Question

Two year old electric cars are worth for their owners20,000 to 60,000(uniform distribution, i.e. informally, each value will occur equally often). Each car worth X for the seller has a potential buyer, for whom it would be worth K*X. Cars are indistinguishable, resulting in adverse selection. Find K if 25% of the cars have been sold.

a.K=6/5 b.K=3/2 c. K=2 d. none of answers is right e.k=1

Answer 1

Answer 2

Solution:

We have a uniform distribution of car values $X$ between 20,000 and 60,000.

• Seller's value: $X\sim \text{Uniform}(20,000,60,000)$

• Buyer's value: $K\cdot X$

• Adverse selection: Only cars where $KX\ge X$ (i.e., $K\ge 1$) will be sold.

• 
  

Step 1: Determine the Range of Sold Cars

Since 25% of cars are sold, the threshold value $X^{\ast }$ satisfies:

$$P(X\le X^{\ast })=0.25$$

For a uniform distribution:

$$\frac{X^{\ast }-20,000}{60,000-20,000}=0.25⟹X^{\ast }=30,000$$

Only cars worth $X\le 30,000$ are sold (since buyers won't pay more than $KX$).

Step 2: Buyer-Seller Equilibrium

At the marginal car (X = 30,000), the buyer's value equals the seller's value:

$$K\cdot 30,000=30,000⟹K=1$$

But this leads to no trade (since $K=1$ implies buyers pay exactly the seller's valuation).

However, since 25% are sold, the correct condition is that buyers pay at least the seller's price:

$$K\cdot X\ge X⟹K\ge 1$$

But since only 25% are sold, the correct $K$ must satisfy:

$$K\cdot 30,000=\text{Expected buyer price for sold cars}$$

But the average sold car value is:

$$\text{Average }X\text{ of sold cars}=\frac{20,000+30,000}{2}=25,000$$

Buyers pay $K\cdot 25,000$, but sellers must be willing to sell at $X\le 30,000$.

The correct $K$ is derived from:

$$K\cdot 25,000=30,000⟹K=\frac{30,000}{25,000}=\frac{6}{5}=1.2$$

Answer:

The correct value of $K$ is $\frac{6}{5}$ (Option a).