I'm reading Advanced Macroeconomics by Romer and having trouble to understand very simple calculation provided in just one sentence.

Define real money market equilibrium condition as $(\frac{M}{P})^s=L(Y,i)=(\frac{M}{P})^d$
Then the price leve is $P=\frac{M}{L(Y,i)}$
If the interest elasticity of money demand is $-0.2$, we need almost $32$ times rise in interest rate to increase price level double.

I can't understand where "$32$ times" comes from.
From the definition of elasticity, and using the fact that
if we want to make price level double then $L(Y,i)$ should be reduced by half,
$\frac{\partial L}{\partial i}\frac{i}{L}=\frac{\partial L}{L}\frac{i}{\partial i}=-\frac{1}{2}\frac{i}{\partial i}=-0.2$ then $\frac{\partial i}{i}=2.3$
Can anyone tell me how to solve this problem?

  • $\begingroup$ Hi! Where did the $-1/2$ come from in your equation $$\frac{\partial L}{\partial i}\frac{i}{L}=\frac{\partial L}{\partial i}\frac{i}{\partial i}=-\frac{1}{2}\frac{i}{\partial i}=-0.2$$? Also, it seems like in this case we would have $\frac{\partial i}{i} = 2.5$? It feels like you are only sharing half of your thoughts with us. $\endgroup$
    – Giskard
    Commented Nov 15, 2023 at 7:17
  • $\begingroup$ @Giskard I edited my post to prove why $\frac{\partial L}{L}=-\frac{1}{2}$ $\endgroup$ Commented Nov 15, 2023 at 7:28
  • $\begingroup$ Can you please tell us a bit about your background, so that someone can answer in an understandable way? Are you familiar with differentials and calculus? $\endgroup$
    – Giskard
    Commented Nov 15, 2023 at 8:28

1 Answer 1


If $L(Y,i)$ has a constant elasticity of $-0.2$ then you can write it as: $$ L(Y,i) = g(Y)i^{-0.2}. $$ Then for the period 0 and period 1 price levels you have: $$ P_0 = \frac{M}{g(Y)i_0^{-0.2}} \text{ and } P_1 = \frac{M}{g(Y)i_1^{-0.2}}. $$ We have that $P_1 = 2 P_0$ so: $$ 2 \frac{M}{g(y) i_0^{-0.2}} = \frac{M}{g(Y)i_1^{-0.2}}. $$ Simplifying gives: $$ 2 i_1^{-0.2} = i_0^{-0.2}. $$ This gives: $$ \begin{align*} \frac{i_1}{i_0} = 2^5 = 32 \end{align*} $$

  • $\begingroup$ Hi @tdm. For my own benefit, could you include any additional steps in going from: $$2i_1^{-0.2} = i_0^{-0.2}$$ to $$\frac{i_1}{i_0} = 2^5$$. Would be appreciated. $\endgroup$
    – EB3112
    Commented Nov 17, 2023 at 14:39
  • $\begingroup$ @ EB3112 It is enough to understand that $0.2=\frac{1}{5}$. $\endgroup$
    – guest
    Commented Nov 18, 2023 at 5:24

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