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Need a math help for the Cagan's model in macroeconomics
The Next CEO of Stack OverflowPresent value of a paymentAre there good step by step math intense books substitute for greene's and woolridge Econometric AnalysisWhat are some books for pricing theory with heavy math?Help understanding expression for continuous discountingModel for simple production chain economyDerive the demand functions: Hotelling-style ModelCan integrals be interpreted simultaneously as aggregates and averages? (Mas-Colell et al. 1995, Proposition 4.C.4)Derivation of demand for intermediate goods in DSGE modelLocal and Central Wage Bargaining: What Is the Difference?Common knowledge in model formulation and solution
$begingroup$
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.
Shouldn't $(p_t+1 - p_t)$ be the log of inflation rate? Why it says just "the inflation rate"?
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.
$ln fracMP = ln (fracp_t+1p_t)^-gamma rightarrow fracMP = (fracp_t+1p_t)^-gamma$
So, if the $(p_t+1 - p_t)$ is just the log of inflation rate, then $fracp_t+1p_t$ is the inflation rate and,
inflation goes up by 1 percentage point
would mean $fracp_t+1p_t$ is going to get +1, right? But I couldn't possibly think it would result the fall of $fracMP$ 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
edited Mar 24 at 10:19
Giskard
13.4k32248
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.
Shouldn't $(p_t+1 - p_t)$ be the log of inflation rate? Why it says just "the inflation rate"?
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.
$ln fracMP = ln (fracp_t+1p_t)^-gamma rightarrow fracMP = (fracp_t+1p_t)^-gamma$
So, if the $(p_t+1 - p_t)$ is just the log of inflation rate, then $fracp_t+1p_t$ is the inflation rate and,
inflation goes up by 1 percentage point
would mean $fracp_t+1p_t$ is going to get +1, right? But I couldn't possibly think it would result the fall of $fracMP$ 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
edited Mar 24 at 10:19
Giskard
13.4k32248
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.
Shouldn't $(p_t+1 - p_t)$ be the log of inflation rate? Why it says just "the inflation rate"?
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.
$ln fracMP = ln (fracp_t+1p_t)^-gamma rightarrow fracMP = (fracp_t+1p_t)^-gamma$
So, if the $(p_t+1 - p_t)$ is just the log of inflation rate, then $fracp_t+1p_t$ is the inflation rate and,
inflation goes up by 1 percentage point
would mean $fracp_t+1p_t$ is going to get +1, right? But I couldn't possibly think it would result the fall of $fracMP$ 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
edited Mar 24 at 10:19
Giskard
13.4k32248
asked Mar 24 at 10:09
dolcodolco
1304
$endgroup$
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.
Shouldn't $(p_t+1 - p_t)$ be the log of inflation rate? Why it says just "the inflation rate"?
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.
$ln fracMP = ln (fracp_t+1p_t)^-gamma rightarrow fracMP = (fracp_t+1p_t)^-gamma$
So, if the $(p_t+1 - p_t)$ is just the log of inflation rate, then $fracp_t+1p_t$ is the inflation rate and,
inflation goes up by 1 percentage point
would mean $fracp_t+1p_t$ is going to get +1, right? But I couldn't possibly think it would result the fall of $fracMP$ 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
mathematical-economics
edited Mar 24 at 10:19
Giskard
13.4k32248
asked Mar 24 at 10:09
dolcodolco
1304
edited Mar 24 at 10:19
Giskard
13.4k32248
asked Mar 24 at 10:09
dolcodolco
1304
edited Mar 24 at 10:19
Giskard
13.4k32248
edited Mar 24 at 10:19
Giskard
13.4k32248
edited Mar 24 at 10:19
Giskard
13.4k32248
13.4k32248
asked Mar 24 at 10:09
dolcodolco
1304
asked Mar 24 at 10:09
dolcodolco
1304
asked Mar 24 at 10:09
dolcodolco
1304
1304
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 24 at 10:19
GiskardGiskard
13.4k32248
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Mar 24 at 10:19
GiskardGiskard
13.4k32248
$endgroup$
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.
answered Mar 24 at 10:19
GiskardGiskard
13.4k32248
$endgroup$
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.
answered Mar 24 at 10:19
GiskardGiskard
13.4k32248
$endgroup$
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.
answered Mar 24 at 10:19
GiskardGiskard
13.4k32248
answered Mar 24 at 10:19
GiskardGiskard
13.4k32248
answered Mar 24 at 10:19
GiskardGiskard
13.4k32248
answered Mar 24 at 10:19
GiskardGiskard
13.4k32248
13.4k32248
add a comment |
add a comment |
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