Supplemental Appendix for Referees
This appendix for referees and interested readers will be posted online if our paper is published in JOP.
1
We denote the investigator’s strategy in the …rst period as
2
Similarly for the second period investigator
( ; s) : [0; 1]
(
I ; s)
: [0; 1]
fG; N Gg ! fT; Dg. Let
executive’s belief about the probability that the incumbent investigator is passive (
the …rst period case to trial and it is acquitted. Likewise let
and let
a
(A) denote the probability that
denote beliefs as
p
(C) ;
n
(C) ;
a
(C) ;
p
>
I
(D) ;
replacement is passive, neutral, or aggressive as
n
I
p
(A) denote the
< ), when she takes
(A) denote the probability that
I
2( ; )
after an acquittal. Similarly for a conviction or drop,
n
p
fG; N Gg ! fT; Dg.
(D) ; and
= F ( );
a
n
(D) : De…ne the probability that a random
= F ( ) F ( ) ; and
a
= 1 F ( ). Finally,
given signal s we denote the di¤erence between the probability that the investigator is retained after trying
and the probability that the investigator is retained after dropping as
r(s)
C
s
[
(1
G)
s
+ (1
)
N G]
+
A
s
[(1
) (1
N G)
+
s
G]
D:
(4)
Note that if r(s) > 0 the investigator has an accountability incentive to try whereas for r(s) < 0 she has an
incentive to drop.
Proof of Lemma 2 An investigator sees either s = G or s = N G. In terms of the utility received from
the case in a given period, which we calculate the same way as in Lemma 1, one of these information sets is
more important to the investigator in the following sense.
A passive investigator cares more about dropping a case when s = N G than she does when s = G, i.e.,
using the reasoning from Lemma 1, for any
G)
I
< ; U (Dropjs = N G)
U (T ryjs = N G) > U (Dropjs =
U (T ryjs = G) because
NG
G
I
(1
G)
NG
[
+ 1
I
(1
NG
G)
(1
+ (1
I ) NG
>
I ) N G]
> 0:
1
G
I
(1
G)
+ 1
G
(1
I ) NG
The last expression holds because
G
>
NG
:
Similarly, an aggressive investigator cares more about trying rather than dropping when s = G than
when s = N G, i.e., for
I
> ; U (T ryjs = G)
U (Dropjs = G) > U (T ryjs = N G)
U (Dropjs = N G).
For a neutral investigator we solve for ~ such that the investigator is indi¤erent in terms of which decision
is more important, i.e., U (T ryjs = G)
G
I
G
+
I
NG
(1
G)
1
(1
G)
+ 2
G
U (Dropjs = G) = U (Dropjs = N G)
(1
NG
I ) NG
G
=
NG
2
I
(1
G)
NG
I
(1
I ) NG
NG
NG
G
G ) + (2
< ~ < : We still need to establish that for
drop in the …rst period when s = N G and for
NG
+ 1
G
2
G
N
G
( +
) (1
~
It is straightforward to con…rm that
NG
=
U (T ryjs = N G) :
I
NG
NG
G)
:
NG
~ it is optimal to
> ~ it is optimal to try in the …rst period when s = G,
regardless of accountability incentives. To do this, we …nd bounds on how much the investigator’s …rst
period actions can a¤ect her utility from second period actions. For
I
~ , the largest possible di¤erence
between a investigator’s expected utility from her own choice of whether to try when retained versus a
replacement investigator’s choice occurs when s = N G in the second period. She can potentially lose up
to U (Dropjs = N G)
U (T ryjs = N G) if the replacement is aggressive. However, the probability of this
occurring is strictly less than 1, because there is some chance that the second period signal is s = G and
there is also some probability that the replacement is not aggressive. Thus a strict upper bound on the
investigator’s expected second period utility loss from choosing x = T when s = N G in the …rst period is
U (Dropjs = N G)
U (T ryjs = N G): Because the investigator’s …rst period utility di¤erence between trying
and dropping is U (Dropjs = N G)
the …rst period. For
I
U (T ryjs = N G) it is thus strictly optimal for her to drop the case in
~ ; a symmetric argument shows that it is strictly optimal to chose x = T when
s = G in the …rst period.
2
1
Proof of Lemma 3 We characterize
1
The argument for
investigator with
I
; the cutpoint for …rst period investigator behavior when s = G:
is essentially similar, except using s = N G: First note that, from Lemma 2, any
~ strictly prefers to try when s = G. There are three cases, based on the di¤erence in
probability of retention from trying versus dropping after a guilty signal: r(G) = 0; r(G) > 0; and r(G) < 0:
Case 1: r(G) = 0. First period actions don’t a¤ect the investigator’s retention probability when
s = G, so
1
= , i.e., she chooses her most preferred action.
Case 2: r(G) > 0: Any investigator with
strictly prefers to try. That’s what she wants to
I
do anyway in the …rst period and doing so increases the chance that she will be retained, which strictly
increases her utility in the second period.
To characterize the behavior of investigators with
I
from trying versus dropping, which we will denote as UT D (
The …rst component of UT D (
I ; G; r(G))
which, as in the proof of Lemma 1 is
I
G
(1
; we …nd an investigator’s utility di¤erence
<
I ; s; r(s)).
is just the …rst period utility di¤erence from the two actions,
G)
(1
I)
1
G
N G:
The second component is the second period e¤ect of her …rst period action. The di¤erence between
her probability of being retained if she tries and her probability of being retained if she drops is r (G) : If a
passive investigator is replaced, there is an increased chance of an incorrect conviction in the second period,
which results in
(1
I)
utility for the investigator. Speci…cally, it may be the case that the replacement
investigator is a neutral type who mistakenly observes s = G when the defendant is innocent and thus brings
the case to trial (which the passive investigator wouldn’t do) and the trial produces a mistaken outcome.
The probability of this happening is
n
(1
q) (1
)
N G.
