Information Acquisition under Persuasive Precedent versus Binding Precedent (Preliminary and Incomplete)

Transcription

1 Information Acquisition under Persuasive Precedent versus Binding Precedent (Preliminary and Incomplete) Ying Chen Hülya Eraslan January 9, 216 Abstract We analyze a dynamic model of judicial decision making. A court regulates a set of activities by allowing or banning them. In each period a new judge is appointed and a new case arises which must be decided by the judge appointed in that period. The judges share the same preferences over whether a case should be allowed or banned, but they are uncertain about the correct ruling until a costly investigation is made. We compare two institutions: persuasive precedent and binding precedent. Under persuasive precedent, a judge is not required to follow previous rulings but can use the information acquired by the investigations of previous judges. Under binding precedent, however, a judge must follow previous rulings when they apply. We show that the incentives to acquire information for the appointed judges are stronger in earlier periods when there are few precedents under binding precedent than under persuasive precedent, but as more precedents are established over time, the incentive to acquire information for the appointed judge becomes weaker under binding precedent than under persuasive precedent. Department of Economics, Johns Hopkins University Department of Economics, Rice University 1

2 1 Introduction We analyze a dynamic model of judicial decision making. A court regulates a set of activities by allowing or banning them. In each period a new judge is appointed and a new case arises which must be decided by the judge appointed in that period. The judges share the same preferences over whether a case should be allowed or banned, but they are uncertain about the correct ruling until a costly investigation is made. Following Baker and Mezzetti [212], we assume that the judge appointed in a given period can either investigate the case before making a ruling or summarily decide without investigation. We compare two institutions: persuasive precedent and binding precedent. Under persuasive precedent, a judge is not required to follow previous rulings but can use the information acquired by the investigations of previous judges. Under binding precedent, however, a judge must follow previous rulings when they apply. We show that the incentives to acquire information for the appointed judges are stronger in earlier periods when there are few precedents under binding precedent than under persuasive precedent, but as more precedents are established over time, the incentive to acquire information for the appointed judge becomes weaker under binding precedent than under persuasive precedent. To see why, note that the cost of making a wrong summary decision is higher under binding precedent than under persuasive precedent since future judges have to follow precedents when they are binding even if they are erroneous. Hence, a judge who faces few precedents is more inclined to investigate to avoid mistakes under binding precedent. As more precedents are established over time, however, the value of information acquired through investigation becomes lower under binding precedent since future judges may not be able to use the information to make rulings, and this discourages judges from acquiring information under binding precedent. We establish these results first in a simple three-period model and then in an infinite-horizon model. In the infinite horizon model, we show there is a unique Markov perfect equilibrium payoff by showing that the Contraction Mapping Theorem applies. In our model, the uniqueness of Markov perfect equilibrium payoff 2

3 implies the uniqueness of Markov perfect equilibrium strategy profile. Given the contraction property, value function iteration converges to the unique equilibrium payoff. The numerical results we obtain are consistent with our finding that the appointed judge in earlier periods investigate more under binding precedent than under persuasive precedent, but as more precedents are established over time, eventually the appointed judge investigates less under binding precedent. Related Literature Landes and Posner [1976], Schwartz [1992], Rasmusen [1994], Talley [1999], Bueno De Mesquita and Stephenson [22], Gennaioli and Shleifer [27], Baker and Mezzetti [212], Ellison and Holden [214], Anderlini, Felli, and Riboni [214], Callander and Hummel [214], Callander and Clark [215]. 2 Model A court regulates a set of activities by allowing or banning them. In each period, a new case arises which must be decided by the judge appointed in that period. We refer to the judge appointed to make the decision in period t as judge t. Each judge prefers to allow activities that she regards as beneficial and ban activities which she regards as harmful. We denote a case by a real number x [, 1]. Judge t has a threshold value [, 1] such that she regards case x as beneficial and would like it to be permitted if and only if x <. Since we assume that does not vary with t, the judges have the same preferences regarding the cases. Following Baker and Mezzetti [212], we assume that the cases are observed by the judges, but the preference parameter is unknown initially. Specifically, we assume is distributed according to cumulative distribution function F with density function f. The support of is [, ] and < < < 1 holds. In period t {1,... }, a case x t randomly arises according to distribution G on [, 1]. We assume that the cases are independent across periods. The precedent at time t is captured by two numbers L t and R t with < L t < R t < 1 where L t is the highest case that was ever allowed and R t is the lowest case that was ever banned 3

4 by time t. We assume L 1 < < < R 1, that is, the precedent at the beginning of period 1 is consistent with the judges preferences. The timing of the events is as follows. In period t, after case x t is brought to the court, judge t chooses whether to investigate the case or not before deciding whether to permit the activity or ban it. 12 An investigation allows judge t to learn what is at a fixed cost z. If the case is decided without an investigation, we say the judge made a summary decision. Let s = ((L, R), x). In what follows, for expositional convenience, we refer to s as the state even though it does not include the information about. Let S p denote the set of possible precedents, i.e. S p = {(L, R), [, 1] 2 : R > L}, and let S denote the set of possible states, i.e. S = S p [, 1]. Denote the ruling at time t by r t {, 1}, where r t = if the case is banned and r t = 1 if the case is permitted. After the judge makes her ruling, the precedent changes to L t+1 and R t+1. If x t was permitted, then L t+1 = max{l t, x t } and R t+1 = R t ; if x t was banned, then L t+1 = L t and R t+1 = min{r t, x t }. Formally, the transition of the precedent is captured by the function π : S {, 1} S p where π(s, ) is the vector (L, min{r, x}) and π(s, 1) is the vector (max{l, x}, R). Judge t observes the investigation decisions by the previous judges as well as the outcome of the investigation and the ruling. We consider two institutions: a binding precedent and a persuasive precedent. Under binding precedent, judge t must permit x t if x t L t and ban it if x t R t. Under persuasive precedent, judge t is free to make any decision. In this case, the role of the precedent is potentially to provide information regarding whether the case is beneficial or not for judge t. We assume the violation of a binding precedent is infinitely costly. The payoff of any judge from the ruling r t on case x t in period t is 1 We assume judges investigate the case when indifferent for expositional simplicity. 2 We assume that a judge learns about his preference parameter through investigation. Alternatively, we can assume that a judge knows about his preferences in terms of the consequences of cases, but does not know how the consequence of a particular case unless he investigates. To illustrate, let c(x) denote the consequence of a case x and assume that c(x) = x + γ. A judge would like to permit case x if c(x) is below some threshold c and would like to ban it otherwise. Suppose that judge knows c and observe x, but γ is unknown until a judge investigates. This alternative model is equivalent to ours. 4

