Supplementary Material to “Pre-election Polls as
Strategic Coordination Devices”
Costel Andonie∗and Christoph Kuzmics†
August 21, 2012
1
Deriving the Pivot Events for the Pure Coordination
Model
In this brief section we derive the pivot events for C types as well as AB types (and, thus,
by symmetry also BA types) in the pure coordination model.
Suppose a C type sends message C. Then the set of states after which candidate C is
elected is given by
WINC = {nA = nB }.
Thus, the set of states in which C is not elected, and, thus, a C type receives a payoff of 0,
is given by
LOSEC = WINC = {nA 6= nB }.
Similarly, if sending message A, these sets are
WINA = {nA = nB − 1}
and
LOSEA = WINA = {nA 6= nB − 1}.
Thus, the set of states in which a C type gains payoff (from 0 to 1) by switching from
∗
Department of Economics and Finance, City U of Hong Kong, Email: [email protected]
†
IMW, Bielefeld University, Bielefeld, Germany, Email: [email protected]
message C to message A is given by
GAIN = WINA ∩ LOSEC ,
which yields
GAIN = {nA 6= nB } ∩ {nA = nB − 1}
= {nA = nB − 1}.
Similarly, we can find the set of states in which a C type loses payoff (from 1 to 0) by
switching from C to A. This set is given by
LOSS = {nA = nB }.
This concludes the derivation of the pivot events for a C type. Now turning to the AB
types, fix attainable strategy profile x ∈ X. An AB type, sending message A receives a
payoff of 1 in the following set of states,
WIN1A = {nA ≥ nB },
of v in this set of states
WINvA = {nA < nB − 1},
and of 0 in this set of states
LOSEA = {nA = nB − 1}.
An AB type, sending message C receives a payoff of 1 in the following set of states,
WIN1C = {nA > nB },
of v in this set of states
WINvC = {nA < nB },
and of 0 in this set of states
LOSEC = {nA = nB }.
Thus, the set of states in which an AB type gains a payoff-difference of 1 by switching
2
from message A to C is given by
GAIN1 = LOSEA ∩ WIN1C = ∅.
Similarly, a payoff-difference of v is achieved in the set
GAINv = LOSEA ∩ WINvC = {nA = nB − 1},
a payoff-difference of 1 − v in
GAIN(1-v) = WINvA ∩ WIN1C = ∅,
a payoff-difference of −1 in
LOSS1 = WIN1A ∩ LOSEC = {nA = nB },
a payoff-difference of −(1 − v) in
LOSS(1-v) = WIN1A ∩ WINvC = ∅,
and, finally, a payoff-difference of −v in
LOSSv = WINvA ∩ LOSEC = ∅.
This concludes the derivation of the pivot events for the pure coordination model.
2
Proof of results from “Voting Stage” section (Pure
Coordination Model)
2.1
Proof of Lemma 1
Because of the symmetry between AB & BA types, it is enough to check the incentives of
an AB type for example. So, consider an AB type voter who contemplates voting for A vs.
voting for B. The pivot events of interest to an AB voter, associated payoffs with choosing
A or B and their differences are given in Table 1. (See Myatt and Fisher (2002) for a more
detailed description of these pivot events, in a more general context.)