Or it may be the case that the replacement is
aggressive, the defendant is innocent, and the trial produces a mistaken outcome. The probability of this
happening is
a
(1
)
N G:
If the passive investigator is replaced, there is also a decreased chance of a correct second period conviction, which counts for
I
utility. Speci…cally, the replacement investigator may be a neutral type who
3
correctly observes s = G when the defendant is guilty, brings the case to trial, and receives a correct trial
outcome. The probability of this happening is
nq
(1
G) :
Or it may be the case that the replacement
is aggressive, the defendant is guilty, and the trial produces a correct outcome. The probability of this
happening is
a
(1
G) :
Combining all of these terms, for a passive investigator, i.e.,
UT D (
I ; G; r(G))
=
I
G
(1
G)
+r (G) (1
r (G)
=
NG
I
(1
I)
I ) (1
(1
I
(1
G)
+
I NG
1
)[
G
n
:
1
G
+
(1
NG
[
NG
G) [ nq
fr (G) (1
+
)
I
n
q) +
nq
I:
a]
+
r (G) (1
Focusing on the last two lines of this expression, we see that for
function of
q) +
a]
a]
r (G) [
G
(1
)[
I
(1
s
)g
a]
n
(1
q) +
2 [0; ]; UT D (
a]
:
I ; s; r(s))
is a linear
Obviously, for r (G) > 0, UT D ( ; G; r(G)) > 0, i.e., an investigator who is indi¤erent between
trying and dropping in terms of …rst period outcomes when s = G strictly prefers to try when doing so
increases the probability that she is retained. Because UT D (
I ; G; r(G))
is linear in
I
this means there are
two possible situations. First, it may be the case that UT D (0; G; r(G)) > 0; in which case all investigators
with
some
I
1
1
2 [0; ] strictly prefer to try when s = G; in this case
2 (0; ) ; UT D (
1
; G; r(G)) = 0, in which case UT D (
I ; G; r(G))
(because UT D ( ; G; r(G)) > 0) and hence all investigators with
and those with
1
I
>
1
strictly prefer to try. Setting UT D (
NG
=
NG
[(1
Note that for r (G) > 0;
G)
1
r (G) (1
1
G
r (G) (1
) [ n (1 q) +
= 0: Second, it may be the case that for
I
<
1
I ; G; r(G))
) [ n (1
a ]] + (1
must be strictly increasing in
I
strictly prefer to drop when s = G
= 0 and solving out yields
q) + a ]
G
r (G) [
G) (
nq
+
a ])
:
(5)
is a continuous function of r (G) :
Case 3: r(G) < 0: In this case, a passive investigator obviously will not try a case. For a neutral
4
investigator, the utility di¤erence between trying versus dropping is
UT D (
I ; G; r(G))
=
I
G
(1
G)
+r (G) (1
+r (G)
=
I ) (1
(1
I
+
I
(1
G)
+
I NG
1
I.
I)
)
G)
r (G) (1
NG
Note that this expression is linear in
(1
)
G
G
1
NG
NG
pq
a
aq
p
p
(1
q)
(1
+ r (G)
G
aq
q)
pq
r (G) (1
(1
G
1
(1
q)
aq
p
a
)
q)
(1
q)
:
Moreover, it is strictly increasing because an investigator at
strictly prefers to drop and, from Lemma 2, an investigator at ~ strictly prefers to try when s = G. Thus
for r (G) < 0, there is a unique solution
1
1
NG
=
G)
(1
NG
r (G) (1
)
1
2 ( ; ~ ), which is a continuous function of r (G):
G
r (G) (1
aq
p (1
)
q)
aq
+ (1
G)
p (1
G
q)
+ r (G)
pq
a (1
q)
(6)
For part (iv) of Lemma 3, note that as r (G) ! 0; the right hand sides of Equations 5 and 6 both converge
to
G
NG 1
G )+(1
(
N G (1
)
G)
G
, i.e.,
, so
1
is a continuous function of r(G): From Equation 4 it is obvious that
r (G) is a continuous function of the executive’s strategy ; so
For part (v) of Lemma 3, we assume that
Assume that
1
1
= 0 and
1
1
is also a continuous function of :
= 1 then derive a contradiction.
= 0; and note that an investigator with
I
= 0 cares only about avoiding mistaken
convictions. If she tries the case in the …rst period when s = G, this will lead to 1
G
NG
mistaken
convictions. On the other hand, by trying the case, she changes her probability of retention by r (G), and if
retained, she will avoid mistaken convictions in two circumstances: her replacement is neutral and receives
an incorrect signal about an innocent defendant who is then mistakenly convicted, or her replacement is
aggressive, the defendant is innocent, and the defendant is mistakenly convicted. For the investigator at
5
:
I
= 0 to try when s = G requires that
G
1
(1
) (1 q)
q + (1
) (1 q)
1 q
q)
q) + a
n (1
1
q + (1
NG
) (1
Substituting in r (G) =
G
C
(1
G)
r (G) [
n
(1
) (1
q)
r (G) [
n
(1
) (1
q) +
+
NG
a
(1
a
(1
)
N G]
)]
r (G) :
G
+ 1
NG
+
A
+
A
1
G
1
(1
N G)
+
G
D
G
from
Equation 4, and rearranging terms this reduces to
C
G
(1
G)
G
+ 1
NG
G
(1
N G)
+
G
G
(7)
D
1 q
n (1 q)+
1
q+(1
Assume also that
1
)(1 q)
= 1: An investigator for whom
I
a
:
= 1 cares only about ensuring conviction of the
guilty, so for her to drop when s = N G the number of foregone correct …rst period convictions must be less
than the expected decrease in the number of correct second-period convictions if she drops the …rst period
case:
NG
(1
(1
1
q) + (1
Substituting in r (N G) =
)q
C
(1
G)
(1 q)
q) + (1
1 q
p + n (1
NG
(1
)q
p
(1
r (N G)
p
+
n
+
A
1
G)
+
(1
n
(1
q) (1
G)
q)
r (N G) :
q)
G)
r (N G)
NG
+ 1
NG
NG
(1
N G)
+
NG
G
D
from Equation 4, and rearranging terms this reduces to
C
NG
(1
G)
NG