5 given by u(x t,, r t ) = { if x t < and r t = P, or x t and r t = B, l x t otherwise, where l x t is the variable cost of making a mistake, that is, permitting a case when it is above or banning a case when it is below. The total payoff of the appointed judge is the sum of her discounted payoffs from the rulings made in each period net of the investigation cost if she carried out an investigation. The discount factor is denoted by δ. We consider stationary Markov perfect equilibria. A Markov strategy depends only on payoff-relevant events, and a stationary Markov strategy does not depend on calendar time. A stationary Markov perfect equilibrium is a subgame perfect Nash equilibrium in stationary Markov strategies. We henceforth refer to a stationary Markov perfect equilibrium simply as an equilibrium. Persuasive precedent In the model with persuasive precedent, the payoff-relevant state in any period is the realized case x [, 1] and the information about. If is known at the time when the relevant decisions are made, then it is optimal not to investigate the case for any s, it is optimal to permit x if x <, and it is optimal to ban x if x >. If is unknown at the time when the relevant decisions are made, a (pure) strategy for the judge is a pair of functions σ = (µ, ρ), where µ : [, 1] {, 1} is an investigation strategy and ρ : [, 1] {, 1} is an uninformed ruling strategy, where µ(x) = 1 if and only if an investigation is made when the case is x; and ρ(x) = 1 if and only if case x is permitted. To each strategy σ = (µ, ρ) we can associate two functions V ( ; σ) and W ( ; σ). The value V (x; σ) represents the dynamic payoff of a judge when she is appointed in the current period and the value W (x; σ) represents the dynamic payoff of a judge when she was appointed in a previous period when the current case is x and is unknown, and the strategy profile σ will be played from the current period 5

6 onwards. In what follows, we suppress the dependence of the dynamic payoffs on σ for notational convenience. A triple (σ, V, W ) is a Markov Perfect Equilibrium if and only if (E1-P) Given (V, W ), the uninformed ruling strategy satisfies ρ (x) = 1 if and ρ (x) = if for any case x. l l max{x,} max{x,} ( x)df () > l ( x)df () < l min{x, } min{x, } (x )df () (x )df () (E2-P) Given (V, W ) and the uninformed ruling strategy ρ, for any state s, the investigation strategy satisfies µ (x) = 1 if and only if z ρ (x)l max{x,} ( x)df () + (1 ρ (x))l (x )df () + δ min{x, } (E3-P) Given σ, for any state s, the dynamic payoffs satisfy W (x) = (1 µ (x)) [ ρ (x)l max{x,} ( x)df () + (1 ρ (x))l (x )df () + δ min{x, } V (x) = µ (x)z + W (x). 1 W (x )dg(x ) Condition (E1-P) says that the the ruling decision depends on only the current period payoff, and in particular, the judge chooses the ruling that minimizes the expected cost of making a mistake in the current period. ] 1 This is because under persuasive precedent, the ruling does not affect the judge s continuation payoff. If a judge investigates case x, then becomes known and no mistake in ruling W (x )dg(x ). 6

7 will be made in the current period as well as in the future. In this case, the dynamic payoff of the appointed judge, V (x), is negative of the cost of investigation, and the dynamic payoff of a previously appointed judge, W (x), is. If a judge does not investigate case x, then her dynamic payoff is the sum of the expected cost of making a mistake in the current period and the continuation payoff. In this case, since the appointed judge does not incur any investigation cost, her dynamic payoff is the same as the dynamic payoff of a previously appointed judge. Condition (E3- P) formalizes this and condition (E2-P) says that the appointed judge chooses to investigate a case if and only if her dynamic payoff from investigating is higher than her expected dynamic payoff from not investigating. Binding precedent In the model with binding precedent, the payoff-relevant state in any period is the precedent pair (L, R) [, 1] 2, the realized case x [, 1] and the information about. If is known at the time when the relevant decisions are made, then it is optimal not to investigate the case for any s, it is optimal to permit x if x < max{l, }, and it is optimal to ban x if x > min{r, }. Let C(L, R) denote the expected dynamic payoff of the judge when the precedent is (L, R), conditional on being known when decisions regarding the cases are made where the expectation is taken over before it is revealed and over all future cases x. Formally C(L, R) = l [ 1 δ L L ( x)dg(x)df () + R R ] (x )dg(x)df (), (1) where L is the (possibly degenerate) interval [, max{l, }] and R is the (possibly degenerate) interval [min{r, }, ]. To see how we derive C(L, R), note that if < L and x (, L], then the judge incurs a cost of l( x) since she has to permit x; similarly, if > R and x [R, ), then the judge incurs a cost of l(x ) since she has to ban x. It follows [ that the expected per-period payoff of a judge conditional on being known is l ( x)dg(x)df () + ] L (x )dg(x)df (), L R R and her dynamic payoff in the infinite horizon model is 1/(1 δ) times the per-period payoff. Note that max{l, } is increasing in L and min{r, } is decreasing in R, and 7

8 therefore C(L, R) is decreasing in L and increasing in R. If is unknown at the time when the decisions regarding the cases are made, a (pure) strategy for the judge is a pair of functions σ = (µ, ρ), where µ : S {, 1} is an investigation strategy and ρ : S {, 1} is an uninformed ruling strategy, where µ(s) = 1 if and only if an investigation is made when the state is s, and ρ(s) = 1 if and only if case x is permitted when the state is s. To each strategy σ = (µ, ρ) we can associate two functions V ( ; σ) and W ( ; σ). The value V (s; σ) represents the dynamic payoff of a judge when she is appointed in the current period and the value W (s; σ) represents the dynamic payoff of a judge when she is not appointed in the current period, when the state is ((L, R), x), is unknown, and the strategy profile σ will be played from the current period onwards. In what follows, we suppress the dependence of the dynamic payoffs on σ for notational convenience. A triple (σ, V, W ) is a Markov Perfect Equilibrium if and only if (E1-B) Given (V, W ), the uninformed ruling strategy satisfies ρ (s) = 1 if either x L or x (L, R) and l > l max{x,} ( x)df () + δ (x )df () + δ 1 1 min{x, } and ρ (s) = if either x R or x (L, R) and l < l for any state s. max{x,} ( x)df () + δ (x )df () + δ 1 1 min{x, } W (π(s, 1), x )dg(x ) W (π(s, ), x )dg(x ), W (π(s, 1), x )dg(x ) W (π(s, ), x )dg(x ), (E2-B) Given (V, W ) and the uninformed ruling strategy ρ, for any state s, the 8

