Eco211A, Autumn 2016 Simon Board

Solutions to Homework 1

October 24, 2016

Solution to Question 1

(a) Suppose w ∼ N(µ, σ2). Then

−E[u(w)] =1√

2πσ2

∫exp[−w] exp

[− 1

2σ2(w − µ)2

]dw

=1√

2πσ2

∫exp

[− 1

2σ2[2wσ2 + (w − µ)2

]]dw

=1√

2πσ2

∫exp

[− 1

2σ2[(w − µ+ σ2)2 − σ4 + 2µσ2

]]dw

= exp

[−µ+

σ2

2

]1√

2πσ2

∫exp

[− 1

2σ2(w − µ+ σ2)2

]dw

= −u(µ− σ2

2

)since the latter term integrates to one.

(b) The principal’s problem can be written as

maxα,β,a

a− (α+ βa)

s.t. α+ βa− β2

2σ2 − g(a) ≥ 0

a ∈ argmaxa′∈< α+ βa′ − β2

2σ2 − g(a′)

First, we can use (IR) to substitute for α. Second, using the first–order approach, (IC) can be

written as β = g′(a). Hence the problem becomes

maxβ,a

a− g(a)− β2

2σ2

s.t. β = g′(a)

The optimal a is then solves the FOC

1− g′(a∗)− g′(a∗)g′′(a∗)σ2 = 0

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Eco211A, Autumn 2016 Simon Board

Rearranging, and using (IC), we have

β∗ =1

1 + g′′(a∗)σ2

(c) The solution would not change if the agent knows x. This can be seen from the (IC)

constraint which is independent of x. Intuitively, this is because the contract is linear: incentives

are unchanged if output increases by a fixed amount x. This is a key insight underlying

Holmstrom and Milgrom (1987).

[Source: Milgrom and Roberts (1992)]

Solution to Question 2

(a) The principal’s problem can be written as

maxα,β,γ,a

a− w

s.t. w − 1

2Var(w)− g(a) ≥ 0

a ∈ argmaxa′∈< w − 1

2Var(w)− g(a′)

where w = α+βa and Var(w) = β2Var(x+γy). Substituting for α, the problem is to maximise

welfare subject to (IC). That is,

maxβ,γ,a

a− β2

2Var(x+ γy)− g(a)

s.t. a ∈ argmaxa′∈< βa′ − g(a′)

Hence γ is chosen to minimise Var(x+ γy). That is, γ∗ = −σxy/σ2y .

If σxy, then γ∗ < 0. Intuitively, a high y means that it is easy to produce high output, and the

agent should not be rewarded as much.

(b) As above, the principal chooses {γj}j 6=i to minimise

Var(qi −∑j 6=i

γjqj) = σ2i −∑j 6=i

γ2j σ2j + (1−

∑j 6=i

γj)2σ2c

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Eco211A, Autumn 2016 Simon Board

The FOC yields for γj yields

γjσ2j = (1−

∑j 6=i

γj)σ2c

Hence γjσ2j = γkσ

2k for all j, k 6= i. Substituting this into the FOC and rearranging yields

γ∗j = σ2c

σ2j + σ2c∑k 6=i

σ2jσ2k

−1

As σ2j increases, qj becomes a noisier estimator of xc and γ∗j falls. As σ2k increases, so their γ∗kfalls and there is more of xc to be explained, leading to an increase in γ∗j .

[Source: Milgrom and Roberts (1992)]

Solution to Question 3

(a) Without an insurance market the agent chooses a to maximise

p(a)

∫u(W0 − x)dG(x) + (1− p(a))U(W0)− ψ(a)

This is a concave maximisation problem with FOC

p′(a)

∫[u(W0 − x)− u(W0)]dG(x) = ψ′(a)

(b) The agent chooses (a,R(x)) to maximise

p(a)

∫u(W0 − x+R(x))dG(x) + (1− p(a))U(W0 +R(0))− ψ(a) (1)

subject to the individual rationality constraint of the insurers:

p(a)

∫R(x)dG(x) + (1− p(a))R(0) ≥ 0 (IR)

The Lagrangian is

p(a)

