If one always spends the same amount, what's one's price elasticity of demand?

Question

Source: p 109, Question 5.9, Principles of Microeconomics, 7 Ed, 2014, by N Gregory Mankiw
= Question 5.7, Principles of Microeconomics, 4 Ed, 2008, by N Gregory Mankiw

9. $\color{green } { \text { Before looking at the price,} }$ Jessie spends $d$ dollars on something (abbreviate this S). What is Jessie’s price elasticity of demand?

Given Answer: Jessie's price elasticity of demand is one, because he spends the same amount on S, no matter what the price, which means $\color{darkred} { \text { his percentage change in quantity } }$ is equal to the percentage change in price.

From p 91: Price elasticity of demand $= \dfrac{ \text{ Percentage change in quantity demanded }} { \text{ Percentage change in price } } $

The green implies Jessie's ignorance of prices, so % change in price = 0. Yet please explain the answer? Especially the red, because the question implies nothing about the % change in quantity demanded?

Footnote: I slightly generalised the question. Mankiw writes 'gas' instead of 'something'.

Answer 1

The intuition is that if Jessie spends the same amount on $S$ no matter the price $p$, then a % change in price of price will reduce the demand by the same % . The thing that might confuse you is that the formula you are given is an approximation for the true price elasticity of demand given by \begin{align} \eta=\frac{dS\Big/ S}{dp\Big/ p} \end{align} where $d$ is the derivative operator. The approximation \begin{align} \frac{\%\Delta S}{\% \Delta p} \end{align} only works for very small changes in $p$ and $S$. To see why the elasticity of demand is $\mid 1 \mid$, let $b$ be the fixed amount spend on $S$. Then the quantity demanded $S$ is given by \begin{align} S=\frac{b}{p} \end{align}

Assume that there is a small price change from $p$ to $p+\epsilon, \, \epsilon>0$. Then we can define $S_1=\frac{b}{p}$ as the demand before the price change and $S_2=\frac{b}{p+\epsilon}$ as the demand after the price change. Thereby we see that \begin{align} \frac{S_2\Big/ S_1 -1}{\frac{p+\epsilon}{p}-1} = \frac{\frac{b}{p+\epsilon}\Big/\frac{b}{p}-1}{\frac{p+\epsilon}{p}-1}=\frac{\frac{-\epsilon}{p+\epsilon}}{\frac{e}{p}}=-\frac{p}{p+\epsilon}\approx -1 \end{align} when $\epsilon$ is small enough.

Answer 2

The Arc Elasticity is the one that will work using the percentage formula (see also this post)

$$\eta_{arc} = \frac {q_b-q_a}{q_b+q_a}\cdot \frac{p_b+p_a}{p_b-p_a} \tag{1}$$

Quantity demanded is $$S = E/p$$

where $E$ is total expenditure.

Assume that price increases from $p_a$ to $p_b = (1+c)p_a$, where $c$ is not "small". The second quotient in the Arc Elasticity formula is

$$\frac{p_b+p_a}{p_b-p_a} = \frac{(1+c)p_a+p_a}{(1+c)p_a-p_a} = \frac {2+c}{c} \tag{2}$$

Total expenditure remains the same, by assumption. So for the quotient involving quantities we have

$$q_b-q_a = \frac {E}{(1+c)p_a} - \frac {E}{p_a} = \frac {E - (1+c)E}{(1+c)p_a} = \frac {-c}{1+c}\cdot \frac {E}{p} \tag{3} $$

and

$$q_b+q_a = \frac {E}{(1+c)p_a} + \frac {E}{p_a} = \frac {E + (1+c)E}{(1+c)p_a} = \frac {2+c}{1+c}\cdot \frac {E}{p} \tag{4} $$

So

$$\frac {q_b-q_a}{q_b+q_a} = \frac{\frac {-c}{1+c}\cdot \frac {E}{p}}{\frac {2+c}{1+c}\cdot \frac {E}{p}} = \frac {-c}{2+c} \tag {5}$$

Combining $(2)$ and $(5)$ we get

$$\eta_{arc} = \frac {-c}{2+c} \cdot \frac {2+c}{c} = -1$$

with the minus sign indicating direction of movement.