3.6 Extensions
3.6.2 Generalization of the Public Vetting Equilibrium
In this subsection, I maintain assumptions 1 thru 7. I also allow the possibility of strong (but not perfect) information about the attributes of the challenger. I show that the behavior of the politician is characterized by a cutoff, with the politician only implementing her policy if the signal is sufficiently low.
There are two main ways in which the analysis differs when policy is only partially informative and valence is a factor in comparison to the analysis in the main text. First, the reputation associated with each outcome is a function of σ. Consequently, as σ changes, not only does the probability of each outcome change, so do the reputations associated with each outcome. This dependency renders the direct approach used for the main text intractable. Second, r+(σ) < 1 and r−(σ) > 0 for any σ ∈ (0,1), so F(r+) < 1 and F(r−) > 0. Even if r = 1 or r = 0, it is no longer necessary that F(0) = 0 or that F(1) = 1. Due to valence shocks, even a politician certain to be competent may lose to a sufficiently well liked challenger, and even a politician certain to be incompetent may win against a sufficiently disliked challenger.
The proof proceeds in two major steps. First, it is shown that for any particularσ and correspondingr−(σ) andr+(σ), there exists a cutoffσ which completely characterizes the politician’s implementation preferences.
The politician prefers to implement her policy if and only ifσ is less than or equal to σ, and chooses inaction otherwise. Second, it is shown thatσ is a continuously decreasing function ofσ. Consequently, there exists a critical value σpub so that the politician prefers to implement her policy if and only ifσ < σpub.
F(˜σ) P(Re-Elect)
r− r+ σ
0 σ˜
E(F(r)|˜σ)
Figure 3.13: A givenr− and r+ and the associated cutoff, σ. Characterizing the Politician’s Preference
As in the main text, the politician may either keep her current reputation for sure by choosing inaction or reveal additional information about herself with probability(1−γ)by implementing her policy. Also, the voter elects the politician in the election with the highest reputation, so that the probability of re-election isF(r).
In the analysis that follows, consider a particular signal σ and the reputations associated with it, r+ and r−. Suppose that the politician has a hypothetical signalσ˜ ∈R, which determines the probability of attaining r− and r+ after implementation, but does not influence the value of r− or r+. Unlike σ, σ˜ may take on any real value. Becauseσ˜ is not bounded there may be implied probabilities greater than 1 or less than 0. For a fixed r+ and r−, let σ ∈ R be the value of σ˜ for which the politician is indifferent between implementation and inaction.
To understand this part of the proof further, consider Figure 3.13. The line represents the expected utility to the politician of implementing her policy, E(F(r)|˜σ). The CDF F(˜σ) represents her utility from choosing inaction. The valueσis the value of ˜σfor which, given a fixedr− andr+, the politician is indifferent between implementing her policy and inaction. Graphically, σ is the value of ˜σ at the intersection of the line through (r−, F(r−)and(r+, F(r+)withF(˜σ). The goal of the first step of the proof is to establish that this intersection exists and is unique, and that it characterizes the politician’s decision to implement her policy.
Recall that, because the expectation of Bayesian posteriors is equal to the prior,
p(σ)r++ (1−p(σ))r−=σ. (3.63)
Hence,p(σ) =rσ−r−
+−r−.Define
D(˜σ;r−, r+)≡
σ˜−r−
r+−r−
F(r+) +
1− σ˜−r−
r+−r−
F(r−)−F(˜σ). (3.64) This expression is positive if and only if the expected utility of implementing the policy, given σ, exceeds the˜ expected utility of the status quo given afixed r− and r+.5 Consequently,σ must solveD(˜σ, r−, r+) = 0.
To identify σ, the analysis begins by deriving properties of D given fixed values for r+ and r−. Because differentiation is with respect to σ, in the following lemmas let˜ D(˜σ) ≡ D(˜σ;r−, r+) as long as r− and r+
remain fixed. It is first established thatDhas exactly two stationary points, r0 and r00, which characterize the intervals of increase and decrease.
Lemma 3.6. There exist values r0 and r00 withr0 < r00 such that
(i) D0(˜σ)<0 if and only if ˜σ∈(r0, r00) (ii) D0(˜σ)>0 if and only if ˜σ /∈(r0, r00) (iii) D0(˜σ) = 0 if and only if ˜σ∈ {r0, r00}
Further,D(˜σ)→ ∞ asσ˜→ ∞ and D(˜σ)→ −∞as ˜σ→ −∞
Proof. Differentiation of Dimplies that D0(˜σ) =
F(r+)−F(r−) r+−r−
−f(˜σ). (3.65)
Hence,D0(˜σ)is positive whenF(rr++)−F−r−(r−) > f(˜σ), negative whenF(rr++)−F−r−(r−) < f(˜σ), and zero whenF(rr++)−F−r−(r−) = f(˜σ). Because F(rr++)−F(r−r− −) is a constant with respect toσ, the intervals of increasing and decreasing for˜ Dare lower and upper sets off.