∫u(W0−x+R(x))dG(x)+(1−p(a))U(W0+R(0))−ψ(a)−λ

[p(a)

∫R(x)dG(x) + (1− p(a))R(0)

]

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Eco211A, Autumn 2016 Simon Board

The FOCs for R(0) and R(x) are

u′(W0 +R(0)) = λ

u′(W0 − x+R(x)) = λ

Thus we have complete insurance: R(0) = −k and R(x) = x − k. The zero profit condition

yields k = p(a)x where x :=∫xdG(x). Utility is

u(W0 − k)− ψ(a) = u (W0 − p(a)x)− ψ(a)

Differentiating with respect to a,

−p′(a∗)u′ (W0 − p(a∗)x)x = ψ′(a∗)

(c) Now consider the problem of maximising (1) subject to (IR) and incentive compatibility:

a ∈ argmaxa′p(a′)

∫u(W0 − x+R(x))dG(x) + (1− p(a′))U(W0 +R(0))− ψ(a′) (IC)

(IC) is convex in a so we can use the first order approach. The first order condition is

p′(a)

∫[u(W0 − x+R(x))− U(W0 +R(0))]dG(x)− ψ′(a) = 0 (ICFOC)

The Lagrangian is

p(a)

∫u(W0 − x+R(x))dG(x) + (1− p(a))U(W0 +R(0))− ψ(a)

− λ

[p(a)

∫R(x)dG(x) + (1− p(a))R(0)

]+ µ

[p′(a)

∫[u(W0 − x+R(x))− U(W0 +R(0))]dG(x)− ψ(a)

]Maximising pointwise with respect to R(0) and R(x) yield

u′(W0 +R(0))

[1− µ p′(a)

1− p(a)

]= λ

u′(W0 − x+R(x))

[1 + µ

p′(a)

p(a)

]= λ

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Eco211A, Autumn 2016 Simon Board

where µ > 0 (else the agent will exert no effort). We thus have

W0 +R(0) = W0 − k1 if x = 0

W0 − x+R(x) = W0 − k2 if x > 0

where k2 > k1.

(d) This is an insurance contract where k1 is the premium and k2 − k1 is the deductable. The

multiplicative specification of the probability of an accident means that an agent can affect

whether an accident occurs, but not the damage when it occurs. Thus the optimal contract

does not depend on how much damage is incurred, only on whether an accident takes place.

What if an agent has to voluntarily report an accident? If x is small then W0−k1−x > W0−k2and the agent will hide the accident.

[Source: Holmstrom (1979, Rand).]

Solution to Question 4

(a) Set w(q) = t(q) = 0.

(b) The principal’s problem is

maxw(q),t(q)

E[q − t(q)|H]

s.t. E[u(w(q))|H]− g(H) ≥ E[u(w(q))|L]− g(L)

q − t(q) ≥ q − t(q)

t(q) ≥ w(q)

w(q) ≥ 0

(c) This follows from (ICP).

(d) We can rewrite (ICA) as∫u(w(q))

[1− f(q|L)

f(q|H)

]f(q|H)dq − g(H) + g(L) ≥ 0

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Eco211A, Autumn 2016 Simon Board

Let the multipliers on (ICA), (FE) and (LL) be µ, φ(q)f(q|H) and λ(q)f(q|H). The FOC with

respect to w(q) is

µu′(w(q))

[1− f(q|H)

f(q|L)

]− φ(q) + λ(q) = 0

Suppose w ∈ (0, t). Then we must have

f(q|H)

f(q|L)= 1

This uniquely defines some output, q∗. When q < q∗, then w = 0. When q > q∗, then w = t.

Finally, we choose t so that (ICA) holds with equality.

Solution to Question 5

Suppose we start with an arbitrary contract {Bi}. Let k be the highest state such that Bk <

min{Bk+1, qk}. Under a debt contract, there is no such k.

Consider the following variation. First, reduce each {Bi}ni=k+1 by ε. Second, increase Bk by[1− F (qk|H)

f(qk|H)

]ε

This variation leaves the wage bill unchanged. It also satisfies (MON) if ε is sufficient small.

