Setup: Say I have a store and I have 50 bottles that I want to sell. Outside this store there are 100 people who want water bottles and each differs in the price they are willing to pay for a water bottle.

Generally, we say the number of people willing to buy the bottle at price $b$ is a monotonically decreasing function of $b$, but let's further say that the distribution in this price is "scale-free" in that it is independent of the number of people outside. In other words, the fraction of people who are willing to pay at least 5 coins for the water bottle is independent of the number of people out there. This means that if 10 people are willing to pay at least 5 coins when there are 100 people outside, then 100 people will be willing to at least 5 coins when there are 1000 people outside. Such an assumption makes sense if you consider each person as independent and to be "drawn" from a distribution of people who want to buy water bottles.

Comparison: This assumption of "scale-free" fraction in demand differs from typical presentations. In particular, I usually see demand represented as a linear function such as $N_{D}(b) = N_D(0) - m_D b$ where $N_D(0)$ is the total number of people in the market and $m_D$ is an arbitrary constant. By this linear equation, when you increase the number of people in the market, you are simply pushing the $N_{D}(b)$ curve to a higher intercept with the same slope. (see examples ....)

What this implies is that the fraction of people willing to pay a certain price,

$$N_D(b)/N_D(0) = 1 - m_D b/N_D(0)$$

is not "scale free" and in particular having more people leads to a greater fraction of people willing to pay above a certain amount. For example, if you have 100 people and 50 of them are willing to pay above 2 coins, then having 1000 people means 950 of them are willing to pay above 2 coins.

Question: This "scale-dependent" pricing distribution makes much less sense to me than a scale-free version, yet the former is often how supply and demand is presented. What am I missing? Is "scale-free"ness not a good general assumption?

(I realize that what I'm missing might simply be that the linear equation is an approximation, but I'm wondering if I'm missing anything more than this.)


1 Answer 1


A (market) demand curve is really the sum of all the individual demand curves. If there are $N$ possible consumers, and demand for agent $i$ is $d_i(p)$ then the market demand curve is $\sum_{i=1}^Nd_i(p)$. Let $D_N(p):=\sum_{i=1}^Nd_i(p)$ be the market demand with $N$ consumers. "Scale free' market demand would be formulated as $\frac{D_N(p)}{N}=\frac{D_M(p)}{M}$ for all $N$ and $M$. This would be the case when each individual consumer has the same demand $d_i(p)=d_j(p)$. If you move outside of this restrictive case of identical individual demand then I think the property of scale-free is very unlikely to hold. In fact, it almost doesn't by definition. As with two consumers $i$ and $j$ this requires $d_i(p)+d_j(p)=2d_i(p)$ which implies $d_j(p)=d_i(p)$.

Therefore, your notion of scale-free market demand is only true when all individuals have the same demand which is unlikely to hold in most markets.

  • $\begingroup$ Thanks for your answer. It seems that you're suggesting that $D_N(p)/N = D_M(p)/M$ implies that $d_{j}(p) = d_{i}(p)$, but this implication is not valid. It is akin to saying that because the average of two quantities are equal, the elements that make up the averages are also equal. $\endgroup$ Dec 13, 2022 at 4:47
  • $\begingroup$ The implication is valid (as is explained in the asnwer for the case of two consumers). Suppose the equality $D_N(p)/N=D_M(p)/M$ holds for all $p$ and for al natural numbers $N,M$. Consider the case where $M=1$. Then $D_N(p)=ND_1(p)$ for all $p$ and for every natural number $N$. This does amount to every consumer having the same demand function. More generally, a market demand that is scale-free could allow some consumers to be of different "sizes", in the sense that consumer $i$ and $j$ might have demands related by $d_i(p)=kd_j(p)$ for some constant $k$. $\endgroup$
    – smcc
    Aug 9, 2023 at 12:19

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