Verifying the constant elasticity of the demand model Q= aP^b

Question

"It is well known" that the elasticity of $Q$ relative to $P$ is the constant $b$ in the equation $Q(P)=aP^b$. We can verify it using the convenient definition of elasticity $= \frac{dQ}{dP} \frac{P}{Q}$, we can verify it in Wikipedia.

Based on this, when I create a spreadsheet with parameters $b=-2$, $a=100$ and P ranging from 1-5 by 1 I expected to recover an elasticity of -2, where elasticity = % change in Q/ % change in P. This is not the case. Instead the elasticity varies substantially as the % change in P varies.

I'm forced to conclude that 'constant elasticity' does not mean a fixed value regardless of the magnitude of the price change, but only invariant with regards to the value of price OR that I have a spreadsheet error.

I've pasted an image of the spreadsheet results below. Obviously, without seeing the formulas, no one can verify that I don't have an error, but I'm hopeful that someone will give this a spin for themselves and get very different numbers and then I'll know my failure to verify the constant elasticity in this function is a spreadsheet error.

Answer 1

The problem with your calculations is that you have very big changes in price

If you made price increases of $1\%$ in price, you would find the quantity fell by about $2\%$ as shown in the table below leading to an elasticity of about $-2$

If you made the price changes even smaller, the calculated elasticity would get even closer to $2$, and in a sense you are looking for the limit of elasticity as the price change approaches $0$, which will indeed be the constant $b=-2$

 a       b      P       Q=a*P^b pct change Q    pct change P   elasticity

100     -2      1       100.00          
100     -2      1.01     98.03      -1.97%          1%          -1.97
100     -2      2        25.00          
100     -2      2.02     24.51      -1.97%          1%          -1.97
100     -2      3        11.11          
100     -2      3.03     10.89      -1.97%          1%          -1.97
100     -2      4         6.25          
100     -2      4.04      6.13      -1.97%          1%          -1.97
100     -2      5         4.00          
100     -2      5.05      3.92      -1.97%          1%          -1.97

Answer 2

Elasticity is the slope after taking a log transform. Throughout this post $\log$ refers to the natural logarithm. That is, the logarithm with base $e$, the ln function in Excel. The logarithm transforms multiplication into addition. Log differences are something akin to percent changes (and they're equivalent for small changes).

Let's say we have:

$$ Q_t = A P_t^b $$

Take the logarithm of both sides:

$$ \log Q_t = \log A + b \log P_t$$

Log differences are conceptually similar to percent changes (for small percent changes). The slope of the line using the log transformed variables is $b$. This is the elasticity.

$$ \frac{d \left( \log Q_t \right) }{\log P_t} = b $$

Why do people treat elasticities as percent changes?

$$\log(Y) - \log(X) \approx \frac{Y-X}{X} \quad \quad \text{ for }\frac{y}{x} \approx 1$$

Log differences are approximately the percent change. Why? Take the first-order taylor expansion of the log to approximate it with a line near 1.

\begin{align*} \log(Y) &\approx \log(1) + \frac{1}{\log(1)}\left(Y - 1 \right) \\ &\approx Y - 1 \end{align*} Hence $\log\left(\frac{Y}{X} \right) \approx \frac{Y-X}{X}$ for $Y \approx X$. This works well for small changes, eg. $\log(1.03) = .0296$, but it gets quite off for big changes $\log(1.5) = .4055$. This is probably where your stuff is getting off...

Example where logs are nice...

Let's say we have $1, 2, 4, 8$. Each is a 100% increase from the previous value. Taking logs we have $0, .693, .1386, 2.079$. The numbers increase linearly by $\log(2)$. Looking at percent increases though, it's less clean: $\frac{2-1}{1} = 100\%$ and $\frac{4-2}{2} = 100\%$ but $\frac{4-1}{1} = 300%$.

$$ \left(1 + R_1 \right) \left(1 + R_2 \right) = 1 + R_1 + R_2 + R_1R_2$$ Dealing with percent changes gets messy when $R_1R_2$ is big enough to matter. Taking logs though, let $r_1 = \log(1+R_1)$ etc...

$$ \log\left((1+R_1)(1+R_2) \right)= r_1 + r_2$$

What I would do if I were you:

Take a log and then it will be clean, and simple. Add a few columns to your spreadsheet.

• $q_t = \log Q_t$.
• $a = \log A$
• $p_t = \log P_t$

Then the equation is: $$ q_t = a + b p_t$$

$$\frac{dq_t}{dp_t} = b $$

$$ \begin{array}{rrrrr} P & Q & \log P & \log Q & \Delta \log P & \Delta \log Q & \frac{\Delta \log P}{\Delta \log Q} \\ 1 & 100 & 0 & 4.605 & \\ 2 & 25 & .6931 & 3.2189 & .6931 & -1.3863 & -2.0\\ 3 & 11.11 & 1.0986 & 2.4079 & .4055 & -.8109 & -2.0\\ 4 & 6.25 & 1.3863 & 1.8326 & .2877 & -.5754 & -2.0\\ 5 & 4 & 1.6094 & 1.3863 & .2231 & -.4463 & -2.0 \end{array} $$

$\Delta$ means "change in", hence $\Delta \log Q_i = \log Q_i - \log Q_{i-1}$.