Wednesday, April 24, 2013

Supply and demand from information transfer

At this point we will take our information transfer process and apply it the the economic problem of supply and demand. In that case, we will identify the information process source as the demand $Q^d$, the information transfer process destination as the supply $Q^s$, and the process signal detector as the price $p$. The price detector  relates the demand signal $\delta Q^d$ emitted from the demand $Q^d$ to a supply signal $\delta Q^s$ that is detected at the supply $Q^s$ and delivers a price $P$.

We translate Condition 1 in [1] for the applicability of our information theoretical description into the language of supply and demand:
Condition 1: The considered economic process can be sufficiently described by only two independent process variables (supply and demand: $Q^d, Q^s$) and is able to transfer information.
We are now going to look for functions $\langle Q^s \rangle = F(Q^d)$ or  $\langle Q^d \rangle = F(Q^s)$ where the angle brackets denote an expected value. But first we assume ideal information transfer $I_{Q^s} = I_{Q^d}$ such that:
$$(4) \space P= \frac{1}{\kappa} \frac{Q^d}{Q^s}$$
$$(5) \space \frac{dQ^d}{dQ^s}= \frac{1}{\kappa} \frac{Q^d}{Q^s}$$
Note that Eq. (4) represents movement of the supply and demand curves where  $Q^d$ is a "floating" information source (in the language of Ref [1]), as opposed to movement along the supply and demand curves where $Q^d =Q^d_0$ is a "constant information source".

If we do take $Q^d =Q^d_0$ to be a constant information source and integrate the differential equation Eq. (5)
$$(6) \space \frac{\kappa }{Q_0^d}\int _{Q_{\text{ref}}^d}^{Q^d}d\left(Q^d\right)'=\int_{Q_{\text{ref}}^s}^{\left\langle Q^s\right\rangle } \frac{1}{Q^s} \, d\left(Q^s\right)$$
We find
(7) \space \Delta Q^d=Q^d-Q_{\text{ref}}^d=\frac{Q_0^d}{\kappa }\log \left(\frac{\left\langle Q^s\right\rangle }{Q_{\text{ref}}^s}\right)
Equation (7) represents movement along the demand curve, and the equilibrium price $P$ moves according to Eq. (4) based on the expected value of the supply and our constant demand source:
\text{(8a) }P=
\frac{1}{\kappa }\frac{Q_0^d}{\left\langle Q^s\right\rangle }
\text{(8b) }
\Delta Q^d=\frac{Q_0^d}{\kappa }\log \left(\frac{\left\langle Q^s\right\rangle }{Q_{\text{ref}}^s}\right)
Equations (8a,b) define a demand curve. A family of demand curves can be generated by taking different values for $Q_0^d$ assuming a constant information transfer index $\kappa$.

Analogously, we can define a supply curve by using a constant information destination $Q_0^s$ and follow the above procedure to find:
$$ \text{(9a) }P= \frac{1}{\kappa }\frac{\left\langle Q^d\right\rangle }{Q_0^s}$$ 
$$ \text{(9b) }\Delta Q^s = \kappa Q_0^s \log \left(\frac{\left\langle Q^d\right\rangle}{Q_{\text{ref}}^d}\right)$$
So that equations (9a,b) define a supply curve. Again, a family of supply curves can be generated by taking different values for $Q_0^s$.

Note that equations (8) and (9) linearize (Taylor series around $Q^x=Q_\text{ref}^x$)
Q^d =Q_{\text{ref}}^d +\frac{Q_0^d}{\kappa }-Q_{\text{ref}}^s P
Q^s = Q_{\text{ref}}^s-\kappa Q_0^s+\frac{Q_0^s{}^2\kappa ^2}{Q_{\text{ref}}^d}P
plus terms of order $(Q^x)^2$ such that
Q^d=\alpha -\beta P
Q^s=\gamma +\delta P
where  $\alpha =Q_{\text{ref}}^d+\left.Q_0^d\right/\kappa$, $ \beta  = Q_{\text{ref}}^s$ ,$ \gamma  = Q_{\text{ref}}^s-\kappa  Q_0^s $ and $ \delta =\kappa ^2 Q_0^s{}^2/Q_{\text{ref}}^d$. This recovers a simple linear model of supply and demand (where you can add a time dependence to the price e.g. $ \frac{dP}{dt} \propto Q^s - Q^d $).

