To equation is

\begin{align} MV=PY \end{align}

where $V=\frac{1}{k}$.

Why $V$ and $Y$ can be taken as fixed or constant?

Why can $V=\frac{1}{k}$ too?

Thanks in advance

  • 1
    $\begingroup$ Assuming that in some future period $Y$ and $V$ are fixed (or more precisely that they are independent of $M$ and $P$, as they are unlikely to be the same as earlier $Y$ and $V$) is an empirical assumption that should be tested (and would typically fail such a empirical test) $\endgroup$
    – Henry
    Apr 23 at 8:10

There are multiple answers to this question.

  1. In any model you can always make a thought experiment where you hold certain variables fixed. So one answer, although not very satisfying one, is that you can view it as a thought experiment. For example, in physics distance traveled equals velocity times time or $D=tv$ and you can always make a thought experiment where you choose 1 or 2 variables that will be fixed and see how others behave.

  2. If you are asking why it would make an economic/intuitive sense to take $Y$, $V$ fixed then:

  • Regarding $Y$: From economic theory we know that $Y$ in a long run depends on productive capacity of an economy that is independent of $M$, $P$ or $V$ (you can learn more about that in any 101 Macro/econ book such as Mankiw Macroeconomics or Mankiw Principles of Economics). So you can consider it to be exogenously given and holding it constant is just performing a thought experiment where you assume you have economy that is not growing so you can see how other variables behave.

  • Regarding $V$, in the past (before 2008), empirically speaking velocity of money was very stable (see the Fed graph below). As you can see between 60s and late 2000s it was always hovering around 1.9 so on empirical grounds many economists considered $V$ stable, although it is worth while pointing that recently it significantly declined. Nonetheless, given that it was historically very stable this might be just transitional, or it is possible there was a structural break and now velocity will remain approximately constant at some lower level. It is also possible that it was wrong to assume it is constant in the long run based on previous empirical observations.

enter image description here

  • 1
    $\begingroup$ I would suggest that, from your FRED chart, fairly stable $V$ stopped being a reasonable assumption after 1992 $\endgroup$
    – Henry
    Apr 23 at 8:13
  • 1
    $\begingroup$ @Henry $V$ certainly was certainly less stable after 1992, but if I would have to put one point after which this would stop being reasonable assumption I would put it in late 2000s where $V$ actually started having a negative trend $\endgroup$
    – 1muflon1
    Apr 23 at 8:30
  • $\begingroup$ Thank you for your answer. $\endgroup$
    – manzanilla
    Apr 23 at 17:44
  • 1
    $\begingroup$ @VerónicaRmz. Well why not? 1. It is perfectly possible to have a country with zero economic growth- in fact in recent years you have some examples of countries with economic growth approximately 0. 2. Holding things constant is the most elementary science 101. For example, in physics if you want to understand how Newtonian gravitational force depends on distance, where gravitational force is given by F=M_1*M_2*G/r^2, you will hold masses M_2 and M_1 constant together with gravitational constant G and see how change in r affects force of gravity $\endgroup$
    – 1muflon1
    Apr 23 at 18:47
  • 1
    $\begingroup$ It does not matter that mass of objects changes over time (for example our sun loses about 4.2*10^12 grams of mass every second), it’s one of the most fundamental ‘tricks’ of a scientist to preform thought experiments where some variables are held constant - in fact that principle also applies to not just thought but real experiments where precautions are made to hold as many things constant as possible $\endgroup$
    – 1muflon1
    Apr 23 at 18:51

To clarify the use of notation this reference shows the definition of slope change where delta notation is used to indicate a finite difference calculation:


As a thought experiment if one holds $V = 1/k$ and $Y$ constant then:

$\frac{\Delta M}{\Delta t} = kY\frac{\Delta P}{\Delta t}$

which means the finite change in the money supply ${\Delta}M$ over a finite period ${\Delta}t$ equals a constant $kY$ times the finite change in the price level ${\Delta}P$ over the same finite period ${\Delta}t$. If one could perform this experiment in reality it would tend to validate or invalidate the monetarist theory of inflation.

Edit: Below I write out the terms for finite difference calculations to check my logic.

$M_{t} = k_{t}Y_{t}P_{t}$

$M_{t-1} = k_{t-1}Y_{t-1}P_{t-1}$

$k = k_{t} = k_{t-1}$

$Y = Y_{t} = Y_{t-1}$

$M_{t} - M_{t-1} = kYP_{t} - kYP_{t-1} = kY(P_{t} - P_{t-1})$

${\Delta}M = M_{t} - M_{t-1}$

${\Delta}P = P_{t} - P_{t-1}$

${\Delta}M = kY{\Delta}P$

  • $\begingroup$ Thanks for your answer. I'm not sure how it answers my question though. $\endgroup$
    – manzanilla
    Apr 23 at 18:02
  • 1
    $\begingroup$ There is no way to hold k and Y constant in the actual economy and therefore no definitive answer to your question. This is because real output Y (price deflated GDP) tends to grow historically over time but it also declines during so-called economic recessions. In practice economists measure YP or nominal output. Then they apply a price deflator to estimate real output Y. They measure the money supply as a statistical aggregate such as M1, M2, which are financial instruments in the banking sector. Then they find velocity as ratio V = YP/M. Monetarists argue change in M causes change in P. $\endgroup$ Apr 23 at 18:16
  • $\begingroup$ @1muflon1 The identity M = kYP is easily derived from the identity MV = YP and the identity V = 1/k which are given in the literature and not referenced in the question or in your answer. Substitute 1/k for V in equation MV = YP to write the equivalent identity M/k = YP. Multiply both sides by k to write an equivalent identity M = kYP. Then if we let M and P vary but hold kY constant from one accounting period to another the delta notation only applies for the change in M and the change in P but not to the constant term kY. $\endgroup$ Apr 23 at 21:40
  • $\begingroup$ @1muflon1 In the context of my answer ${\Delta}M$ is the change in the money supply stated in terms of money units {mu} and ${\Delta}P$ is the change in price level stated in terms of money units {mu}. This means the dimensions are the same for $M = kYP$ and for ${\Delta}M = kY{\Delta}P$. This is the delta notation one learns in Calculus. Perhaps for accuracy both sides of the equation should have ${\Delta}t$ in the denominator to recognize the change occurs over time from one period to another. Monetarists make a weak argument that changes in money cause changes in price level. $\endgroup$ Apr 23 at 23:32
  • 1
    $\begingroup$ If one comprehends the math of finite differences then this is exactly the model one gets when holding k and Y constant during a thought experiment. This model is the rational or reason for holding k and Y constant as a thought experiment. The fact that k and Y are only constant as a thought experiment is given in my answer, my comments, and in the other answer. $\endgroup$ Apr 26 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.