tag:blogger.com,1999:blog-6837159629100463303.post6094295199392582535..comments2023-06-18T01:25:08.748-07:00Comments on Information Transfer Economics: I. Quantity theory and effective field theoryJason Smithhttp://www.blogger.com/profile/12680061127040420047noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-6837159629100463303.post-7660161943886939192016-01-18T12:53:53.373-08:002016-01-18T12:53:53.373-08:00Ah! Ok. That's what you meant by "it'...Ah! Ok. That's what you meant by "it's not really a Taylor expansion" ... I thought perhaps that was a comment on my attempt with substituting u=D/S.<br />Thanks!Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-21318979334886711072016-01-18T12:25:32.924-08:002016-01-18T12:25:32.924-08:00As I said, it's not really a Taylor expansion....As I said, it's not really a Taylor expansion. A Taylor expansion around D = S is just a conceptual jumping-off point for people not familiar with effective field theory. It's an effective field theory expansion (the Lagrangian that goes with a Feynman diagram expansion) -- you write down all the terms consistent with the symmetry principle (invariant under D→αD, S→αS). But neither the coefficients nor the terms are necessarily "small" ... unless you have good physical reasons for it.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-26191767310706802932016-01-18T11:25:06.111-08:002016-01-18T11:25:06.111-08:00Thanks Jason.
You write:
"In reality it'...Thanks Jason.<br /><br />You write:<br /><br />"In reality it's not a Taylor expansion and you need terms like (D/S) dD/dS ... but I mention those as generalized 2nd order terms immediately after."<br /><br />This is the part immediately after, right?:<br /><br />"Note that higher order derivatives by themselves are not consistent with homogeneity. If we take D→α D, S→α S means that d²D/dS² → (1/α) d²D/dS². Terms like D d²D/dS² would be necessary, which we'll subsume into "generalized" second order terms."<br /><br />Sorry to be so slow on the uptake here. What if g(D/S) = (D/S-1)^3? (I'm trying to make a super simple example, but if you've got a better one...)<br /><br />I don't see any partials, so it's not a 2D Taylor series:<br />http://www.math.ubc.ca/~feldman/m200/taylor2dSlides.pdf<br /><br />So you're just expanding about either D or S then (rather than the ratio as a whole)? E.g., I need to replace my<br /><br />g'(u) with g'(u)*du/dD<br /><br />?Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-78524678562311404102016-01-17T16:35:21.370-08:002016-01-17T16:35:21.370-08:006. I call both "kappa" and k the IT inde...6. I call both "kappa" and k the IT index, but I now try to be consistent in using k for a general index that is in the numerator on the RHS of the information transfer equation.<br /><br />But it really doesn't matter as the map 1/x : R⁺ → R⁺ is a diffeomorphism :)Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-20440362106406530972016-01-17T16:28:40.091-08:002016-01-17T16:28:40.091-08:003. Yes.
4. Yes.
5. Yes, it's an alternative ...3. Yes.<br /><br />4. Yes.<br /><br />5. Yes, it's an alternative approach based on "effective field theory".Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-59088928855237169352016-01-17T16:27:21.836-08:002016-01-17T16:27:21.836-08:002. That is expanding around the equilibrium where ...2. That is expanding around the equilibrium where demand = supply.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-73265810749283558942016-01-17T16:26:39.345-08:002016-01-17T16:26:39.345-08:001. I'm not sure what your point is here. If it...1. I'm not sure what your point is here. If it is that the form doesn't look exactly the same -- you can expand all the terms (u-1)^n and collect powers of u. In reality it's not a Taylor expansion and you need terms like (D/S) dD/dS ... but I mention those as generalized 2nd order terms immediately after.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-53762322281186831822016-01-16T23:17:26.972-08:002016-01-16T23:17:26.972-08:008) "Our process with equation (3) was analogo...8) "Our process with equation (3) was analogous to writing down every consistent with long run neutrality of money (analogous to a symmetry)."<br /><br />I think you forgot the word "term."Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-91974761507896315702016-01-16T23:10:26.851-08:002016-01-16T23:10:26.851-08:001) "A Taylor expansion of g around D=S result...1) "A Taylor expansion of g around D=S results in an equation of the form..."<br /><br />So we could consider g(u) with u=D/S, and thus a Taylor expansion around D=S would be an expansion around u=1:<br /><br />g(u-1) = g(1) + g'(1)*(u-1) + [g''(1)*(u-1)^2]/2 + ...<br /><br />Where g' = dg/du and g'' = d(dg/du)/du, etc.<br /><br />Correct?<br /><br />2) Why about D=S (equivalently u=1)?<br /><br />3) Also, what do you mean by MB here? Base money or base money minus reserves (currency)? (i.e. which did you use to make the plots of your model results with? It looks pretty good, so I'd assume you meant currency). <br /><br />4) You use M0 in the title to a plot above, but I assume that you don't mean currency by that, but instead a normalization parameter?<br /><br />5) I read this before (I think), but it's a fairly mind blowing post. I'd forgotten about it I guess. Basically it's an alternative derivation (alternative to the information theory approach) of the price level formula based on long run neutrality of money, and the kinds of differential equations that implies.<br /><br />6) "κ (which I'll just call the IT index for now)"<br />I know we've been over this before, but 1/κ = k you now call the IT index, right? (I just checked your draft paper, and yes that seems true)<br /><br />7) I had all kinds of questions about the plots, but I'm pretty sure I resolved them all by staring at them long enough... my 1st attempt to comment (w/ those questions) was lost by me pressing the wrong key... it may have benefited other readers.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-28410713611163494702014-02-16T16:39:07.151-08:002014-02-16T16:39:07.151-08:00I did an update to convert the equations to mathja...I did an update to convert the equations to mathjax ... hopefully that didn't mess too much up.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.com