I was reading: This article on money velocity (see the summary section)

Where two formulas are presented:

first an equality:

$$\text{Quantity of Money} (M) \times \text{Velocity of Money} (V) = \text{Real GDP} (Y) \times \text{Prices} (P) $$

or just stated as

$$ MV = YP$$

then its derivative is stated in the article as:

$$ \Delta M + \Delta V = \Delta Y + \Delta P $$

This second statement makes no sense to me, since if I apply the product rule I should get

$$ M \Delta V + V\Delta M = Y \Delta P + P \Delta Y$$

And I see no way to easily drop the non differential terms. Where did I go wrong here?


In regard to the question about the meaning of $\Delta$ notation I agree substantially with the answer given by @1muflon1. However I could only become convinced of this answer by expanding the "logic" implied in the notation with reference to the information below.

Notation Used in the Article (See Question)


"Equation of Exchange"

$MV = YP$

"It can also be derived from the equation of exchange, for small changes in the factors

$\Delta M + \Delta V = \Delta P + \Delta Y$

The Quantity Theory of Money (1 page)


The quantity theory of money (QTM) asserts that aggregate prices (P) and total money supply (M) are related according to the equation P = VM/Y, where Y is real output and V is velocity of money. With lower-case letters denoting percentage changes (growth rates), the QTM can be expressed as p = v + my, with p as the rate of inflation and y, v, and m as growth rates of output, velocity, and money stock, respectively.

This reference expresses QTM (Equation of Exchange) in terms of percent change but without any mention of the reasoning, derivation, use of natural logarithms, or the method for computing percent change.

Use of Logarithms in Economics


Thus if you earn r% interest that is compounded continuously, at the end of the year your money will have grown by $P_{2}/P_{1}=e^{r}$. For example, if r = 4%, continuous compounding would actually give you 4.08% at the end of the year, a little more than the 4.06% from quarterly compounding or 4.0% with no compounding.

Taking natural logarithms is just the inverse of the above operation: $ln(P_{2}/P_{1})=r$, or since the log of a ratio is the difference of the logs, $ln(P_{2})-ln(P_{1})=r$. In other words, taking the difference between the log of a stock price in year 2 and the log of the price in year 1 is just calculating a rate of return on the holding, quoted in terms of a continuously compounded rate.

For low values of r, the continuously compounded return is almost the same as the non-compounded return, so that the log difference is almost the same number as the percentage change. For the example just given, the percentage change is $(P_{2}-P_{1})/P_{1}=0.0408$ whereas the log change is $ln(P_{2})-ln(P_{1})=0.04$. So any time that you see a graph that is measured in logs, an increase of 0.01 on that scale corresponds very closely to a 1% increase. A graph that is a straight line over time when plotted in logs corresponds to growth at a constant percentage rate each year.

A Monetarist Model of the Inflationary Process (11 pages)


Page 2 (Nominal page 14, final paragraph)

Three other features of the model should be mentioned at the outset. First, all relations are linear and are expressed in logarithmic form. There is a specific reason for this formulation. Modern monetarist analysis is usually stated in terms of percentage rates of change of the relevant variables. And since the percentage change of any variable over a given interval of time can be represented mathematically by the first time difference of its logarithm, it follows that a log-linear formulation facilitates the analysis.

Page 3 (Nominal page 15, second full paragraph)

As for notation, the model employs the following symbols. Let m, be the money stock, y actual real income, ... and p the price level - with all variables expressed as logarithms.

The symbols $\Delta$ and $\Delta^2$ appearing before a variable denote first and second time differences, respectively, so that the model is effectively expressed in terms of proportional rates of change and rates of acceleration or deceleration of those rates of change.


That is not a differential form of the equation. $\Delta$ there stands just for changes not differential. You can derive the equation in such form by log-linearization.