It is first proven that the upper set,Lu ={˜σ ∈R|F(rr+)−F(r−)
+−r− < f(˜σ)}, is nonempty. SupposeLuwere empty.
Then it must be the case that F(rr+)−F(r−)
+−r− ≥maxσ˜f(˜σ). Becausef is unimodal, the maximum is attained at m. Consequently, it must be that F(r+r)−F(r−)
+−r− ≥ f(m). This implies that F(r+)−F(r−) ≥ f(m)(r+ −r−) or, equivalentlyRr+
r− f(˜σ)d˜σ ≥Rr+
r− f(m)d˜σ. Because f(m) > f(˜σ) for every ˜σ 6=m, this is impossible. Hence,
F(r+)−F(r−)
r+−r− < f(m) and, therefore, Lu is nonempty. By the strict quasiconcavity of f, the upper set is an open interval, denoted(r0, r00). Hence, for anyσ˜ ∈(r0, r00),D0(˜σ)<0. Strict quasiconcavity of f implies that
F(r+)−F(r−)
r+−r− > f(˜σ) for σ /˜ ∈ [r0, r00]. Hence, by continuity, F(rr++)−F(r−r− −) = f(˜σ) at r0 and r00. Consequently, D0(˜σ) = 0 ifσ˜ ∈ {r0, r00} andD0(˜σ)>0 whenever σ /˜∈(r0, r00).
Finally, to compute the limits of Dasσ˜ goes to infinity, note that (3.64) may be rewritten as D(˜σ) =F(r−) +
σ˜ r+−b−
[F(r+)−F(r−)]− r−
r+−r−
[F(r+)−F(r−)]−F(˜σ). (3.66) Taking the limit asσ˜ → ∞, the second term becomes arbitrarily large, whileF(˜σ)→0. Hence, D(˜σ) becomes arbitrarily large. Taking the limit asσ˜ → −∞, the second term becomes arbitrarily negative, whileF(˜σ) = 0.
Hence,D(˜σ) becomes arbitrarily negative.
The parameter σ is now precisely defined. Let σ be a value of σ˜ such that D(σ) = 0 and either (i) σ ∈ {r/ −, r+} or (ii) σ ∈ {r−, r+} and D0(σ) = 0. That is, σ is the signal for which a politician would be indifferent between implementing the policy and the status quo. It is distinct from r− and r+ except in knife-edge cases. In those knife edge cases, eitherr− orr+ is a stationary point of D, and that value is chosen to beσ. It is now shown thatσ exists, is unique, and that the politician prefers implementation if and only ifσ≤σ.
Lemma 3.7. Given σ, and the correspondingr− and r+, there exists a unique σ such that implementation is preferred if and only if σ≤σ and inaction is preferred if and only ifσ > σ.
Proof. Note thatD(r−) =D(r+) = 0. By Rolle’s Theorem, there must exist a stationary point ofDin(r−, r+). That is, at least one of r0 orr00 must lie in(r−, r+). Becauser0< r00, there are three cases to consider:
1. r0∈(r−, r+) and r00≥r+. 2. r0≤r− and r00∈(r−, r+).
3. r0∈(r−, r+) and r00∈(r−, r+).
BecauseD(σ) = 0 wheneverσ∈ {r/ −, r+}, Rolle’s Theorem also implies the following necessary conditions:
(a) Ifσ> r+, there must be a stationary point of Din the open interval(r+, σ).
(b) Ifσ< r−, there must be a stationary point of Din the open interval(σ, r−).
(c) Ifr−< σ < r+, thenr0 ∈(r−, σ) andr00∈(σ, r+).
(d) There is at most one σ∈ {r/ −, r+} such thatD(σ) = 0.
Recall thatr0 andr00are the only stationary points ofD. In Case 1, it cannot be thatσ 6=r−andσ< r+ because neither (b) nor (c) hold. Further D0(r−) 6= 0 because r− < r0 and, by Lemma 3.6, this implies that D0(r−)<0, so it cannot be the case thatσ=r−. Ifr00 =r+, thenr+=σ. There cannot be anotherσ> r+
by requirement (a), so σ is unique. If r00 > r+, then D0(r+) < 0. Thus, for σ > r˜ + but arbitrarily close to r+, D(˜σ) < 0. Further, Lemma 3.6 implies that as σ˜ → ∞, D(˜σ) → ∞. Hence, by the Intermediate Value Theorem, there must existσ∈(r+,∞)such thatD(σ) = 0. By (d), there cannot be anotherσ ∈ {r/ −, r+}, and neither D0(r+) nor D0(r−) is 0. Hence, σ must be unique. Now consider the preferences of a politician with signalσ. Note that D(σ)>0 for allσ ∈(r−, r+) becauseD0(r−)>0,D(σ) is continuous, and there are no other roots ofD(σ) in (r−, r+). Hence, in Case 1 σ≤σ for everyσ ∈(r−, r+), and the politician prefers to implement her policy.