We can explicitly show the supply and demand curves using equations (8a,b) and (9a,b) and plotting price $P$ vs change in quantity $\Delta Q^x=\text{$\Delta $Q}^s $ or $\text{$\Delta $Q}^d$. Here we take $\kappa = 1$ and $Q_{\text{ref}}^x=1$ and show a few curves of $Q_0^x = 1 \pm  0.1$. For example, for $x = s$ and +0.1, we are shifting the supply curve to the right. In the figure we show a shift in the supply curve (red) to the right and to the left (top two graphs) and a shift in the demand curve (blue) to the right and to the left (bottom two graphs). The new equilibrium price is the intersection of the new colored (supply or demand) curve and the unchanged (demand or supply, respectively) curve.

[1] Information transfer model of natural processes: from the ideal gas law to the distance dependent redshift P. Fielitz, G. Borchardt


  1. Jason, in equation 7, you start out on the left with "Q_superscript_d = " but should that be "Delta_Q_superscript_d = " ?

    1. Yes, you are correct. I will edit it.

    2. I finally made the change -- I think I forgot to save the result 3 months ago ...

  2. I'm having trouble digesting this... I can get parts of it, but not all. I looked at versions 1 and 3 of your 1st reference... it's good to see some of the same stuff in there, so that lets me know you're sticking with their derivation I think, but it'll take some study on my part.

    A couple of quick questions:

    What does the ' indicate on the left hand side of equation 6 (inside the integral)?

    Where did the expected value (angle brackets) come from? Is there an underlying probability distribution here that I'm unaware of? I'm used to this kind of expected value (where X is a random variable with probability density function px()):

    E{X} = integral from -inf to inf of px(u)*u*du

    Are we talking about the same thing? Of course I realize it might be a probability mass function instead: Px()) taking on values at integer values of u and nowhere else (thus giving you a summation instead of an integral), but that's basically the same thing.

    1. I actually used version 2 because that is the one that was available at the time I wrote this. It has the most overlap, especially in notation, with Fielitz and Borchardt's paper.

      The ' mark is there to denote a dummy variable in the integral.

      Regarding the angle bracket, all of the variables should be considered to be expected values of stochastic processes. However in version 2 of the reference, there is "condition 8" that takes < x > = x. The reason is best explained with an analogy: in the model, they take < V > = V for the volume of an ideal gas. Individual atoms don't have volume, so volume doesn't really make sense except as an expected value of an operator ... you could say that volume is emergent.

      Individual transactions don't really have "supply" and "demand", however a market has an expected value of the quantity supplied.

      Also, your E{X} is essentially what the partition function approach does explicitly. The thermodynamic expected value is given in terms of a sum (or integral) over states of energy E with the probability weighting factor exp(-βE).

    2. Thanks Jason. OK, version 2 it is then. :D.... any glaring prerequisites I might want to bone up on 1st? My background: ECE masters over a decade ago in control systems. I never had a proper class in thermo... just noisy systems (noise figure, power spectral densities, basic probability theory, etc), and two undergrad communication courses that touched on information theory. I'm best with state space stuff, linear algebra, Kalman filters, etc. Do you think I can handle your reference, or should I do some background reading on information theory first? Well, I'll dig in and see how far I get this time.

      BTW, that reference: How would you describe it to someone? As a general theory of macro level things that sometimes does a good job? It seems like a one-size-fits all approach in some regards to "processes." I realize that their are special conditions that must apply, etc., but I was looking for a simple description. Thanks.