Combining the pivot probabilities and their associated gains/losses, we can compute the
3
Table 1: Pivot Events and their associated Gains/ Losses for an AB voter
Pivot Event
Payoff if
Payoff if
Gains/ Losses associated
voter chooses A voter chooses B with choosing A over B
1
v
1−v
NA = NB = NC − 1
2
2
2
NA = NB ≥ NC
1
v
1−v
1+v
1−v
NA = NC = NB − 1
v
2
2
1+v
2−v
NA = NC = NB + 1
1
3
3
1
1
NA = NC > NB + 1
1
2
2
1+v
1−v
NB = NC = NA − 1
1
2
2
1+v
1−2v
NB = NC = NA + 1
v
3
3
v
v
NB = NC > NA + 1
v
−
2
2
1
1
NA = NC − 1 > NB
0
2
2
1+v
1−v
NA = NB + 1 > NC + 1
1
2
2
v
v
NB = NC − 1 > NA
0
−
2
2
1+v
1−v
NA = NB − 1 > NC
v
2
2
expected gains/losses when a voter chooses to vote A over B as:
1−v
[P (NA = NB = NC − 1) + P (NA = NC = NB − 1) + P (NB = NC = NA − 1)
2
+ P (NA = NB + 1 > NC + 1) + P (NA = NB − 1 > NC )]
E[uA − uB ] =
+ (1 − v)P (NA = NB ≥ NC )
2−v
+
P (NA = NC = NB + 1)
3
1
[P (NA = NC > NB + 1) + P (NA = NC − 1 > NB )]
+
2
1 − 2v
+
P (NB = NC = NA + 1)
3
v
−
[P (NB = NC > NA + 1) + P (NB = NC − 1 > NA )]
2
where E[uA − uB ] are his expected gains/ losses when switching from B to A.
Consider first the equilibrium in which σ(AB) = σ(BA) = A; σ(C) = C. This implies
πA = 2q, πB = 0 and πC = 1 − 2q, so NB = 0 and the AB’s incentive reduces to:
1
E[uA − uB ] = [P (NA = NC > 1) + P (NA = NC − 1 > 0)] > 0
2
Therefore the AB prefers to vote for A as required.
Consider now the equilibrium in which σ(AB) = σ(BA) = B; σ(C) = C. This implies
4
πA = 0, πB = 2q and πC = 1 − 2q, so NA = 0 and the AB’s incentive reduces to
v
E[uA − uB ] = − [P (NB = NC > 1) + P (NB = NC − 1 > 0)] < 0
2
Therefore the AB prefers to vote for B as required.
Finally, consider the equilibrium in which σ(AB) = A; σ(BA) = B; σ(C) = C. This
implies πA = q, πB = q and πC = 1 − 2q. Because πA = πB , we have:
P (NA = NC = NB + 1) = P (NB = NC = NA + 1)
P (NA = NC > NB + 1) = P (NB = NC > NA + 1)
P (NA = NC − 1 > NB ) = P (NB = NC − 1 > NA )
Therefore, AB’s incentives can be simplified to
1−v
[P (NA = NB = NC − 1) + P (NA = NC = NB − 1) + P (NB = NC = NA − 1)
2
+ P (NA = NB + 1 > NC + 1) + P (NA = NB − 1 > NC )]
E[uA − uB ] =
+ (1 − v)P (NA = NB ≥ NC )
3 − 3v
P (NA = NC = NB + 1)
+
3
1−v
+
[P (NA = NC > NB + 1) + P (NA = NC − 1 > NB )] > 0
2
Therefore the AB prefers to vote for A as required. QED
2.2
Proof of Lemma 2
Let (1AB , 1BA , 1C ) denote the outcome of a trinomial draw with probabilities (q, q, 1 − 2q),
where 1AB indicates an AB type was drawn etc. Let N be the total number of voters.
Consider first the equilibrium in which AB and BA’s coordinate on A. In this case
NA =
N
X
(1AB + 1BA )
i=1
NB = 0
N
X
NC =
1C
i=1
By the Law of Large Numbers
1
N
N A
→ 2q and
1
N
N C
→ 1 − 2q. Because 2q > 1 − 2q, we
have P (NA > NC ) → 1 as N → +∞. So A wins with probability one in the limit.
5
A similar argument applies to the equilibrium involving coordination on B.
Finally, consider the equilibrium in which the majority support is evenly split between
A and B. Because q < 1 − 2q, by the Law of Large Numbers, C wins with probability one in
the limit.
3
Proof of Lemma 8
Write the an and bn for the even and odd cases as
a2n =
n
X
k=0
a2n+1 =
n
X
k=0
b2n =
n−1
X
k=0
b2n+1 =
n
X
k=0
(2n)!
ρ2k (1 − 2ρ)2n−2k
k!k!(2n − 2k)!
(2n + 1)!
