Supplementary Appendix A: a mixed-strategy equilibrium in the
2-seller case
Suppose that the disclosure decisions of each seller are limited by the extreme cases: a seller may
either reveal full information about his product or no information at all. Consider a symmetric
mixed-strategy pro…le, in which each seller discloses information with probability and does
not disclose with probability 1
. Under the non-disclosure policy, setting pn = 0 is not
optimal, and there is no a pure-strategy equilibrium price pn > 0. (Otherwise, any seller can
increase the probability of selling, and, hence, his pro…t by setting p0n = pn ", where " # 0.) By
the same logic, if F ( ) is the equilibrium distribution of prices under the non-disclosure policy,
it may not contain mass points. Also, the lower and upper bounds of the support of F ( ), s
and a, respectively, must satisfy the following conditions: s > 0 and a v e (setting s = 0 or
> a brings zero pro…ts to the seller that is not optimal).
Consider the seller’s pro…t function under the non-disclosure policy. Given that the
competitor discloses information and charges price p with probability , and does not disclose
information and chooses a random price z according to F (z), setting a price results in the
seller’s pro…t
e
+ p) + (1
) (1 F ( ))) :
(1)
n ( ) = ( G (v
Here, G (v e
+ p) = Pr fv e
v pg is the probability of selling the product if the rival
discloses information, which happens with probability . With probability 1
, the seller
competes with the “non-disclosing”rival. In this case, the chance of selling the product is given
by Pr fv e
v e zg = Pr fz
g = 1 F ( ). Because the seller randomizes over prices,
each price must bring the same payo¤:
n
Denote a = arg max G (v e
( ) = K > 0; 2 [s; a] :
+ p) . Due to Condition 1, G ( ) has an increasing hazard
rate. That is, G (v e
+ p) is strictly pseudo-concave, which implies that a is unique. Then,
the following statements hold.
Claim 1 a = a .
Proof By contradiction, if a < a , then
n
(a ) = G (v e
a + p) a > G (v e
since F (a) = F (a ) = 1. Similarly, if a > a , then 1
n
(a ) = G (v e
a + p) a + (1
a + p) a =
(a) ;
F (a ) > 0 implies
F (a )) a > G (v e
) (1
n
a + p) a =
n
(a) :
Claim 2 The support of F ( ) is an interval.
Proof By contradiction, let S be the support of F ( ) and contain a ‘hole’ ( 0 ; 00 ) 2
= S,
such that 0 > s; 00 < a, and both 0 and 00 are in S. This implies F ( 0 ) = F ( 00 ) and
0
00
G (v e
+ p) 0 < G (v e
+ p) 00 , which lead to the contradiction n ( 0 ) < n ( 00 ).
1
Then, (1) gives the expressions for the distribution
G (v e
(1
G (v e
K
F( )=1
=1
+ p)
)
a + p) a
G (v e
=1
G (v e
1
a + p) a
G (v e
(1
)
+ p)
; 2 [s; a] ;
+ p)
and the density
f( )=
G (v e
a + p) a
g (v e
2
1
+ p) ; 2 [s; a] :
Now, consider the seller’s pro…ts in the case of disclosing information. If such a seller sets
price x, his payo¤ is given by
d
(x) = ((1
(2)
) Pn (x) + Pd (x)) x.
Here,
Pd (x) = G (p) (1
G (x)) +
Z1
(1
G (y
p + x)) dG (y)
p
is the probability of selling the product under the non-disclosure policy if the rival discloses
information. Similarly,
Za
Pn (x) = 1 G (v e
+ x) dF ( )
s
is the probability of selling the product if the rival does not disclose information. This is obtained
by integrating Pr fv x v e
g = 1 G (v e
+ x) over .
e
, and
For the uniform G (x), we have v e = 21 ; a = v 2+p = 1+2p
4
K = G (v e
a + p) a = a2 ;
(3)
This gives the condition on s:
K=
n
(s) = ( G (v e
s + p) + 1
)s =
1
2
s+p +1
y+p
x) dy;
s.
