<p>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:</p>
<p><a href="https://www.mathsisfun.com/calculus/derivatives-introduction.html" rel="nofollow noreferrer">https://www.mathsisfun.com/calculus/derivatives-introduction.html</a></p>
<p>As a thought experiment if one holds <span class="math-container">$V = 1/k$</span> and <span class="math-container">$Y$</span> constant then:</p>
<p><span class="math-container">$\frac{\Delta M}{\Delta t} = kY\frac{\Delta P}{\Delta t}$</span></p>
<p>which means the finite change in the money supply <span class="math-container">${\Delta}M$</span> over a finite period <span class="math-container">${\Delta}t$</span> equals a constant <span class="math-container">$kY$</span> times the finite change in the price level <span class="math-container">${\Delta}P$</span> over the same finite period <span class="math-container">${\Delta}t$</span>. If one could perform this experiment in reality it would tend to validate or invalidate the monetarist theory of inflation.</p>
<p>Edit: Below I write out the terms for finite difference calculations to check my logic.</p>
<p><span class="math-container">$M_{t} = k_{t}Y_{t}P_{t}$</span></p>
<p><span class="math-container">$M_{t-1} = k_{t-1}Y_{t-1}P_{t-1}$</span></p>
<p><span class="math-container">$k = k_{t} = k_{t-1}$</span></p>
<p><span class="math-container">$Y = Y_{t} = Y_{t-1}$</span></p>
<p><span class="math-container">$M_{t} - M_{t-1} = kYP_{t} - kYP_{t-1} = kY(P_{t} - P_{t-1})$</span></p>
<p><span class="math-container">${\Delta}M = M_{t} - M_{t-1}$</span></p>
<p><span class="math-container">${\Delta}P = P_{t} - P_{t-1}$</span></p>
<p><span class="math-container">${\Delta}M = kY{\Delta}P$</span></p>

CC BY-SA 4.0
2021-04-23T18:02:59.027Z
Thanks for your answer. I'm not sure how it answers my question though.
Verónica Rmz.
CC BY-SA 4.0
2021-04-23T18:16:26.727Z
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.
SystemTheory
CC BY-SA 4.0
2021-04-23T21:40:48.000Z
@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.
SystemTheory
CC BY-SA 4.0
2021-04-23T23:32:14.827Z
@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.
SystemTheory
CC BY-SA 4.0
2021-04-24T00:52:18.850Z
Oh ok, I guess you are not actually trying to have a rates of changes here but just differences but then your terminology in text was wrong but I will retract the post notice
1muflon1
CC BY-SA 4.0
2021-04-25T23:20:50.117Z
This does not even try to answer the question
csilvia
CC BY-SA 4.0
2021-04-26T00:24:57.640Z
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.
SystemTheory
CC BY-SA 4.0
2021-04-23T00:54:24.853Z
2021-04-24T01:22:33.730Z
2021-04-24T01:22:33.730Z
SystemTheory
<p>There are multiple answers to this question.</p>
<ol>
<li><p>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 <span class="math-container">$D=tv$</span> and you can always make a thought experiment where you choose 1 or 2 variables that will be fixed and see how others behave.</p>
</li>
<li><p>If you are asking why it would make an economic/intuitive sense to take <span class="math-container">$Y$</span>, <span class="math-container">$V$</span> fixed then:</p>
</li>
</ol>
<ul>
<li><p>Regarding <span class="math-container">$Y$</span>: From economic theory we know that <span class="math-container">$Y$</span> in a long run depends on productive capacity of an economy that is independent of <span class="math-container">$M$</span>, <span class="math-container">$P$</span> or <span class="math-container">$V$</span> (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.</p>
</li>
<li><p>Regarding <span class="math-container">$V$</span>, 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 <span class="math-container">$V$</span> 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.</p>
</li>
</ul>
<p><a href="https://i.sstatic.net/gBQA8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gBQA8.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-04-23T08:13:35.407Z
I would suggest that, from your FRED chart, fairly stable $V$ stopped being a reasonable assumption after 1992
Henry
CC BY-SA 4.0
2021-04-23T08:30:19.683Z
@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
1muflon1
CC BY-SA 4.0
2021-04-23T18:12:23.290Z
Regarding $Y$ in the short run, would it also be possible to consider it somehow fixed?
Verónica Rmz.
CC BY-SA 4.0
2021-04-23T18:38:53.633Z
@VerónicaRmz. Yes you can consider it fixed if you assume economic growth of economy is zero.
1muflon1
CC BY-SA 4.0
2021-04-23T18:40:29.363Z
And why could be considered as 0?
Verónica Rmz.
CC BY-SA 4.0
2021-04-23T18:47:05.047Z
@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
1muflon1
CC BY-SA 4.0
2021-04-23T18:51:13.883Z
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
1muflon1
CC BY-SA 4.0
2021-04-23T20:03:56.757Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/123362/discussion-between-veronica-rmz-and-1muflon1).
Verónica Rmz.
CC BY-SA 4.0
2021-04-22T20:21:40.940Z
2021-04-22T21:25:07.867Z
2021-04-22T21:25:07.867Z
1muflon1
<p>To  equation is</p>
<p><span class="math-container">\begin{align}
MV=PY 
\end{align}</span></p>
<p>where <span class="math-container">$V=\frac{1}{k}$</span>.</p>
<p>Why <span class="math-container">$V$</span> and <span class="math-container">$Y$</span> can be taken as fixed or constant?</p>
<p>Why can <span class="math-container">$V=\frac{1}{k}$</span> too?</p>
<p>Thanks in advance</p>

