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Random Preorders Random preorders Peter Cameron and Dudley Stark School of Mathematical Sciences Queen Mary, University of London Mile End, London, E1 4NS U.K. Submitted to Combinatorica Abstract A random preorder on n elements consists of linearly ordered equiva- lence classes called blocks. We investigate the block structure of a preorder chosen uniformly at random from all preorders on n elements as n ∞. ! 1 Introduction Let R be a binary relation on a set X. We say R is reflexive if (x;x) R for all x X. We say R is transitive if (x;y) R and (y;z) R implies (x;2z) R.A partial2 preorder is a relation R on X which2 is reflexive and2 transitive. A relation2 R is said to satisfy trichotomy if, for any x;y X, one of the cases (x;y) R, x = y, or (y;x) R holds. We say that R is a preorder2 if it is a partial preorder2 that satisfies trichotomy.2 The members of X are said to be the elements of the preorder. A relation R is antisymmetric if, whenever (x;y) R and (y;x) R both hold, then x = y. A relation R on X is a partial order if it2 is reflexive,2 transitive, and antisymmetric. A relation is a total order, if it is a partial order which satisfies trichotomy. Given a partial preorder R on X, define a new relation S on X by the rule that (x;y) S if and only if both (x;y) and (y;x) belong to R. Then S is an equivalence relation.2 Moreover, R induces a partial order x on the set of equivalence classes of S in a natural way: if (x;y) R, then (x;y) R, where x is the S-equivalence class containing x and similarly2 for y. We will2 call an S- equivalence class a block. If R is a preorder, then the relation R on the equivalence 1 classes of S is a total order. See Section 3.8 and question 19 of Section 3.13 in [4] for more on the above definitions and results. Preorders are used in [6] to model the voting preferences of voters. (A different but equivalent definition of preorders is used in [6], where they are called weak orders.) We suppose that there are n candidates and m voters. Suppose that X is a finite set representing a collection of candidates. Let Ri, i = 1;2;:::;m, be a set of weak orders on X. Then (x;y) R means that the ith voter prefers candidate y 2 i to candidate x. The Ri blocks correspond to sets of candidates to which voter i is indifferent. Let p(n) denote the number of preorders possible on a set of n elements. The assumption is made in [6] that each voter chooses his voting preference uniformly at random from all of the p(n) possibilities independently of the other voters. An algorithm for generating a random preorder is given in [6] and the ideas behind the algorithm are used to derive a formula for the probability of the occurrence of “Condorcet’s paradox”. See [5] for a survey of assumptions on voter preferences used in the study of Condorcet’s paradox. We are interested in the size of the blocks in a random preorder. Let B1 be the size of the first block, let B2 be the size of the second, and let Bi be the size of the ith block. If the preorder has N blocks we define Bi = 0 for i > N. It is an identity that ∞ ∑ Bi = n: (1.1) i=1 We can represent a preorder on the set X by the sequence (B1;B2;:::), where the Bi are disjoint and i Bi = X. A related combinatorial object to preorders is set partitions, for whichS the blocks are not ordered. The block structure of random set partitions has been studied in [8]. In this paper we look at the block structure of a random preorder. We give asymptotic estimates of the number of blocks, the size of a typical block, and the number of blocks of a particular size. We are able to show that the maximal size of a block asymptotically takes on one of two values. Let S(n;k) denote the Stirling number of the second kind. Note that the number of preorders with exactly r blocks is p(n;r) = r!S(n;r) and therefore n p(n) = ∑r=1 r!S(n;r). The