tag:blogger.com,1999:blog-6837159629100463303.post731216590931378020..comments2023-06-18T01:25:08.748-07:00Comments on Information Transfer Economics: Information equilibrium: a common language for multiple schoolsJason Smithhttp://www.blogger.com/profile/12680061127040420047noreply@blogger.comBlogger26125tag:blogger.com,1999:blog-6837159629100463303.post-8252780869970578532016-03-26T19:44:30.156-07:002016-03-26T19:44:30.156-07:00...plus it was reassuring to see you write it out.......plus it was reassuring to see you write it out.Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-60789674307985542772016-03-26T19:42:07.476-07:002016-03-26T19:42:07.476-07:00Thanks, that's a big help. Brings back memorie...Thanks, that's a big help. Brings back memories of second year electromagnetics class. I'd already concluded the expression you give above based on your earlier introduction of $\Delta D$ so I was going to stop worrying about it, but that Stokes' theorem link cleared up my remaining questions.Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-65804077992503261852016-03-26T15:10:08.238-07:002016-03-26T15:10:08.238-07:00The notation means "boundary". The integ...The notation means "boundary". The integral isn't saying much more than the fundamental theorem of calculus.<br /><br /><a href="https://en.wikipedia.org/wiki/Stokes%27_theorem" rel="nofollow">https://en.wikipedia.org/wiki/Stokes%27_theorem</a><br /><br />It's not so much the path integral as the differential forms. If you'd like you can take:<br /><br />$$<br />\int_{\Gamma} dD = \int_{t_{0}}^{t} dt' \frac{d}{dt'} D(t') = D(t) - D(t_{0}) \equiv \Delta D \equiv B \equiv G - T<br />$$<br /><br />All of physics is easier with differential forms.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-46179112863227086292016-03-26T12:30:38.895-07:002016-03-26T12:30:38.895-07:00... Ah, but I then I forgot the bijectve requireme...... Ah, but I then I forgot the bijectve requirement for r... so for 1D cases, r can't reverse I guess.Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-82544025588100307142016-03-26T12:25:37.236-07:002016-03-26T12:25:37.236-07:00... I should have said for monotonic r it doesn...... I should have said for monotonic r it doesn't much matter, but if r should reverse directions on its path from a to b one or more times, it does matter (I think).Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-36196179357056452202016-03-26T12:02:03.476-07:002016-03-26T12:02:03.476-07:00Or I guess instead of f=1 we could have dD/ds = f....Or I guess instead of f=1 we could have dD/ds = f. I'm at my worst with math when I can't imagine some concrete examples to generalize from. So let me know if this makes sense as an example: going back to f=1 (for simplicity), we could have a=0, b=1, r(t)=2t then (1) above would evaluate to 2. We get the same answer for r(t) = +/- 2t^n, which makes sense... r just being a map to get us from 0 to 2 (i.e. draw out the 1-D "curve" (in this case) $\Gamma$): it doesn't much matter how we get there. What I'm not seeing is the advantage of introducing the path integral here... I'm not saying their isn't one, I'm just not seeing it. Do you have an example of a case where there's an advantage to the path integral concept?Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-13692299692219745732016-03-25T23:23:55.344-07:002016-03-25T23:23:55.344-07:00Ok, thanks. I'm matching up this:
$$\int_{\G...Ok, thanks. I'm matching up this: <br /><br />$$\int_{\Gamma} dD \tag 1$$<br /><br />With what I found in <a href="https://en.wikipedia.org/wiki/Line_integral#Definition" rel="nofollow">this section of the Wikipedia article on line integrals</a>:<br /><br />$$\int\limits_{C} f\, ds = \int_{a}^{b} f(\mathbf{r}(t))\,|\mathbf{r}' (t)|\,dt \tag 2$$<br /><br />It seems $\Gamma$ corresponds to the curve (or path) $C,$ the constant $1$ corresponds to the integrand $f,$ and $dD$ corresponds to what they call an "elementary arc length" $ds.$ That all seems clear enough. Apparently $\mathbf{r}$ get's you from one end of the curve $C$ to the other via its parameter $t$ as $t$ varies from $a$ to $b$ with $a < b.$ The result should be independent of the parameterization $\mathbf{r}$ of $C.$ That all makes sense. However, I'm not familiar with this notation:<br /><br />$$\int_{\partial \Gamma} D \tag 3$$<br /><br />What is the meaning of that exactly? Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-75212969552351053702016-03-25T19:19:46.226-07:002016-03-25T19:19:46.226-07:00Yes, that is one way you'd parameterize the pa...Yes, that is one way you'd parameterize the path from one value of D to another.<br /><br />I tried to get away from imposing a coordinate system.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-58736604577660608602016-03-25T15:36:52.963-07:002016-03-25T15:36:52.963-07:00OK, I see. I'm reviewing path integrals. Would...OK, I see. I'm reviewing <a href="https://en.wikipedia.org/wiki/Line_integral" rel="nofollow">path integrals</a>. Would it be fair to characterize Γ as generally a function of time, between two sample times (nTs and (n+1)Ts say)?Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-88606202747837719692016-03-25T15:14:30.760-07:002016-03-25T15:14:30.760-07:00It's not the integral of B, it's the integ...It's not the integral of B, it's the integral of dD.<br />$$<br />\int_{\Gamma} dD = \int_{\partial \Gamma} D = \Delta D = B = G - T<br />$$Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-17760446458656235412016-03-25T15:05:23.161-07:002016-03-25T15:05:23.161-07:00I'd do the sum (integrate) the other way aroun...I'd do the sum (integrate) the other way around, like this:<br />$$D_n = \sum_{i=2}^n B_i \tag 3$$<br />Assuming $D_1 = 0$ and $B_1 = 0$ (<a href="https://3.bp.blogspot.com/-4qdQVxSJbTw/VtjXZe4PW0I/AAAAAAAAJEI/QPBSmHG_o_g/s1600/numerical%2Bgodley%2B1.png" rel="nofollow">as G&L apparently do</a>)<br />Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-37099323176494446982016-03-25T14:58:32.859-07:002016-03-25T14:58:32.859-07:00Sounds about right. Not sure what you are missing....Sounds about right. Not sure what you are missing.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-38977003264152157012016-03-25T13:51:39.260-07:002016-03-25T13:51:39.260-07:00Sorry Jason, I don't get it. I would sum all t...Sorry Jason, I don't get it. I would sum all the deficits (B[n]) to get the total debt at period n (D[n]). What am I missing?Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-44497766207839159622016-03-25T12:38:54.340-07:002016-03-25T12:38:54.340-07:00B is the amount of bonds issued in a time period; ...B is the amount of bonds issued in a time period; D is the stock of debt.<br /><br />D is defined by that integral.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-58598313207238546402016-03-25T11:58:01.226-07:002016-03-25T11:58:01.226-07:00So B is defined both as G - T ((government spendin...So B is defined both as G - T ((government spending over one time step) - (taxes collected over one time step)) and path integral of D? That 2nd part actually defining D I guess (I don't see D defined until that integral). I'm unclear on what D is in words.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-27386526484680831982016-03-24T21:13:24.827-07:002016-03-24T21:13:24.827-07:00Nope. But I will fix it when I have time.Nope. But I will fix it when I have time.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-6514511180373285742016-03-24T21:12:26.933-07:002016-03-24T21:12:26.933-07:00Ah, so then it should be
$$
\begin{align}
\Delta H...Ah, so then it should be<br />$$<br />\begin{align}<br />\Delta H & \rightleftarrows B \tag 1\\<br />& \text{and}\\<br />\Delta H & \sim B^k \tag 2<br />\end{align}<br />$$<br />Which will still mean that $\lim\limits_{t \to \infty} \Delta H = 0$ if $\theta \neq 0$, but that may be OK.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-49823446452676454392016-03-24T17:00:16.817-07:002016-03-24T17:00:16.817-07:00Typo in equation. Will fix when I get a chance. No...Typo in equation. Will fix when I get a chance. Not terribly important as the general idea is there. Linear transformation plus one dynamical equation.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-35749333045454572282016-03-24T09:14:28.968-07:002016-03-24T09:14:28.968-07:00Should be "$\Delta H$ makes more sense."...Should be "$\Delta H$ makes more sense." Also, if theta=0 the the steady state gain for T (and integrated T) are both 0 of course, but I think that's the only exception.Tom Brownhttp://www.google.comnoreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-35764383295793897422016-03-24T08:43:45.816-07:002016-03-24T08:43:45.816-07:00Back to B≡G−T ... eventually T=G (since it's s...Back to B≡G−T ... eventually T=G (since it's steady state gain is 1 no matter what the parameters are), so the steady state gain for B is 0, no matter what the parameters are. Again, interpreting all of them as defined over one period. So eventually H goes to zero if H goes as B^k (unless k=0, then H stays at 1). It seems to me Delata_H makes more sense there.