ρ2k (1 − 2ρ)2n+1−2k
k!k!(2n + 1 − 2k)!
(2n)!
ρ2k+1 (1 − 2ρ)2n−2k−1
k!(k + 1)!(2n − 2k − 1)!
(2n + 1)!
ρ2k+1 (1 − 2ρ)2n−2k
k!(k + 1)!(2n − 2k)!
6
Then
a2n =
=
=
=
n
X
k=0
n−1
X
k=0
n−1
X
k=0
n−1
X
k=0
=
n−1
X
k=0
(2n)!
ρ2k (1 − 2ρ)2n−2k
k!k!(2n − 2k)!
(2n)!
(2n)! 2n
ρ2k (1 − 2ρ)2n−2k +
ρ
k!k!(2n − 2k)!
n!n!
(2n − 1)!(2n − 2k + 2k) 2k
(2n)! 2n
ρ (1 − 2ρ)2n−2k +
ρ
k!k!(2n − 2k)!
n!n!
n−1
X
(2n − 1)!(2k) 2k
(2n)! 2n
(2n − 1)!(2n − 2k) 2k
2n−2k
ρ (1 − 2q)
+
ρ (1 − 2ρ)2n−2k +
ρ
k!k!(2n − 2k)!
k!k!(2n
−
2k)!
n!n!
k=0
n−1
X
(2n − 1)!
(2n − 1)!(2k) 2k
(2n)! 2n
2k
2n−2k
ρ (1 − 2ρ)
+
ρ (1 − 2ρ)2n−2k +
ρ
k!k!(2n − 1 − 2k)!
k!k!(2n
−
2k)!
n!n!
k=1
= (1 − 2ρ)
= (1 − 2ρ)
= (1 − 2ρ)
n−1
X
k=0
n−1
X
k=0
n−1
X
k=0
n
X
(2n − 1)!
(2n − 1)!(2k) 2k
ρ2k (1 − 2ρ)2n−1−2k +
ρ (1 − 2ρ)2n−2k
k!k!(2n − 1 − 2k)!
k!k!(2n
−
2k)!
k=1
n−1
X
(2n − 1)!
(2n − 1)!(2)
ρ2k (1 − 2ρ)2n−1−2k +
ρ2k+2 (1 − 2ρ)2n−2
k!k!(2n − 1 − 2k)!
k!(k
+
1)!(2n
−
2k
−
2)!
k=0
n−1
X
(2n − 1)!
(2n − 1)!
2k
2n−1−2k
ρ (1 − 2ρ)
+ 2ρ
ρ2k+1 (1 − 2ρ)2n
k!k!(2n − 1 − 2k)!
k!(k + 1)!(2n − 2k − 2)!
k=0
= (1 − 2ρ)a2n−1 + 2ρb2n−1
7
The probability of a near-tie can similarly be computed as
b2n =
=
=
−
n−1
X
k=0
n−1
X
k=0
n−1
X
k=0
n−1
X
k=0
=
n−1
X
k=0
(2n)!
ρ2k+1 (1 − 2ρ)2n−2k−1
k!(k + 1)!(2n − 2k − 1)!
(2n − 1)!(2n − 2k − 1 + 2k + 2 − 1) 2k+1
ρ
(1 − 2ρ)2n−2k−1
k!(k + 1)!(2n − 2k − 1)!
n−1
X
(2n − 1)!(2n − 2k − 1) 2k+1
(2n − 1)!(2k + 2)
2n−2k−1
ρ
(1 − 2ρ)
+
ρ2k+1 (1 − 2ρ)2n−2k
k!(k + 1)!(2n − 2k − 1)!
k!(k
+
1)!(2n
−
2k
−
1)!
k=0
(2n − 1)!
ρ2k+1 (1 − 2ρ)2n−2k−1
k!(k + 1)!(2n − 2k − 1)!
n−1
X
(2n − 1)!
(2n − 1)!(2)
2k+1
2n−2k−1
ρ
(1 − 2ρ)
+
ρ2k+1 (1 − 2ρ)2n−2k−1
k!(k + 1)!(2n − 2k − 2)!
k!k!(2n
−
2k
−
1)!
k=0
n−1
−
(2n)!
