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

CC BY-SA 4.0
2023-11-17T14:39:09.213Z
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.
EB3112
CC BY-SA 4.0
2023-11-18T05:24:05.200Z
@ EB3112 It is enough to understand that $0.2=\frac{1}{5}$.
guest
CC BY-SA 4.0
2023-11-16T06:45:39.917Z
2023-11-16T06:45:39.917Z
tdm
<p>I'm reading Advanced Macroeconomics by Romer and having trouble to understand very simple calculation provided in just one sentence.</p>
<p>Define real money market equilibrium condition as <span class="math-container">$(\frac{M}{P})^s=L(Y,i)=(\frac{M}{P})^d$</span><br />
Then the price leve is <span class="math-container">$P=\frac{M}{L(Y,i)}$</span><br />
If the interest elasticity of money demand is <span class="math-container">$-0.2$</span>, we need almost <span class="math-container">$32$</span> times rise in interest rate to increase price level double.</p>
<p>I can't understand where &quot;<span class="math-container">$32$</span> times&quot; comes from.<br />
From the definition of elasticity, and using the fact that<br />
if we want to make price level double then <span class="math-container">$L(Y,i)$</span> should be reduced by half,<br />
<span class="math-container">$\frac{\partial L}{\partial i}\frac{i}{L}=\frac{\partial L}{L}\frac{i}{\partial i}=-\frac{1}{2}\frac{i}{\partial i}=-0.2$</span> then <span class="math-container">$\frac{\partial i}{i}=2.3$</span><br />
Can anyone tell me how to solve this problem?</p>

CC BY-SA 4.0
2023-11-15T07:17:45.007Z
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.
Giskard
CC BY-SA 4.0
2023-11-15T07:28:17.163Z
@Giskard I edited my post to prove why $\frac{\partial L}{L}=-\frac{1}{2}$
postasguest
CC BY-SA 4.0
2023-11-15T08:28:31.927Z
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?
Giskard
<macroeconomics><elasticity>
Calculation of simple elasticity problem
postasguest
# Calculation of simple elasticity problem

(Asked by: username_0 on 2023-11-15)

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?

**Tags:** macroeconomics, elasticity
> **Comments:**
> **username_1**: 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.  
> **username_0**: @Giskard I edited my post to prove why $\frac{\partial L}{L}=-\frac{1}{2}$  
> **username_1**: 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?  

---

## 1 answers

### Answer by username_2 on 2023-11-16. Score: 2

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:
$$
\begin{align\*}
\frac{i\_1}{i\_0} = 2^5 = 32
\end{align\*}
$$

> **Comments:**
> **username_3**: 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.  
> **username_4**: @ EB3112 It is enough to understand that $0.2=\frac{1}{5}$.  

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