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 equivalence 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,∈z) R.A partial∈ preorder is a relation R on X which∈ is reflexive and∈ transitive. A relation∈ 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 preorder∈ if it is a partial preorder∈ that satisfies trichotomy.∈ 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 it∈ is reflexive,∈ 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.∈ 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 similarly∈ for y. We will∈ 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 ∈ 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 θr, 1 r n, ≤ ≤ n ∞ n z n ∑ θr p(n,r) = n![z ] ∑ θn(e 1) !. r=1 n=0 − Proof We observe that ∞ n zn ∞ ∞ p(n,r)zn ∑ ∑ θr p(n,r)! = ∑ ∑ θr n=0 r=1 n! r=1 n=r n! ∞ ∞ S(n,r)zn = ∑ ∑ r!θr r=1 n=r n! ∞ z r = ∑ θr(e 1) . r=1 − The lemma follows immediately.

If we take θr = 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 2πi) 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 ε)− ), where r = log2+2πi ; that is, exponentially small. An − z 1 | | 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 λ to be n n λ = . (2.6) n 2log2 k We will show that E(Xn) λ 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 > ε) = 0 for all ε > 0. → | − | 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) λ (2.8) k ∼ n and that a.a.s. Xn λn, ∼ where λn 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 θr = (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 + 2πi . Singularity analysis (Sec- tion 11 of [7]) implies that | |

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) λ , 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 − − λn + o(λn ) + (λn + o(λn)) (λn + o(λn)) = λn(1 + o(λn)). The conclusion that a.a.s. − Xn λn is a consequence of ∼ 2 P( Xn/λn 1 > ε) = P( Xn λn > ελn) Var(Xn)/(ελn) = o(1). | − | | − | ≤

As a random variable Z with Poisson(λn) distribution has falling moments k exactly equal to E(Z) = λ , and (Z λn)/√λn converges weakly to the standard k n − normal distribution if λn ∞, (2.8) indicates that (Xn λn)/√λn should have a distribution that is approximately→ normal. Asymptotic− normality could be a subject for future research.

5 3 The size of a typical block

Because the blocks in a random preorder are linearly ordered, we may take B1 as the size of a typical block. Given a preorder (B1,B2,...) on X, we may deﬁne a new preorder on X B by the sequence (B ,B ,...). This operation can be \ 1 2 3 reversed: given a preorder on X B1, (B2,B3,...), we can insert B1 to get the original preorder on X. The above\ correspondence implies

n p(n k) P(B = k) = − 1 k p(n) and for ﬁxed k the asymptotic (1.4) gives

n (n k)! (log2)k P(B = k) − (log2)k = . (3.10) 1 ∼ k n! k! It is easily checked that the distribution deﬁned by the right hand side of (3.10) is the same as the distribution of the conditioned random variable (Z Z > 0), where Z is Poisson(log2) distributed. | We will use an argument similar to the one above and the results of Section 2 to show that the distribution of ﬁxed block sizes are asymptotically i.i.d. and distributed as (Z Z > 0). |

Theorem 2 Let a ﬁnite set of indices i1,i2,...,iL and a sequence of nonnegative integers a1,a2,...,aL be given. Then

L (log2)ai P(B = a ,B = a ,...,B = a ) ∏ . i1 1 i2 2 iL L ∼ i=1 ai! That is, the distribution of the B converges weakly to an i.i.d. sequence of random i j variables distributed as (Z Z > 0), where Z is Poisson(log2) distributed. | Proof Given a preorder with blocks B ,B ,...,B on X, we can form a new pre- i1 i2 iL L order by (B1,...,Bi 1,Bi 1+1,...,Bi 1,Bi +1 ...) on X l=1 Bi . On the other 1 2 2 S l − − − \ L hand, a preorder (B1,...,Bi 1,Bi 1+1,...,Bi 1,Bi +1 ...) on X l=1 Bi forms 1 2 2 \S l a valid preorder (B ,B ,...) −on X −by the insertion− of the blocks B ,B ,...,B if 1 2 i1 i2 iL and only if (B1,...,Bi 1,Bi 1+1,...,Bi 1,Bi +1 ...) is a preorder with at least 1− − 2− 2

6 i L nonempty blocks. Therefore, with b deﬁned as b = ∑L a , L − l=1 l P(B = a ,B = a ,...,B = a ) i1 1 i2 2 iL L ∞ n ∑r=i L p(n b,r) = L− − a ,a ,...,a ,n b p(n) 1 2 L − n p(n b) = − P(X a L). (3.11) a ,a ,...,a ,n b p(n) n b ≥ L − 1 2 L − − The probability in (3.11) approaches 1 because of Theorem 1. The other factors have asymptotics that give the theorem.

