general equilibrium solution and write P as a function of either J or L alone.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-90102966235856213762016-01-26T00:45:22.870-08:002016-01-26T00:45:22.870-08:00Well since your 2nd equation in the post is the so...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:<br /><br />P = (1/kappa)*J/L<br /><br />It seems P is a function of both L and J together.<br /><br />Do I know ahead of time what the numerical value P is or something?Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-9373168756432430332016-01-25T22:55:10.336-08:002016-01-25T22:55:10.336-08:00You need to use the price equation
P = f(L)
P = ...You need to use the price equation<br /><br />P = f(L)<br /><br />P = f(J)Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-38645945966039530142016-01-25T21:13:55.853-08:002016-01-25T21:13:55.853-08:00I should have written:
"Now divide both side...I should have written:<br /><br />"Now divide both sides of (3) by the right side of (3):"Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-28451383942091806682016-01-25T21:10:40.154-08:002016-01-25T21:10:40.154-08:00Jason, starting from:
(1) J/Jref = (L/Lref)^(1/ka...Jason, starting from:<br /><br />(1) J/Jref = (L/Lref)^(1/kappa)<br /><br />raise both sides to the power of kappa:<br /><br />(2) (J/Jref)^kappa = L/Lref<br /><br />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:<br /><br />(3) (J/Jref)^(kappa-1) = (L/Lref)^(1-1/kappa)<br /><br />Now divide the left side of (3) by the right side of (3):<br /><br />(4) [(J/Jref)^(kappa-1)]*[(L/Lref)^(1/kappa-1)] = 1<br /><br />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).Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-60582373520164852192014-03-09T11:36:39.170-07:002014-03-09T11:36:39.170-07:00I forgot to explicitly say what the matching funct...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<br />$$<br />M(U, V) = (E+U)^{1-1/\kappa}(E+V)^{\kappa - 1} - \lambda_{ref}<br />$$<br />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 Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-69944644655866471072014-03-08T18:27:26.964-08:002014-03-08T18:27:26.964-08:00The equation after (1) is the solution to the diff...The equation after (1) is the solution to the differential equation<br />$$<br />\frac{dJ}{dL} = \frac{1}{\kappa} \; \frac{J}{L}<br />$$Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.com