42 W
.V. Gehrlein, D. Lepelley Mathematical Social Sciences 41 2001 39 –50
WSR with l51. However, the definition that we use, with weights W, 1, 0, is equivalent to the usual definition, and it is particularly useful in simplifying derivations
that follow. BR is the WSR which has W 5 2. Van Newenhizen 1992 showed that BR is the WSR that maximizes Condorcet
efficiency for a class of distributions, not including IAC, that define the probability that each possible profile is observed. Saari 1990 gives a number of other properties of BR
to give strong additional support to its use as a WSR. Given this background, the development of a representation for the Condorcet efficiency of BR under IAC is of
significant interest.
2. A representation for the Condorcet efficiency of Borda Rule
A
Let J n denote the number of IAC profiles with n voters for which A is both the
A
Condorcet winner and the winner by BR. To develop a representation for J n, we first
B . A
let J n denote the number of IAC profiles with n voters for which A is the
A
Condorcet winner and for which B beats A by Borda Rule. Let Con n denote the number of IAC profiles with n voters for which A is the Condorcet winner. Lhuilier
1793 in Sommerlad and McLean, 1989 showed that the Condorcet winner can never be beaten by both of the other candidates with BR. It then follows that:
A A
B . A C . A
J n 5 Con n 2 J n 2 J
n By the symmetry of IAC with respect to candidates:
A A
B . A
J n 5 Con n 2 2J n
A
We know the representation for Con n from Gehrlein and Fishburn 1976:
3
n 1 1n 1 3 n 1 5
A
]]]]]] Con n 5
384
B . A
so we only need to develop a representation for J n.
The conditions to have A as the Condorcet winner, with B the BR winner over A are given by:
n 1 n 1 n . n 1 n 1 n
1 2
3 4
5 6
n 1 n 1 n . n 1 n 1 n
1 2
4 3
5 6
2n 1 n 1 n 1 n . 2n 1 n 1 n 1 n
3 5
1 6
1 2
3 4
The restrictions that are needed to require consistency among the n ’s, given all three
i
of these inequalities are:
W .V. Gehrlein, D. Lepelley Mathematical Social Sciences 41 2001 39 –50
43
n 2 1 ]]
0 n
56
2 0 n n
5 56
n 2 1 ]]
0 n 2
n
3 56
2 n 2 1
]] 0 n
2 n
4 56
2 2n 1 n 1 2n
2 n
3 5
56
0 n Min
H J
2
n 2 n 2 n 2 n
3 4
56 a
Here, Min h j is the minimum of a and b, and n 5 n 1 n .
b 56
5 6
When A is the Condorcet winner, it is easily shown that we must have: 2n 1 n 1 2n
2 n , n 2 n 2 n 2 n
3 5
56 3
4 56
Thus, the Min argument in the n summation counter can be removed. In addition, to
2
have consistency between the upper and lower summation limits of the n counter, we
2
must be assured that: 2n 1 n 1 2n
2 n 0
3 5
56
So, we must add the restriction that:
1 ]
n [n 2 n 2 2n ]
3 5
56 2
To then have consistency between the upper and lower summation limits of the n
3
counter, we must be assured that: n 2 1
1 ]]
] 2
n [n 2 n 2 2n ]
56 5
56
2 2
which requires n 1, so that n 1.
5 56
The resulting limitations on the n ’s after accounting for all of these consistency
i
requirements are given by: n 2 1
]] 1 n
56
2 1 n n
5 56
n 2 1 1
]] Max
n 2
n
3 56
H J
2 ] n 2 n 2 2n
f g
5 56
2 n 2 1
]] 0 n
2 n
4 56
2 0 n 2n 1 n 1 2n
2 n
2 3
5 56
a
Here, Max h j is the maximum of a and b.
