tag:blogger.com,1999:blog-6837159629100463303.post4864722235080499929..comments2023-06-18T01:25:08.748-07:00Comments on Information Transfer Economics: The IS-LM model as an effective theory at low inflationJason Smithhttp://www.blogger.com/profile/12680061127040420047noreply@blogger.comBlogger31125tag:blogger.com,1999:blog-6837159629100463303.post-91813308066701253792016-02-06T15:59:57.182-08:002016-02-06T15:59:57.182-08:00... nobody cares, but my LPF() above should take a...... nobody cares, but my LPF() above should take all the higher order terms as an argument, not t.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-68407922545974633432016-02-06T14:44:21.931-08:002016-02-06T14:44:21.931-08:00The fiscal stimulus argument is different: there y...The fiscal stimulus argument is different: there you propose that the CB exogenously fix r and MB in a liquidity trap (i.e. when dMB/dY >> 1), so it's plain to see that's not going to be very effective monetary offset.Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-78255791852218998972016-02-06T14:31:25.883-08:002016-02-06T14:31:25.883-08:00...the nice thing about the above argument for me ......the nice thing about the above argument for me is it doesn't involve having to consider fluctuations: i.e. anything not "happening instantly." Fluctuations could cause Y to either "move first" (r up, then down) or "move last" (r down, then up, as you describe) in relation to MB and r, and you arrive at the same place: liquidity trap.Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-39500232100170291332016-02-06T14:08:50.378-08:002016-02-06T14:08:50.378-08:00We'll have
Y ~ MB^k1
and
r ~ p^k2
It seem...We'll have <br /><br />Y ~ MB^k1<br /><br />and<br /><br />r ~ p^k2<br /><br />It seems like you're assuming 0 < k1 < 1 and 0 < k2, which seems reasonable.<br /><br />Then another way to say what you do in the blue curve case is to say dY/dMB goes to 0 as MB goes to infinity, or alternatively as r goes to 0 (since k3 < 0, where r ~ MB^k3, and k3 = (k1 - 1)*k2).<br /><br />I understand you're relating the argument to IS-LM, so you break it down accordingly.Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-63245109754882045252016-02-05T17:24:48.102-08:002016-02-05T17:24:48.102-08:00E[V] = V is not quite the same thing. Ideal point ...E[V] = V is not quite the same thing. Ideal point particles have zero volume, so the expectation value of the volume operator is an interesting concept ... especially since real atoms do have volume (or at least a thermodynamic size from the thermal wavelength). That is essentially a statement that our macro concept of volume is "emergent".<br /><br />"But it sounds like there is a little bit of time dynamics assumed then?"<br /><br />No, there is no specific dynamics assumed. It just doesn't happen instantly. How "not instantly"?<br /><br />And in economics it would probably be impossible to separate the "equilibrium fluctuations" (around eq values) from the non-equilibrium transition fluctuations (transition from one eq to another). The former are perhaps even larger than the latter ...Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-87996134300655680972016-02-05T16:52:19.966-08:002016-02-05T16:52:19.966-08:00Shoot, I went and pestered you again. Oh well. You...Shoot, I went and pestered you again. Oh well. You can ignore me.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-43113963071734944972016-02-05T16:46:33.642-08:002016-02-05T16:46:33.642-08:00"Last I heard 0 ≈ 0"
Ha!
"M0 and ..."Last I heard 0 ≈ 0"<br /><br />Ha!<br /><br />"M0 and one is MB"<br /><br />Ah, excellent point! When you used "M" instead of "M0" it kind of threw me, but that makes sense, and I should have figured that out since your inflation models all work best when M = M0. (The casual reader might have even less chance than I do of catching onto that, however). Also, doesn't classic IS-LM work just fine in an all cash society away from a liquidity trap (i.e. away from r ≈ 0)? Don't you just require π ≈ 0?<br /><br />Well, maybe it doesn't if we're setting MB = M0. I don't know. <br /><br />"In general, these are all time averages or equilibrium solutions. Don't treat them as Newtonian orbits."<br /><br />I guess I usually have been assuming that F&B's "ideal point particles" condition holds (condition C8 in their 1st draft, which justifies assuming E[V] = V)<br /><br />"it will approach the thermodynamic equilibrium isothermal expansion result"<br /><br />Ah, I almost used that myself. I was thinking of an imperfect thermal coupling between the bath and the gas cylinder, so a short time delay would be required to bring the gas back to the bath's temperature after an exogenous change in volume, but you're thinking of something much much faster I think.<br /><br />But it sounds like there is a little bit of time dynamics assumed then? (especially with 10^9 "particles")? After all "approach" implies it's not truly at "thermodynamic equilibrium" during each and every 5.4e-44 second interval when reacting to an exogenous change in volume (or to a change in MB).<br /><br />OK, thanks. I'll stew on all that a bit.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-74636613788266625382016-02-05T16:14:40.850-08:002016-02-05T16:14:40.850-08:00Last I heard 0 ≈ 0