    3. I would say you don't need additional pre-reqs. I knew very little about information theory when I started this. The basic idea is that this information theory motivates a 1st order ODE, then everything comes from the ODE.

      Regarding the description of the paper -- I like the first line of their abstract: "Information theory provides shortcuts which allow [one] to deal with complex systems." The specific thrust of the paper is that it looks at how far you can go without specifying "constraints" -- this refers to partition function constraints:

      that define the "Lagrange multipliers". In thermodynamics terms, basically the method allows you to look at thermodynamic systems without having defined a temperature (Lagrange multiplier) and without having the related constraint (thermal equilibrium).

      In economics, there is no real concept of "temperature" or (sorry, economists) economic equilibrium. That is to say, more generally, there are no Lagrange multipliers or constraints on an economic system.

      Rambling on a bit, what the partition function approach seems to say is that there may be a Lagrange multiplier/temperature that's a function of log M0 and the "constraint" may be something that reduces to the quantity theory of money in certain limits (an economy in equilibrium is one that satisfies the quantity theory to some approximation).

    4. Wow, that's an interesting comment. Thanks! BTW, I emailed one of the paper's authors (Peter) with some suggestions for small grammar improvements to their abstract. I'm guessing English is not their 1st language. :D

    5. I thought it was interesting too ... ha! I turned it into a new post.

      Regarding the paper, I believe they are in the process of submitting it to a journal. Things like that usually get cleaned up during the peer review process.

  3. Of course we can combine equations 8a and 8b to obtain (I'm using [] instead of angle brackets):

    Eq. 8:

    P = (1/k)*[Qd]/Qs0 = (1/k)*(Qdref/Qs0)*exp(deltaQs/(k*Qs0)

    And we can combine equations 9a and 9b to obtain:

    Eq. 9:

    P = (1/k)*Qd0/[Qs] = (1/k)*(Qd0/Qsref)*exp(-k*deltaQd/Qd0)

    Then it's crystal clear that for Eq. 8, if we change Qs0 to Qs0+0.1 the curve of P vs deltaQs moves down (or to the right), etc. (for the other three plots).

    Here's the bit that never sunk into my thick skull:

    1. Where did Qsref and Qdref come from?

    2. Focusing primarily on Eq. 8, what is the meaning of holding Qs constant at Qs = Qs0, but then developing an equation (Eq. 8) that relates [Qd] to Qs, Qsref and Qs0, as if Qs =/= Qs0.

    3. Why do you need the downward sloping demand curve in the upper two plots? Without it, you'd still have P = f([Qd], Qs0) = g(Qs, Qs0, Qsref)... in other words, [Qd] is there in P even without the demand curve. I think you're going to say "To demonstrate how the equilibrium point (where demand and supply curves cross) shifts when Qs0 goes to Qs0 +/- 0.1." I guess since you've got delta quantity on the x-axis, and the supply & demand curves nominally cross at x = 0, you're also demonstrating how you're starting at P = (1/k)*Qdref/Qsref, i.e. the "floating" or "general equilibrium solution," true?

    3. Putting aside perturbing Qs0 for the moment, we can then simplify Eq. 8 to be
    P = f([Qd]) = g(Qs-Qsref)

    or better yet

    P ~ [Qd] ~ exp(Qs-Qsref)

    But in what sense does [Qd] go as exp(Qs-Qsref) when Qs is fixed at a constant Qs0? Is it Qsref that's varying (I wouldn't think so)? Assuming Qsref isn't varying, that seems like a logical contradiction. Do you have a reference that explains this general idea for someone with a thick skull such as myself?

    5. Likewise for Eq 9 and Qd.

    1. OK, ***maybe*** the light is ever so slowly starting to dimly flicker on...

      When you normally have an equation like:

      dD/D = (1/k)*dS/S

      and you claim that D is "held constant" at D0 (or is "slowly varying" around D0), and you instead write it like:

      dD/D0 = (1/k)*dS/S

      What you're saying is that you're writing an equation assuming that D is changing so slowly that for the purposes of solving this differential equation, we might as well replace the D part of the equation with a constant. The mind bending part for me, is that the solution to such an equation produces a variable D again. There's no such assumption on S. Once we solve it and find how [S] varies with D, it's really only valid in a small region for D around D0, over which there are no such restrictions on S. Is that kind of it? It seems like cheating somehow.