+ 1
+
NG
1
(1 q)+(1
+
A
1
NG
1 q
p + n (1 q)
)q
6
(1
N G)
+
NG
G
D:
(8)
Because the same value of
C
+
A
NG
(1
+
G)
NG
1
NG
+ 1
(1
N G)
+
NG
G
C
NG
+
G
(1
G
1
A
)q
q)
q+(1
N G)
)(1 q)
(1
Using the fact that
p
p + n (1
q)
1 q
n (1 q)+
)(1 q)
G
1 q
q)+
NG
G
G
a
NG
(1
G
N G) ( C
)q
n
A) :
a
1
= F ( );
+
n (1
Note that the right hand side is strictly less than 1, so a necessary condition for
1
q) + (1
= 1 requires that
1 q
)q
1
q+(1
(1
1
G
+ 1
1
p + n (1
1
(1 q)+(1
= 0 and
G)
1 q
1
(1 q)+(1
+
1
must satisfy Equations 7 and 8, to have
D
q
p + n (1
1
q)
=F( )
F ( ) ; and
+
q + (1
) (1
a
= 0 and
1 q
q) +
n (1
q)
=1
1
< 1:
a
because
(1
q) [(1
q) (F ( )
)
(1
F ( )) <
+ (1
F ( ))] < q [
1
q
q
7
<
(1
q) (
)
F ( )]
)
(9)
< 1:
Note that under our assumptions in the main text about the distribution F we can show that
F ( ) + (1
= 1 is
F ( ) ; this can be re-written as
1
1 q
(1 q) + (1
) q F ( ) + (1 q) (F ( ) F ( ))
1
1 q
+
q + (1
) (1 q) (1 q) (F ( ) F ( )) + 1 F ( )
1 q
F ( )+(1 q)(F ( ) F ( )) ;
1
F( )
(1 F ( ))
1 q
+(1 q)(
)
<
and that
(1
1 q
(1 q)(
<
)+1
q) (F ( )
1 q
(1 q)(F ( ) F ( ))+(1 F ( )) ,
F ( )) + (1
(1
because
F ( )) < (1
q) (
q +1
(1
q) [
F ( )] <
(1
q) [
F ( )] < q [(1
F( )
(1 F ( ))
)
<
)
q
1
)+1
q
1+F ( )
(1
F ( ) + qF ( )
F ( ))]
:
Thus, for Equation 9 to hold requires that
1
1
(1 q) + (1
)q
1
+
q + (1
) (1 q) (1
+
(1
1
q) + (1
) (1
q) q (1
To simplify Equation 10, we work on the terms q
)
< 1
1 q
) q q + (1 q)
1 q
) + (1 q) (1
)
1
q + (1
+ (1
1
q) (
q
q) (
q
)+1
1 q
+(1 q)
and
q(1
1 q
)+(1 q)(1
< 1:
),
using the expressions derived
in the proof of Lemma 1:
=
=
=
1
G
(1
G
NG
G)
G ) + (1
NG
(1
)(1 q)
q+(1
)(1 q) N G
(1
)(1 q)
q
G ) + q+(1
q+(1
)(1 q) (1
)(1 q) N G
(1
q (1
) (1
)
+
(1
G
q) N G
) (1 q)
8
;
NG
(10)
and
NG
=
Substituting for
and
NG
1
=
NG
N G)
G ) + (1
NG
(1
) q NG
:
q) (1
) q NG
G ) + (1
(1
(1
and simplifying yields
1 q
q + (1 q)
=
1
)q
(1
NG
1
G )+(1
q(1
)(1 q)
NG
1
+
;
1
G )+(1
(1 q)(1
)q
(11)
NG
and
q (1
1 q
) + (1 q) (1
1
=
)
q (1
G)
(1 q)(1
1
G )+(1
)q
NG
1
+
q(1
1
G )+(1
:
)(1 q)
(12)
NG
Substituting in Equations 11 and 12 into Equation 10 yields
h
1
(1 q)+(1
)q
1
)q
(1
NG
1
q)(1
(1
G )+(1
)q N G
1
+
1
+
q+(1
q (1
1
q(1
)(1 q)
1
G )+(1
)(1
G)
<
1
(1 q)(1
G )+(1
)q
NG
+
1
q(1
G )+(1
)(1 q)
NG
i
< 1:
q) N G
Multiplying out the second term on the left hand side, this requires that
1
q(1
G)
1
(1 q)+(1
)q (1
1
)q
NG
(1
(1
q) (1
1
)q
NG
(1
+ (1
)q
NG
> [ (1
q) (1
)q
G)
) q (1
G)
(1
(1
>
+1 >
NG
9
1
q+(1
)(1 q)
: However, breaking this into two separate in-
equalities, we see that the inequality cannot hold. Speci…cally,
1
q) + (1
+
(1
q) (1
1
G ) + (1
q) + (1
) q] (1
q) + (1
)q
)q
)q
NG
NG
and
1
q + (1
q (1
1
) (1
G)
q)
q (1
+ (1
) (1
q)
(1
) (1
q (1
q)
G)
1+
1
)
+
(1
G
) (1
> [ q + (1
) (1
q)] q (1
>
) (1
q) ;
>
G)
NG
NG
q (1
q + (1
where the last line of each of these inequalities holds because
(1
) q and 1 > q + (1
) (1
q)
NG
G)
2 (0; 1) and q 2 (0; 1) ; so 1 >
(1
q). Thus we have reached a contradiction.
Proof of Lemmas 4 and 5 The proof of these lemmas is based on the fact that for any cutpoints
1
q) +
1
and
an executive who is either passive or aggressive has a strict incentive to retain or remove the investigator,
based solely on her decision to try or drop the case in the …rst period.
There are four cases to consider: (i)
1
1
> ; (iv)
>
and
1
1
<
and
1
<
; (ii)
1
>
and
1
>
; (iii)
1
<
and
< : We show below that in each of these four cases, after the …rst period policy
outcome is revealed, executive beliefs about the probability that the incumbent investigator is passive can be
ordered as follows:
p (D)
is a weighted average of
and
p (D)
show that
>
a
p
<
>
p
>
p (A);
p
p
(C)
p
(C) ; and
(A). Because the …rst period outcome must be A; C or D,
p
(D). Thus it is su¢ cient to prove that
p
>
p
(C)
p
p
(A)
follows. Similarly for beliefs about the probability that the investigator is aggressive, we
a
(C)
a
(A) so that
a (D)
<
a
<
a
(C)
a
(A).
For a passive executive, a passive investigator produces the highest expected utility and an aggressive
investigator produces the lowest expected utility in the second period. If D is the …rst period outcome then
the probability of the best type is greater than the prior and the probability of the worst type is lower than
the prior. Thus it is strictly optimal to retain, setting
outcome then
C
p
= 0. Likewise
>
p
A
(C) and
a
<
a
D
= 1. On the ‡ip side, if C is the …rst period
(C) ; so it is strictly optimal to remove the investigator, setting
= 0 is optimal. Because our analysis allows for any
10
1
and
1
, except for the case
of
1
1
= 0 and
= 1; which we ruled out in Lemma 3(v), we thus establish a unique equilibrium for the
case of a passive executive. A similar argument establishes that for an aggressive executive there is a unique
1
equilibrium because for any
1
and
it is optimal to set
= 0, and
D
=
C
A
= 1:
We now give the details of the executive’s beliefs in cases (i)-(iv).