Case 2 is symmetric to the first.
In Case 3, by (a) and (b), it must be thatσ∈(r−, r+) because there are no stationary points ofDoutside of that interval. Note that D0(r−) >0 because r− < r0, which implies for ˜σ > r− but arbitrarily close to r−
that D(˜σ) > 0. Similarly, D0(r+) > 0, which implies that for σ < r˜ + but arbitrarily close to r+, D(˜σ) <0. Hence, by the Intermediate Value Theorem, there must exist σ ∈ (r−, r+) such that D(σ) = 0. Because neither D0(r+) nor D0(r−) equal 0, and by (d) there cannot be another σ ∈ {r/ −, r+}, σ is unique. Now consider the preferences of a politician with signal σ. Because D0(r−)>0,D(σ) is continuous, and there are no other roots ofD in (r−, σ), it must be that D(σ)>0 for all σ ∈(r−, σ). Similarly, D0(r+) >0 implies thatD(σ)<0for all σ ∈(σ, r+). Consequently, implementation is preferred if and only ifσ ≤σ.
The three cases presented are exhaustive. Hence,σis always defined and unique, and for anyσ∈(r−, r+), the politician prefers to implement her policy if and only ifσ ≤σ.
In the proof of Lemma 3.7, the locations of r+ and r− relative tor0 and r00, as well as σ, were derived. In Case 1,r−< r0 < r+≤r00 ≤σ. In Case 2, σ≤r0 ≤r−< r00< r+. In Case 3, r−< r0< σ< r00< r+. The Critical Cutoff
What has shown in Lemma 3.7 is that for each value ofσ and its associated beliefsr−and r+, one can define a cutoff,σ, that characterizes the politician’s behavior. It is now shown that asσincreases, this point decreases.
Consequently, there is a cutoffσpub such that σ ≤σpub implies a preference for action and σ > σpub implies a preference for the status quo for the policy lottery consistent with σ.
Consider
D(σ, rˆ +(σ), r−(σ)) =
σ−r−(σ) r+(σ)−r−(σ)
F(r+(σ)) +
1− σ−r−(σ) r+(σ)−r−(σ)
F(r−(σ))−F(σ). (3.67) The functionDˆ is the utility difference between policy implementation and the status quo for the lottery faced by a politician with reputation σ. By Lemma 3.7, there exists a unique σ for this lottery such that the politician prefersx1=θ1 if and only ifσ≤σandx1=s1 if and only ifσ > σ. The following theorem shows that σ is a decreasing function of σ, so that if a lottery is accepted at reputation σ, it will also be accepted for any smaller value ofσ and, similarly, if the status quo is preferred at anyσ, then it is also preferred for any larger value ofσ. Hence, there exists a value of the signal,σpub∈[0,1], that divides these cases.
Theorem 3.5. There exists σpub ∈ [0,1] such that for σ ≤ σpub policy implementation is preferred an for d σ > σpub implies the status quo is preferred.
Proof. Becauseσ must satisfyD(σˆ )≡0, the Implicit Function Theorem is used. An increase inσ increases bothr+ andr−. In general, the increase inr+and r− may have a positive or negative effect on the value ofD.ˆ
However, it will be shown that the effect of an increase inσ is always of the same sign, so that the conditions of the Implicit Function Theorem required to show that σ is a decreasing function ofσ are met.
