# Scale-Dependent Demand Curve

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.)

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 $$d_j(p)=d_i(p)$$.
• 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. Dec 13, 2022 at 4:47
• 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 $k$.