Asymptotics of certain coagulation-fragmentation processes and invariant Poisson-Dirichlet measures

We consider Markov chains on the space of (countable) partitions of the interval $[0,1]$, obtained first by size biased sampling twice (allowing repetitions) and then merging the parts with probability $\beta_m$ (if the sampled parts are distinct) or splitting the part with probability $\beta_s$ according to a law $\sigma$ (if the same part was sampled twice). We characterize invariant probability measures for such chains. In particular, if $\sigma$ is the uniform measure then the Poisson-Dirichlet law is an invariant probability measure, and it is unique within a suitably defined class of ``analytic'' invariant measures. We also derive transience and recurrence criteria for these chains.

By size-biased sampling according to a point p ∈ Ω 1 we mean picking the j-th part p j with probability p j . The starting point for our study is the following Markov chain on Ω 1 , which we call a coagulation-fragmentation process: size-bias sample (with replacement) two parts from p. If the same part was picked twice, split it (uniformly), and reorder the partition. If different parts were picked, merge them, and reorder the partition.
We call this Markov chain the basic chain. We first bumped into it in the context of triangulation of random Riemann surfaces [5]. It turns out that it was already considered in [15], in connection with "virtual permutations" and the Poisson-Dirichlet process. Recall that the Poisson-Dirichlet measure (with parameter 1) can be described as the probability distribution of (Y n ) n≥1 on Ω 1 obtained by setting Y 1 = U 1 , Y n+1 = U n+1 (1 − n j=1 Y j ), and reordering the sequence (Y n ) n , where (U n ) n is a sequence of i.i.d. Uniform[0,1] random variables. Tsilevich showed in [15] that the Poisson-Dirichlet distribution is an invariant probability measure for the Markov chain described above, and raised the question whether such an invariant probability measure is unique. While we do not completely resolve this question, a corollary of our results (c.f. Theorem 3) is that the Poisson-Dirichlet law is the unique invariant measure for the basic chain which satisfies certain regularity conditions.
Of course, the question of invariant probability measure is only one among many concerning the large time behavior of the basic chain. Also, it turns out that one may extend the definition of the basic chain to obtain a Poisson-Dirichlet measure with any parameter as an invariant probability measure, generalizing the result of [15]. We thus consider a slightly more general model, as follows.
For any nonnegative sequence x = (x i ) i , let |x| = i x i , the ℓ 1 norm of x, and |x| 2 2 = i x 2 i . Set and Ω ≤ = {p ∈ Ω : |p| ≤ 1}. Let 0 = (0, 0, . . . ) and defineΩ = Ω ∪ {0} and Ω ≤ = Ω ≤ ∪ {0}. Unless otherwise stated, we equip all these spaces with the topology induced from the product topology on R N . In particular,Ω ≤ is then a compact space. For a topological space X with Borel σ-field F we denote by M 1 (X) the set of all probability measures on (X, F) and equip it with the topology of weak convergence. M + (X) denotes the space of all (nonnegative) measures on (X, F).
Define the following two operators, called the merge and split operators, onΩ, as follows: Note that the operators M ij and S u i preserve the ℓ 1 norm. Let σ ∈ M 1 ((0, 1/2]) be a probability measure on (0, 1/2] (the splitting measure). For p ∈Ω ≤ and β m , β s ∈ (0, 1], we then consider the Markov process generated inΩ ≤ by the kernel K σ,βm,βs (p, ·) := 2β m i<j p i p j δ M ij p (·) + β s i p 2 i δ S u i p (·) dσ(u) It is straightforward to check (see Lemma 4 below) that K σ,βm,βs is Feller continuous. The basic chain corresponds to σ = U (0, 1/2], with β s = β m = 1. It is also not hard to check (see Theorem 6 below) that there always exists a K σ,βm,βs -invariant probability measure µ ∈ M 1 (Ω 1 ). Basic properties of any such invariant probability measure are collected in Lemma 5 and Proposition 7. Our first result is the following characterization of those kernels that yield invariant probability measures which are supported on finite (respectively infinite) partitions. To this end, let S := {p ∈ Ω 1 | ∃i ≥ 2 : p i = 0} be the set of finite partitions. On the left side a part of size p i has been chosen twice and is split with probability β s . On the right side two different parts of sizes p i and p j have been chosen and are merged with probability β m .
