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2

$begingroup$

From the appendix after the chapter 4 in Macroeconomics 7th edition by Gregory Mankiw.

To keep the math as simple as possible, we posit a money demand function that is linear in the natural logarithms of all the variables. The money demand function is

$m_t − p_t = −gamma( p_{t+1} − p_t)$,

where $m_t$ is the log of the quantity of money at time t, $p_t$ is the log of the price level at time t, and $gamma$ is a parameter that governs the sensitivity of money demand to the rate of inflation. By the property of logarithms, $m_t − p_t$ is the log of real money balances, and $p_{t+1} − p_t$ is the inflation rate between period t and period t+1. This equation states that if inflation goes up by 1 percentage point, real money balances fall by $gamma$ percent.

1. Shouldn't $(p_{t+1} - p_t)$ be the log of inflation rate? Why it says just "the inflation rate"?



2. 
   

   
   

   This equation states that if inflation goes up by 1 percentage point, real money balances fall by $gamma$ percent.

   
   

   
   
   

   My math level is like that of a high school. Would anyone be so nice and explain this for me? To me, it doesn't make sense at all.

   
   
   

   $ln frac{M}{P} = ln (frac{p_{t+1}}{p_t})^{-gamma} rightarrow frac{M}{P} = (frac{p_{t+1}}{p_t})^{-gamma}$

   
   
   

   So, if the $(p_{t+1} - p_t)$ is just the log of inflation rate, then $frac{p_{t+1}}{p_t}$ is the inflation rate and,

   
   
   

   
   

   inflation goes up by 1 percentage point

   
   

   
   
   

   would mean $frac{p_{t+1}}{p_t}$ is going to get +1, right? But I couldn't possibly think it would result the fall of $frac{M}{P}$ by the $gamma$ point. What am I missing?

   
   
   

   And secondly, if the $(p_{t+1} - p_t)$ is just the inflation rate,(not the log of any) then it bugs me more than the former. So, +1 change to the inflation rate is like nothing but that we would get "$−gamma(1 + p_{t+1} − p_t)$" at the right side, right? How could this be the case?

   
   

mathematical-economics

share|improve this question

edited Mar 24 at 10:19

Giskard

13.6k32348

asked Mar 24 at 10:09

dolcodolco

1304

$endgroup$

add a comment | 

2

$begingroup$

From the appendix after the chapter 4 in Macroeconomics 7th edition by Gregory Mankiw.

To keep the math as simple as possible, we posit a money demand function that is linear in the natural logarithms of all the variables. The money demand function is

$m_t − p_t = −gamma( p_{t+1} − p_t)$,

where $m_t$ is the log of the quantity of money at time t, $p_t$ is the log of the price level at time t, and $gamma$ is a parameter that governs the sensitivity of money demand to the rate of inflation. By the property of logarithms, $m_t − p_t$ is the log of real money balances, and $p_{t+1} − p_t$ is the inflation rate between period t and period t+1. This equation states that if inflation goes up by 1 percentage point, real money balances fall by $gamma$ percent.

1. Shouldn't $(p_{t+1} - p_t)$ be the log of inflation rate? Why it says just "the inflation rate"?



2. 
   

   
   

   This equation states that if inflation goes up by 1 percentage point, real money balances fall by $gamma$ percent.

   
   

   
   
   

   My math level is like that of a high school. Would anyone be so nice and explain this for me? To me, it doesn't make sense at all.

   
   
   

   $ln frac{M}{P} = ln (frac{p_{t+1}}{p_t})^{-gamma} rightarrow frac{M}{P} = (frac{p_{t+1}}{p_t})^{-gamma}$

   
   
   

   So, if the $(p_{t+1} - p_t)$ is just the log of inflation rate, then $frac{p_{t+1}}{p_t}$ is the inflation rate and,

   
   
   

   
   

   inflation goes up by 1 percentage point

   
   

   
   
   

   would mean $frac{p_{t+1}}{p_t}$ is going to get +1, right? But I couldn't possibly think it would result the fall of $frac{M}{P}$ by the $gamma$ point. What am I missing?

   
   
   

   And secondly, if the $(p_{t+1} - p_t)$ is just the inflation rate,(not the log of any) then it bugs me more than the former. So, +1 change to the inflation rate is like nothing but that we would get "$−gamma(1 + p_{t+1} − p_t)$" at the right side, right? How could this be the case?

   
   

mathematical-economics

share|improve this question

edited Mar 24 at 10:19

Giskard

13.6k32348

asked Mar 24 at 10:09

dolcodolco

1304

$endgroup$

add a comment | 

$begingroup$

From the appendix after the chapter 4 in Macroeconomics 7th edition by Gregory Mankiw.