CC BY-SA 4.0
2021-04-23T08:10:12.540Z
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)
Henry
<macroeconomics><inflation><money-supply><quantity-theory-of-money>
Why in the quantitative equation: $MV=PY$, $V$ and $Y$ can be taken as fixed?
Verónica Rmz.
# Why in the quantitative equation: $MV=PY$, $V$ and $Y$ can be taken as fixed?

(Asked by: username_0 on 2021-04-22)

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

**Tags:** macroeconomics, inflation, money-supply, quantity-theory-of-money
> **Comments:**
> **username_1**: 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)  

---

## 2 answers

### Answer by username_3 on 2021-04-22. Score: 3 | Accepted answer

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](https://i.sstatic.net/gBQA8.png)](https://i.sstatic.net/gBQA8.png)

> **Comments:**
> **username_1**: I would suggest that, from your FRED chart, fairly stable $V$ stopped being a reasonable assumption after 1992  
> **username_3**: @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  
> **username_0**: Regarding $Y$ in the short run, would it also be possible to consider it somehow fixed?  
> **username_3**: @VerónicaRmz. Yes you can consider it fixed if you assume economic growth of economy is zero.  
> **username_0**: And why could be considered as 0?  
> **username_3**: @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  
> **username_3**: 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  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/123362/discussion-between-veronica-rmz-and-1muflon1).  

---

### Answer by username_2 on 2021-04-23. Score: -1

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:

<https://www.mathsisfun.com/calculus/derivatives-introduction.html>

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.

$M\_{t} = k\_{t}Y\_{t}P\_{t}$

$M\_{t-1} = k\_{t-1}Y\_{t-1}P\_{t-1}$

$k = k\_{t} = k\_{t-1}$

$Y = Y\_{t} = Y\_{t-1}$

$M\_{t} - M\_{t-1} = kYP\_{t} - kYP\_{t-1} = kY(P\_{t} - P\_{t-1})$

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

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

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

> **Comments:**
> **username_0**: Thanks for your answer. I'm not sure how it answers my question though.  
> **username_2**: 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.  
> **username_2**: @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.  
> **username_2**: @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.  
> **username_3**: Oh ok, I guess you are not actually trying to have a rates of changes here but just differences but then your terminology in text was wrong but I will retract the post notice  
> **username_4**: This does not even try to answer the question  
> **username_2**: 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.  

---