For case (i),
p
p
(A) =
[F (
[1 F (
[F (
(C) =
) F(
1
1
)] Pr(s=G) Pr(T =Ajs=G)
1 )] Pr(s=N G) Pr(T =Ajs=N G)
F(
)] Pr(s=G) Pr(T =Cjs=G)
1 )] Pr(s=G) Pr(T =Cjs=G)+[1
[1 F (
) F(
1 )] Pr(s=G) Pr(T =Ajs=G)+[1
F(
1 )] Pr(s=N G) Pr(T =Cjs=N G)
and
: We show that
p
(A) <
p
(C), by
multiplying out these two expressions, and cancelling terms to get
Pr(T = Cjs = N G) Pr(T = Ajs = G) < Pr(T = Ajs = N G) Pr(T = Cjs = G):
(13)
Expanding out Equation 13, we need
NG
(1
G
NG
G
NG
G)
+ 1
+ 1
G
(1
G)
NG
+ 1
(1
G
1
NG
NG
NG
<
G
N G)
(1
G
N G)
(1
NG
<
NG
+ 1
G
0 <
1
NG
+ 1
G
NG
G)
+ 1
G
1
The …rst term in brackets is strictly greater than zero because
is strictly greater than zero because
To show
p
(C) =
p
[1
(C) <
F(
p
NG
< 1=2 and
G
= F ( ) note that in case (i),
1 )] Pr(s
G
1
G
NG
NG
N G)
NG
N G)
G
G
(1
G
(1
>
(1
G)
G
[(1
NG
and the second term in brackets
N G ) (1
G)
N G G] :
< 1=2:
1
<
so the second term in the denominator of
F ( ) F 1 Pr(s = G) Pr(T = Cjs = G)
= G) Pr(T = Cjs = G) + 1 F 1 Pr(s = N G) Pr(T = Cjs = N G)
11
(14)
is strictly greater than zero, and it’s su¢ cient to show that:
1
F( ) F
[1 F (
Pr(s = G) Pr(T = Cjs = G)
= G) Pr(T = Cjs = G)
F( )
1 )] Pr(s
F
1
F( )
F ( )F
1
F
F( )
1
F ( )F
1
:
(15)
Now we turn to beliefs about the probability that the investigator is aggressive in case (i). Here
[1 F (
a (A)
[1 F ( )][Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]
1 ) F ( 1 )] Pr(s=G) Pr(T =Cjs=G)
1 )][Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]+[F (
=
[1 F (
[1 F ( )][Pr(s=G) Pr(T =Ajs=G)+Pr(s=N G) Pr(T =Ajs=N G)]
1 ) F ( 1 )] Pr(s=G) Pr(T =Ajs=G) :
For
a
<
a (C);
above expression for
[1
F(
1)
a (C);
)
F
1
Straightfor-
a (A):
Pr(s = N G) Pr(T = Cjs = N G) to the denominator of the
cancel terms and note that
a (C)
>
[1 F ( )][Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]
1 ) F ( 1 ) [Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]
]
][Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]+[F (
1 F( )
1 F ( 1)
1
F ( ):
1
For case (ii), because
p
1
we add F (
a (C)
=
and
1 )][Pr(s=G) Pr(T =Ajs=G)+Pr(s=N G) Pr(T =Ajs=N G)]+[F (
ward though tedious algebra shows that
a (C)
(C)
p
>
no passive type ever tries so
p
(C) =
p
(A) = 0 and thus
p (D)
>
p
>
(A) :
In case (ii),
a
(A) =
[1
[1 F (
F(
1
)] Pr(s=N G) Pr(T =Ajs=N G)+[1
1 )] Pr(s=N G) Pr(T =Ajs=N G)+[1
F ( )] Pr(s=G) Pr(T =Ajs=G)
F ( )] Pr(s=G) Pr(T =Ajs=G)+[F ( ) F ( 1 )] Pr(s=G) Pr(T =Ajs=G)
and
a
(C) =
show that
[1
[1 F (
a
F(
1
)] Pr(s=N G) Pr(T =Cjs=N G)+[1
1 )] Pr(s=N G) Pr(T =Cjs=N G)+[1
(C)
a
F ( )] Pr(s=G) Pr(T =Cjs=G)
:
F ( )] Pr(s=G) Pr(T =Cjs=G)+[F ( ) F ( 1 )] Pr(s=G) Pr(T =Cjs=G)
(A) ; we multiply out and cancel several terms to get Pr(T = Ajs = G) 1
Pr(s = N G) Pr(T = Cjs = N G)+ Pr(T = Ajs = G) [1
Pr(T = Cjs = G) 1
F
1
Pr(s = N G) Pr(T = Ajs = N G)+ Pr(T = Cjs = G) [1
Pr(T = Ajs = N G); a condition that we already checked above as Equation 13.
a
<
a
(C) we need
12
1
F ( )] Pr(s = G) Pr(T = Cjs = G)
Pr(T = Ajs = G), which reduces to Pr(T = Ajs = G) Pr(T = Cjs = N G)
For
F
To
F ( )] Pr(s = G)
Pr(T = Cjs = G)
=
1 F( )<
1
F
[1
F(
1
)] Pr(s=N G) Pr(T =Cjs=N G)+[1
1 )] Pr(s=N G) Pr(T =Cjs=N G)+[1
[1 F (
F ( )] Pr(s=G) Pr(T =Cjs=G)
:
F ( )] Pr(s=G) Pr(T =Cjs=G)+[F ( ) F ( 1 )] Pr(s=G) Pr(T =Cjs=G)
Adding
Pr(s = G) Pr(T = Cjs = G) to the denominator decreases the right hand side, so it is su¢ cient to
show that
1
F ( )] 1
[1
1
F
1
1
1
F
1
F( )
F
1
Pr(s = N G) Pr(T = Cjs = N G) + [1 F ( )] Pr(s = G) Pr(T = Cjs = G)
F 1 Pr(s = N G) Pr(T = Cjs = N G) + Pr(s = G) Pr(T = Cjs = G)
:
(16)
1
This inequality holds because F ( ) 2 (0; 1) and F
For case (iii), the argument for
p (D)
>
p
>
2 [0; 1].
p
(C)
p
1
only di¤erence is that we need to allow for the possibility that
(A) is almost identical to case (i). The
= 1, in which case the second term in the
denominator of Equation 14 is zero. So we need the inequality in Equation 15 to hold strictly, but this is
1
guaranteed because when
The argument for
a (D)
= 1 Lemma 3(v) tells us that
<
a
<
a
(C)
a
1
that we need to allow for the possibility that
1
1
1
< 1 and thus F
= 0, in which case F
1
Pr(s = G) Pr(T = Cjs = G) = 0:
a (D)
<
a
<
a
(C)
1
= 0 Lemma 3(v) tells us
2 (0; 1) :
For case (iv), the argument for
for
> 0:
(A) is almost identical to case (ii). The only di¤erence is
So we need Equation 16 to hold strictly, but this is guaranteed because when
that
1
> 0 and hence F
a
p (D)
>
p
>
p
(C)
p
(A) is identical to case (ii). The argument
(A) is identical to case (i).
