This scene asks whether high effort is worth buying. Making someone work hard is never free.
As the effort cost rises, the profit from high effort falls. The profit from low effort stays flat. Where the two profits meet is the cutoff. Incentives pay only on the cheap side of it. The effort cost is exactly the risk premium you must cover.
For example, a parent sets the 💰 allowance for an errand. A trip to the nearby 🏪 shop takes 1,000 won. A far shop is tiring and needs more. If the price gets too high, going in person is cheaper. Hard effort is worth buying only while it stays cheap.
Π1(c)=p1xH+(1−p1)xL−C1(c)
Π0=p0xH+(1−p0)xL−uˉ1/α
Π1(c∗)=Π0
Examples & Exercises
Let p1=0.8, p0=0.2, xH=2000, xL=0, uˉ=20, α=1/2, c=10.
Compute the principal's profit from implementing high effort and from settling for low effort, and determine which to choose.
Π1=0.8×2000−C1,Π0=0.2×2000−uˉ2
c=10, α=1/2 — the current point lies in the "implement high effort" region.Show proof
Implementing high effort gives expected output 0.8×2000=1600, and the expected wage from the previous problem is C1=944.44. Hence Π1=1600−944.44=655.56.
Settling for low effort needs only the flat wage uˉ1/α=202=400, so Π0=0.2×2000−400=0.
Since Π1=655.56>0=Π0, the principal implements high effort. The expected-output gain from effort, (0.8−0.2)×2000=1200, comfortably exceeds the extra incentive cost 944.44−400=544.44.
With α=1/2, find the threshold effort cost c∗ at which the principal is indifferent between implementing high effort and settling for low effort.
Use the signs at c=18 and c=19 to show c∗ lies between them.
Π1(c)−Π0=1200−(C1(c)−400),c∗≈18.13
c=18, α=1/2 — just left of the crossing, still in the "implement high effort" region.Show proof
Π1(c)−Π0=1200−(C1(c)−400). Since C1(c) is strictly increasing in c (a higher effort cost widens the utility spread uH−uL=c/0.6 and raises the risk premium), the gap is strictly decreasing and the crossing is unique.
At c=18, uL=14 and uH=44, so C1=0.8×442+0.2×142=1588 and Π1−Π0=1600−1588=12>0.
At c=19, uL=13.667 and uH=45.333, so C1≈1681.44 and Π1−Π0=1600−1681.44=−81.44<0. The sign flips, so c∗∈(18,19); numerically c∗≈18.13.