To keep the math as simple as possible, we posit a money demand function that is linear in the natural logarithms of all the variables. The money demand function is

$m_t − p_t = −gamma( p_{t+1} − p_t)$,

where $m_t$ is the log of the quantity of money at time t, $p_t$ is the log of the price level at time t, and $gamma$ is a parameter that governs the sensitivity of money demand to the rate of inflation. By the property of logarithms, $m_t − p_t$ is the log of real money balances, and $p_{t+1} − p_t$ is the inflation rate between period t and period t+1. This equation states that if inflation goes up by 1 percentage point, real money balances fall by $gamma$ percent.

1. Shouldn't $(p_{t+1} - p_t)$ be the log of inflation rate? Why it says just "the inflation rate"?



2. 
   

   
   

   This equation states that if inflation goes up by 1 percentage point, real money balances fall by $gamma$ percent.

   
   

   
   
   

   My math level is like that of a high school. Would anyone be so nice and explain this for me? To me, it doesn't make sense at all.

   
   
   

   $ln frac{M}{P} = ln (frac{p_{t+1}}{p_t})^{-gamma} rightarrow frac{M}{P} = (frac{p_{t+1}}{p_t})^{-gamma}$

   
   
   

   So, if the $(p_{t+1} - p_t)$ is just the log of inflation rate, then $frac{p_{t+1}}{p_t}$ is the inflation rate and,

   
   
   

   
   

   inflation goes up by 1 percentage point

   
   

   
   
   

   would mean $frac{p_{t+1}}{p_t}$ is going to get +1, right? But I couldn't possibly think it would result the fall of $frac{M}{P}$ by the $gamma$ point. What am I missing?

   
   
   

   And secondly, if the $(p_{t+1} - p_t)$ is just the inflation rate,(not the log of any) then it bugs me more than the former. So, +1 change to the inflation rate is like nothing but that we would get "$−gamma(1 + p_{t+1} − p_t)$" at the right side, right? How could this be the case?

   
   

mathematical-economics

share|improve this question

edited Mar 24 at 10:19

Giskard

13.6k32348

asked Mar 24 at 10:09

dolcodolco

1304

$endgroup$

share|improve this question

edited Mar 24 at 10:19

Giskard

13.6k32348

asked Mar 24 at 10:09

dolcodolco

1304

share|improve this question

edited Mar 24 at 10:19

Giskard

13.6k32348

asked Mar 24 at 10:09

dolcodolco

1304

edited Mar 24 at 10:19

Giskard

13.6k32348

edited Mar 24 at 10:19

Giskard

13.6k32348

asked Mar 24 at 10:09

dolcodolco

1304

asked Mar 24 at 10:09

dolcodolco

1304

1 Answer
1

active

oldest

votes

3

$begingroup$

The answer to both your questions is that for small $x$ values
$$
ln(1+x) approx x,
$$
the difference being less than $x^2/2$. (Proof by Taylor-approximation.)

So if inflation is around 10%, then the absolute error from this type of approximation is less then 0.5%, which is pretty good.

This should also answer your second question, as the approximation
$$
gamma x approx ln(1+ gamma x),
$$
works as well.

It may also be worthwhile to look into elasticity.

share|improve this answer

answered Mar 24 at 10:19

GiskardGiskard

13.6k32348

$endgroup$

add a comment | 

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3

$begingroup$

The answer to both your questions is that for small $x$ values
$$
ln(1+x) approx x,
$$
the difference being less than $x^2/2$. (Proof by Taylor-approximation.)

So if inflation is around 10%, then the absolute error from this type of approximation is less then 0.5%, which is pretty good.

This should also answer your second question, as the approximation
$$
gamma x approx ln(1+ gamma x),
$$
works as well.

It may also be worthwhile to look into elasticity.

share|improve this answer

answered Mar 24 at 10:19

GiskardGiskard

13.6k32348

$endgroup$

add a comment | 

3

$begingroup$

The answer to both your questions is that for small $x$ values
$$
ln(1+x) approx x,
$$
the difference being less than $x^2/2$. (Proof by Taylor-approximation.)

So if inflation is around 10%, then the absolute error from this type of approximation is less then 0.5%, which is pretty good.

This should also answer your second question, as the approximation
$$
gamma x approx ln(1+ gamma x),
$$
works as well.

It may also be worthwhile to look into elasticity.

share|improve this answer

answered Mar 24 at 10:19

GiskardGiskard

13.6k32348

$endgroup$

add a comment | 

$begingroup$

The answer to both your questions is that for small $x$ values
$$
ln(1+x) approx x,
$$
the difference being less than $x^2/2$. (Proof by Taylor-approximation.)

So if inflation is around 10%, then the absolute error from this type of approximation is less then 0.5%, which is pretty good.

This should also answer your second question, as the approximation
$$
gamma x approx ln(1+ gamma x),
$$
works as well.

It may also be worthwhile to look into elasticity.

share|improve this answer

answered Mar 24 at 10:19

GiskardGiskard

13.6k32348

$endgroup$

share|improve this answer

answered Mar 24 at 10:19

GiskardGiskard

13.6k32348

answered Mar 24 at 10:19

GiskardGiskard

13.6k32348

answered Mar 24 at 10:19

GiskardGiskard

13.6k32348