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-2438710887462071232016-03-24T08:38:27.755-07:002016-03-24T08:38:27.755-07:00Also, it seems there is no "price" conce...Also, it seems there is no "price" concept in SIM originally, true? Is that generally true of SFC models?Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-8348872967663548562016-03-24T08:36:04.738-07:002016-03-24T08:36:04.738-07:00Regarding SFC models. With H⇄B, B is defined as ov...Regarding SFC models. With H⇄B, B is defined as over one period (B≡G−T), since G and T are wrt one period (rather than being integrated values). But H is an integrated value over all periods.<br /><br />Also, I had a different thought, which I suppose isn't just SFC specific, and that is the idea of assuming the agents explore the whole space allowed them. In this case it would be letting alpha1 and alpha2 range over all possible values they are allowed... and then taking the "center of mass" of those results as the maximum entropy answer.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-82551914936064728622016-03-23T11:14:16.964-07:002016-03-23T11:14:16.964-07:00Cheers, Cameron.
One thing I like about the Becke...Cheers, Cameron.<br /><br />One thing I like about the Becker approach is that if agents clump up in that opportunity set (everyone wants to push consumption to period 2), you can get a breakdown of 'supply and demand' and (emergent) rationality. I don't think traditional econ identifies failure modes like that. I hope more people follow your lead! If starting that small change in teaching econ is all I end up contributing, that will be worth it!<br /><br />I completely agree that there needs to be an even more fundamental definition of what the units of economic discourse are.<br /><br />FWIW, in the IT framework, those variables represent probability distributions over some domain ... e.g. space, time. You have equilibrium when e.g. the spatial probability distribution of 'demand events' for X is equal to the spatial probability distribution of 'supply events' of X -- and there are a large number of units of X (so that the probability distribution comes close to being the actual distribution). In that case, the information required to construct both distributions is equal I(Pd(X)) = I(Ps(X)).<br /><br />Since the whole thing simplifies if you talk about uniform distributions, you can think of e.g. NGDP as the total number of 'demand events' (measured in e.g. dollars). When uniformly distributed (not true, but works to leading order), the information in a string of 'demand events' drawn from that distribution is just proportional to the number of events ... I(Pd(D)) ~ NGDP.<br /><br />However, since this definition is pretty malleable, it could easily fit distributions of property rights, accounts, etc. you discuss in your follow-up.<br /><br />For example, I've used it for the distribution of price states in a time series. The distribution over nominal interest rate states is equal to the distribution over "the price of money", which is proportional to velocity in this very basic model.<br /><br />The IT framework is just one example of a possible way to address this issue, but I agree that more needs to be done in this regard.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-55966643697560274632016-03-22T23:01:48.283-07:002016-03-22T23:01:48.283-07:00Yes, I'm glad he posted it too. It's late,...Yes, I'm glad he posted it too. It's late, so I'll have to take a detailed look later though.Tom Brownhttps://www.blogger.com/profile/17654184190478330946noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-44200945277335782502016-03-22T19:38:03.768-07:002016-03-22T19:38:03.768-07:00"The fact that I criticized an aspect of SFC ..."The fact that I criticized an aspect of SFC analysis upset the MMT and PK tribes (see the post and comments) led me to not end up posting the work I'd done."<br /><br />Don't let the critics put you off. The MMT and PKE people have been taking criticism for years and they are still going strong. Criticism and debate are necessary for the development and dissemination of knowledge. This is a great contribution of blogging.Tom Hickeyhttps://www.blogger.com/profile/08454222098667643650noreply@blogger.com