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Proof of Proposition 1
The value ρd =
τ
τ +z
follows immediately from equating V Ad |{αd = 0} and V Ad |{αd = 1}, where V Ad is
given by (1). Similarly, (4) is derived.
Because the agent is made indifferent between complying and not complying, V Ak follows after
evaluating these functions in αk = 1 and solving for V Ak for each k ∈ {w, d, t}.
Substituting V Ad =
solutions: ρt =
τ +z 1−δA
c
δA
−
τ
τ +z
τ
τ +z
1−τ −ρd c
1−δA ,
and ρt =
V At =
τ +z 1−δA
c
δA .
1−τ −ρt c
1−δA ,
and ρd =
τ
τ +z
into (4) and solving for ρt yields two
Note that the solution ρt =
τ +z 1−δA
c
δA
is decreasing in δA . Solving
= 0 yields δ ? .
The agent’s strategy as formulated in (6)-(8) follows from equating V Rk |{ρk = 0} and V Rk |{ρk = 1},
where V Rk is given by (2). Because the regulator is made indifferent between auditing and not auditing,
V Rk follows after evaluating these functions in ρk = 0 and solving for V Rk .
Now suppose that αd > αt . Then V Rd > V Rt , which however would imply from (6) and (7) that
αd < αt . Similarly, a contradiction is found for the alternative assumption αd < αt . Therefore, the only
solution is αd = αt . The proof for αw is similar. Indeed, when solving for αt in the way we did for ρt , we
again find two solutions: αt = αd and αt =
τ +z−δR z
δR τ
> 1. This second solution is therefore not feasible.
Proof of Proposition 2
Assume that the discount factor δA is drawn from some distribution function, the regulator learns the
agent’s δA after the first audit, and sets the audit probability based on the following three conditions:
A. If the agent complies and δA < δ ? , then the agent faces ρd in the next period.
B. If the agent complies and δA > δ ? , then the agent faces ρt in the next period.
C. If the agent does not comply, the agent faces ρd in the next period.
After the first audit, the regulator knows the agent’s discount factor, so the equilibria of Proposition 1
apply. Because ρt =
τ
τ +z+δA (V At −V Ad ) ,
learning that δA < δ ? would make the regulator increase the audit
probability in the trusted state (i.e., ρt > ρd ), but this contradicts the principle of trust. Proposition 1
shows that the regulator is indifferent between the games with and without trust. The regulator’s payoffs
are not affected by auditing an agent with δA < δ ? with probability ρd instead of ρt , so the regulator can
credibly commit to applying condition A.
Now define ρ̃ as the audit probability that makes an agent with δA = δ̃ indifferent between complying
and not complying. Given ρ̃, the agent’s pure strategy is to comply if δA > δ̃, and not to comply when
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δA < δ̃. Only when ρ̃ = ρd , an agent with δA < δ̃ is indifferent between complying and not complying.
In the waiting state, the value function of the regulator is given by
V Rw = ατ + ρw [(1 − α)(τ + z − δR (V Rt − V Rd )) − s + δR V Rt ] + (1 − ρw )V Rw ,
After the first audit, the equilibria of Proposition 1 apply, so V Rt = V Rd , and the value function simplifies
to
V Rw =
ατ + ρw [(1 − α)(τ + z) − s + δR V Rt ]
,
1 − δR (1 − ρw )
where α is a function of ρ as explained above.
Taking the derivative with respect to ρ yields:
sign
∂V Rw
∂ρ
s
− α(ρ)) + (1 − δR + δR ρ)[τ − ρ(τ + z)]α0 (ρ)
= sign [τ + z(1 − δR )](1 −
τ +z
(1)
Now let us analyze the following two cases:
i. The regulator’s assessment of δA is such that Pr {δA > δ ? } > α? . Then, when ρ ≥ ρd =
τ
τ +z ,
the
second term of the right-hand side of (11) is less than or equal to zero and the first term is negative.
The latter holds because Pr {δA > δ ? } > α? implies that α(ρ) > 1 −
only be equal to zero for ρ < ρd . However, if α(ρ) ≤ 1 −
s
τ +z ,
s
τ +z .
The derivative can thus
the first term is greater than or equal
to zero, and this can only follow from ρ < ρd , which implies that the second term is also positive.
Therefore V R is maximized for some ρ < ρt that yields α(ρ) > α? .
ii. The regulator’s assessment of δA is such that Pr {δA > δ ? } ≤ α? . When ρ = ρd , the second term of
the right-hand side of (11) equals zero and the first term is positive unless an agent with δA < δ ? uses
a mixed strategy. Now let ρw = ρd . When δA > δ ? , the waiting agent has a pure strategy to comply.
When δA ≤ δ ? , let the waiting agent comply with probability α̂ =
α? −Pr{δA >δ ? }
Pr{δA <δ ? } .
From the regulator’s
point of view, the compliance probability is equal to Pr {δA > δ ? } · 1 + Pr {δA < δ ? } · α̂ = 1 −
s
τ +z ,
which indeed makes the regulator indifferent between auditing and not auditing (as shown in
Proposition 1). Indeed, in such case
∂V Rw
∂ρ
= 0.
For the agent, the optimal strategy is to comply when δA > δ ? because it results in entering the
trust state, and to comply with probability α̂ when δA < δ ? because she is indifferent between
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complying and not complying (in such case, ρw = ρt = ρd and therefore V At = V Ad ).
Finally, we note that when Pr {δA > δ ? } = 0, solution ii is equivalent to the deterrence-based
equilibrium, so that a trust-based solution does not exist.
From the value functions of the regulator, which are given in Proposition 1, it follows that the regulator
has the same expected payoffs with and without trust-based regulation.
Proof of Proposition 3
This proof goes along the same lines as the proof of Proposition 1. Let us start with the compliance
probabilities. The value for the regulator is again given by V Rk =
αk τ
1−δR .
For the compliance probabilities
αk , αj , with k, j ∈ {w, d, t}, it follows from these value functions (V Rk =
αk τ
1−δR )
that αk > αj would
imply that V Rk > V Rj . However, (10) would imply that αk < αj , so the only possible solution is
s
αw = αd = αt . The value 1 − τ +z
follows from the fact that the term V Rk − V Rj ∀ k, j ∈ {w, d, t} equals
zero.
The values for ρt and ρd are obtained by solving the system of equations given by (10), while expressing
the values using αk and solving for V Ak . Now, solving V Ad for αd = 0 and V At for αt = 1 yields:
V Ad =
1 − ρd (τ + z + c)
1 − τ − ρt c
and V At =
.
1 − δA
1 − δA
Substituting these expressions in (10) yields the following two equations:
ρd
=
τ ρt
uτ +ρt (τ +z)(1−u)
ρd
=
τ +ρt c
τ +z+c
+
(1−δA )τ −ρt (τ +z)
δA ρt (τ +z+c)
Solving for ρt yields the two solutions given in the proposition. The third root is
h
i
p
1
F − F 2 + 4cδA (1 − u)τ u(1 − δA ) ,
2cδA (1 − u)
which is not a feasible solution as this root is strictly negative.
Finally, it can be verified for this solution that ρt = ρd if and only if
u=
cδA τ − (1 − δA )(τ + z)2
.
δA τ (τ + z + c)
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Because u? = 0 for δA = δ ? , u? is strictly increasing in δA , and u? =
τc
τ (c+τ +z)
< 1 for δA = 1, it follows
that u? ∈ (0, 1) so that the trust-based solution satisfies ρd > ρt iff u < u? . This completes the proof.