9 investigation strategy µ satisfies µ(s) = 1 if and only if ρ (s)l max{x,} z + 1 L (x)l x ( x)df () + (1 ρ (s))l ( x)df () + 1 R (x)l min{x, } where 1 A (x) takes the value 1 if x A and otherwise. (x )df () + δ (E3-B) Given σ, for any state s, the dynamic payoffs satisfy W (s) = µ (s) [ 1 L (x)l + (1 µ (s)) [ x ρ (s)l ( x)df () + 1 R (x)l max{x,} ( x)df () + (1 ρ (s))l (x )df () + δ min{x, } V (s) = µ (s)z + W (s). 1 x (x )df () + δc(l, R) x 1 (x )df () + δc(l, R) W (π(s, ρ (s)), x )dg(x ) Under binding precedent, the ruling decision may change the precedent, which in turn may affect the continuation payoff. As such, condition (E1-B) says the ruling decision depends on both the current period payoff and the continuation payoff. In particular, the judge chooses the ruling that maximizes the sum of the current period payoff and the continuation payoff, taking into consideration how her ruling affects the precedent in the next period. If a judge investigates case x, then becomes known. When the precedents are binding, however, mistakes in ruling can still happen if < L or if > R. this case, the dynamic payoff of the appointed judge, V (s), is the expected cost of making mistakes in the ruling, both in the current period and in future periods, minus the cost of investigation. The difference between V (s) and the dynamic payoff of a previously appointed judge, W (s), is still the cost of investigation. If a judge does not investigate case x, then her dynamic payoff is the sum of the expected W (π(s, ρ (s)), x )dg(x ) ] ] In 9

10 cost of making a mistake in the current period and the continuation payoff. In this case, since the appointed judge does not incur any investigation cost, V (s) is the same as W (s). Condition (E3-B) formalizes this and condition (E2-B) says that the appointed judge chooses to investigate a case if and only if her dynamic payoff from investigating is higher than her expected dynamic payoff from not investigating. 3 A three-period model Before we analyze the infinite horizon model, we discuss a three-period model to illustrate some of the intuition. For simplicity, we assume that F is uniform distribution on [, ] and G is uniform distribution on [, 1]. 3.1 Persuasive Precedent Consider the judge in period t. If a previous judge has investigated, then is known and judge t permits or bans case x t according to. Suppose no previous judge has investigated, then judge t s belief about is the same as the prior. If judge t investigates, his payoff is z in period t and in future periods. The following result says that if judge t does not investigate, then he permits x t if x t < E() and bans x t if x t > E(). For notational convenience, let e = E() and α =. Lemma 1. Under persuasive precedent, for any x t e, if no previous judge has investigated, then E [u(x t,, B) u(x t,, P )](x t e ) >. Lemma 1 implies that if no previous judge has investigated and judge t does not investigate case x t, then he permits it if x t < e and bans it if x t > e. Now we analyze the judges investigation decisions. The following lemma says that when the investigation cost is sufficiently low, the judge appointed in each period investigates with positive probability and the cases that judge t investigates forms an interval. 1

11 Lemma 2. Suppose z < αl. In the three-period model under persuasive precedent, 8 judge t {1, 2, 3} investigates x t with positive probability in equilibrium. Specifically, there exist x L t and x H t > x L t such that judge t investigates x t if and only if x t [x L t, x H t ]. 3.2 Binding precedent We first show that in equilibrium, in each period t, the cases that judge t investigates also form a (possibly degenerate) interval under binding precedent. Lemma 3. In the three-period model under binding precedent, the set of cases that judge t investigates in equilibrium is either empty or convex for any precedent (L t, R t ). In the next proposition, we show that judge 1 investigates more under binding precedent than under persuasive precedent, but judges 2 and 3 investigate less under binding precedent than under persuasive precedent. Proposition 1. Judge 3 investigates less under binding precedent than under persuasive precedent. Specifically, for any precedent (L 3, R 3 ), judge 3 investigates case x 3 if and only if x 3 (L 3, R 3 ) [x L 3, x H 3 ]. Judge 2 also investigates less under binding precedent than under persuasive precedent. Specifically, if the precedent in period 2 satisfies [, ] [L 2, R 2 ], then the set of cases that judge 2 investigates under binding precedent is the same as [x L 2, x H 2 ]; otherwise the set of cases he investigates under binding precedent is a subset of [x L 2, x H 2 ]. Judge 1 investigates more under binding precedent than under persuasive precedent, that is, given any precedent (L 1, R 1 ) (, ), the set of cases that judge 1 investigates under binding precedent contains the set [x L 1, x H 1 ]. The reason for judge 3 to investigate less under binding precedent is that investigation has no value if x 3 L 3 or if x 3 R 3 since the judge must permit any x 3 L 3 and must ban any x 3 R 3 no matter what the investigation outcome; moreover, since period 3 is the last period, the information about has no value for the future either. For x 3 (L 3, R 3 ), judge 3 faces the same incentives under binding and persuasive precedent and therefore investigate the same set of cases. 11

12 For judge 2, if the precedent in period 2 satisfies [, ] [L 2, R 2 ], then investigation avoids mistakes in ruling in the current period as well as the future period even under binding precedent. In this case, judge 2 faces the same incentives under binding and persuasive precedent and therefore investigate the same set of cases. However, if the precedent in period 2 does not satisfy [, ] [L 2, R 2 ], then even if x 2 (L 2, R 2 ) and judge 2 investigates, mistakes in ruling can still happen in period 3 under binding precedent if / [L 2, R 2 ] since judge 3 has to follow the precedent. In this case, the value of investigation is lower under binding precedent than under persuasive precedent and therefore judge 2 investigates less under binding precedent. For judge 1, since the precedent in period 1 satisfies [, ] [L 1, R 1 ], investigation avoids mistakes in ruling in the current period as well as in future periods even under binding precedent. However, for x 1 (, ), if the judge does not investigate x 1 and makes a summary ruling, then she changes the precedent in a way that [, ] [L 2, R 2 ]. As discussed in the previous paragraph, this diminishes judge 2 s incentive to investigate, which in turn lowers judge 1 s dynamic payoff. Hence, judge 1 s payoff from not investigating is lower under binding precedent than under persuasive precedent, and therefore she has a stronger incentive to investigate under binding precedent. 4 Infinite-horizon model We now consider the infinite-horizon model, that is, T =. 4.1 Persuasive precedent If a previous judge already investigated, then judge t knows the value of and would permit or ban a case according to. We next show that if no previous judge has investigated, then the cases that judge t investigates form a convex set. Proposition 2. Under persuasive precedent, if judge t investigates x 1 and x 2 > x 1 in equilibrium, then she investigates any x [x 1, x 2 ] in equilibrium. 12