Proof of Lemma 6 We …rst establish existence of the cutpoints. We do this for
for
C
and
A
are essentially similar. For
D
D
: The arguments
, note that the di¤erence in the executive’s expected utility
di¤erence from retaining versus removing the investigator is a linear, and hence monotonic, function of
Speci…cally, the utility di¤erence is
p (D) E
a (D) [ E
n (D) [ E
G
+ (1
(q
E )(1
G
+ (1
)
q)) + (1
E )(1
N G]
13
) (1
q)
N G]
E.
p E
n
[
E
(q
+ (1
G
q)) + (1
E )(1
) (1
q)
N G]
a
[
E
+ (1
G
E )(1
)
N G]
;
which equals
E
p (D)
p
+(1
E )(1
+[
) f[
n (D)] (q G
n
n (D)] (1
n
q)
+ (1
NG
q)) + [
+[
the investigator when she drops the …rst period case and an executive with
with
E
<
D
a (D)
there exists a cutpoint
prefers to retain whereas an executive with
D
To show that
and
E
1
is a continuous function of
1
are functions of
p (D)
=
and
1
E
1
and
D
>
=
D
=
E
E
=
strictly prefers to retain
strictly prefers to remove
2 ( ; ) such that an executive
prefers to remove the investigator.
; we …rst note that voter beliefs
p (D);
n (D);
:
min F ( ); F ( 1 ) + Pr(s = N G) F ( )
F ( 1 ) + Pr(s = N G) F 1
a (D)
(17)
a (D)] N G g
a
Also, as established in the proof of Propositions 4 and 5 an executive with
her. Thus because Equation 17 is linear in
a (D)] G
a
min F
Pr(s = N G) F 1
1
F ( ) + Pr(s = N G) F 1
min F ( ); F (
F ( 1)
1
;F( )
F ( 1)
1
)
;
;
and
n (D)
Next, we explicitly solve for
D
D
=
D
)f[
n
n (D)](1
q)
N G +[ a
is a continuous function of
p (D)
a (D):
by setting Equation 17 equal to zero, yielding:
(1
(1
=1
)f[
n
n (D)](1
a (D)] N G g
p (D);
n (D);
q)
f[
and
N G +[ a
p (D)
p
a (D)
]+[
a (D)] N G g
n
n (D)](q G +(1
q))+[
a
a (D)] G
so it is a continuous function of
1
g
: Note that
and
1
:
We now order the cutpoints relative to each other. First we note that it’s impossible to have both
C
<
D
and
A
<
D
: If this were the case then any executive type with
E
2 (max
C
;
A
;
D
)
would strictly prefer to retain the incumbent investigator after all possible …rst period outcomes. This is a
contradiction because the replacement is drawn from the same pool as the incumbent. A similar contradiction
14
C
results if
D
>
A
and
D
>
.
The …nal part of the argument is to show that
A
: To prove that
C
A
C
A
, which enables us to conclude that
C
D
, we show that if an executive’s expected utility from retaining the investigator
after an acquittal is greater than his utility from retaining after a conviction then his utility from retaining
after a conviction is greater than his utility from a new randomly drawn investigator. We denote these utilities
as U (oldjC), U (oldjA) ; and U (rndm) : We also will use U ( > x) to denote an executive’s expected utility
from a investigator randomly drawn from the portion of the investigator type distribution F that is greater
than x: Similarly U ( 2 (x; y)) denotes expected utility from a investigator drawn from the distribution F
restricted to the interval (x; y) :
First note that if
1
>
then, as shown in the proof of Lemmas 4 and 5,
passive investigators never choose x = T when
always have U (oldjC)
Step 1. We …rst show that Pr
>
1
jC ,
[1
[1 F (
> ;
p (C)
=
p (A)
(C)
a
(A) and because
= 0; so for a neutral executive we
U (oldjA) :
1
The argument is more complicated when
Pr
1
a
F(
>
1
1
jA
. We proceed in four steps.
> Pr
>
1
jC
> 1
F
1
: For Pr
>
1
jA
)][Pr(s=G) Pr(T =Ajs=G)+Pr(s=N G) Pr(T =Ajs=N G)]
1 )][Pr(s=G) Pr(T =Ajs=G)+Pr(s=N G) Pr(T =Ajs=N G)]+[F (
1)
F(
1 )] Pr(s=G) Pr(T =Ajs=G)
>
must
be strictly greater than
[1
[1 F (
F(
1
)][Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]
1 )][Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]+[F (
1)
F(
1 )] Pr(s=G) Pr(T =Cjs=G)
. After multiplying out
and cancelling, this reduces to Pr(T = Cjs = G) Pr(T = Ajs = N G) > Pr(T = Ajs = G) Pr(T = Cjs = N G),
which we already checked as Equation 13.
For Pr
1
>
[1
[1 F (
jC > 1
F(
1
F
1
, we need
)][Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]
1 )][Pr(s=G) Pr(T =Cjs=G)+Pr(s=N G) Pr(T =Cjs=N G)]+[F (
1)
F(
1 )] Pr(s=G) Pr(T =Cjs=G)
plying out and canceling, this reduces to
1
F
F
[Pr(s = G) Pr(T = Cjs = G) + Pr(s = N G) Pr(T = Cjs = N G)] >
1
F
1
Pr(s = G) Pr(T = Cjs = G); i.e.,
15
>1
F
1
: Multi-
1
F
Pr(s = N G) Pr(T = Cjs = N G) >
1
F
Pr(s = G) Pr(T = Cjs = G):
Step 2. We show that if U (oldjA) > U (oldjC) then U
Pr
+ Pr
2
Because Pr
1
Pr
Pr
1
>
1
>
1
<
1
jA U
1
jA U
>
1;
jA = Pr
jC for Pr
jA
U
+U
>
2
Pr
U
>
>
1
1
;
Pr
+ Pr
2
U
>
1
1;
1
>U
1
>
1;
Pr
1
2
jC U
1
jC U
1
>
1
;
:
1
>
2
1;
jA for Pr
1
1
2
1
;
1
1
;
1
>
1;
2
Pr
>
jC
>
1
jC
U
+U
>
2
1
U
1;
1
1
2
2
1
;
1
For the …rst case, if
1
1
>
1
>
:
Step 3. We show that U
<
;
> 0:
1
From Step 1 we know that the …rst term in brackets is strictly greater than zero, so U
U
jA and
jC ; to get
Pr
1
U (oldjC)
1
2
1
jA
U (oldjA) >
jC = 0, we substitute 1
1
U
1;
1
1
<
2
>
2
>
1
2
1
2
;
1
;
1
>U
1
<
: There are two cases:
prefers neutral over passive investigators so U
2 we know that U
>
1
1
>
2
1
;
1
1
2
implies that if
>U
<
and
1
> :
then the investigator is either a neutral type or a passive type, and if
1
the investigator is a passive type with probability 1, because
For the second case,
1
>
1
1
;
>U
<
. A neutral executive strictly
1
:
then the investigator is surely aggressive. From Step
: Note that the region
1
1
;
includes some investigators
who are passive, some who are neutral, and some who are aggressive. Also, a neutral executive most
prefers a neutral investigator so the only way that U
is the executive’s least preferred type. Because
U
2
1
;
1
>U
<
1
>
<
:
16
1
1
>U
2
1
;
1
is if a passive investigator
implies that the investigator is passive for sure,
Step 4. We show that U (oldjC) > U (rndm), i.e.,
Pr
+ Pr
2
>
1;
1
jC U
1
jC U
1
1
>
2
1;
>
1
1
+ F
2
1
;
1
> U
1
2
1
;
1
jC U
Pr
2
1
;
1
jC = 1 because no investigator with
1
;
> 1
1
>
U
U
1
F
U
I
>
1
<
1
+F
1
<
U
1
1
2
so the inequality will hold if Pr
Pr
2
1
<
U
1
F
+F
From Step 3, U
1
F
2
1
1
;
:
1
>
1
;
jC U
1
>
1
+
: Note that Pr
will bring a case to trial. Substituting in
and collecting terms, we need
Pr
>
1
jC
1
F
1
U
>
1
U
2
1
;
1
> 0:
Step 1 and Step 2 establish that each term is strictly greater than zero.