However this increases the slack,

S =n∑i=1

(qi −Bi)[f(qi|H)− f(qi|L)

],

in the (IC) constraint. To see this observe that,

dS = −[

1− F (qk|H)

f(qk|H)

]ε[f(qk|H)− f(qk|L)

]+

n∑i=k+1

ε[f(qk|H)− f(qk|L)

]=

ε

f(qk|H)

[(1− F (qk|H))f(qk|L)− (1− F (qk|L))f(qk|H)

]Hence dS ≥ 0 if and only if MHRP holds. This proves that MHRP is sufficient (and that it is

also necessary).

Here’s an intuition. Consider random variables X ∼ f(·) and Y ∼ g(·). One can think of

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Eco211A, Autumn 2016 Simon Board

X as output given low effort, and Y as output given high effort. Y ≥lr X in the likelihood

ratio order if g(v)/f(v) > g(u)/f(u) for v > u, so a higher output is indicative of higher effort.

Y ≥hr X in the hazard rate order if [1 − G(v)]/[1 − F (v)] > [1 − G(u)]/[1 − F (u)] for v > u,

so knowing output is above ’v’, rather than output being above ’u’ is more indicative of high

effort. Crucially, this is a statement about increasing sets, i.e. {x|x > u}.

The (MON) constraint implies that payments Bq are increasing in output q. Observe that any

increasing function can be decomposed into the sum of step functions. These step functions are

exactly of the form I{x|x > u}, where I is an indicator function. Hence the hazard rate order

is the correct concept, rather than the likelihood order.

Solution to Question 6

(a) Suppose effort is observable. The principal’s problem is to solve

maxw(q),a

E[q − w(q)|a] s.t. E[u(w(q)− g(a))|a] ≥ u(0)

The Lagrangian is∫[q − w(q)] dF (q|a) + λ

∫[u(w(q)− g(a))− u(0)] dF (q|a)

If λ = 0 then the optimal solution is w(q) = −∞. This breaks (IR), so we must have λ > 0.

Pointwise optimisation yields

λu′(w(q)− g(a)) = 1 (∀q)

Hence the optimal wage is constant. Using (IR) this wage is given by w∗ = g(a). The principal

thus chooses effort to maximise

E[q|a]− g(a)

(b) Suppose effort is unobservable. The principal’s problem is to solve

maxw(q),a

E[q − w(q)|a]

s.t. E[u(w(q)− g(a))|a] ≥ u(0)

a ∈ argmaxa′∈AE[u(w(q)− g(a′))|a′]

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Eco211A, Autumn 2016 Simon Board

(c) Suppose effort is observable. The agent solves the problem:

maxw(q),a

E[u(w(q)− g(a))|a] s.t. E[q − w(q)|a] ≥ 0

The Lagrangian is ∫[u(w(q)− g(a))] dF (q|a) + λ

∫[q − w(q)] dF (q|a)

If λ = 0 then the optimal solution is w(q) = ∞. This breaks (IR), so we must have λ > 0.

Pointwise optimisation yields

u′(w(q)− g(a)) = λ (∀q)

Hence the optimal wage is constant. Using (IR) this wage is given by w∗ = E[q|a]. The agent

thus chooses effort to maximise

u(E[q|a]− g(a))

Since u(·) is increasing, this is the same effort as in part (a).

(d) The principal’s problem is to solve

maxw(q),a

E[u(w(q)− g(a))|a]

s.t. E[q − w(q)|a] ≥ 0

a ∈ argmaxa′∈AE[u(w(q)− g(a′))|a′]

Write the optimal utility from this program by u(x). The dual can then be written as

maxw(q),a

E[q − w(q)|a]

s.t. E[u(w(q)− g(a))|a] ≥ u(x)

a ∈ argmaxa′∈AE[u(w(q)− g(a′))|a′]

Using the fact utility is CARA, u(w+ x) = u(w)e−x. Letting w(q) := w(q)− x, we can rewrite

the program as

maxw(q),a

E[q − w(q)|a]− x

s.t. E[u(w(q)− g(a))|a] ≥ u(0)

a ∈ argmaxa′∈AE[u(w(q)− g(a′))|a′]

This is identical to the program in (b), and thus has the same solution.

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