tag:blogger.com,1999:blog-6837159629100463303.post4474377311233552583..comments2023-06-18T01:25:08.748-07:00Comments on Information Transfer Economics: I have no idea what you are talking aboutJason Smithhttp://www.blogger.com/profile/12680061127040420047noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6837159629100463303.post-87065175518228604252015-01-19T10:17:57.690-08:002015-01-19T10:17:57.690-08:00Hi Roger,
In the equations above, k in the ITM is...Hi Roger,<br /><br />In the equations above, k in the ITM is essentially a model for velocity in the quantity theory -- I will do a check of the math, but k does seem to be similar to the exponent in your decay model since tax receipts are roughly proportional to GDP:<br /><br /><a href="http://research.stlouisfed.org/fred2/graph/?g=XAJ" rel="nofollow">http://research.stlouisfed.org/fred2/graph/?g=XAJ</a>Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-79946056916118474052015-01-18T17:14:12.093-08:002015-01-18T17:14:12.093-08:00I look at MV = PY as being two descriptions (each ...I look at MV = PY as being two descriptions (each equal to the other) of an economy which lives in a container of size one-year.<br /><br />Yes, this would be the equivalent of equating PY as being the sum of the velocity of each gas particle in a container where there are many particles and many velocities. <br /><br />Yes, this would be the equivalent of equating MV as being the average velocity (V) of a single representative particle (M), with the limit that the representative particle represents less than the total number of particles in the container.<br /><br />Now we notice that MV changes from year to year. Something is happening to the entire container so that one year varies from the adjacent year.<br /><br />It seems to me that this is equivalent to accelerating the entire container of gas. If we accelerate the entire container of gas, during acceleration, the particles will concentrate on one side of the container and de-concentrate on the opposite side. A thermometer would measure that the concentrated side has become warmer, the de-concentrated side has become cooler.<br /><br />The above description can be applied to an economy. It seems to me that adding money (which is an increase in term M) imbalances the part of the economy that first receives the added money. This added money does not reach the later-reached sectors of the economy until much later, maybe not until the money addition stops. This would correspond with the visual image of a container of gas where the particles would remain concentrated on one side until the acceleration stopped.<br /><br />It is time to turn to the use of logarithmic notation to represent this process. I don't see any need for logs up to this point. There is, however, another potential justification. Taxes tend to be uniformly applied at every monetary exchange so a repetitive transfer to government occurs. <br /><br />Logarithmic notation can be used to learn the number of money supply turnovers based on a tax rate influenced money supply. I wrote a post “Finding the Exponent in the Fiat Decay Model” which can be found at <br /><br />http://mechanicalmoney.blogspot.com/2014/11/finding-exponent-in-fiat-decay-model.html<br /><br />It seems to me that the exponent here is equivalent to the K (or velocity) of your<br /><br /> k ~ log base M PY<br /><br /> equation.<br /><br />The Fiat Decay Model is based on the theory that a one-time injection of fiat money into an economy can be completely recovered by government using taxation. A one-time injection will create a predictable maximum GDP as it slowly decays with each succeeding transaction.<br /><br />Perhaps my understanding of your IT theory is improving but I cannot confidently compare your result to my work. You may find that the parallels and analogies here in this comment are not too good. Please remember that I am trying, and be patient.<br />Roger Sparkshttps://www.blogger.com/profile/01734503500078064208noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-34961290649594317512015-01-17T08:59:59.301-08:002015-01-17T08:59:59.301-08:00Another instance where my physics background leads...Another instance where my physics background leads to a lack of clarity. The ~ symbol is <a href="http://en.wikipedia.org/wiki/Asymptotic_analysis" rel="nofollow">from here</a>. It really just allows you to write down the essence of an equation. If y = m0*x + b0 for a line then you can write y ~ x.<br /><br />In the case above, I left off some constants and lower order terms (they change much more slowly than the first terms in the equation):<br /><br />log PY = k log M + k log m0 - log(p0 y0)<br />log P = (k - 1) log M + (k - 1) log m0 + log k - log(p0)<br /><br />As far as what the P, Y, M and k mean, they are from the <a href="http://en.wikipedia.org/wiki/Equation_of_exchange" rel="nofollow">equation of exchange</a> or the <a href="http://en.wikipedia.org/wiki/Cambridge_equation" rel="nofollow">Cambridge equation</a> where P is the price level, Y is output, M is the 'money supply' and k is sort of like the Cambridge k -- except the information transfer model represents a new model for it.<br /><br />What these equations all do is describe a 'quantity theory of money' where monetary expansion ('printing money') leads to increased inflation (P goes up) and output (PY goes up).<br /><br /><a href="http://en.wikipedia.org/wiki/Quantity_theory_of_money" rel="nofollow">http://en.wikipedia.org/wiki/Quantity_theory_of_money</a><br /><br />Pure quantity theories have mostly been rejected by looking at the data (unless k or V = velocity of money are pure fudge factors which makes the quantity theory unfalsifiable) so there are various modifications -- for example, market monetarists have something like an expectations-augmented quantity theory of money in their minds. There is also debate about which <a href="http://en.wikipedia.org/wiki/Money_supply" rel="nofollow">monetary aggregate</a> to use for M.Jason Smithhttps://www.blogger.com/profile/12680061127040420047noreply@blogger.comtag:blogger.com,1999:blog-6837159629100463303.post-75377897278176445432015-01-17T07:47:14.285-08:002015-01-17T07:47:14.285-08:00Can you give a simpler explanation of what
logPY∼...Can you give a simpler explanation of what<br /><br />logPY∼klogM<br /><br /><br />logP∼(k−1)logM<br /><br />means ?<br /><br />You lost me at that point.Anonymousnoreply@blogger.com