Geography Reference
In-Depth Information
H
2
(
0.5
)
H
1
;0.5
1
0.5 1
1
N
<
p
2
(
0.5
)
$ p
1
a
N
b4
i
5 1, . . .,
N
(24A.3)
Since the two locations are identical and
x
is the share of i rms choosing location 1,
by symmetry it holds that p
1
(
x
)
5 p
2
(
1 2
x
)
. Using this relation we can rewrite both
(24A.2) and (24A.3) as:
H
1
(0.5) H
1
;
0.5
1
0.5 1
1
p
1
(
0.5
)
$ p
1
a
N
b4
i
5 1, . . .,
N
N
<
The latter is satisi ed if and only if the function p
1
(
x
)
is not increasing at
x
= 0.5. Direct
computation of
d
p
1
(
x
)
/
dx
0
x
50.5
(see the following proofs for the explicit expression)
shows that this is never the case, implying that the symmetric distribution is never a
PSNE.
Proof of Proposition 3
Consider the Taylor expansion up to the i rst order of each term in (24.17) as a function
of
z
=
x
− 0.5:
H
l
(
z
)
H
l
(
0.5
)
z
(
)
l
;29;
1
t
K1
1 2
t
s21
1 1
t
s21
b
2
2
p
1
(
z
)
5 p
1
(
0.5
)
1
z
(
2
)
l
29<
O
(
z
2
)
,
l
1, 2
1
O
(
z
2
)
, l 5 1, 2
a
2a
a
1
t
K1
<
2 2a
b
Using the expressions above to linearize expp
l
in (24.19), and writing them in terms of
the number of i rms
n
l
,
l
=1, 2 we obtain the expressions of the linearized exponential
payof
c
l
,
(
1 2
t
s21
)
2
(
1 1
t
s21
)
2
2 1
n
1
N
2
1
c
1
5 1 2 2a
a
ba
2
b
(
1 2
t
s21
)
2
(
1 1
t
s21
)
2
2 1
2
2
n
2
1
c
2
5 1 1 2a
a
ba
N
b
This shows that
a
and
b
are given by:
(
1 2
t
s21
)
2
(
1 1
t
s21
)
2
2 1
a
5 1 1 a
a
b
(
1 2
t
s21
)
2
(
1 1
t
s21
)
2
2 1
b
52
2a
N
a
b
which, using Assumption 4 to eliminate
N
, correspond to (24.21).
Proof of Proposition 5
From (24.22) it follows that the distribution is symmetric around
N
/2, that is
p
(
N
/2 1
n
)
5 p
(
N
/2 2
n
)
for every
n
5 0, . . .,
N
/2. Consequently it sui ces to analyse
the set
{
0,. . .,
N
/2
}
.
When
a
=
b
, q
n
(
a
,
b
)
reduces to
n
!. As a result the probability density becomes:
p
(
n
)
5
N
!
C
(
N
,
a
,
b
)
Z
(
N
,
a
,
b
)
4
n
,