    2. The other mind bending part for me is the introduction of both Dref and D0. Off hand it almost seems like you can make Dref do the job of both, and in fact you set them both to 1 in your example above (as you do for both Sref and S0).

    3. I guess this business shouldn't be so shocking to me.. I'm used to constructing extended Kalman filters (EKFs) for non-linear DE systems, by linearizing around a nominal operating point and then construction a 1st order ODE to approximate the system. That seems similar to what you're doing here. Is that true?

      Is it fair to call the resulting supply and demand curves (from using D0 and S0) approximations?

      For example, I could imagine a non-linear DE with a slowly varying non-linear time dependence in the parameters, perhaps like:

      x' = f(t)*x

      Then I could linearize around some nomial t=t0:

      x' = K*x

      with K = f(t0)

      then for t close to t0, that equation should be approximately correct, and the solution of course is:

      x ~ exp(K*t)

      That seems to be exactly what's going on here, only you're letting both Qd and Qs (in turn) take the place of the time variable in my example.

      Is it just the absence of the word "approximation" which is causing me so much grief? Lol.

    4. Actually, my equation above should be:

      delta_x' = K*delta_x


      delta_x = x - x0

      Where x0 is the solution at t0 to the original true equation. That would explain the D0 and S0. Now the only thing missing here is Dref and Sref. I'm still sure where they came from. Is it safe to say that Dref and Sref represent two values at which we know the original differential equation was solved?

    5. Should be: "I'm still not sure where they came from." ... but I have an inkling from your source's gravitational example (if I'm correct about that (see comment in your recent post))..

    6. I think this post may be helpful:

      When we hold D constant at D0 we are doing the analogous thing to keeping temperature constant in an isothermal expansion. You can change the temperature of the heat bath -- i.e. changing D0 -- which just changes which isotherm you are on. You can also move along the isotherm (an isothermal expansion or compression).

      In real life, if you expand a gas, it cools (in general equilibrium, i.e. floating process variable). But if you keep it in contact with a thermal bath, you get isothermal expansion (which is the equivalent of a demand curve with diminishing marginal utility and lower prices/pressure).

    7. OK, I'll take a look. So the words "approximately true for small movements in X around X0" are not actually required then? (regarding your solutions, such as to the DE dD/D0 = (1/k)*dS/S)

      I don't have much of a thermodynamics intuition built up, but maybe it's time I acquire one... it's almost certainly more concrete that econ stuff!

    8. Yes, the "approximately true for small movements" isn't required. We're imposing a restriction on X not making an approximation X ~ X0 in the equations.

      In real life (in economics), we occasionally are making the approximation that X is restricted (i.e. partial equilibrium). But that isn't the same thing as saying X is approximately equal to X0. X0 just chooses which "isotherm" ... which isosupply or isodemand curve.

    9. "When we hold D constant at D0 we are doing the analogous thing to keeping temperature constant in an isothermal expansion."

      I guess what's less confusing about the isothermal expansion is that there's a separate symbol for temperature (T). We don't hold T = T0 = constant, and then find a curve over which delta_T = T - Tref varies, right? Everywhere on the isotherm T = T0.

      But in these examples, we set D = D0, and we call the solution we get from that an "isodemand," but then we find a curve which takes on different values of delta_D = D - Dref, and I don't think it's Dref that's varying, right? If Dref doesn't vary that implies that D does (for each point on the curve).