(4)
Also,
Pd (x) = p (1
x) +
minf1;1
Z x+pg
(1
(5)
p
and
Pn (x) =
1
Za
1
2
x+
s
2
a2
2
1 d :
(6)
Thus, the values of and p are determined by the FOC for (2) at x = p and the seller’s
indi¤erence between the disclosure and non-disclosure policies:
d
dx
(7)
(x = p) = 0,
d
d
(p) = K.
Taking the derivative and simplifying the expressions leads to:
Za
1
2
2p +
a2
1
p2 + 2p
2
d +
2
1
= 0;
(8)
a2
1 + 2p
;a =
.
p
4
(9)
s
Za
1
2
p+
a2
1
d +
2
p2
1
2
=
s
Summing the equations results in
Za
a2
1
a2
;
p (a)2
d =1
2
s
where p (a) =
4a 1
,
2
(10)
and
s)2
(a
s
=1
a2
:
p (a)2
The right-hand side (RHS) of (10) is equal to:
HR (a) = 1
a2
=1
p (a)2
1 + 2p (a)
4p (a)
2
1
1
+
4p (a) 2
=1
2
,
1
, or, equivalently, in p for p
0. Also, HR (a) takes values
which is increasing in a for a
4
1
1
from 1 to 0 as a changes from 4 to 2 .
The left-hand side (LHS) of (10) is a function of two variables:
HL (s; a) =
Za
1
a2
2
d =
s)2
(a
s
.
s
Note that for all a > 0, HL (s; a) in increasing in s and takes values from 1 to 0 as s changes
from 0 to a. Therefore, for any a 2 41 ; 21 there is a unique continuous solution s0 (a) 2 (0; a)
to (10). Also, as a increases from 14 to 21 , s0 (a) takes all values between 0 and 12 . Finally, s0 (a)
is increasing in a. To show this, the implicit function theorem results in
ds
js=s0 (a) =
da
@H (s; a) =@a
js=s0 (a) ;
@H (s; a) =@s
3
where H (s; a) = HL (s; a)
HR (a). Then,
@H (s; a)
js=s0 (a) =
@a
1
4
Because a
ds0 (a)
da
2
a
s0 (a)
s0 (a)
0.
Thus,
Now, consider (8). It can be expressed as:
3
4
+
4a
1
a
a2
2
1)3
(4a
@H(s;a)
js=s0 (a)
@a
and s0 (a) 2 [0; a], it follows that
Za
8
.
@H(s;a)
@s
0. Also,
a2
s20 (a)
=
1
7
= 0:
8
d + 2a2 + a
0.
(11)
s0 (a)
The LHS of (11) is continuous in a; a 2 ( 41 ; 12 ]. If a =
Ra
1
1
1
0;
+ 34 4a
1
0;
2
(0;
],
and
lim
2
4
16
4
1
1
,
4
it follows that s0
a2
1
d =
2
a! 4 s0 (a)
then s0 12 = 12 . Thus, the LHS of (11) is equal to 2a2 + a
solution a0 2 14 ; 12 . In addition, we use (4) to …nd :
0
=
7
8
=
1
8
1
4
=
1. If a = 12 ,
> 0. Thus, (11) has a
s0 (a0 )
.
s0 (a0 ) (s0 (a0 ) 2a0 + 1) + a20
The numerical analysis shows that there is a solution to (11), such that a0 = 0:455 that gives
s0 = 0:225; p0 = 0:409; 0 = 0:81, and K = 0:167. To verify that the seller’s strategies with the
parameters a0 ; s0 (a0 ) ; p0 = p (a0 ), and 0 constitute an equilibrium, we need to show that there
is no a pro…table deviation to another policy-price pair. Consider …rst the case of non-disclosure
policy. The seller can deviate by setting the price 2 [0; s) or 2 (a; 12 ] (a > 12 = v e is not
optimal). If 2 [0; s), his pro…t is given by:
n
G (v e
( )=
) < s G (v e
+ p) + (1
where the inequality holds, since ( G (v e
Similarly, if 2 (a; 21 ], we have
n
( )=
G (v e
s + p) + s (1
n
(s) ;
)) is strictly increasing in ; < a.
+ p) + (1
+ p) < a G (v e
)=
a + p) =
n
(a) ,
because G (v e
+ p) is strictly decreasing in ; > a.