13 Fix an equilibrium. Suppose the set of cases the judge investigates is nonempty. Let a = inf{x : µ (x) = 1} and b = sup{x : µ (x) = 1}. We next show that if a judge faces a case such that there is no uncertainty about what the correct ruling is for that case (that is, if x or if x ), then the judge does not investigate the case in equilibrium, even though the information from investigation is valuable for future rulings. Proposition 3. In equilibrium, we have < a b <. Let W e = 1 W (x )dg(x ). To find the equilibria, note that δw e if x, or if x, l x W (x) = ( x)df () + δw e if < x < a, if x [a, b], l e (x )df () + δw if b < x <. x (2) Hence, we have W e = δw e [G(a) + 1 G(b)]+l a x ( x)df ()dg(x)+l b x (x )df ()dg(x). Since the appointed judge is indifferent between investigating and not investigating when x = a, we have z = l a (3) ( a)df () + δw e. (4) Similarly, since the appointed judge is indifferent between investigating and not investigating when x = b, we have z = l b (b )df () + δw e. (5) We can solve for W e, a, b from equations (3), (4), and (5). Plugging these in (2), 13

14 we can solve for W (x). We can also solve for V (x) since V (x) = z for x [a, b] and V (x) = W (x) for x / [a, b]. 4.2 Binding precedent We now consider binding precedent. Let F denote the set of bounded measurable functions on S taking values in R. For f F, let f = sup{ f(s) : s S}. An operator Q : F F satisfies the contraction property for if there is a β (, 1) such that for f 1, f 2 F, we have Q(f 1 ) Q(f 2 ) β f 1 f 2. For any operator Q that satisfies the contraction property, there is a unique f V such that Q(f) = f. Recall L is the (possibly degenerate) interval [, max{l, }] and R is the (possibly degenerate) interval [min{r, }, ]. Let A(s) denote the appointed judge s dynamic payoff if he investigates in state s = ((L, R), x), not including the investigation cost. Formally, A(s) = 1 L (x)l x ( x)df () + 1 R (x)l x (x )df () + δc(l, R). Also, let g p (s) be the appointed judge s current period payoff if he permits the case without investigation in state s and g b (s) be his current period payoff if he bans the case without investigation in state s. Formally, l max{x,} ( x)df () if x < R, g p (s) = if x R, l (x )df () if x > L, g b min{x, } (s) = if x L. Note that for any s S, µ (s) as defined in (E2-B) satisfies µ (s) arg max µ[ z + A(s)] µ {,1} + (1 µ) max{g p (s) + δ 1 W (max{x, L}, R)dG(x ), g b (s) + δ 14 1 W (L, min{x, R})dG(x )}.

15 For W F and any s S, define T W (s) = µ(s)z + max{ z + A(s), g p (s) + δ 1 W (max{x, L}, R)dG(x ), g b (s) + δ Note that W as defined in (E3-B) satisfies W = T W. We next show that T is a contraction. Proposition 4. T : F F is a contraction. In contrast to a dynamic single-agent decision problem with discounting, value functions in a dynamic game often fail the contraction mapping property because a player s payoff depends not only on her own action but other players actions as well. In the game we analyze, however, the judges have the same preference over the rulings and the only divergence of interest comes from the private cost a judge incurs if she carries out an investigation. We are able to exploit this to establish the contraction mapping property. Since T is a contraction, there is a unique W that satisfies T W = W and therefore a unique W that satisfies (E3-B). Once we solve for W, we can solve for the equilibrium strategies ρ and µ from (E1-B) and (E2-B). We can then solve for V from (E3-B). We next establish that the equilibrium value functions W and V are decreasing in L and increasing in R and the equilibrium investigation strategy µ is also decreasing in L and increasing in R. This result says that as the precedent gets tighter, the appointed judge investigates less and her payoff also becomes lower. 1 W (L, min{x, R})dG(x )}. Proposition 5. Suppose the precedent (ˆL, ˆR) is tighter than (L, R), that is, L ˆL < ˆR R. Under binding precedent, when z is sufficiently low, for any case x [, 1], if an appointed judge investigates x under precedent (ˆL, ˆR), then she also investigates x under precedent (L, R), that is, µ (L, R, x) is decreasing in L and increasing in R; moreover, the value functions W (L, R, x) and V (L, R, x) are also decreasing in L and increasing in R. 15

16 We next show that the set of cases that an appointed judge investigates is either empty or convex, and the appointed judge does not investigate any case for which she must follow the precedent in ruling. Proposition 6. Under binding precedent, for any precedent (L, R) S p, (i) if the appointed judge investigates case x 1 and x 2 > x 1 in equilibrium, then she also investigates any x [x 1, x 2 ], and (ii) the appointed judge does not investigate any case x / (L, R). For any (L, R) S p such that {x : µ (L, R, x) = 1}, let a(l, R) = inf{x : µ (L, R, x) = 1} and b(l, R) = sup{x : µ (L, R, x) = 1}. Proposition 6 implies that when facing the precedent (L, R), the appointed judge investigates x (a(l, R), b(l, R)) where a(l, R) L and b(l, R) R. We refer to (a(l, R), b(l, R)) as the investigation interval under (L, R). We conjecture that if L < a(l, R) < b(l, R) < R, then under precedent (L, R), the judge is indifferent between investigating or not when x = a(l, R) and when x = b(l, R), which implies that the set of cases that the appointed judge investigates is the closed interval [a(l, R), b(l, R)]. Suppose that given the initial precedent (L 1, R 1 ), the set of cases that judge 1 investigates is nonempty (if it is empty, then no judge will investigate any case). Specifically, judge 1 investigates x 1 if and only if x 1 [a(l 1, R 1 ), b(l 1, R 1 )]. For notational simplicity, let a 1 = a(l 1, R 1 ) and b 1 = b(l 1, R 1 ). If x 1 [a 1, b 1 ], judge 1 investigates x 1. In this case, since becomes known, no future judge will investigate. If x 1 / [a 1, b 1 ], then judge 1 makes a summary ruling without any investigation and changes the precedent to (x 1, R 1 ) if he permits the case and to (L 1, x 1 ) if he bans the case. Note that when judge 1 makes a summary ruling, the resulting new precedent satisfies L 2 < a 1 and b 1 < R 2. Monotonicity of µ in Proposition 5 implies that the investigation interval in period 2 satisfies a(l 2, R 2 ) a 1 and b(l 2, R 2 ) b 1 and therefore we have L 2 < a(l 2, R 2 ) b(l 2, R 2 ) < R 2. An iteration of this argument shows that on any realized equilibrium path, the investigation interval is a strict subset of the precedent in any period and is either closed or empty. Denote a nonempty investigation interval on an equilibrium path by [a(l e, R e ), b(l e, R e )]. By Propositions 5 and 6, we have 16