:)
But in the two markets (in...Last I heard 0 ≈ 0<br /><br />:)<br /><br />But in the two markets (inflation and interest rate models) don't have the same money supply -- one is M0 and one is MB.<br /><br />In general, these are all time averages or equilibrium solutions. Don't treat them as Newtonian orbits.<br /><br />If you do an isothermal expansion on a gas, it will induce some fluctuations in the density and temperature. After some time (quickly in physics with 10^23 particles, but slowly in economics with 10^9) it will approach the thermodynamic equilibrium isothermal expansion result.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-50816006005808284662016-02-05T15:40:01.220-08:002016-02-05T15:40:01.220-08:00Your scope conditions say this works for π ≈ 0. So...Your scope conditions say this works for π ≈ 0. So why not π = 0? In that case:<br /><br />r = i<br />k = 1<br />Y = N = (Nref/MBref)*MB (from the opening section of the post and assuming M = MB)<br /><br />In which case Y reacts just as fast as MB always. Unless there's some hidden time dynamics on Y which temporarily retards its reactions to changes in MB so it's temporarily out of G.E. for the expression resulting from<br /><br />p : Y ⇄ MB<br /><br />But Y ~ MB seems to contradict your assumption that either Y or MB could "move last" in relation to the other. Is π = 0 outside the scope conditions? Where did I go wrong?<br /><br />I must be driving you nuts by now. I won't pester you about this anymore today! (ever?). I was just surprised to see that my assumption of unstated time dynamics (the "LPF" in the comment above) was apparently not what you were thinking of, which got me thinking about simplifying by setting k=1 precisely.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-68980173983443012822016-02-05T15:34:41.867-08:002016-02-05T15:34:41.867-08:00There is no time dimension in these, so the only t...There is no time dimension in these, so the only time is implicit: Y(t) and MB(t). MB moves "quickly" only relative to Y. It depends on the size of the MB move, not its "speed".Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-86075765253335913952016-02-05T14:18:57.271-08:002016-02-05T14:18:57.271-08:00Which brings up an interesting question: what if M...Which brings up an interesting question: what if MB was a large step function in time? Your sin(dx) analogy seems to indicate Y should move quickly (be a step itself actually), simply because MB is large, and thus thus the higher order terms immediately make a big contribution. Or do you think it would rise slowly?Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-23989533758473408582016-02-05T13:55:43.354-08:002016-02-05T13:55:43.354-08:00Ah, Ok.
To continue your analogy, I was imaginin...Ah, Ok. <br /><br />To continue your analogy, I was imagining something different. Something like this:<br /><br />f(x(t)) = x(t) + LPF(t)*[-x(t)^3/3! + x(t)^5/5! - ...]<br /><br />Where LPF(t) is some unspecified low pass filter of its argument with unity gain at frequency=0.<br /><br />So that if x(t) is a step function (say), then f(x) -> sin(x) as t -> infinity, but<br />f(x) ≈ x for t << 1 <br /><br />So then in the post, the unspecified LPF would be what causes Y to react more slowly (to say a step change in MB).<br /><br />OK, thanks.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-47155474345524635802016-02-05T13:28:46.887-08:002016-02-05T13:28:46.887-08:00"Restored" is not a technical mathematic..."Restored" is not a technical mathematical term.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-64301224190322850302016-02-05T13:26:36.333-08:002016-02-05T13:26:36.333-08:00If dx << 1, then
sin(dx) ≈ dx
But if dx ~ ...If dx << 1, then<br /><br />sin(dx) ≈ dx<br /><br />But if dx ~ 1, then<br /><br />dx → sin(dx)<br /><br />It is "restored". The sin(dx) form always applied; for a brief period the form dx was valid.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-50729310868470590622016-02-05T13:19:17.249-08:002016-02-05T13:19:17.249-08:00Just one last thing bothers me. In the comments ab...Just one last thing bothers me. In the comments above you write:<br /><br />"But the general equilibrium solution always applies..."<br /><br />But then why do you state in the post:<br /><br />"As Y rises, interest rates will come back up and equation (1) will hold -- general equilibrium restored."<br /><br />What is general equilibrium "restored" from if it already "always applies?"Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-65769717250982069722016-02-05T13:07:54.091-08:002016-02-05T13:07:54.091-08:00TL:DR version: I'm satisfied! (you'll be g...TL:DR version: I'm satisfied! (you'll be glad to know!)<br /><br />"The result is as given in the paper."<br /><br />From the paper (pg 7, section 2.1), equations (2.14) and (2.15) produce a function for P in terms of ΔD, with the same form as my eq (2) expressing r in terms of ΔY. Likewise eqs (2.16) and (2.17) express P in terms of ΔS with the same form as my eq (3) expressing r in terms of ΔMB. Namely, exponentials.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-29395684123833641492016-02-05T12:04:16.787-08:002016-02-05T12:04:16.787-08:00I guess I don't understand your question.