Take the original equation:


Apply natural log to both sides:

$$ \ln M + \ln V = \ln Y + \ln P$$

And now you are done, since in an equation that is in log-log form you can interpret everything in terms of changes.


$$ \ln M + \ln V = \ln Y + \ln P \implies \Delta M + \Delta V = \Delta Y + \Delta P$$

  • $\begingroup$ I think the notation is ambiguous in the original article. Also based on the most voted answer to this question: stats.stackexchange.com/questions/244199/… it is not clear to me how for example ln M is a percent change unless one expands the symbol ln M = ln M(t) - ln M(t-1) for a relatively small change in M over time. Of course this logic would apply for all the variables. $\endgroup$ – SystemTheory Mar 3 at 0:42
  • $\begingroup$ @SystemTheory no this is not based on that, that is different property of logs, here I am using fact that if you have $\ln y (t) = \ln x (t) \implies \dot{y}/y= \dot{x}/x \implies \Delta y= \Delta x$ - this is not the same as $\ln y_t- \ln y_{t-1} \approx \%\Delta$ just have a look at some undergraduate math textbook to refresh your knowledge on this. $\endgroup$ – 1muflon1 Mar 3 at 8:59
  • $\begingroup$ I will look into my old textbook(s) and further online research. The article in the question states "[F]or small changes in the factors, that the change in money growth (ΔM) plus the change in the velocity of money (ΔV) equals the change in prices (ΔP) plus the change in real GDP (ΔY)." Are you saying this only makes sense if the delta is interpreted as a natural log transformation and so the the "units" of the equation should be stated as "percent change" rather than merely the "change" in each variable? Also is this an exact solution or an approximation for small percent changes? $\endgroup$ – SystemTheory Mar 3 at 17:37
  • $\begingroup$ @SystemTheory the result is based on a time derivative so it holds with equality $\frac{d}{dx}[f(x)] = f'(x)$. e.g. for $d/dt[\ln y(t)=\ln x(t)]$ you directly get $\dot{y}/y=\dot{x}/x$. It holds for small changes in $t$ since $\dot{y}$ is slope of function at the limit $t$ goes to 0, but that does not matter since t can be any unit Also, I probably did not phrased it properly, it is common to also multiply both sides of equation by 100 which turns changes into percent changes, but it is true that I did not show that step so I removed the % from the answer $\endgroup$ – 1muflon1 Mar 3 at 17:53
  • $\begingroup$ I have seen the identity ln $M$ + ln $V$ = ln $Y$ + ln $P$ used in some economic papers and power point presentations but cannot find a derivation of this use. Simply transforming a product term to its log or natural log representation is helpful to avoid multiplication (video lesson youtu.be/cMA8m6upFPM) but natural log transformation alone does not give a slope line or percent change. This reference shows percent change with log transformation for price only flypapereffect.wordpress.com/2013/05/31/…. $\endgroup$ – SystemTheory Mar 4 at 22:05

The author is using notation which is non-standard within mathematics, but might be normal in economics.

The $\Delta$ operator is presumably defined as: $$ \Delta x = \frac{\frac{dx}{dt}}{x}. $$

This is equivalent to a percentage change (continuous time).

Then: $$ \Delta MV = \frac{\frac{d MV}{dt}}{MV}, $$ $$ = \frac{V \frac{dM}{dt} + M \frac{dV}{dt}}{MV} = \Delta M + \Delta V. $$ Same algebra for the other side of the equation.

  • $\begingroup$ Where do you get that definition of $\Delta x$? Typical notation $\Delta x = x_2 - x_1$. Why divide by $x$ or $MV$? Most voted answer here stats.stackexchange.com/questions/244199/… says For $x_2$ and $x_1$ close to each other the percent change ($x_2$ - $x_1$)/$x_1$ is approximated by the log difference log $x_2$ - log $x_1$. I can follow the logic applied by @1muflon1 with reference to the link. Solution requires ability to drop any term with zero change and integrate to find change in the desired variable. $\endgroup$ – SystemTheory Mar 3 at 1:01
  • $\begingroup$ It’s not standard notation, but the standard definition won’t work. It’s the equivalent of a percentage change. The issue with the log interpretation is that it scales with magnitudes, which does not appear to be desired, $\endgroup$ – Brian Romanchuk Mar 3 at 1:44
  • $\begingroup$ Rocket scientists define momentum $p = mv$ and set change in momentum equal to force times time. This expands via the product rule as $dp/dt = m(dv/dt) + v(dm/dt) = Ft$. Reasoning by analogy one could write a differential equation in the form $M(dV/dt) + V(dM/dt) = PY$ where I think the Monetarist assumption is that dV/dt = 0 and/or dM/dt >> dV/dt to state, "Inflation is always and everywhere a monetary phenomenon." In reality changes in both money supply M and velocity of money V are coupled to credit/debt deals. Bank credit (leverage) increases M and nonbank credit (leverage) increases V. $\endgroup$ – SystemTheory Mar 3 at 2:27

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