identity ∞ S(n;r)zn (ez 1)r ∑ = − (1.2) n=0 n! r! is proven in Proposition (5.4.1) of [4] using inclusion-exclusion. The following 2 ∞ n consequence of (1.2) will be useful. If f (z) = ∑n=0 anz is a power series, then n we use the notation [z ] f (z) to denote an. Lemma 1 For any sequence qr, 1 r n, ≤ ≤ n ∞ n z n ∑ qr p(n;r) = n![z ] ∑ qn(e 1) !: r=1 n=0 − Proof We observe that ∞ n zn ∞ ∞ p(n;r)zn ∑ ∑ qr p(n;r)! = ∑ ∑ qr n=0 r=1 n! r=1 n=r n! ∞ ∞ S(n;r)zn = ∑ ∑ r!qr r=1 n=r n! ∞ z r = ∑ qr(e 1) : r=1 − The lemma follows immediately. If we take qr = 1 in Lemma 1, then we find that n z 1 p(n) = n![z ](2 e )− ; − z 1 an identity proved in [2]. The singularity of smallest modulus of (2 e )− occurs at z = log2 with residue − z log2 1 1 lim − = lim = ; (1.3) z log2 2 ez z log2 ez −2 ! − ! − by l’Hopital’sˆ rule. So the function 1 1 + 2 ez 2(z log2) − − z 1 is analytic in a circle with centre at the origin and the next singularities of (2 e )− − (at log2 2pi) on the boundary. Thus ± n! 1 n+1 p(n) ; (1.4) ∼ 2 log2 and indeed it follows from Theorem 10.2 of [7] that the difference between the n two sides is o((r e)− ), where r = log2+2pi ; that is, exponentially small. An − z 1 j j exact expression for (2 e )− is given in [1] in terms of its singularities and the truncation error from using− only a finite number of singularities is estimated. 3 2 The number of blocks We denote the number of blocks of a random preorder on n elements by Xn. In ∑∞ terms of the block sizes Bi we may express Xn as Xn = i=1 I[Bi > 0], where I[Bi > 0] is the indicator variable that the ith block has positive size. In this section we give asymptotics for Xn. The kth falling factorial of a real number x is defined to be (x) = x(x 1)(x k − − 2) (x k + 1) and the kth falling moment of Xn to be ··· − E(Xn) = EXn(Xn 1)(Xn 2) (Xn k + 1): (2.5) k − − ··· − Define l to be n n l = : (2.6) n 2log2 k We will show that E(Xn) l for each fixed k 0, where we use the notation k ∼ n ≥ an bn for sequences an, bn to mean limn ∞ an=bn = 1. By a standard argument using∼ Chebyshev’s inequality, the asymptotics! of the first two moments implies a:a:s: n a:a:s: that Xn , where we write Xn an (Xn converges to an asymptotically ∼ 2log2 ∼ almost surely) to mean limn ∞ P( Xn=an 1 > e) = 0 for all e > 0. ! j − j Theorem 1 The kth falling moment of the number of blocks of a random preorder equals k!n! (ez 1)k E(X ) = [zn] − : (2.7) n k p(n) (2 ez)k+1 − It follows that for fixed k k E(Xn) l (2.8) k ∼ n and that a:a:s: Xn ln; ∼ where ln is defined by (2.6). Proof In order to prove (2.7) it suffices to note that n p(n;r) 1 n E (Xn)k = ∑ (r)k = ∑ p(n;r)(r)k; r=1 p(n) p(n) r=1 to apply Lemma 1 with qr = (r)k, and to observe that ∞ k!xk ∑ (n) xn = : k (1 x)k+1 n=0 − 4 We now proceed to show (2.8). An analysis similar to (1.3) shows that (z log2)k+1(ez 1)k 1 k+1 lim − − = z log2 (2 ez)k+1 −2 ! − and that (ez 1)k ( 1=2)k+1 − − (2 ez)k+1 − (z log2)k+1 − − is analytic on any disc of radius less than log2 + 2pi . Singularity analysis (Sec- tion 11 of [7]) implies that j j z k k+1 n (e 1) 1 n k 1 [z ] − [z ](1 z=log2)− − (2 ez)k+1 ∼ 2log2 − − 1 k+1 nk ; (2.9) ∼ 2log2 Γ(k + 1)(log2)n ∞ where Γ(x) = e ttx 1dt. Therefore, R0 − − k+1 k n! 1 n k E(Xn) l ; k ∼ p(n) 2log2 (log2)n ∼ n where we have used (1.4) and Γ(k + 1) = k!. 2 The variance of Xn is asymptotically Var(Xn) = EXn(Xn 1)+EXn (EXn) = 2 2 2 − − ln + o(ln ) + (ln + o(ln)) (ln + o(ln)) = ln(1 + o(ln)). The conclusion that a:a:s: − Xn ln is a consequence of ∼ 2 P( Xn=ln 1 > e) = P( Xn ln > eln) Var(Xn)=(eln) = o(1): j − j j − j ≤ As a random variable Z with Poisson(ln) distribution has falling moments k exactly equal to E(Z) = l , and (Z ln)=pln converges weakly to the standard k n − normal distribution if ln ∞, (2.8) indicates that (Xn ln)=pln should have a distribution that is approximately! normal.
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