b
It simplifies matters in later discussion to treat the specific case of n 51 as a special
56
case. For this special case, the n ’s must have: n 5 0, n 5 1, n 5 n 2 3 2, n 5 0,
i 6
5 3
2
and 0 n n 2 3 2. It is easily shown that there are n –1 2 possible profiles of this
4
type, and we are then interested in determining the number of IAC profiles in the remainder of the space covered by the summation indexes with:
44 W
.V. Gehrlein, D. Lepelley Mathematical Social Sciences 41 2001 39 –50
n 2 1 ]]
2 n
56
2 1 n n
5 56
n 2 1 1
]] Max
n 2
n
3 56
H J
2 ] n 2 n 2 2n
f g
5 56
2 n 2 1
]] 0 n
2 n
4 56
2 0 n 2n 1 n 1 2n
2 n
2 3
5 56
1 ]
The [n 2 n 2 2n ] term in the Max argument on the n counter poses a particular
5 56
3 2
problem, since it is constrained to have an integer value that is consistent with that limit. To account for this, it is necessary to deal with odd and even values of n as separate
5
cases. The first step in the procedure is to account for odd and even values of n as
56
separate cases. We assume that the term n 2 1 4 is integer valued, so that n [ 9, 21, h
33, . . . , 189, . . . . These values on n will then allow for direct comparison with the j
Condorcet efficiency values that are reported in Table 1 for the other voting rules. For any given n 2 1 2 upper limit of n , there are n 2 1 4 different even values
56
of n to be enumerated, with n
5 2, 4, 6, 8, . . . , n 2 1 2. For each n 2 1 2 upper
56 56
limit of n , there are n 2 5 4 odd values of n to be enumerated, with n 53, 5,
56 56
56
7, . . . , n 23 2, since n 51 was treated as a special case. The space of n ’s that is
56 i
shown above is then partitioned into two subspaces to account for the differences that result from the odd and even n
terms.
56
The case of even values of n defines n 52n
with:
56 56
56
n 2 1 ]]
1 n
56
4 1 n 2n
5 56
n 2 1 1
]] Max
n 2
2n
3 56
H J
2 ]
n 2 n 2 4n f
g
5 56
2 n 2 1
]] 0 n
2 2n
4 56
2 0 n 2n 1 n 1 4n
2 n
2 3
5 56
The case of odd values of n defines n 52n
1 1 with:
56 56
56
n 2 5 ]]
1 n
56
4 1 n 2n
1 1
5 56
n 2 3 1
]] Max
n 2
2n
3 56
H J
2 ]
n 2 2 2 n 2 4n f
g
5 56
2 n 2 3
]] 0 n
2 2n
4 56
2 0 n 2n 1 n 1 4n
2 n 1 2
2 3
5 56
Once that we have established the odd–even status of the n counter, it is then
56
possible to deal with the odd–even status of n .
5
W .V. Gehrlein, D. Lepelley Mathematical Social Sciences 41 2001 39 –50
45
Consider the subspace that accounts for even values of n , and the corresponding
56
value for the upper limit of the n index. For each even upper limit of n , there are n 2
5 5
56
even values of n to be enumerated, with n 5 2, 4, 6, . . . , n . Similarly, for each even
5 5
56
upper limit value of n , there are n 2 odd values of n to be enumerated, with n 5 1,
5 56
5 5
3, 4, . . . , n 2
1. Once the odd–even status of both of the n and n indexes have been
56 56
5 1
]
established, the associated integer value of the [n 2 n 2 2n ] term to be used in the
5 56
2
Max argument on the n counter can be determined. The summation that is given above
3
for even n is then partitioned into two subspaces to account for odd and even values of
56
n , when n is even.
5 56
The case of even n with even values of n , with n 5 2n
[denoted as Subspace
56 5
5 5
Even–Even] has limits given by: n 2 1
]] 1 n
56
4 1 n n
5 56
n 2 1 n 1 1
]] Max
n 2
2n
3 56
H J
2 ]]
2 n 2 2n
5 56
2 n 2 1
]] 0 n
2 2n
4 56
2 0 n 2n 1 2n 1 4n
2 n
2 3
5 56
The case of even n with odd values of n , with n 5 2n 2 1 [denoted as Subspace
56 5
5 5
Even–Odd] has limits given by: n 2 1
]] 1 n
56
4 1 n n
5 56
n 2 1 n 1 1
]] Max
n 2
2n
3 56
H J
2 ]]
2 n 2 2n
5 56
2 n 2 1
]] 0 n
2 2n
4 56
2 0 n 2n 1 2n 1 4n
2 n 2 1
2 3
5 56
The same general procedure can then be used for the subspace that accounts for odd values of n , and the corresponding upper limit of the n index. For each possible odd
56 5
upper limit value of n , there are n 21 2 even values of n to be enumerated, with
5 56
5
n 52, 4, 6, . . . , n 21. Similarly, for each odd upper limit value of n , there are
5 56
5
n 1
1 2 odd values of n to be enumerated, with n 5 1, 3, 4, . . . , n . Once the
56 5
5 56
odd–even status of both of the n and n
indexes have been established, the
56 5
1 ]
corresponding integer value of the [n 2 n 2 2n ] term to be used in the Max
5 56
2
argument on the n counter can be determined. The summation that is given above for
3
odd n is partitioned into two subspaces to account for odd and even values of n when
56 5
n is odd.