4 The number of blocks of ﬁxed size

(s) Deﬁne Xn to be the number of blocks of size s = s(n) in a random preorder on n (s) elements. Deﬁne λn to be (log2)s 1n λ (s) = − . (4.12) n 2s! Theorem 3 The kth falling moment of the number of s-blocks of a random preorder equals

k z k j k!n! (k) (e 1) − E(X(s)) = [zn ks] ∑ j − . (4.13) n k p(n)(s!)k − j!(2 ez)k j+1 j=0 − − It follows that for ﬁxed k and s = o(n) such that λ (s) ∞, n → E(X(s)) (λ (s))k n k ∼ n and that a.a.s. X(s) λ (s), (4.14) n ∼ n (s) where λn is deﬁned by (4.12).

Proof Let ps(n,k) be the number of preorders on n elements with exactly k blocks (s) of size s. The kth falling moment of Xn is 1 ∞ E (s) (Xn )k = ∑(r)k ps(n,r). p(n) r=0

7 ∑∞ The quantity r=0(r)k ps(n,r) counts the number of preorders with k labelled s- blocks, where each of the labelled s-blocks is given a unique label from the set 1,2,...,k . This number is also counted by: ﬁrst, choosing k s-blocks to be the {ones marked;} second, forming a preorder on the n ks remaining elements with r blocks; third, inserting the s-blocks into the preorder− in the order they were chosen k+r in one of k ways; fourth, marking the inserted s-blocks in one of k! ways. We therefore have 1 ∞ n n s n (k 1)s k + r E (s) (Xn )k = ∑ − − − p(n ks,r) k! p(n) r=1 s s ··· s − k n! ∞ = ∑ p(n ks,r)(k + r) p(n)(s!)k(n ks)! − k − r=1 n! ∞ = [ n ks] ( + ) ( z )n k z − ∑ k n k e 1 p(n)(s!) n=0 − (4.15) where we have made use of Lemma 1 at (4.15). We use the identity

∞ k k k k j d x k (k j)!(k) jx − ∑ (k + n) xn = = ∑ − k dxk 1 x j (1 x)k j+1 n=0 − j=0 − −

k j k j d ∑k k d d − in (4.15), which follows from the formula k uv = j=0 j j u k j v for func- dx dx dx − tions u(x) and v(x). After substitution of the identity and simpliﬁcation (4.15) becomes (4.13). In (4.13), the singularity of largest degree occurs at z = log2 when j = 0. The E (s) asymptotics of (Xn )k are given by

z k (s) n!k! n ks (e 1) E(X ) [z − ] − n k ∼ p(n)(s!)k (2 ez)k+1 − 2k!(log2)n+1 1 k+1 (n ks)k k Γ − n ks ∼ (s!) 2log2 (k + 1)(log2) − s 1 k (log2) − n , ∼ 2s! where we have used (1.4), (2.9), and the assumption s = o(n). The almost sure convergence result (4.14) is an application of Chebyshev’s inequality as in the

8 proof of Theorem 1.

(s) X λn(s) One would expect from Theorem 3 that the distribution of n − converges √λn(s) weakly to a standard normal distribution as long as λn(s) ∞, where λn(s) = s 1 → (log2) − n 2s! . This could be the subject of further investigations. The method of the proof of Theorem 3 can be used to derive asymptotics of (s ) (s ) (s ) (s ) E 1 2 1 k1 2 k2 joint falling moments. For example, ((Xn )k (Xn )k ) (λn ) (λn ) for 1 2 ∼ ﬁxed s1, s2, k1, k2. ∑∞ (s) ∑∞ (s) Observe that s=1 sλn = n and s=1 λn = λn, showing that Theorem 4.13 agrees with (1.1) and Theorem 1, respectively, and indicating that Theorem 3 gives a good picture of the block structure of a random preorder.

