Job Assignments under Moral Hazard - School of Economics and ...

fh(ē, ˜ θ)−Ch(ē, ˜ θ) = fl(ē)−Cl(ē) ⇔ fh(ē, ˜ θ)−fl(e)−Ch(ē, ˜ θ) = fl(ē)−fl(e)−Cl(ē). The
right-hand side is nonnegative because Condition 1 holds for job l. Using fh(e, ˜ θ) < fl(e)
therefore yields fh(ē, ˜ θ)−fh(e, ˜ θ) > Ch(ē, ˜ θ) (Condition 1). Thus, the maximized second-
best profit in job h at θ = ˜ θ is Π SB
h (ē, ˜ θ), and by the definition of ˜ θ it follows that
θ SB = ˜ θ. Furthermore, we know that θ SB = ˜ θ < θ F B = ˆ θ(ē, ē). We now show that
θ SB is the only job assignment threshold below θ F B , i.e., Π SB (ē, θ) never falls back to
or below fl(ē) − Cl(ē) for θ ∈ (θ SB , ˆ θ(ē, ē)]. By way of contradiction, suppose that
there exist values θn ∈ (θSB , ˆ θ(ē, ē)] for which ΠSB h (ē, θn) = fl(ē) − Cl(ē), and let θ1
denote the smallest of these. Because ΠSB h (ē, θSB ) = fl(ē) − Cl(ē) and ΠSB h θ (ē, θSB ) > 0
(by Lemma 1) it follows that ΠSB h θ (ē, θ1) < 0. This, however, implies that Condition 1
is violated at θ1, i.e., ΠSB h (ē, θ1) < fh(e, θ1), which leads to a contradiction because
fl(ē) − Cl(ē) ≥ fl(e) > fh(e, θ1).
C Continuous effort model
First best
F B
The efficient effort level given θ solves maxe fj(e, θ) − c(e). For job l this yields e
l
= α
F B
B
and for job h we have eh (θ) = α θ. Hence, the expected profits are ΠFl α2 = 2 and
F B Πh (θ) = α2θ2 2
B
− k, respectively. Setting profits equal defines the efficient cutoff θF
implicitly by α2θ2 − 2 k − α2 = 0. Solving for θF B yields
θ F B √
α2 + 2 k
≡ .
α
Second best
To implement effort e in job j the wage after success must be set at ¯wj(e, θ) = ce(e)
fj e(e,θ) ,
where subscript e denotes the partial derivative with respect to effort. That is,
¯wl(e) = e
α and ¯wh(e, θ) = e . Denote the expected cost of implementing effort in job j by
α θ
Cj(e, θ) = fj(e, θ) ¯wj(e, θ). Maximizing the principal’s expected profit fj(e, θ) − Cj(e, θ)
over e gives the second-best effort level for a type-θ agent. In job l this yields e SB
l
half the efficient effort level. In job h we have 17
e SB
h (θ) = 1
α θ +
2
k
.
α θ
α = 2 ,
17 SB
Note that an effort of zero yields an expected utility of zero. So for eh (θ) > 0 to be optimal the
incentive constraint requires that EUh(eSB h (θ), θ) ≥ 0. Equation (5) and the derivations below show
that EUh(eSB h (θ), θ) ≥ 0 ⇔ θ ≥ √ 3 k
α . This explains A4.
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