56
The case of odd n with even values of n , with n 5 2n
[denoted as Subspace
56 5
5 5
Odd–Even] has limits given by:
46 W
.V. Gehrlein, D. Lepelley Mathematical Social Sciences 41 2001 39 –50
n 2 5 ]]
1 n
56
4 1 n n
5 56
n 2 3 n 2 1
]] Max
n 2
2n
3 56
H J
2 ]]
2 n 2 2n
5 56
2 n 2 3
]] 0 n
2 2n
4 56
2 0 n 2n 1 2n 1 4n
2 n 1 2
2 3
5 56
The case of odd n with odd values of n , with n 5 2n 2 1 [denoted as Subspace
56 5
5 5
Odd–Odd] has limits given by: n 2 5
]] 1 n
56
4 1 n n
1 1
5 56
n 2 3 n 2 1
]] Max
n 2
2n
3 56
H J
2 ]]
2 n 2 2n
5 56
2 n 2 3
]] 0 n
2 2n
4 56
2 0 n 2n 1 2n 1 4n
2 n 1 1
2 3
5 56
At this point, the odd–even factor has been removed from each of the subspaces, and it is now possible to use standard procedures to further partition the subspaces, in order
to remove the Max argument from the n index limit, as in Gehrlein 1982.
3
Subspace [Even–Even] is partitioned into three subspaces: Subspace No. 1
Subspace No. 2 n 1 3
n 2 1 n 2 3
]] ]]
]] n
1 n
56 56
6 4
6 n 1 3
]] 2
2n n n
1 n n
56 5
56 5
56
2 n 2 1
n 1 1 n 2 1
]] ]]
]] 0 n
2 2n
2 n 2 2n
n 2
2n
3 56
5 56
3 56
2 2
2 n 2 1
n 2 1 ]]
]] 0 n
2 2n
0 n 2
2n
4 56
4 56
2 2
0 n 2n 1 2n 1 4n 2
n 0 n 2n 1 2n 1 4n
2 n
2 3
5 56
2 3
5 56
Subspace No. 3 n 1 3
n 2 1 ]]
]] n
56
6 4
n 1 1 ]]
1 n 2
2n
5 56
2 n 1 1
n 2 1 ]]
]] 2
n 2 2n n
2 2n
5 56
3 56
2 2
n 2 1 ]]
0 n 2
2n
4 56
2 0 n 2n 1 2n 1 4n
2 n
2 3
5 56
W .V. Gehrlein, D. Lepelley Mathematical Social Sciences 41 2001 39 –50
47
To describe the partitioning of Subspace [Even–Even], suppose that we eliminate the Max argument in the n counter by requiring 0 . n 1 1 2 2 n 2 2n . This then leads
3 5
56
to the additional consistency restriction that: n 1 3
]] 2
2n
56
Max n n
2
5 56
H J
1 For consistency between the upper and lower limits of this modified n
counter, we
5
must have n n 1 3 6. Given the limits on the n
counter, it is easily shown that
56 56
n 1 3 2 2 2n 1, to eliminate the Max argument in this modified n
counter. All of
56 5
this leads to the summation limits of Subspace No. 1. The remaining partition of Subspace [Even–Even], with 0 n 1 1 2 2 n 2 2n
to
5 56
eliminate the Max argument in the n counter, leads to the summation limits for
3
Subspace No. 2 and Subspace No. 3. In the same fashion, Subspace [Even–Odd] is partitioned into three subspaces:
Subspace No. 4 Subspace No. 5
n 1 3 n 2 1
n 1 3 n 2 1
]] ]]
]] ]]
n n
56 56
6 4
6 4
n 1 3 n 1 1
]] ]]
2 2n
n n 1 n
2 2n
56 5
56 5
56
2 2
n 2 1 n 1 1
n 2 1 ]]
]] ]]
0 n 2
2n 2
n 2 2n n
2 2n
3 56
5 56
3 56
2 2
2 n 2 1
n 2 1 ]]
]] 0 n
2 2n
0 n 2
2n
4 56
4 56
2 2
0 n 2n 1 2n 1 4n 2
n 2 1 0 n 2n 1 2n 1 4n
2 n 2 1
2 3
5 56
2 3
5 56
Subspace No. 6 n 2 3
]] 1 n
56
6 1 n n
5 56
n 1 1 n 2 1
]] ]]
2 n 2 2n
n 2
2n
5 56
3 56
2 2
n 2 1 ]]
0 n 2
2n
4 56
2 0 n 2n 1 2n 1 4n
2 n 2 1