      I realize that D represents "quantity demanded" not "demand" (where demand =. a demand curve)

    10. Confusing, right?

      What we are looking at is analogous to an isothermal expansion/compression (in contact with a heat bath) so that

      ΔW ~ nRT log(V/Vref)

      It's "isothermal" so the temperature doesn't change. But if the energy of a gas is nRT and we do work on the gas (by compressing it) shouldn't the energy change -- and therefore the temperature? If you do work on a point mass, it acquires kinetic energy so what is happening here?

      Well, the temperature change is "absorbed" by the heat bath, so what we're looking at is work done by us (ΔW) -- it's exogenous work in economics terminology.

      That is to say the ΔD is an exogenous change in demand where the system maintains a level of demand (in contact with a "demand bath" at "constant" demand D₀).

    11. I understand your gas description (I think), but I'm wondering if something about the symbols in the econ case is killing me, so let me try to line it all up.

      So delta_W is a change in energy and nRT is energy.

      delta_D is a change in quantity demanded and D0 is a quantity demanded.

      Am I on the right track here? Thus:

      delta_W is analogous to delta_D?


      nRT is analogous to D0 (with n, R and T all constants)?

      It's too bad we can't use different symbols in the economics case (that is something different for the D component of delta_D that we use in the expression D=D0), because the isothermal case avoids that confusion with different symbols. Well, it does, if I'm correct about what's analogous to what above.

    12. Yes, you have that right.

      Work is a form of energy and they are measured in the same units, e.g. J. An economist would say you have exogenous changes in energy (work) and energy (nRT -- n is moles, R is gas constant and T temp).

      In economics, you have exogenous changes in demand (ΔD) and demand (D₀).

      If I was to write the work equation in the form of the ITM, I'd have

      ΔE = E₀ log(V/Vref)

      But physicists call exogenous changes in energy "work".

    13. "But physicists call exogenous changes in energy "work"."

      The work in this case was done on the bath, right? Now what if there was no bath, and the vessel holding the gas was perfectly insulated. Then as we compress the gas the temperature of the gas goes up, correct? And thus it's energy goes up. I know it's probably a different formula, but qualitatively that's what I expect. Would a physicist call the change in energy of the gas W as well? It's not exogenous in the sense of being outside our vessel, but in some sense we've done work on the gas haven't we?

      Economists should call exogenous changes in demand something other than "demand." I think that would clear this whole business up. Or at least find some other way to distinguish exogenous from endogenous (when referring to demand or supply, or more correctly I guess, quantity demanded or quantity supplied). If somebody had done that it would have saved poor old Sraffa a lot of ink. :^)

      I did read in one of your posts that one important thing about the expected value angle brackets < > you use is that it's a sign that the variable inside is endogenous. So if we're holding endogenous demand at a constant value D0, does it make sense to signify this by saying you're going to hold expected quantity demanded constant at D0, like this:

      < D > = D0

      It would probably be OK, if you started from this (like you do somewhere, perhaps at least once in your draft paper):

      P = dD/dS = (1/kappa) * < D > / < S >

      Also, you don't have a problem with me changing your "demand" to "quantity demanded" in any of the cases I did that?

    14. "The work in this case was done on the bath, right?"

      No, work is done on the gas in the cylinder.

      "Now what if there was no bath, and the vessel holding the gas was perfectly insulated. Then as we compress the gas the temperature of the gas goes up, correct?"

      Yes, that is called an isentropic process and represents the general equilibrium solution in the information transfer model. It is still called work if you act against pressure, though.

      "Also, you don't have a problem with me changing your 'demand' to 'quantity demanded' in any of the cases I did that?"

      That's fine. I probably should be more careful about that language, though I don't really know what the confusion that would result is. I can refer to the isothermal process curve as 'the pressure' or a point on it as 'the pressure' much the same way that NGDP is both a time series and a specific value. But maybe I am missing something.


      RE: dD/dS = (1/kappa) * < D > / < S >

      I most definitely don't want to do this because the differential equation is an operator relationship, not an expected value relationship -- we have

      < dD/dS > ~ < D/S >

      but < D > / < S > ≠ < D/S > in general.