1 X
ρ2k+1 (1 − 2ρ)2n−2k−1
2n k=0 k!(k + 1)!(2n − 2k − 1)!
= (1 − 2ρ)
n−1
X
k=0
−
1
2n
n−1
X
k=0
n−1
X
(2n − 1)!
(2n − 1)!
ρ2k+1 (1 − 2ρ)2n−2k−2 + 2ρ
ρ2k (1 − 2ρ)2n
k!(k + 1)!(2n − 2k − 2)!
k!k!(2n
−
2k
−
1)!
k=0
(2n)!
ρ2k+1 (1 − 2ρ)2n−2k−1
k!(k + 1)!(2n − 2k − 1)!
= (1 − 2ρ)b2n−1 + 2ρa2n−1 −
1
b2n
2n
A similar relationship holds for the odd-indexed terms.
4
Proof of Lemma 10
Writing out an+2 and bn+2 in terms of an and bn , we have
an+2 = (1 − 2ρ)an+1 + 2ρbn+1
n+1
[(1 − 2ρ)bn + 2ρan ]
n+2
n+1
n+1
= [(1 − 2ρ)2 + (2ρ)2
]an + [(1 − 2ρ)2ρ + (1 − 2ρ)2ρ
]bn
n+2
n+2
n+1
2n + 3
= [(1 − 2ρ)2 + (2ρ)2
]an + (1 − 2ρ)2ρ
bn
n+2
n+2
= (1 − 2ρ)[(1 − 2ρ)an + 2ρbn ] + 2ρ
8
and
n+2
[(1 − 2ρ)bn+1 + 2ρan+1 ]
n+3
n+2
n+1
=
[(1 − 2ρ)
[(1 − 2ρ)bn + 2qan ] + 2ρ[(1 − 2ρ)an + 2ρbn ]]
n+3
n+2
n+2
n+1
n+2
n+1
=
[(1 − 2ρ)2ρ
+ (1 − 2ρ)2ρ]an +
[(1 − 2ρ)2
+ (2ρ)2 ]bn
n+3
n+2
n+3
n+2
2n + 3
n+2
n+1
= (1 − 2ρ)2ρ
an +
[(1 − 2ρ)2
+ (2ρ)2 ]bn
n+3
n+3
n+2
bn+2 =
Then
an+2 − bn+2 = αn an − βn bn
where
n+1
2n + 3
− (1 − 2ρ)2ρ
n+2
n+3
n
+
1
n
+
2
2n + 3
= (1 − 2ρ)2
+ (2ρ)2
− (1 − 2ρ)2ρ
n+3
n+3
n+2
αn = (1 − 2ρ)2 + (2ρ)2
βn
To show αn > 0, write
αn = 4(1 +
2n + 3
n + 1 2n + 3 2
+
)ρ − 2(2 +
)ρ + 1
n+2
n+3
n+3
Its discriminant is
2n + 3 2
n + 1 2n + 3
) − 16(1 +
+
)
n+3
n+2
n+3
2n + 3 2
2n + 3
n + 1 2n + 3
4(4 + (
) +4
) − 16(1 +
+
)
n+3
n+3
n+2
n+3
2n + 3 2
n+1
4(
) − 16
n+3
n+2
4
[(2n + 3)2 (n + 2) − 4(n + 1)(n + 3)2 ]
(n + 3)2 (n + 2)
4
[−8n2 − 27n − 18] < 0
2
(n + 3) (n + 2)
∆ = 4(2 +
=
=
=
=
Since there are no real roots and at ρ = 0, αn > 0, it follows that αn > 0 for all ρ and
n ≥ 0.
9
For the second claim, write
1
1
2
+
)ρ +
n+3 n+2
n+3
1
1
1
+
)(
= 2(
1 − ρ)
n + 3 n + 2 2 + n+2
αn − βn = −2(
5
Proof of Observation 1
Let
a(ρ) =
X
e
k
k
−ρm (ρm) −ρm (ρm)
k!
k≥0
e
k!
the probability of having a tie, and
b(ρ) =
X
k≥0
e−ρm
(ρm)k −ρm (ρm)k+1
e
k!
