Offers on an asset arrive one per day, independent and uniform on [0, 100]. Each additional day of waiting costs you $2. What reservation price maximises your expected net proceeds, and what net do you end up with?

Offers on an asset arrive one per day, independent and uniform on [0, 100]. Each additional day of waiting costs you $2. What reservation price maximises your expected net proceeds, and what net do you end up with?

Approach: A stationary problem has a stationary threshold, so set the cost of one more search equal to the expected improvement over that threshold and solve the resulting quadratic.

80. The stationary threshold R satisfies c = E[(X - R)^+], since one more draw costs c and only the expected excess above R is gained. For X uniform on [0, 100], E[(X - R)^+] = (100 - R)^2/200, so setting it equal to the search cost of 2 gives (100 - R)^2 = 400 and the reservation price R = 80. The accounting closes: acceptance probability 0.2 means 5 days of waiting on average at a cost of $10, and E[X | X > 80] is 90, leaving a net of 80, exactly the threshold. That equality is general, since the reservation price is by definition the value of the whole search. Raising the cost to $8 moves the threshold to 60.

Follow-up: How does the threshold change if the offer distribution is exponential with mean 50 instead of uniform?

Key concepts: reservation price, search cost, stationary threshold, expected excess.