Differentiating (3.67) with respect to r+ and r−,
∂Dˆ
∂σ
σ=σ =
F(r+)−F(r−) r+−r−
−f(σ); (3.68)
∂Dˆ
∂r+
σ=σ =−
σ−r−
(r+−r−)2
F(r+) +
σ−r−
r+−r−
f(r+) +
σ−r−
(r+−r−)2
F(r−); (3.69)
∂D dr−
σ=σ =
σ−r+ (r+−r−)2
F(r+) +
r+−σ
(r+−r−)2
F(r−)−
r+−σ
r+−r−
f(r−). (3.70) The derivative with respect toσ, (3.68), is positive if and only if
F(r+)−F(r−) r+−r−
> f(σ) (3.71)
or, equivalently, if
dD dσ
σ=σ >0. (3.72)
The derivative with respect tor+, (3.69), is positive if and only if
−
σ−r−
r+−r−
F(r+) + (σ−r−)f(r+) +
σ−r−
r+−r−
F(r−)>0. (3.73)
Factoring out(σ−r−) simplifies (3.73) to (σ−r−)
f(r+)−F(r+)−F(r−) r+−r−
>0. (3.74)
Factoring out a −1 from the bracketed term and substituting the definition of D, it follows that (3.74) is equivalent to
(r−−σ) ∂D
dσ σ=r+
>0. (3.75)
Inequality (3.75) holds ifσ> r− and ∂Ddσ
σ=r+ <0, or ifσ< r− and ∂Ddσ
σ=r+ >0.Hence, ∂r∂Dˆ+
σ=σ >0. The derivative with respect tor−, (3.70), is positive if and only if
σ−r+ (r+−r−)
F(r+) +
r+−σ
(r+−r−)
F(r−) + (r+−σ)f(r−)>0. (3.76) Dividing on both sides of (3.76) by −1, it is equivalent to
r+−σ
(r+−r−)
F(r+)−
r+−σ
(r+−r−)
F(r−)−(r+−σ)f(r−)<0. (3.77) Factoring out(r+−σ) simplifies (3.77) to
(r+−σ)
F(r+)−F(r−) r+−r−
−f(r−)
<0. (3.78)
Substituting the definition ofD, it follows that (3.78) is equivalent to (r+−σ)
∂D
∂σ σ=r−
<0. (3.79)
Inequality (3.79) satisfied ifσ> r+ and ∂D∂σ
σ=r− >0or ifσ< r+ and ∂D∂σ
σ=r− <0.Hence, ∂r∂Dˆ−
σ=σ>0.
Each possible case for the location ofσ is considered, following the cases from Lemma 3.7.
In Case 1, it has been shown that r− < r0 < r+ ≤ r00 ≤ σ. Therefore, by Lemma 3.6, ∂D∂σ
σ=r+ ≤ 0 and ∂D∂σ
σ=r− >0. By (3.75) and (3.79), this implies that ∂r∂Dˆ+
σ=σ ≥ 0 and ∂r∂Dˆ−
σ=σ >0. Further, because σ < r0 in case (i), (3.72) implies that ∂∂σDˆ
σ=σ >0. Hence, by the Implicit Function Theorem, dσdσ <0, and σ varies continuously with σ.
In Case 2, it has been shown that σ ≤r0 ≤r− < r00 < r+. Therefore, by Lemma 3.6, ∂D∂σ
σ=r+ > 0 and
∂D
∂σ
σ=r− ≤0. By (3.75) and (3.79), this implies that ∂r∂Dˆ+ >0 and ∂r∂Dˆ− ≥0. Further, because σ > r00, (3.72) implies that ∂D∂σ
σ=σ >0. Hence, by the Implicit Function Theorem, dσdσ <0, andσ(σ) varies continuously withσ.
In Case 3, it has been shown that r− < r0 < σ < r00 < r+. Therefore, by Lemma 3.6, ∂D∂σ
σ=r− >0, and
∂D
∂σ
σ=r− < 0. By (3.75) and (3.79), this implies that ∂r∂Dˆ+ < 0 and ∂r∂Dˆ− < 0. Because σ ∈ (r−, r+), (3.72) implies that dDdσ
σ=σ <0. Hence, by the Implicit Function Theorem, dσdσ <0, andσ(σ) varies continuously withσ.
Hence, σ(σ) is a decreasing, continuous function of σ. If σ(0)≤0, the politician prefers the status quo for anyσ, andσ∗ = 0. Ifσ(1)≥1, then the politician prefers to implement her policy for anyσ, andσ∗ = 1.
Ifσ(0)>0andσ(1)<1, there is a unique pointσpub∈(0,1)such thatσ(σpub) =σ∗pub, and for allσ < σpub it holds thatσ < σ(σ) and the politician prefers to implement her policy, while for allσ > σ∗pub,it holds that σ > σ(σ), and the politician prefers the status quo.
Theorem 3.5 shows that even if policy implementation becomes less informative of competence and additive valence plays a factor, the result that the politician prefers to implement her policy only when the expected outcome is sufficiently low remains. Further, σpub∈(0,1).
Corollary 3.2. The cutoff σpub is strictly between0 and 1
Proof. As σ → 1, r− → 1, and consequently, both r− and r+ must be on the concave portion of F. Because σ is the expected value of r− and r+, by Jensen’s inequality, F(σ) must be greater than the expectedF from implementation. Hence for sufficiently highσ, inaction is preferred. Therefore,σpub <1. Conversely, asσ →0, r+→0, and therefore bothr−and r+are on the convex portion ofF. Again by Jensen’s inequality,F(σ)must be less than the expected F from implementation. Hence for sufficiently low σ, implementation is preferred.
Therefore, σpub>0.
This corollary implies that, as long as F is strictly unimodal, there are some high signals for which the politician will not implement her policy, despite it being extremely likely to generate good outcomes, and simultaneously, low signals that induce the politician to implement a policy despite it being extremely likely to generate bad outcomes.