(k + 1)!
the probability of having a near-tie.
Then
da(ρ)
= −2m[a(ρ) − b(ρ)]
dρ
which is negative from our results in the Poisson case. So a(ρ) is decreasing in ρ, and C’s as a
group would like to make ρ as small as possible. Since ρ = qxAB (A)+qxBA (A)+(1−2q)xC (A),
C’s would choose xC (A) = 0.
6
Proof of Observation 2
Let
m
am (ρ) =
b2c
X
k=0
m!
ρ2k (1 − 2ρ)m−2k
k!k!(m − 2k)!
the probability of having a tie, and
b m−1
c
2
bm (ρ) =
X
k=0
m!
ρ2k+1 (1 − 2ρ)m−2k−1
k!(k + 1)!(m − 2k − 1)!
10
the probability of having a near-tie.
Then
dam (ρ)
= −2m[am−1 (ρ) − bm−1 (ρ)]
dρ
which is negative from our results in the odd case. So am (ρ) is decreasing in ρ, and C’s as a
group would like to make ρ as small as possible. Since ρ = qxAB (A)+qxBA (A)+(1−2q)xC (A),
C’s would choose xC (A) = 0.
7
Polling equilibria with alternative equilibrium maps
(Pure Coordination Model)
Consider the case of a more general equilibrium map:
sAB


 vote A, if − k ≤ mA − mB ≤ k
=
vote A, if mA − mB > k


vote B, if mA − mB < −k
sBA


 vote B, if − k ≤ mA − mB ≤ k
=
vote B, if mA − mB < −k


vote A, if mA − mB > k
and
where k ≥ 0 is a parameter. In words, AB and BA voters will coordinate in the voting
stage only if either A’s lead over B or B’s lead over A is at least (k + 1) messages. In this
context, similarly to the case of k = 0, the incentives of C types and AB types to deviate
can be computed as:
∆C = P (nB = nA + k + 1) − P (nA = nB + k)
= P (nA = nB + k + 1) − P (nA = nB + k)
and
∆AB = vP (nB = nA + k + 1) − P (nA = nB + k)
= vP (nA = nB + k + 1) − P (nA = nB + k)
Again we note that if v < 1 then ∆AB < ∆C . For simplicity we consider the Poisson case
11
i.e. the sample size follows a Poisson distribution.
Proposition S1 Consider the pure coordination model with the equilibrium map as described
above. Suppose the poll size follows a Poisson distribution. Then the polling game has a
unique attainable equilibrium in which AB types send message A; BA types send message B
and C types send message C.
The proof of the proposition follows from the following lemma.
Lemma S1 Suppose nA , nB are two iid random variables distributed as P oisson(n). Then
for any k ≥ 0 we have
P (nA = nB + k) > P (nA = nB + k + 1)
This lemma implies that, for any attainable strategies of its opponents, a C type would
always want to send a truthful message. Moreover, the same holds true for an AB type.
Therefore, the unique equilibrium has all three types sending a truthful message.
We already know the lemma is true for k = 0, but would like to have a general result.
For any k ≥ 0, let us denote by
uk = P (nA = nB + k) =
X
e−n
j≥0
−2n
= e
nj −n nj+k
e
j!
(j + k)!
X
j≥0
n2j+k
j!(j + k)!
To simplify notation, in what follows we ignore the constant e−2n and use
uk =
X
j≥0
n2j+k
j!(j + k)!
The next lemma gives a recurrence equation relating the k, k + 1 and k + 2 terms
Lemma S2 For any k ≥ 0, we have
k+1
uk+1 = uk − uk+2
n
12
Proof: Write
uk =
X
j≥0
=
n2j+k
j!(j + k)!
X (j + k + 1)n2j+k
j≥0
=
X
=
X
j≥0
j≥1
j!(j + k + 1)!