  4. "No, work is done on the gas in the cylinder."

    Yes, but ultimately the energy (nRT) of the gas stays fixed, right? So that rather than increasing the energy of the gas in the cylinder, that energy is dissipated in the bath. You don't call that doing work on the bath? Pressure also rises inside the cylinder, correct? (Pressure is analogous to price here?).

    So as we move the piston to compress the gas, we do work, and some of that work goes to increasing the pressure inside the cylinder and some of it goes to heat lost into the bath. Where have I gone wrong?


    Still think I saw this somewhere in one of your posts (without the differentials perhaps ... which I guess makes all the difference?):

    P = (1/kappa) * < D > / < S >

    Was it your draft paper perhaps? (I can't get to your paper from the machine I'm on now, so I can't check).


    Now for this case:

    dD/D = (1/kappa)*dS/S0

    Analogous to constant volume? No bath anymore (or is there?). I think I saw a post on this somewhere... I'll go look it up.

    1. "Yes, that is called an isentropic process and represents the general equilibrium solution in the information transfer model."

      That is the other case you mentioned here:

      So if that's general equilibrium, what's analogous to moving along an iso-quantity-supplied curve?

      Also, I mentioned elsewhere that I learned to be careful around Nick Rowe in distinguishing between a quantity demanded and demand (he'll yell at you if you get those mixed up! Lol). Nick takes demand to be the whole curve. Similar for quantity supplied and supply. Any comment on that?

    2. I asked an ME about some of this, and I think I confused him. Lol.

    3. Shoot, I forget I've got two PhDs in physics from CalTech 3 and 6 doors down the hall (a husband and wife, BTW). They both studied quantum computing I think. Oh well, I can exercise their thermo brains... maybe at lunch today.

    4. Regarding the work, it is more critical that you are doing the work, not what you are doing the work on. I guess the energy ultimately ends up in the bath. In a simple mechanics problem, you do work against friction. The energy ends up as heat that is radiated away into the environment. I wouldn't say you were doing work on the environment when you are acting against friction, though.

      If I did write that equation, I will have to go back and take a look at it. I believe I wrote it after I solved the differential equation so that integrating (k = 1 WOLOG):

      dD/D0 = dS/S0

      (D - Dref)/D0 = (S - Sref)/S0

      ΔD/ΔS = D0/S0

      So I probably should have written D0/S0 instead of < D > / < S >.

      But the key issue is the differential equation.

      There isn't a good physical analogy for the case of constant volume AFAICT. It would involve being in contact with a "volume bath". For economics, it makes sense (a "supply bath" where you can exchange goods analogous to a heat bath where you exchange heat).

      Nick is a real economist, so if he says you need to be careful I'll take his advice.

    5. LOL!... well, the wife escaped (she had a conf. call in 3 min), but I did confuse the husband. This is great fun!

      So in our isothermic bath set up, if we dial up the force on the piston to compress the gas (decrease it's volume) w/o raising the temperature of the gas, and then we dial the force on the piston back down to its starting value, will the piston eventually end up back where it started? Would that make the cylinder+bath an isoentropic system?

      It kind of seems like it should since the bath will heat the gas back to the fixed bath temperature as it expands and cools and re-acquires it's previous pressure. However, my physics PhD colleague was left scratching his head on that. He mumbled something about never finding a good thermo text book... Lol. (I didn't want to take up too much of his time). His wife overheard us when she was done with her call and poked her head in the door, but when she saw we were talking thermo... she hesitated for a moment and went on her way.

      He did bring up the point that the "energy of the gas" is

      U = (some other constant having to do with degrees of freedom of the gas) * nRT

      He showed me the formula right there on a Wikipedia page. But he agreed that this was a constant as well and wouldn't change. (that's probably why you used a "~" in your W expression rather than a "=" come to think of it).

      I thought that was a good time to bring up "enthalpy" ... but no light-bulbs went off (for him) when I did so. That they didn't go off for me goes without saying.