17 L e < a(l e, R e ) b(l e, R e ) < R e and given the precedent (L e, R e ), the judge is indifferent between investigating x or not if x = a(l e, R e ) or if x = b(l e, R e ). On the equilibrium path, the investigation intervals either converge to or to some nonempty set [â, ˆb] such that if the precedent is (L, R) = (â, ˆb), then a(l, R) = â and b(l, R) = ˆb. We conjecture that in period 1 when the precedent is (L 1, R 1 ), the appointed judge investigates more under binding precedent than under persuasive precedent. But as more precedents are established over time, the appointed judge has less freedom in making her ruling when precedents are binding, and eventually she investigates less than if the precedent is persuasive. Recall that [a, b] denotes the set of cases that an appointed judge investigates under persuasive precedent. Conjecture 1. We have (â, ˆb) [a, b] [a(l 1, R 1 ), b(l 1, R 1 )]. Conjecture 1 is analogous to Proposition 1 in the 3-period model, which says that judge 1 investigates more under binding precedent than under persuasive precedent, but judges 2 and 3 investigate less under binding precedent. We next provide a numerical example to illustrate the results. Since T is a contraction as shown in Proposition 4, value function iteration converges to the unique equilibrium value function W. Example 1. Suppose that is uniformly distributed on [.2,.8], x is uniformly distributed on [, 1], δ =.95, z =.5, l = 1. Under persuasive precedent, if no previous judge had investigated, then the appointed judge investigates x if and only if x [.429,.571], and he does not investigate x and permits it if x [,.429) and bans it if x (.571, 1]. We now turn to binding precedent. Figure 1 illustrates the investigation decision of the appointed judge for different values of L when R = 1 when no previous judge has investigated. As the figure shows, the cases that the appointed judge investigates is an interval for L sufficiently low and the investigation interval shrinks as L increases and eventually becomes empty, consistent with Propositions 5 and 6. Figure 2 also illustrates the investigation of the appointed judge for different values of L, but now R is fixed at.6. Similar to Figure 1, Figure 2 shows that 17

18 the investigation interval shrinks as L increases and eventually becomes empty. We can also compare the investigation intervals in the figures with the same value of L. Indeed, the investigation interval is larger when R = 1 than when R =.6 when L is fixed. (Note that since L < R, Figure 2 has fewer graphs displayed than Figure 1.) Under binding precedent, given the initial precedent (L 1, R 1 ) = (, 1), judge 1 investigates x 1 [a(l 1, R 1 ), b(l 1, R 1 )] where a(l 1, R 1 ) =.37 and b(l 1, R 1 ) =.63. We also calculate that â =.45 and ˆb =.55. Since under persuasive precedent, the appointed judge investigates x [a, b] where a =.429 and b =.571, we have (â, ˆb) [a, b] [a(l 1, R 1 ), b(l 1, R 1 )], consistent with Conjecture 1. Figure 3 illustrates the ruling decisions for different values of L when R = 1. As the figure shows, the ruling decision follows a cut-off rule: without investigation, there is a threshold such that the appointed judge permits any case below it and bans any case above it. 1. R= µ x L=. L=.2 L=.4 L=.6 L=.8 Figure 1: Investigation decision for different values of L when R = 1 18

19 1. R= µ x L=. L=.2 L=.4 Figure 2: Investigation decision for different values of L when R =.6 1. R= ρ x L=. L=.2 L=.4 L=.6 L=.8 Figure 3: Ruling decision for different values of L when R = 1 19

20 5 Appendix Proof of Lemma 1: First note that [ max{,xt} ] E [u(x t,, B) u(x t,, P )] = l ( x t )df () (x t )df (), min{x t, } (6) where l min{x t, } t)df () is judge t s expected payoff when he bans x t but x t <, and l max{,x t} (x t )df () is judge 3 s expected payoff when judge t permits x t but x t >. If x t, then clearly E [u(x t,, B) u(x t,, P )] > ; and if x t, then clearly E [u(x t,, B) u(x t,, P )] <. We consider x t (, ). Note that = = x t ( x t )df () xt (x t )df () x t df () x t [ F ( ) F (xt ) ] x t [F (x t ) F ()] + = e x t. df () x t xt df () It follows that E [u(x t,, B) u(x t,, P )](x t e ) = l(x t e ) 2 > for x t e. Proof of Lemma 2: Consider judge 3 first. If x 3 <, then judge 3 knows that > x 3, and therefore he permits the case without investigation. If x 3 >, then judge 3 knows that > x 3, and therefore he bans the case without investigation. For x 3 [, e ), if judge 3 does not investigate, he permits the case and his expected payoff is x 3 l(x 3 ) d α 3 = l (x 2α 3 ) 2. For x 3 ( e, ), if judge 3 does not investigate, he bans the case and his expected payoff is d 2 = l x 3 l( x 3 ) x 3 α = e, judge 3 s expected payoff if he does not investigate is αl 8 z < αl 8, then judge 3 investigates some cases. Specifically, let xl 3 = + ( x 2α 3 ) 2. When. Hence, if 2αz l and 2

21 x H 3 2αz =. If x l 3 [x L 3, x H 3 ], then judge 3 investigates case x 3. Now consider judge 2. Suppose judge 1 has not investigated. If judge 2 chooses to investigate in period 2, then his payoff in period 2 is z and his period-3 expected payoff is. If judge 2 chooses not to investigate in period 2, then by Lemma 1, he permits any case x 2 < e and bans any case x 2 > e. Note that if x 2 or if x 2, then judge 2 s payoff is if he does not investigate since he makes the correct decision. Consider < x 2 < e and suppose judge 2 does not investigate the case. Since judge 2 permits such a case, his expected payoff in period 2 is 1 α x2 l(x 2 )d = l(x 2 ) 2. 2α Similarly, for e < x 2 <, if judge 1 does not investigate the case, he bans it and in this case, his expected payoff in period 2 is 1 α l( x 2 )d = l( x 2) 2 x 2 2α Recall that judge 3 investigates x 3 if and only if x 3 [x L 3, x H 3 ] where x L 2αz 3 = + l and x H 2αz 3 =. l For x 3, judge 3 permits x 3 and judge 2 s payoff in period 3 is ; for x 3, judge 3 bans x 3 and judge 2 s payoff in period 3 is. Hence, the expected payoff of judge 2 in period 3 is x L 3 l 2α (x 2 ) 2 dx 2 + l x H 2α ( x 2) 2 dx 2 3 = l [ 1 2α 3 (xl 3 ) ] 3 ( xh 3 ) 3 2αz. = 2z 3 l 21

