Akerlof's 1970 paper models the utility of two trading groups as

$$ U_1 = M + \sum_{i=1}^n x_i \\ U_2 = M + \sum_{i=1}^n \frac{3}{2} x_i $$ where $M$ is the consumption of good other than cars, $x_i$ is the quality of the $i$th car, and $n$ is the number of cars.

The quality of the cars held by group one have a uniformly distributed quality $0 \leq x \leq 2$ and the price of goods other than the cars is unitary.

Income for the two groups is denoted $Y_1$ and $Y_2$.

The paper goes on to give the demand for cars for type one traders: $$ D_1 = Y_1/p \quad \quad \mu/p > 1 \\ D_1 = 0 \quad \quad \mu/p < 1 $$ The supply of cars from type one is $$ S_1 = pN/2 \quad \quad p \leq 2 $$ and their quality is $\mu = p/2$. The paper states that to drive the expressions for supply and quality, the uniform distribution of car quality is used. How is this done exactly?

up vote 2 down vote accepted

A type-1 seller will trade her car only if the car's measure of quality $x$ (privately known) is less than or equal to the average car quality in the market, $\mu$. $x \in [0,2]$ since no buyer values a car more than $2$ and a car cannot be sold for less than $0$. To find the probability of a car having quality less than or equal to $\mu$, we consider the CDF of $x$, $F(x)$, assuming $x$ is uniformly distributed over the support $[0,2]$:

$$ F(x)= \begin{cases} 0 && x \leq 0 \\ \frac{x}{2} && x \in [0,2] \\ 1 && x \geq 2. \end{cases} $$

So the probability that a car has $x \leq \mu$ is $P(x \leq \mu) = F(\mu) = \frac{\mu}{2}$. A seller facing a price $p$ for her car becomes indifferent towards selling at $\mu = p \implies F = \frac{p}{2}$. Scaling up by a total of $N$ cars possessed by type-1 sellers implies that the market supply from type-1 sellers is $S_1(p) = \frac{p}{2}N$.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.