The...I guess I don't understand your question.<br /><br />The curves I plotted in the graphs are notional curves. The result is as given in the paper. I probably should have drawn them free hand like here:<br /><br /><a href="http://informationtransfereconomics.blogspot.com/2015/10/info-eq-101.html" rel="nofollow">http://informationtransfereconomics.blogspot.com/2015/10/info-eq-101.html</a>Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-1882444996353773242016-02-05T11:13:14.348-08:002016-02-05T11:13:14.348-08:00Also, this seems rather appropriate as well...Also, <a href="http://krugman.blogs.nytimes.com/2011/09/18/ive-never-actually-seen-the-resemblance/?_r=0" rel="nofollow">this</a> seems rather appropriate as well...Todd Zorickhttps://www.blogger.com/profile/10976192775890569092noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-17997819902231379372016-02-05T11:08:35.737-08:002016-02-05T11:08:35.737-08:00All hail Krugtron the InvincibleAll hail <a href="http://noahpinionblog.blogspot.com/2013/04/krugtron-invincible.html" rel="nofollow">Krugtron the Invincible</a> Todd Zorickhttps://www.blogger.com/profile/10976192775890569092noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-31975546907389977602016-02-05T10:55:15.044-08:002016-02-05T10:55:15.044-08:00Does the function you plotted to create both blue ...Does the function you plotted to create both blue curves, i.e. your f in <br /><br />r = f(ΔY)<br /><br />Have the same form as my eq (2)? If not, what form?Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-19704095383293910782016-02-05T10:45:45.435-08:002016-02-05T10:45:45.435-08:00"r = f(ΔY)"
My eq. (2) has that form.
..."r = f(ΔY)"<br /><br />My eq. (2) has that form.<br /><br />"But the general equilibrium solution always applies, the partial equilibrium solutions are approximations to it for small values of ΔY."<br /><br />In the ideal gas law, it's not an approximation for the isothermal case, is it? ΔW (the energy transferred from the gas in the cylinder to the bath when the piston is moved (i.e. when the volume is changed), so as to keep the gas at a constant temperature) is described exactly by an equation like (2) for any sized ΔW, isn't it?Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-67557725731236542422016-02-05T10:12:48.502-08:002016-02-05T10:12:48.502-08:00It doesn't matter the order in which you apply...It doesn't matter the order in which you apply the "approximately constant": either to the differential equation itself or its solution.<br /><br />As a *function*, the curves are just a function of ΔY (or ΔMB), i.e. r = f(ΔY). The ⟨MB⟩ or ⟨Y⟩ parameterizes length along the curve.<br /><br />But the general equilibrium solution always applies, the partial equilibrium solutions are approximations to it for small values of ΔY.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-33247837791700072092016-02-05T09:34:32.639-08:002016-02-05T09:34:32.639-08:00If (2) and (3) do indeed describe those curves, th...If (2) and (3) do indeed describe those curves, then your statements about "moves last" and G.E. being "restored" make sense, since they imply a transition from satisfying (2) and (3) respectively to satisfying (1).Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-48693591034522873382016-02-05T09:26:16.711-08:002016-02-05T09:26:16.711-08:00"doesn't inform me how to change both ΔY ..."doesn't inform me how to change both ΔY and Yref such that their sum (Y) stays constant as either r or MB change"<br /><br />Should be<br /><br />"doesn't inform me how to change both ΔY and Yref such that their sum (Y) stays constant as <b>both</b> r <b>and</b> MB change"Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-88083908462285246452016-02-05T09:24:46.747-08:002016-02-05T09:24:46.747-08:00Sorry, I feel like a block head, but say I wanted ...Sorry, I feel like a block head, but say I wanted to replicate your blue curves by using eq (1) and by holding Y constant. I can change either r or MB, but eq (1) doesn't inform me how to change both ΔY and Yref such that their sum (Y) stays constant as either r or MB change. And even if it did, then the result would still satisfy G.E. (eq (1)), so when you write that Y "moves last" and<br /><br />"As Y rises, interest rates will come back up and equation (1) will hold -- general equilibrium restored."<br /><br />I'm wondering why it moves at all, since by satisfying eq. (1) the whole time we were in G.E. before it moved. It didn't need to be "restored" to G.E., it was already there.<br /><br />Instead, as per your draft paper, I'd expect the actual formulas you used to plot the blue curves to look like the partial equilibrium solutions:<br /><br />(2) r = constant1 * exp(-ΔY * constant2)<br /><br />and the formulas you used to draw the red curves to look like the other partial equilibrium solution:<br /><br />(3) r = constant3 * exp(ΔMB * constant4)<br /><br />If not those exact expressions, then approximations to them. Those come from your partial equilibrium solutions, not the G.E. (eq. (1)), and they tell me exactly how to draw those curves.<br /><br />In fact the bottom curves (especially) look just like what those exponential P.E. formulas describe. I can clearly see the curves crossing the vertical ΔY=0 and ΔMB=0 lines (respectively) at r=1 just like exp(-ΔY*constant2) and exp(ΔMB*constant4), respectively, would do. Are you sure you didn't use (2) and (3) to draw those?<br /><br />I'm probably missing something obvious.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.com