22 Now consider judge 2 s optimal strategy in period 2. For x 2 / [, ], if 2z 3 2αz l > z, then it is optimal for judge 2 not to investigate x 2 in period 2. Since z < αl, this is 8 satisfied. Hence judge 2 does not investigate x 2 if x 2 / [, ]. For < x 2 < e, if z < l 2α (x 1 ) 2 + 2z 2αz 3 l, then it is optimal for judge 2 not to investigate x 2 in period 2. Let x 2 = e. Then 2αz l RHS = αl + 2z 8 3 (, e ) that satisfies < z since z < αl 8. Hence, by continuity, there exists xl 2 z = l 2α (xl 2 ) 2 + 2z 2αz 3 l such that if x 2 (x L 2, e ], judge 2 investigates x 2 in period 1. Similarly, there exists x H 2 which satisfies z = l 2α ( xh 2 ) 2 + 2z 2αz 3 l such that if x 2 [ e, x H 2 ], judge 2 investigates x 2 in period 2. So the set of x 2 such that judge 2 investigates x 2 in period 2 is convex. It is [x L 2, x H 2 ]. Note that [x L 3, x H 3 ] [x L 2, x H 2 ]. Now consider judge 1. If judge 1 chooses to investigate in period 1, then his dynamic payoff is z. If he does not investigate in period 1, then he permits x 1 if x 1 < e and bans x 1 if x 1 > e and his expected continuation payoff is [ x L 2 l 2α (x 2 ) 2 2z ] 2αz dx l = l [ 1 2α 3 (xl 2 ) ] 3 ( x H 2 ) 3 2z 3 x H 2 [ l 2α ( x 2 ) 2 2z 3 2αz l ( x H 2 + x L 2 ). ] 2αz dx 2 l 22

23 For x 1 [, e ), judge 1 permits x 1 if he does not investigate it. Hence, his dynamic payoff is l 2α (x 1 ) 2 l [ 1 2α 3 (xl 2 ) ] 3 ( x H 2 ) 3 2z 2αz 3 l ( x H 2 + x L 2 ) and x L 1 satisfies z = l 2α (xl 1 ) 2 + l [ 1 2α 3 (xl 2 ) ] 3 ( x H 2 ) 3 + 2z 2αz 3 l ( x H 2 + x L 2 ). Similarly, for x 1 ( e, ], judge 1 bans x 1 if he does not investigate it and his dynamic payoff is l 2α ( x 1 ) 2 l [ 1 2α 3 (xl 2 ) ] 3 ( x H 2 ) 3 2z 2αz 3 l ( x H 2 + x L 2 ) and x H 1 satisfies z = l 2α ( x 1 ) 2 + l [ 1 2α 3 (xl 2 ) ] 3 ( x H 2 ) 3 + 2z 2αz 3 l ( x H 2 + x L 2 ). Substituting for x L 2 and x H 2, we can solve for when judge 1 investigates x 1. Proof of Lemma 3: Consider judge 3 first. Suppose no previous judge has investigated. Recall that under persuasive precedent, judge 3 investigates x 3 if and only if x 3 [x L 3, x H 3 ]. Since under binding precedent, investigation has no value if x 3 L 3 or if x 3 R 3, judge 3 investigates x 3 if x 3 [x L 3, x H 3 ] (L 3, R 3 ). Hence, the set of cases that judge 3 investigates is either empty or convex. Let k(l, R) denote a judge s expected payoff in period t under binding precedent when the precedents are (L, R) in period t conditional on being known where the expectation is taken over before it is revealed and over all possible cases x. Formally [ k(l, R) = l L L ( x)dg(x)df () + R R ] (x )dg(x)df () 23

24 where L is the (possibly degenerate) interval [, max{l, }] and R is the (possibly degenerate) interval [min{r, }, ]. Note that k(l, R) and k(l, R) < if L > or if R <. Note also that k(l, R) is decreasing in L and increasing in R. Let EV t (L, R) denote the expected equilibrium continuation payoff in period t given that the precedent in period t is (L, R). Claim 1. If EV t (L, R) is decreasing in L and increasing in R, then the set of cases that judge t 1 investigates in period t 1 is either empty or convex for any precedent in period t 1. Proof: Suppose that EV t (L, R) is decreasing in L and increasing in R. Fix the precedent in period t 1 and denote it by (L t 1, R t 1 ). Suppose that the judge investigates cases x and x > x in period t 1. We next show that the judge also investigates case ˆx [x, x ]. Let g p (L, R, x) be the appointed judge s current period payoff if he permits the case without investigation in state s = (L, R, x) and g b (s) be his current period payoff if he bans the case without investigation in state s. Note that for any (L, R), g p (L, R, x) is decreasing in x and g b (L, R, x) is increasing in x. Suppose ˆx (L t 1, R t 1 ). If the judge investigates ˆx, then his payoff is z + δk(l t 1, R t 1 ). Suppose the judge does not investigate ˆx and without loss of generality, suppose it is optimal for him to permit ˆx if he does not investigate it. Since the judge investigates x under precedent (L t 1, R t 1 ), we have z + δk(l t 1, R t 1 ) g p (L t 1, R t 1, x ) + δev (max{x, L t 1 }, R t 1 ). Since g p is decreasing in x, we have g p (L t 1, R t 1, x ) > g p (L t 1, R t 1, ˆx). Moreover, since ˆx > max{x, L t 1 } and EV t is decreasing in L, we have we have EV t (max{x, L t 1 }, R t 1 ) > EV (ˆx, R t 1 ). Hence, we have z + δk(l t 1, R t 1 ) g p (L t 1, R t 1, ˆx) + δev t (ˆx, R t 1 ), which implies that the judge investigates case ˆx. Suppose ˆx L t 1. Then judge t 1 has to permit ˆx regardless of whether he investigates it or not. Hence, judge t 1 investigates ˆx if z + δk(l t 1, R t 1 ) δev t (L t 1, R t 1 ). Since the judge investigates x < ˆx, we have z + δk(l t 1, R t 1 ) δev t (L t 1, R t 1 ), implying that the judge investigates ˆx. A similar argument shows that judge t 1 investigates ˆx if ˆx R t 1 as well. Hence, the set of cases that judge 24