Proof of Lemma 7 Lemmas 4 and 5 characterize equilibrium behavior for passive and aggressive executives. Here we handle the case of neutral executives.
For the type of equilibrium in Lemma 7(i), set
cutpoint
C
D
= 1;
=
C
A
= 0 and from Lemma 6 …nd the
that arises from the resulting …rst investigator behavior. The executive behavior in part (i) is
optimal for any
E
C
:
For the type of equilibrium in Lemma 7(ii), set
C
Lemma 6 to …nd the cutpoint
optimal to play
D
= 1 and
A
D
= 1;
A
= 0; and for each value
C
implied by the resulting …rst investigator behavior. For
2 (0; 1) apply
E
=
C
it is
= 0 and because this executive type is indi¤erent after observing a conviction
in the …rst period, he can mix using the particular
C
C
2 (0; 1) that was used to generate
:
The construction of equilibria for parts (iii)-(vii) of the lemma is similar.
Note that no other type of equilibrium can exist for any
strategy) executive strategy
. Given
E
2 [ ; ]. Consider any (possibly mixed
, Lemma 3 implies that there exist cutpoints
17
1
and
1
for …rst
>
1
jC +
period investigator behavior. Given these cutpoints, Lemma 6 characterizes cutpoints for executive behavior.
It is straightforward to check that the only executive strategies
that are compatible with these cutpoints
are the 7 types listed in Lemma 7.
Finally, we establish existence. To do this, we construct a function
(z) : [0; 7] ! [ ; ]. Each possible
executive strategy , whether a pure strategy or a mixed strategy, in parts (i)-(vii) of Lemma 7 is speci…ed
by some value of z, and we use the intermediate value theorem to show that for any
z such that
(z) =
E,
E
2 [ ; ] there is some
and thus there is an equilibrium with one of these 7 types of executive behavior.
For any executive strategy
=(
D;
C;
A) ;
1
let
1
( )=
1
( );
( ) represent the cutpoints for op-
timal …rst period investigator behavior from Lemma 3, given that the executive’s strategy is . Given any cutpoints for …rst period investigator behavior,
1
1
and
, let
CDA
be the cutpoints for executive behavior from Lemma 6. Let
;
1
=
C
(1)
=
C
1
1
C
1
(1; 1; 0) ;
1
(1; 1; 0) ;
(3)
=
D
1
(1; 1; 0) ;
1
(1; 1; 0) ;
A
1
(0; 1; 0) ;
1
(0; 1; 0) ;
(6)
=
A
1
(0; 1; 1) ;
1
(0; 1; 1) : De…ne
(z) =
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
+z
2[ ; ]:
2[ ; ]:
D
2[ ; ]:
D
= 1;
C
(2)
+ (z
=1
D
(z
= 0;
C
(6)
A
1
A
1
1
;
1
(1; 0; 0) ;
1
(0; 1; 0) ;
for z 2 [1; 2]
(2)
for z 2 [2; 3]
= 0 is an equilibrium
for z 2 [3; 4] :
(5)
(4)
for z 2 [4; 5]
=z
5 is an equilibrium
= 1;
4)
D
;
(0; 1; 0) ;
= 0 is an equilibrium
(3)
C
= 1;
+ (z
D
1
for z 2 [0; 1]
1;
2)
3) ;
+ (z
=
;
(1; 0; 0) ;
(1)
=z
(4)
(4)
1
6)(
18
A
(6) )
for z 2 [5; 6]
for z 2 [6; 7]
;
A
1
(2)
=
(5)
=
;
1
And let
~ (z) =
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
(1; 0; 0)
(1; z
for z 2 [0; 1]
1; 0)
for z 2 [1; 2]
(1; 1; 0)
for z 2 [2; 3]
(1 (z 3) ; 1; 0)
>
>
>
>
>
>
(0; 1; 0)
>
>
>
>
>
>
>
>
(0; 1; z 5)
>
>
>
>
>
:
(0; 1; 1)
for z 2 [3; 4] :
for z 2 [4; 5]
for z 2 [5; 6]
for z 2 [6; 7]
Note that ~ (z) is a continuous function of z: Thus, by part (iv) of Lemma 3, the investigator cutpoints given
1
by
(~ (z)) are continuous in z, which in turn implies, by part 2 of Lemma 6, that cutpoints for executive
CDA
behavior
1
(~ (z)) are a continuous function of z: In particular, we care that
a continuous function of z for z 2 [1; 2];
A
1
D
1
1
(~ (z)) is
(~ (z)) is a continuous function of z for z 2 [3; 4]; and
(~ (z)) is a continuous function of z for z 2 [5; 6]:
Thus by construction,
(z) : [0; 7] ! [ ; ] is a continuous function where
the intermediate value theorem implies that for each
that
C
E
=
(z) : By construction of
E
(0) =
and
(7) =
so
2 [ ; ] there exists at least one z 2 [0; 7] such
(z) this implies that there exists an equilibrium. If z 2 [0; 1] [ [2; 3] [
[4; 5] [ [6; 7] this equilibrium is a pure strategy equilibrium from part (i), (iii), (v), or (vii) of Lemma 7 and
if z 2 [1; 2] [ [3; 4] [ [5; 6] it is a mixed strategy equilibrium from part (ii), (iv), or (vi) of Lemma 7.
Proof of Lemma 8 The proof of this lemma is straightforward. Here we state the argument for part
(i), when the executive is either passive or passive-neutral. The arguments for other types of executives are
essentially identical.
For
1
> ; note that in the equilibrium that we characterize for passive and passive-neutral executives,
r (G) < 0, i.e., when s = G the investigator is strictly more likely to be retained if she drops than is she tries.