2 3
5 56
Subspace [Odd–Even] is partitioned into three subspaces:
48 W
.V. Gehrlein, D. Lepelley Mathematical Social Sciences 41 2001 39 –50
Subspace No. 7 Subspace No. 8
n 1 3 n 2 5
n 1 3 n 2 5
]] ]]
]] ]]
n n
56 56
6 4
6 4
n 2 1 n 2 3
]] ]]
2 2n
n n 1 n
2 2n
56 5
56 5
56
2 2
n 2 3 n 2 1
n 2 3 ]]
]] ]]
0 n 2
2n 2
n 2 2n n
2 2n
3 56
5 56
3 56
2 2
2 n 2 3
n 2 3 ]]
]] 0 n
2 2n
0 n 2
2n
4 56
4 56
2 2
0 n 2n 1 2n 1 4n 2
n 1 2 0 n 2n 1 2n 1 4n
2 n 1 2
2 3
5 56
2 3
5 56
Subspace No. 9 n 2 3
]] 1 n
56
6 1 n n
5 56
n 2 1 n 2 3
]] ]]
2 n 2 2n
n 2
2n
5 56
3 56
2 2
n 2 3 ]]
0 n 2
2n
4 56
2 0 n 2n 1 2n 1 4n
2 n 1 2
2 3
5 56
Subspace [Odd–Odd] is partitioned into three subspaces: Subspace No. 10
Subspace No. 11 n 1 3
n 2 5 n 2 3
]] ]]
]] n
1 n
56 56
6 4
6 n 1 1
]] 2
2n n n
1 1
1 n n 1
1
56 5
56 5
56
2 n 2 3
n 2 1 n 2 3
]] ]]
]] 0 n
2 2n
2 n 2 2n
n 2
2n
3 56
5 56
3 56
2 2
2 n 2 3
n 2 3 ]]
]] 0 n
2 2n
0 n 2
2n
4 56
4 56
2 2
0 n 2n 1 2n 1 4n 2
n 1 1 0 n 2n 1 2n 1 4n
2 n 1 1
2 3
5 56
2 3
5 56
Subspace No. 12 n 1 3
n 2 5 ]]
]] n
56
6 4
n 2 1 ]]
1 n 2
2n
5 56
2 n 2 1
n 2 3 ]]
]] 2
n 2 2n n
2 2n
5 56
3 56
2 2
n 2 3 ]]
0 n 2
2n
4 56
2 0 n 2n 1 2n 1 4n
2 n 1 1
2 3
5 56
When considering odd values of n that are not included in 9, 21, 33, . . . , 189, . . . , h
j
W .V. Gehrlein, D. Lepelley Mathematical Social Sciences 41 2001 39 –50
49
the upper and lower limits on the n index in these subspaces would simply be modified
56
to account for the restriction that each must have an integer value. By sequentially using known algebraic relations for sums of powers of integers, a
simple representation, [S , can be obtained for the number of IAC profiles in each of the
i
twelve subspaces. Including the special case with n 5
1, we can then obtain a
56 A
representation for J n from:
12
n 2 1
A A
]] J n 5 Con n 2 2
1
O
[S
H J
i
2
i 51
After significant algebraic reduction, we obtain the representation:
5 4
3 2
123n 1 1785n 1 9970n 1 27270n 1 38547n 1 24705
A
]]]]]]]]]]]]]]]] J n 5
51840
A
This representation for J n has been verified by computer enumeration. It follows from the symmetry of IAC with respect to candidates that the joint
probability that a candidate is both the Condorcet winner and the BR winner is
A
3J n Nn. This joint probability is then divided by P n, IAC to obtain the
Con
Condorcet efficiency, CEn, BR, IAC, of BR for n voters with three candidates under IAC. After substitution and reduction we obtain:
4 3
2
123n 1 1416n 1 5722n 1 10104n 1 8235 ]]]]]]]]]]]]
CEn, BR, IAC 5
2
135n 1 1n 1 3 n 1 5 with n [ 9, 21, 33, . . . , 189, . . . . Computed values of CEn, BR, IAC are given in
h j
Table 1. The computed values of CEn, BR, IAC in Table 1 are consistent with results obtained from computation by enumeration in Gehrlein 1995.
3. Conclusion
