1
$\begingroup$

everyone. I am studying Advanced Macroeconomics, by Derek Leslie, and I am having some troubles in understanding the result of a differentiation in the Chapter 1, section 5. This section approaches the concepts of seigniorage and inflation tax.

The author starts by a simple statement of a real money demand function:

$(1)$ $\frac{M_t}{P_t} = A + bY_t - cr$

Which represents the fact that real money demand responds positively from an increase in output ($Y_t$) and negatively from an increase in the nominal interest rate ($r$).

Then, $Z_t$ is introduced as the inflation tax:

$(2)$ $Z_t = \frac{[π /(1 + π)] M_t}{P_t}$

$Z_t$ is differentiated by $π$ in order to get the maximum possible value of seigniorage when $\frac{d Z_t}{d π} = 0$. There is an additional statement made by the author to support this calculation:

$(3)$ $\frac{M_t}{P_t} = A_0 - e π$

Where $A_0 = A + bY_t - c\rho$ and is considered as a constant for the purpose of the argument. $e$ is equal to $c(1 + \rho)$. $\rho$ is considered as the real interest rate.

After this new argument, the differentiation is calculated and it results in:

$(4)$ $\frac{d Z_t}{d π} = \frac{M_t/P_t}{1+π} - e π$

The author states that the fact that $\frac{d\frac{M_t}{P_t}}{d π} = - e$ in $(3)$ was necessary to calculate it. I see that, but in every way I try to use this info to help me in this differentiation, I don't get the right answer.

The closest I could get from it was when I thought $(2)$ as

$Z_t = \frac{π}{1 + π} * \frac{M_t}{P_t}$

And used the product rule ($u' * v + u * v'$), but I could only get to

$\frac{d Z_t}{d π} = \frac{M_t/P_t}{(1 + π)^2} - \frac{e π}{1 + π}$

I am probably miscalculating somewhere, but I do not know where exactly.

$\endgroup$
4
$\begingroup$

Your calculation is correct. There's probably just a mistake in the book and (4) should really be $(1+π)\frac{d Z_t}{d π} = \frac{M_t/P_t}{1+π} - e π$. Since presumably the RHS is then set to zero, this doesn't change the maximizer, so the mistake is innocent.

$\endgroup$
1
  • $\begingroup$ Thank you, VARRulle. I was wondering if it could be a mistake in the book as you said, but even considering this I was stuck because I wasn't able to get the rationale behind that $(1 + π) \frac{dZ_t}{d π}$ would be "neutral" to the equation since the right hand side is set to zero. Now that you explained this, it is so clear to me. $\endgroup$
    – PGabriel96
    Jun 28 at 15:10

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.