In terms of second period policy, any investigator is better o¤, in expectation, when retained, because there
19
is a strictly positive probability that her replacement will choose a di¤erent action than the one she would
have chosen. In terms of …rst period outcomes, an investigator with
weakly prefers to drop when
I
s = G: Thus, because r (G) < 0, when considering both …rst and second period outcomes any investigator
with
strictly prefers to drop, and hence
I
For
1
1
> :
> ; note that in terms of …rst period outcomes any investigator with
weakly prefers to
I
drop when s = N G: Thus, because r(N G) < 0, when both …rst and second period outcomes are taken into
account any investigator with
1
must strictly prefer to drop when s = N G; so
I
ER
Proof of Lemma 9 First we solve for
> :
, the cutpoint between executives who prefer passive versus
random replacement investigators. Let U (pass) denote utility from a passive replacement and U (rndm)
denote utility from a random replacement, where U (pass) =
U (rndm) =
p E
n
[
E
(q
G
+ (1
q)) + (1
and
E
E )(1
) ((1
q)
N G )]
a
[
E
G
+ (1
E )(1
Combining terms, we get
U (pass)
U (rndm)
=
(1
)
Note that this is strictly decreasing in
ER
To see that
with
E
ER
=
(1
)
NG
[
n
E
(1
[
1
E
E
NG
(1
n
(1
p
)
q) +
n
NG
[
(q
n
G
(1
a]
+ (1
q) +
q))
a G
a] :
so
(1
) N G [ n (1 q) + a ]
q) + a ] + 1
p
n (q G + (1
q))
:
a G
> , note that a passive investigator will act optimally from the perspective of an executive
= , so an executive at
We now prove that
ER
<
strictly prefers a passive investigator over a random replacement.
E:
An executive at
E
is indi¤erent between a random draw and an
investigator who dropped when behaving according to …rst period cutpoints
20
1
=
and
1
2 ( ; 1), i.e.,
)
N G] :
he’s at
E
=
D
from Lemma 6 given these cutpoints. The executive at
ER
is indi¤erent between a
random draw and a passive investigator, which also means that he’s indi¤erent between a random draw and
an investigator who dropped when playing according to …rst period cutpoints
1
=
and
1
= 1. Because
aggressive investigators are the worst type from this executive’s perspective, he must strictly prefer to retain
an investigator, rather than replace her with a random replacement, if she drops when behaving according to
1
…rst period cutpoints
=
and
1
ER
2 ( ; 1) : Thus by Lemma 6,
is strictly less than the
D
generated
by these cutpoints for investigator behavior.
To solve for
U (agg) =
E
ER
; we take a similar approach, setting the utility from an aggressive replacement, i.e.,
(1
G
ER
E ) (1
=
)
N G,
(1
(1
)
Arguments similar to the ones for
N G [1
ER
equal to U (rndm). Solving out, we get
)
(1
n
NG
[1
q)
n (1
]
+
(1
a
q)
establish that
E
<
ER
Proof of Lemma 10 First, we prove part 1 of the lemma, for
D
2 (0; 1), and
C
=
A
a]
G)
p
+
n
(1
q)
:
< :
E
2
;
ER
. Suppose the executive plays
= 0, which means that r (G) = r (N G) < 0: We need to show that it is optimal
for the investigator to behave according to cutpoints
If s = G then any investigator with
I
<
1
=
when s = G and
1
2 ( ; 1) when s = N G:
strictly prefers to drop. In terms of …rst-period policy, she
is better o¤ dropping than trying. And accountability incentives have no e¤ect on a passive investigator
because the replacement chosen by the executive will be passive.
If s = G, then a neutral investigator with
I
2 ( ; ) strictly prefers to try. In terms of …rst-period
policy she is better o¤ trying. In the second period, the signal will be either s = G or s = N G: If the
investigator tries and loses o¢ ce by doing so and the second period signal is s = G then she is no worse o¤
as a result of having tried. On the other hand if the second period signal is s = N G, the dogmatic passive
replacement will do exactly what the neutral investigator would do in the second period, so she winds up
21
being strictly better o¤ as a result of trying in the …rst period.
If s = G, then an investigator with
better o¤ trying. And because
U (T ryjs = N G)
I
strictly prefers to try. In terms of …rst-period policy she is
I
> ~ , for an investigator at
I;
U (T ryjs = G)
U (Dropjs = G) >
U (Dropjs = N G): The worst-case scenario if the investigator tries is that by trying in
the …rst period she loses o¢ ce and the second period signal is s = G but her dogmatic passive replacement
drops the case. However, it’s also possible that s = N G in the second period, in which case she would have
been strictly better o¤ trying in the …rst period.
We now turn to the case of s = N G: If s = N G then an investigator with
I
<
strictly prefers to
drop. Doing so makes her strictly better o¤ in terms of …rst period utility and at least weakly better o¤ in
terms of second-period utility.
For an investigator with
^T D (
U
I ; N G; r(N G)),
I
the investigator’s utility di¤erence from trying versus dropping is
where we put a hat over the U because with a passive replacement the utility di¤erence
is not the same as it was for a random replacement in the proof of Lemma 3. The …rst period utility di¤erence
is the same, based the reasoning in Lemma 1. In the second period, the passive replacement will always
drop, whereas the incumbent investigator with
will always try if retained. Thus, being replaced
I
avoids some mistaken convictions but also results in some failures to convict, i.e., for
^T D (
U
I ; N G; r(N G))
=
I
NG
(1
G)
r (N G) (1
+r (N G)
=
I
+
I
(1
G)
+
I NG
1
22
I)
I ) (1
(1
)
1
NG
NG
NG
G)
NG
1
NG
(1
I
+ r (N G) (1
NG
NG
)
+ r (N G)
+ r (N G) (1
) :
(18)
Note that Equation 18 is linear in
s = G, i.e., U^T D (
I ; N G; r(N G))
I:
Also for r (N G) < 0, an investigator at
strictly prefers to drop when
< 0, because she is indi¤erent in terms of …rst period actions and strictly
prefers to be retained rather than to have a dogmatic passive investigator choose second period actions.
^T D (1; N G; r(N G))
There are two possibilities. First, it may be the case that U
0; so
1
= 1 and all
1
investigator types strictly prefer to drop when s = N G: Second, it may be the case that for some
^T D (
U
1
; N G; r(N G)) = 0, in which case all investigator types with
with
I
>
1
I
1
<
2 ( ; 1) ;
strictly prefer to drop and those
strictly prefer to try when s = N G: Solving out for this case, we get
1
=
NG
1
+ r (N G) (1
)
N G + r (N G) ] +
N G ) + r (N G) (1
)
[
[(1
G
NG
NG
(1
1
Note that as r (N G) ! 0,
!