Now, consider the policy of full disclosure. We need to show that the FOC (7) is su¢ cient
for maximizing d (x). Using (5) and (6) results in:
P (x)
=
=
(1
Za
) Pn (x) + Pd (x)
max
1
2
x + ;0
=
a2
2
1
Pn (x) + Pd (x)
1 d + p (1
x) +
Z1
p
s
4
max f1
y+p
x; 0g dy,
or
P (x)
=
8
>
>
>
>
>
>
>
>
>
>
>
>
<
Ra
1
2
s
Ra
1
2
s
Ra
>
>
>
>
>
>
x 12
>
>
>
>
>
>
:
1
2
a2
x+
2
a2
x+
1 d + p (1
2
a2
x+
1 d + p (1
x) +
1 d + p (1
2
x) +
R1
p
1 Rx+p
p
1 Rx+p
x) +
(1
y+p
x) dy; 0
(1
y+p
x) dy; p < x
(1
y+p
x) dy; s +
p
p (1
x) +
1 Rx+p
(1
y+p
x) dy; a +
p
1
2
<x
x
1
2
p
s+
<x
1
2
a+
1.
If x 2 [0; p], P (x) is linear and decreasing in x. Thus, d (x) = xP (x) is concave in x and
(7) is su¢ cient for maximizing d (x) on [0; p]. By tedious, but routine computations, one can
show that d (x) < d (p) ; x > p, which implies that there are no pro…table deviations from the
derived strategies.
Supplementary Appendix B: no bound on N0
Consider G (v) = v and g (v) = v 1 ; > 0. Under full information revelation by all sellers,
the pro…t function of a single seller is determined by
(x; p) = P (x; p) x,
where P (x; p) is the probability of selling the product:
P (x; p) = p
(N 1)
(1
x )+
minf1;1
Z x+pg
(1
(y
p + x) ) dy
(N 1)
.
(12)
p
The FOC function is:
(x; p) = p
(N 1)
(1
x
x )+
minf1;1
R x+pg
1 (y
p + x)
x (y
1
p + x)
dy
(N 1)
p
If x
0
x
p, then
(x; p) =
p
(N 1)
x
1
Z1
( + 1)
(y
p + x)
(2 (y
p) + (1 + ) x) dy
(N 1)
< 0,
p
so that
(x; p) is strictly concave in x. If x 2 (p; 1], then
0
x
(x; p) =
p
2
(N 1)
(N
x
1)
1
( + 1)
1 Zx+p
(y
p + x) (2 (y
p
+x
2
(N
1) (1
x + p)
(N 1) 1
5
.
p) + (1 + ) x) y
(N 1) 1
dy
(13)
1
2
The …rst two components in (13) are negative, and the last one is positive. Thus, for …xed N
and p > 0, it follows that 0x (x; p) < 0; x 2 (p; 1] as ! 0, since the last two components
have the order of 2 , whereas the …rst one has the order of . Therefore, for any given N , the
solution to the equation
(p) = (p; p) = 0;
(14)
is bounded from 0 as ! 0, then (14) is the necessary and su¢ cient condition for p to be the
market price.
The expression for (p) is given by:
(p) =
If
2
(1
N
N) p
N (N
N( N
1)
1)
2
N (N
1)
2
p
(1
N)
(1
N) ;
.
! 0, in particular, N < 1, then
2
H (p) = 1
N
N p
p
is a concave in p, H (0) = (1
N ) < 0, and H (1) = N (1
N ) > 0.
Thus, there is a unique solution p to H (p) = 0. We will focus on the asymptotic properties
of p when ! 0. First, note that
2
H (p) = 1
= 1
N p
O
2
where O (x) has an order x. That is, p
p
N)
1
N
! e
!0
1
N (N
N
2
O
2
1)
p
p
(1
(1
N)
N) ,
! p^ , where p^ is a solution to:
!0
^ (p) = p
H
Then, p^ = (1
N
N
(1
N ) = 0.
' 0:368, which results in the seller’s pro…ts:
(p ; p ) =
1
GN (p )
p
N
!
!0
1
e
Ne
N
!
!0
e
.
Now, if a single seller deviates by not revealing information and setting price x = v e =
his pro…t increases:
d
= GY (p ) x = (p )
(N 1)
ve = e
6
(N 1)
1+
!
!0
>
e
.
1+
,