5 Maximal block size

We are also able to estimate closely the the maximum size of a block in a random preorder. Deﬁne µn to be

(s) µn = max s : λ 1 n n ≥ o and deﬁne µn if λn(µn) √µn, νn = ≥ (5.16) µn 1 if λn(µn) < √µn. − Theorem 4 Let Mn = maxi 1 Bi be the maximal size of a block in a random pre- ≥ order. Let νn be deﬁned by (5.16). Then νn logn/loglogn and ∼ P(Mn νn,νn + 1 ) 1 as n ∞. (5.17) ∈ { } → → (s) Proof Clearly, λn is monotone decreasing in s for s 2. Taking the logarithm (s) ≥ of λn produces

logλ (s) = logn + (s 1)loglog2 logs! log2 n − − − = logn slogs + O(s). (5.18) − logn Plugging s = loglogn into (5.18) gives logn lognlogloglogn logn logλn = + O ∞, loglogn loglogn loglogn →

9 from which it follows that for large enough n, µn > logn/loglogn. On the other logn + 2logloglogn hand, if we plug loglogn 1 loglogn into the right hand side of (5.18) we get logn 2logloglogn logλn 1 + loglogn loglogn logn 2logloglogn logloglogn = logn 1 + loglogn logloglogn + O − loglogn loglogn − loglogn logn +O loglogn lognlogloglogn logn(logloglogn)2 = + O ∞, − loglogn (loglogn)2 → −

< logn + 2logloglogn so that µn loglogn 1 loglogn for large enough n. We have shown that logn logn 2logloglogn < µn < 1 + loglogn loglogn loglogn

logn logn for large enough n and, in particular, that µn loglogn and νn loglogn . Deﬁne the index sets ∼ ∼

N = n 1 : λn(µn) √µn 1 { ≥ ≥ } and N = n 1 : λn(µn) < √µn . 2 { ≥ } We prove (5.17) first for indices going to infinity in N1 and then for indices going to infinity in N2. ∞ ∞ ∞ When n in N1, λn(νn) √µn as n , so the proof of Theorem 3 (ν→) ≥ → → (µ +1) (µ +2) gives P(X n > 0) 1 and so P(Mn < νn) 0. The ratio λ n /λ n = n → → n n ∞ (µn+1) (µn+2) (µn + 2)/log2 and λn < 1 imply λn 0. We will show, further- more, that → → (s) ∑ E(Xn ) = o(1), (5.19) s µn+2 ≥ which implies P(Mn > νn + 1) 0. By Theorem 3, after some simplification, for all s [1,n] → ∈ (s) n! n s z 2 E(X ) = [z − ](2 e )− n p(n)s! −

10 Kn! n n s (5.20) ≤ p(n)s! (log2) − s 1 (log2) − n K0 ≤ s! for constants K,K0 > 0, where we have used the O( ) version of singularity anal- n(log2)s+1/(s+1)! log2 · ysis [7] at (5.20). The ratios n(log2)s/(s)! = s+1 are less than some ﬁxed ρ < 1 for all s µn + 2 for large enough n, so that for n large enough, ≥ s 1 (µn+2) (s) n(log2) − K0λn ∑ E(X ) K0 ∑ 0. (5.21) n ≤ s! ≤ 1 ρ → s µn+2 s µn+2 ≥ ≥ − (µ ) (µ 1) (µ ) (ν ) When n ∞ in N , λ n 1 and λ n /λ n = µn/log2 ∞ give λ n → 2 n ≥ n − n → → ∞ P (µn) (µn) (µn+1) , hence (Mn < νn) 0. On the other hand, λ < √µn and λn /λn = → (µ +1) 1/2 (µn + 1)/log2 ∞ give λ n = O(µn ) = o(1) and an argument like the → − one showing (5.21) results in P(Mn > νn + 1) 0. → Asymptotic two-point concentration theorems are well known from random graph theory. See Theorem 7, page 260 of [3] for such a result regarding clique number.

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