48 E
. Stensholt Mathematical Social Sciences 37 1999 45 –57
is uniformly distributed. This gives an alternative way to generate P. Instead of the root extractions in Eq. 4 one then needs a sorting routine to find
s.
Proof of Theorem: For t 51, this is just Eq. 5. Without loss of generality, let
1 n 1 n 21
1 n 112t
Y 5 X 1 X 1 . . . 1 X 5 1 2 ´
? ´
? . . . ? ´
. 7
t 1
2 t
1 2
t
The proof now proceeds by induction on t, using the formula 12Y 512Y ?
t 11 t
1 n 2t
´ . Then, since
´ is rectangularly distributed on [0,1] and is also stochastically
t 11 t 11
independent of Y , we get
t
d d
n 2t t 2n
] ]
ProbY x 5
ProbY x and ´
1 2 x ? 1 2 Y
t 11 t
t 11 t
dx dx
x
d n 2 1
n 2t t 21
n 2t
] 5
E
S
n ?
S D
? 1 2 r ? r
? 1 2 1 2 x dx
t 2 1
t 2n
? 1 2 r
D
dr 5 see illustration
x
d n 2 1
n n 2 1
n 2t t 21
n 2t t
] n ?
S D
?
E
1 2 r ? r
dr 2
S D
? 1 2 x ? x
5 n ?
S D
dx t 2 1
t t
1 2
n 2t 21 t
? 1 2 x ? x j
3. A comparison of two ‘‘cultures’’
In election theory, particularly in simulation work as in Gehrlein 1997, Nurmi 1992 and Van Newenhizen 1992, various probabilities are introduced by assuming a
‘‘culture’’, i.e. a certain stochastic behaviour of the electorate. For each voter there is one ranking of a number of candidates, say N. The result is a ‘‘profile’’, i.e. a point Eq.
1 in the n-dimensional simplex S with n 115N corners. Then X is the fraction of the
i
electorate which selects ranking no. i in some fixed ordering of the N possible rankings. Two frequently considered types of cultures are the following:
The impartial cultures IC: Each voter independently picks a ranking at random, with probability 1 N for each. The profile Eq. 1 has a discrete essentially multinomial
distribution which depends on the number of electors.
The impartial anonymous cultures IAC: Each possible profile has the same probability. The individual action of the voters is disregarded; they become ‘‘anonymous’’. Here
we consider only the continuous case, i.e. the limit case when the number of voters tend to infinity. We thus assume the profile Eq. 1 to be uniformly and continuously
distributed in S.
Consider now a fixed triple ha,b,cj in a set of N.3 candidates. To each of the 6
rankings, i.e.
E . Stensholt Mathematical Social Sciences 37 1999 45 –57
49
Fig. 2. The densities for the variables abc, abc 1acb, abc 1acb 1cab.
a . b . c, a . c . b, c . a . b, c . b . a, b . c . a, b . a . c corresponds one sixth of the n 115N rankings of all candidates. Let xyz be the fraction
of the electorate who choose a ranking with x .y .z. For our triple we get a profile vote vector
abc, acb, cab, cba, bca, bac 9
Under the continuous IAC, the components of Eq. 9 become stochastic variables which are disjoint sums of t 5n 11 6 variables X in Eq. 1. With e.g. N 58
i
candidates, the densities for e.g. abc, abc 1acb, abc 1acb 1cab are given by the theorem with n 5821 and t 58 6, 8 3, 8 2 respectively:
7892 33599
6719 11148
26879
0.13844214 ? 10 ? 1 2 x
? x , 0.26288619 ? 10
? 1 2 x ?
13439 12140
20159 20159
x , 0.13553854 ? 10
? 1 2 x ? x
. The integral of the last expression from 0.49 to 0.51 is approx. 0.9999409; this is the
probability that the outcome of a given pairwise contest like a vs. c is in the 0.49–0.51 range. See Fig. 2.
We are now ready to compare the assumptions of IAC and IC: Already for 8 candidates the IAC-profiles cluster around 1 6?1,1,1,1,1,1[S like the IC-profiles with
a large number of voters. The IAC-variance of abc 1acb 1cab, i.e. of Y with n 115N
t
and t 5n 11 2, is calculated from Eqs. 6 and 7 as 1 4?N11, which equals the IC-variance with 11N voters, when 11Nabc 1acb 1cab is binomial 0.5, 11N.
As in Eq. 3, all IC-correlations between different components are also 20.2.
4. IAC and the Condorcet paradox. What is impartiality?