    6. Whoops, I wrote the above w/o seeing your latest comment. So let me read that.

  5. This has been great fun. I really appreciate your comments here Jason.

    Yes, a "volume bath" is where my intuition hit a brick wall too. I mentioned to the ME what this had to do with supply and demand curves (which he seemed to get), hoping he'd fill me in on how to visualize this volume bath... (he didn't), but the fact that you don't find that useful either is good enough for me.


    Maybe this is leading me astray, but according to this:

    Enthalpy is measured in joules (so an energy measure?), and it's

    H = U + pV

    So in terms of our cylinder of gas in a bath, we didn't change U but we did change p (it went up) and V (it went down)... and by the same fractions (since PV = nRT = constant), so that term (pV) didn't change either. Yes, enthalpy is a dead end. (c:

    Haw, this has been quite an education. I really want to figure out a more satisfying physical example though... one in which an analogy to the "bath" concept holds for both D and S. I suppose I might be able to just mentally switch roles for the two and do it that way with the cylinder+bath (as long as I made all the other appropriate changes). I'll have to think about that.

    1. Each term in a thermodynamic potential can be thought of as an information equilibrium relationship, something I use to build an "economic potential" here:

      While the PV term may be bad for physical analogies with supply and demand, I have thought maybe the μN term might be more fruitful -- chemical potential and particle number. Energy (demand) ~ μN with μ as the price and N as supply.

    2. Great. I missed that post the 1st time around. I look forward to digging into it later.

      One last quick question for today:

      If you increase the initial force on the piston from Fref to F and thus decrease the volume of the cylinder from Vref to V, the temperature in the cylinder stays constant due to the bath. Now let's say we add a perfect insulator to the cylinder and remove it from the bath, and after that we then return the force on the cylinder to Fref. Once our system reaches an equilibrium, I would expect the final volume (V1) such that:

      V < V1 < Vref

      I would also expect the final temperature (T1) to decrease so that

      T1 < T

      Where T is the constant temperature of the (now missing) bath. Is that correct?

      If that's the case then both U and pV will have decreased from their starting values, although p will be restored to the original value. So there was a negative change in enthalpy which I would guess equals the energy dissipated into the bath (delta W).

      Did I get any of that right?

    3. And the final equilibrium we reach at V1, T1, p, etc, will be a general equilibrium, because once we remove the insulated cylinder from the bath, it's an isoentropic process. That's the way I explained it to both the ME and the physicist, and they both nodded their heads politely waiting for the noise to stop*, which I assume means they both agree. ;^)

      *I have to credit Scott Adams (Dilbert) for "nod your head waiting for the noise to stop." I wish I could find the strip it came from.

    4. I believe what you are describing is part of what is called a Carnot cycle:

      It's how heat engines work -- isothermal (economic partial eq) and isentropic (economic gen eq) expansions and compressions.

    5. Adiabatic and isentropic are effective synonyms.

    6. I'll assume I didn't make any major errors there then. Thanks! Sometime I might try to write out in my own words exactly what it means to move along the demand or supply curve. Not today... but hopefully I have enough understanding of this now to work my way through it.

      Also, it occurs to me that since the isotherm example also relies on statistics (the statistics of all those gas molecules and bath molecules bouncing around), it could make perfect sense to include expectations there as well (i.e. < > brackets).

      BTW, re: my comment below, will you then change Eq. 8 in your paper, or do you think you'll keep it as is?

    7. The thermodynamic limit is basically < X > = X, so I may just drop all mentions of < X > and relegate it to a footnote as it's all rather pedantic. I do like to use "ensemble average" instead of "expectation value" because the latter has an economic connotation (e.g. rational expectations).

  6. BTW, it's equation 8 in your draft paper (pg. 5):

    P = k * < D > / < S >


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Also, try to avoid the use of dollar signs as they interfere with my setup of mathjax. I left it set up that way because I think this is funny for an economics blog. You can use € or £ instead.

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