25 t 1 investigates in period t 1 is either empty or convex for any precedent in period t 1. Now consider judge 2. Suppose judge 1 did not investigate. Since judge 3 investigates x if x (L 3, R 3 ) [x 3 1, x H 3 ], we have EV 3 (L 3, R 3 ) = k(max{l 3, x L 3 }, min{r 3, x H 3 }). Since k(l, R) is decreasing in L and increasing in R, it follows that EV 3 is decreasing in L 3 and increasing in R 3. By Claim 1, the set of cases that judge 2 investigates is either empty or convex. Now consider judge 1. To establish that EV 2 (L 2, R 2 ) is decreasing in L 2 and increasing in R 2, we show that if judge 2 investigates x 2 given the precedent (L 2, R 2 ), then he also investigates x 2 given the precedent (ˆL 2, ˆR 2 ) if ˆL 2 L 2 and ˆR 2 R 2 Suppose that judge 2 investigates x 2 given the precedent (L 2, R 2 ). First consider the case x 2 (L 2, R 2 ). Then we have the following inequality. z + δk(l 2, R 2 ) max{ x2 ( x 2 )df () + δev 3 (x 2, R 2 ), (x 2 )df () + δev 3 (L 2, x 2 )}. x 2 Since EV 3 (x 2, ˆR 2 ) = k(max{x 2, x L 3 }, min{x H 3, ˆR 2 }) and EV 3 (x 2, R 2 ) = k(max{x 2, x L 3 }, min{x H 3, R 2 }), we have k(ˆl 2, ˆR 2 ) k(l 2, R 2 ) EV 3 (x 2, ˆR 2 ) EV 3 (x 2, R 2 ). k(ˆl 2, ˆR 2 ) k(l 2, R 2 ) EV 3 (ˆL 2, x 2 ) EV 3 (L 2, x 2 ). It follows that z + δk(ˆl 2, ˆR 2 ) max{ x2 Similarly, we have ( x 2 )df () + δev 3 (x 2, ˆR 2 ), (x 2 )df () + δev 3 (ˆL 2, x 2 )} x 2 and therefore judge 2 investigates x 2 under precedent (ˆL 2, ˆR 2 ). Next consider the case x 2 L 2. Since judge 2 investigates x 2 under precedent (L 2, R 2 ), we have z + δk(l 2, R 2 ) EV 3 (L 2, R 2 ). Since k(ˆl 2, ˆR 2 ) k(l 2, R 2 ) EV 3 (ˆL 2, ˆR 2 ) EV 3 (L 2, R 2 ) and x 2 ( x 2)dF () <, it follows that z+δk(ˆl 2, ˆR 2 ) δev 3 (ˆL 2, ˆR 2 ) > x 2 ( x 2)dF ()+δev 3 (ˆL 2, ˆR 2 ). Hence, judge 2 investigates x 2 L 2 under precedent (ˆL 2, ˆR 2 ). A similar argument shows that if x 2 R 2 and judge 2 investigates x 2 under precedent (L 2, R 2 ), then judge 2 also investigates x 2 under precedent (ˆL 2, ˆR 2 ). Since the investigation set is smaller with a tighter precedent in period 2 and in 25

26 period 3, it follows that EV 2 (L 2, R 2 ) is decreasing in L 2 and increasing in R 2. By Claim 1, the set of cases that judge 1 investigates is either empty or convex. Proof of Proposition 1: Consider judge 3 first. Suppose no previous judge has investigated. As shown in the proof of Lemma 3, under binding precedent, judge 3 investigates x 3 if x 3 [x L 3, x H 3 ] (L 3, R 3 ). Now consider judge 2. Recall that under persuasive precedent, judge 2 investigates x 2 if and only if x 2 [x L 2, x H 2 ] where [x L 2, x H 2 ] [x L 3, x H 3 ]. First suppose [, ] [L 2, R 2 ]. Then the incentive of judge 2 is the same under binding precedent as under persuasive precedent. In this case, under binding precedent, judge 2 investigates x 2 if and only if x 2 [x L 2, x H 2 ]. Next suppose [, ] [L 2, R 2 ]. We show below that under binding precedent, judge 2 does not investigate case x L 2. Recall that judge 2 is indifferent between investigating and not investigating x L 2 under persuasive precedent. Hence, we have z = x L 2 ( x L 2 )df () + δk(x L 3, x H 3 ). Consider binding precedent. Suppose x L 2 (L 2, R 2 ). Then the difference in judge 2 s dynamic payoff between investigating and not investigating x L 2 is [ ] x L z + δk(max{l 2, x L 3 }, min{r 2, x H 2 3 }) ( x L 2 )df () + δk(max{x 2, x L 3 }, min{r 2, x H 3 } z x L 2 ( x L 2 )df () = δk(x L 3, x H 3 ) <. Hence, judge 2 does not investigate x L 2 in this case. Suppose x L 2 / (L 2, R 2 ). Then judge 2 has to follow the precedent when ruling on x L 2 and the precedent in period 3 does not change, that is, (L 3, R 3 ) = (L 2, R 2 ). Hence, the difference in judge 2 s dynamic payoff between investigating and not investigating x L 2 is z + δk(max{l 2, x L 3 }, min{r 2, x H 3 }) δk(max{l 2, x L 3 }, min{r 2, x H 3 }) = z <. Hence, judge 2 does not investigate x L 2 in this case either. A similar argument establishes that under binding precedent, judge 2 does not 26

27 investigate x H 2 for any precedent. Given the convexity of the set of cases that judge 2 investigates under either binding or persuasive precedent, this implies that the set of cases that judge 2 investigates under binding precedent is contained in the set of cases he investigates under persuasive precedent. Now consider judge 1. We show below that judge 1 investigates x L 1 under binding precedent. Recall that judge 1 is indifferent between investigating and not investigating x L 1 under persuasive precedent. Hence, we have z = x L 1 ( x L 1 )df () + δk(x L 2, x H 2 ) + δ 2 [G(x L 2 ) + 1 G(x H 2 )]k(x L 3, x H 3 ). Let a 2 (L, R) = inf{x : judge 2 investigates x under binding precedent (L, R)}, b 2 (L, R) = sup{x : judge 2 investigates x under binding precedent (L, R)}. Define a 3 (L, R) and b 3 (L, R) analogously. Since L 1 < < < R 1, if judge 1 investigates x L 1, then his dynamic payoff is equal to z under binding precedent. If judge 1 does not investigate x L 1 and permits it, then the precedent in period 2 becomes (x L 1, R 1 ) and judge 2 investigates cases in ((a(x L 1, R 1 ), b(x L 1, R 1 )). Note also that judge 1 s expected payoff in period 3 is weakly lower than k(x L 3, x H 3 ). It follows that if judge 1 does not investigate x L 1, his dynamic payoff is lower than x L 1 ( x L 1 )df ()+δk((a(x L 1, R 1 ), b(x L 1, R 1 ))+δ 2 [G(a(x L 1, R 1 ))+1 G(b(x L 1, R 1 )]k(x L 3, x H 3 ). As shown previously, a(x L 1, R 1 ) > x L 2 and b(x L 1, R 1 ) < x L 2. Since k(l, R) is decreasing in L and increasing in R, it follows that k((a(x L 1, R 1 ), b(x L 1, R 1 )) < k(x L 2, x H 2 ). Also, since k(x L 3, x H 3 ) <, we have [G(a(x L 1, R 1 )) + 1 G(b(x L 1, R 1 )]k(x L 3, x H 3 ) < [G(x L 2 ) + 1 G(x H 2 )]k(x L 3, x H 3 ). Hence, z > x L 1 ( x L 1 )df ()+δk((a(x L 1, R 1 ), b(x L 1, R 1 ))+δ 2 [G(a(x L 1, R 1 ))+1 G(b(x L 1, R 1 )]k(x L 3, x H 3 ), implying that judge 1 investigates x L 1 under binding precedent. A similar argument establishes that under binding precedent, judge 1 investigates 27