X (k + 1)n2j+k
jn2j+k
+
j!(j + k + 1)! j≥0 j!(j + k + 1)!
X
n2j+k
n2j+k
+ (k + 1)
(j − 1)!(j + k + 1)!
j!(j + k + 1)!
j≥0
= uk+2 +
k+1
uk+1
n
QED
The proof of the lemma follows from the following two results.
Lemma S3 For any k such that k(k + 1) < n, we have
uk − uk+1 > 0
Proof: The proof is by induction. For k = 0 and k = 1 the statement is:
If 0 < n, then u0 − u1 > 0.
If 2 < n, then u1 − u2 > 0. (to be shown later)
Suppose it holds for some k, and we want to show it holds for k + 2:
If (k + 2)(k + 3) < n, then uk+2 − uk+3 > 0.
We can write
uk+2 − uk+3 = uk+2 − [uk+1 −
k+2
uk+2 ]
n
k+2
)uk+2 − uk+1
n
k+2
k+1
= (1 +
)[uk −
uk+1 ] − uk+1
n
n
k+2
k+1
n
= (1 +
)[uk − (
+
)uk+1 ]
n
n
n+k+2
= (1 +
Since k(k + 1) < (k + 2)(k + 3) < n, we can use the induction hypothesis: uk − uk+1 > 0.
Also since (k + 1)(k + 2) < (k + 2)(k + 3) < n, we have 1 >
two inequalities, we obtain uk > ( k+1
+
n
n
)uk+1
n+k+2
13
k+1
n
+
n
.
n+k+2
Combining these
which is equivalent to uk+2 − uk+3 > 0.
Claim S1 If 2 < n, then u1 − u2 > 0.
Proof: Taking the derivative wrt n we have
d
2
1
2
(u1 − u2 ) = (2 + )u0 − (2 + + 2 )u1
dn
n
n n
Since n > 2, we have 2 + n2 > 2 + n1 + n22 . Combine this inequality with u0 > u1 to obtain
d
(u1 − u2 ) > 0
dn
Therefore (u1 − u2 ) is increasing over n ∈ (2, +∞). Since at n = 2 :
u1 − u2 =
X
j≥0
j
22j+1
>0
j!(j + 1)! j + 2
it follows that u1 − u2 > 0 for any n > 2.
QED
Lemma S4 For any k such that k(k + 1) > n, we have
uk − uk+1 > 0
Proof: Again, we do the proof by induction. We note that lemma holds for any k such that
k ≥ n − 1:
uk − uk+1 =
X
j≥0
=
X
=
X
j≥0
j≥0
X
n2j+k
n2j+k+1
−
j!(j + k)! j≥0 j!(j + k + 1)!
n2j+k
n
[1 −
]
j!(j + k)!
j+k+1
n2j+k j + k + 1 − n
>0
j!(j + k)! j + k + 1
Suppose the claim is true for some k and we want to show it holds for k − 2:
If (k − 2)(k − 1) > n, then uk−2 − uk−1 > 0.
14
We can write
k−1
uk−1 − uk−1
n
n−k+1
= uk −
uk−1
n
n−k+1
k
[uk+1 + uk ]
= uk −
n
n
n−k+1
n−k+1k
]uk −
uk+1
= [1 −
n
n
n
uk−2 − uk−1 = uk +
If n + 1 < k, then we immediately obtain uk−2 − uk−1 > 0. Suppose now k ≤ n + 1.
Since k(k + 1) > (k − 2)(k − 1) > n we can use the induction hypothesis uk > uk+1 . At the
k
same time (k − 1)k > (k − 2)(k − 1) > n implies 1 − n−k+1
>
n
n
inequalities, we immediately obtain uk−2 − uk−1 > 0.
n−k+1
.
n
Combining these two
QED
References
Myatt, D.P., Fisher, S.D., 2002. Everything is Uncertain and Uncertainty is Everything:
Strategic Voting in Simple Plurality Elections. University of Oxford, Discussion Paper
Series, Number 115.
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