(using the expression for
)]
:
1
in the proof of Lemma 1), and
is a
continuous function of r(N G), and hence of : Also, from Equation 18 we can solve for the largest value of
r (N G) such that an investigator with
I
= 1 will drop in the …rst period
^T D (1; N G; r(N G))
U
(1
NG
G)
+ r (N G)
+
NG
1
NG
+ r (N G) (1
)
0
NG
1
NG
+ r (N G) (1
NG
r (N G)
Let
1 q
q) + (1
(1
C
= 0:
Having characterized the investigator’s best response, we now solve for the executive type
E
2
D
such that r (N G) solves this expression with equality when
)q
=
D
be the value of
=
A
who is indi¤erent between retaining and replacing the investigator when x = D , given cutpoints
and
1
)
:
;
ER
1
=
2 [ ; 1] for …rst period executive behavior.
1
For
= , any investigator who drops a case is either passive or neutral, so an executive at
E
=
is
indi¤erent between retaining and replacing.
For
ability
1
p
= 1, because
1
=
p Pr(s=G)
Pr(s=G)+Pr(s=N G) ),
any investigator who drops is either a passive type who saw s = G (with probor a random draw who saw s = N G (with probability
23
p
Pr(s=N G)
Pr(s=G)+Pr(s=N G) ).
ER
By the de…nition of
1
For
ER
; an executive at
is indi¤erent between this lottery and a passive replacement.
2 ( ; 1), as in the proof of Lemma 6, the executive’s expected utility di¤erence from retaining is
p (D) E
n (D) [ E
a (D) [ E
G
+ (1
(q
G
+ (1
E )(1
)
q)) + (1
N G] :
E )(1
) (1
q)
N G]
His expected utility from a passive replacement is
: Thus his
E
expected utility di¤erence between retaining and removing is
1
E
p (D)
n (D)(q G
+ (1
Note that this is strictly increasing in
E
=
(1
)[
So we have shown that, holding
denote as
E
(
D) ;
1
We have also shown that
E
of
D
and
E
1
A
E )(1
)[
n (D) (1
q)
NG
+
a (D) N G ] :
Setting it equal to zero yields the executive who is indi¤erent:
)[
NG +
=
(1
C
n (D) (1
q)
a (D) N G ] +
= 0; for any
NG
+
1
D
2 [0;
a (D) N G ]
p (D)
D]
n (D)(q G
+ (1
q))
:
(19)
there is an executive type, which we
who is indi¤erent between retaining and replacing the investigator after she drops, given
and
the cutpoints,
E.
(1
q)
n (D) (1
q))
=
2 [ ; 1] ; for …rst period investigator behavior that is a best response given
and
(0) =
E
is a continuous function of
(
1
D)
ER
=
: Moreover, because
, the composition
intermediate value theorem implies that for each
E
2
;
ER
E
(
D)
1
D.
is a continuous function
is continuous as well. Thus the
there exists some
D
2 (0;
D)
such that
there exists an equilibrium as stated in part 1 of Lemma 10.17
A similar argument proves part 2 of the lemma. The only complexity is that whereas for
needed to vary
D
we now need to consider both
A
and
C.
ER
we only
What makes this fairly straightforward is the
fact that, in contrast to the case of a random replacement, it is possible for an executive with a given
E
to mix both after convictions and after acquittals. The reason for this is that after either a conviction or
an acquittal, because
17
1
<
and
1
=
the executive’s beliefs can be written as a convex combination of
Note that the intermediate value theorem only ensures that for some D 2 [0; 1] there exists an equilibrium:
The strict set inclusion comes from the fact that D = 0 cannot be an equilibrium for E > and D = 1
cannot be an equilibrium for E < ER :
24
(i) a belief that the executive is aggressive and saw s = N G and (ii) a belief that the executive is randomly
drawn from
1
; 1 and saw s = G: The only di¤erence is that after convictions and acquittals the executive
will put di¤erent weights on these two beliefs.
Thus, if after observing a conviction the executive is indi¤erent between retaining the investigator and
replacing her with an investigator who is surely aggressive, he must be indi¤erent between an aggressive
1
type and a random draw from
; 1 : But this in turn means that after an acquittal he must likewise be
indi¤erent about whether to retain the investigator or replace her with an aggressive type. By the same
argument, if the executive is indi¤erent after an acquittal he must be indi¤erent after a conviction.
Thus we can have
C
and
A
both strictly between zero and 1. For any
continuum of equilibria, using di¤erent mixing probabilities
1
to the same investigator behavior, as characterized by
equilibrium with
A
=
A
and
2 (0; 1) and
1
ER
E
2
;
there exists a
C
2 (0; 1), all of which lead
. For simplicity, in the paper we state the
C:
Finally, we sketch technical details supporting footnote 16 in the main text, which mentions the counterintuitive fact that an executive who prefers a dogmatic replacement over a random draw may nonetheless
prefer to pick a random draw over a dogmatist in the …rst period. There are two e¤ects to consider: …rst
period policy considerations and selection e¤ects for the second period. The executive types who prefer to
do this are PN1 types with
case of PN1 types close to
E
close to
ER
and AN2 types with
E
close to
ER .
Here we consider the
ER :
Note that given the equilibrium behavior in part 2 of Lemma 10, any executive with
E
2
;
ER
is
indi¤erent, in terms of selection e¤ects for the second period, between appointing a …rst-period dogmatist
and a …rst-period random draw.
First-period policy considerations are more subtle. In the absence of accountability incentives, a PN1
type is strictly better o¤ having policies chosen by a dogmatist rather than a random draw. However, the
magnitude of this preference is arbitrarily small as
E
!
25
ER :
On the ‡ip side, the accountability incentive
stemming from the executive’s behavior in part 2 of Lemma 10 increases congruence by a …rst-period random
draw. This increases the executive’s utility from appointing a random-draw in the …rst period.
The next step is to show that this utility increase is bounded away from zero. For this step, we need
to show that the equilibrium
D
in part 2 of Lemma 10 does not converge to zero as
this, suppose to the contrary that
period behavior because
by cutpoints
1
=
A
and
=
1
C
D
E
!
ER :
To see
0: Thus accountability incentives have essentially no e¤ect on …rst-
= 0; which means that …rst period investigator behavior is characterized
: This, in turn, implies that an executive with
E
close to
ER
must
strictly prefer to retain the investigator when she drops a case, which is a contradiction with the fact that
the executive mixes in this information set.
Because
from zero as
D
E
does not converge to zero, in the equilibrium in part 2 of Lemma 10,
!
ER ;
1
is bounded away
and thus the magnitude of the accountability-induced increase in the executive’s
utility from appointing a random-draw in the …rst period is bounded away from zero.
Thus, there exists some neighborhood of
ER
such that an PN1 executive within this neighborhood
would choose random draw over a dogmatist in the …rst period if (contrary to the assumptions of our model)
he were allowed to make that choice.
26