28 x H 1. Given the convexity of the set of cases that judge 1 investigates under either binding or persuasive precedent, this implies that the set of cases that judge 1 investigates under binding precedent contains the set of cases he investigates under persuasive precedent. Proof of Proposition 2: Since judge t investigates x 1 (E2-P), we have z max { l for x {x 1, x 2 }. max{x,} ( x)df (), l min{x, } (x )df () and x 2, by (E1-P) and } + δ 1 W (x )dg(x ). Suppose ˆx [x 1, x 2 ]. Since l max{x,} ( x)df () is decreasing in x, we have l max{ˆx,} ( ˆx)dF () l max{x1,} Since l (x )df () is increasing in x, we have min{x, } It follows that { z max l ( x 1 )df (). l (ˆx )df () l (x 2 )df (). min{ˆx, } min{x 2, } max{ˆx,} ( ˆx)dF (), l min{ˆx, } and therefore the judge investigates ˆx in equilibrium. (ˆx )df () } + δ 1 W (x )dg(x ) Proof of Proposition 3: We prove the result by contradiction. Suppose that there exists an equilibrium in which a <. Let W e = 1 W (x )dg(x ). Since a <, the appointed judge s dynamic payoff is z if he investigates x = a, and δw e if he does not investigate x = a. Since µ(a) = 1, we have z δw e. Since for any x >, the judge s dynamic payoff is z if he investigates, and δw e if he does not investigate, 28

29 it must be the case that the judge investigates any case x >. It follows that b = 1. Moreover, since a <, any appointed judge makes the correct decision for x < a. It follows that W e = a δw e dg(x) = δg(a)w e, which implies that W e =. Since z δw e, this is impossible. Proof of Proposition 4: For any s S, let HW (s) = max{ z + A(s), g p (s) + δ 1 W (π(s, 1), x )dx, g b (s)+δ 1 W (π(s, ), x )dx } and KW (s) = µ(s)z. So T W (s) = HW (s) + KW (s). Note that µ(s) = 1 if and only if HW (s) = z + A(s). Suppose that W 1, W 2 F and consider any s S. Without loss of generality, suppose that HW 1 (s) HW 2 (s). For notational convenience, define µ 1 and µ 2 relative to W 1 and W 2. There are three cases to consider. (i) Suppose that HW 1 (s) = z + A(s). Since HW 1 (s) HW 2 (s), we have HW 2 (s) = z + A(s). We also have that µ 1 (s) = 1 and µ 2 (s) = 1. It follows that T W 1 (s) T W 2 (s) =. (ii) Suppose that HW 1 (s) = g p (s) + δ 1 W 1 (π(s, P ), x )dg(x ). Then µ 1 (s) =. We have T W 1 (s) T W 2 (s) g p (s) + δ δ δ W 1 (π(s, 1), x )dg(x ) g p (s) δ [ W 1 (π(s, 1), x ) W 2 (π(s, 1), x ) ] dg(x ) [ W 1 (π(s, 1), x ) W 2 (π(s, 1), x ) ] dg(x )] δ W 1 W 2 1 W 2 (π(s, 1), x )dg(x ) (iii) Suppose that HW 1 (s) = g b (s) + δ 1 W 1 (π(s, ), x )dg(x ). Then a similar argument as in case (ii) shows that T W 1 (s) T W 2 (s) δ W 1 W 2. Since either T W 1 (s) T W 2 (s) = or T W 1 (s) T W 2 (s) δ W 1 W 2 for any s S in all three cases, we have T W 1 T W 2 δ W 1 W 2 and therefore T is a contraction. 29

30 Proof of Proposition 5: Recall that for any s S, we define HW (s) = max{ z + A(s), g p (s) + δ 1 W (π(s, 1), x )dx, g b (s) + δ 1 W (π(s, ), x )dx }, KW (s) = µ(s)z, and T W (s) = HW (s)+kw (s), where µ(s) = 1 if HW (s) = z +A(s) and µ(s) = otherwise. Let EW (L, R) = 1 W (L, R, x)dg(x). To prove the proposition, we first establish the following lemma. Lemma 4. If W F satisfies the following properties: (i) W is decreasing in L and increasing in R, (ii) EW (L, R) EW (ˆL, ˆR) C(L, R) C(ˆL, ˆR), and (iii) KW is decreasing in L and increasing in R, then, for z sufficiently low, T W also satisfies these properties, that is, (i) T W is decreasing in L and increasing in R, (ii) ET W (L, R) ET W (ˆL, ˆR) C(L, R) C(ˆL, ˆR), and (iii) KT W is decreasing in L and increasing in R. We first show that if W F satisfies properties (i), (ii), and (iii), then T W is decreasing in L and increasing in R. Note that A(s) is decreasing in L and increasing in R, g p (s) is constant in L and increasing in R, g b (s) is constant in R and decreasing in L. Suppose W F is decreasing in L and increasing in R. Then 1 W (π(s, 1), x )dx = 1 W ((max{l, x}, R), x )dx and 1 W (π(s, ), x )dx = 1 W ((L, min{r, x}), x )dx are decreasing in L and increasing in R. Hence, HW is decreasing in L and increasing in R. Since KW is decreasing in L and increasing in R, it follows that T W = HW + KW is decreasing in L and increasing in R. Let ŝ = (ˆL, ˆR, x). We next show that if W F satisfies properties (i), (ii), and (iii), then ET W (L, R) ET W (ˆL, ˆR) C(L, R) C(ˆL, ˆR). Consider the following cases. (a) Suppose µ(ŝ) = 1 so that T W (ŝ) = A(ŝ). Then, since KW is increasing in L and decreasing in R, we have µ(s) = 1 and therefore T W (s) = A(s). Hence T W (s) T W (ŝ) = A(s) A(ŝ). (b) Suppose µ(ŝ) = and T W (ŝ) = g p (ŝ) + EW (max{x, ˆL}, ˆR) > A(ŝ) z. Then we must have x < ˆR and g p (s) + EW (max{x, ˆL}, ˆR) > z + A(ŝ). Note that A(s) g p (s) + δ 1 W (π(s, 1), x )dx and A(s) g b (s) + δ 1 W (π(s, ), x )dx since A(s) is the appointed judge s dynamic payoff if she investigates in state s, not including the investigation cost. It follows that T W (s) A(s) and therefore T W (s) T W (ŝ) < A(s) + z A(ŝ). (c) 3