We now turn to the case σ = U (0, 1/2]. In order to define invariant probability measures in this case, set π : Ω → Ω 1 , p := π(p) = (p i /|p|) i≥1 . For each θ > 0 consider the Poisson process on R + with intensity measure ν θ (dx) = θx −1 e −x dx which can be seen either as a Poisson random measure N (A; ω) on the positive real line or as a random variable X = (X i ) ∞ i=1 taking values in Ω whose distribution shall be denoted by µ θ , with expectation operator E θ . (Indeed, and thus X ∈ Ω a.s.). A useful feature of such a Poisson process is that for any Borel subset A of R + with 0 < ν θ (A) < ∞, and conditioned on {N (A) = n}, the n points in A are distributed as n independent variables chosen each according to the law ν θ ( · |A). The Poisson-Dirichlet measure µ θ on Ω 1 is defined to be the distribution of ( X i ) i≥1 . In other words, µ θ = µ θ • π −1 . In the case θ = 1 it coincides with the previously described Poisson-Dirichlet measure. See [9], [10] and [3] for more details and additional properties of Poisson-Dirichlet processes.
We show in Theorem 3 below that, when σ = U (0, 1/2], for each choice of β m , β s there is a Poisson-Dirichlet measure which is invariant for K σ,βm,βs . We also show that it is, in this case, the unique invariant probability measure in a class A, which we proceed to define. Set . + x k < 1} and denote by A k the set of real valued functions onΩ k < that coincide (leb k -a.e.) with a function which has a real analytic extension to some open neighborhood of Ω k < . (Here and throughout, leb k denotes the k-dimensional Lebesgue measure; all we shall use is that real analytic functions in a connected domain can be recovered from their derivatives at an internal point.) For any µ ∈ M 1 (Ω 1 ) and each integer k, define the measure µ k ∈ M + Ω k < by . An alternative description of µ k is the following one: pick a random partition p according to µ and then sample sizebiased independently (with replacement) k parts p i 1 , . . . , p i k from p. Then, µ k (B) = P (the i j -s are pairwise distinct, and (p i 1 , . . . , p i k ) ∈ B) .
Part of the proof of part (b) of Theorem 3 below will consist in verifying that these measures (µ k ) k≥1 characterize µ (see [12,Th. 4] for a similar argument in a closely related context).
Set for k ∈ N, Our main result is part (b) of the following: The Poisson-Dirichlet law of parameter θ belongs to A := ∞ k=1 A k , and is invariant (in fact: reversing) for the kernel K σ,βm,βs . (b) Assume a probability measure µ ∈ A is K σ,βm,βs -invariant. Then µ is the Poisson-Dirichlet law of parameter θ.
The structure of the paper is as follows: In Section 2, we prove the Feller property of K σ,βm,βs , the existence of invariant probability measures for it, and some of their basic properties. Section 3 and 4 are devoted to the proofs of Theorems 1 and 2 respectively, Section 5 studies the Poisson-Dirichlet measures and provides the proof of Theorem 3. We conclude in Section 6 with a list of comments and open problems.
Lemma 4. The kernel K σ,βm,βs is Feller, i.e. for any continuous function f : One may assume that f (p) is of the form F (p 1 , . . . , p k ) with k ∈ N and F ∈ C(Ω k ≤ ), since any f ∈ C(Ω ≤ ) can be uniformly approximated by such functions, and denote accordingly p k the R k norm of p's first k components. We shall prove the lemma in this case by showing that both sums in (2) contain finitely many nonzero terms, this number being uniformly bounded on some open neighborhood of a given q, and that g i and h i are continuous for every i.
For the second sum these two facts are trivial: S u i p and p coincide in their first k components ∀u ∈ (0, 1/2], ∀i > k, since splitting a component doesn't affect the ordering of the larger ones, and thus h i ≡ 0 for i > k. Moreover, h i 's continuity follows from equicontinuity of (S u i ) u∈(0,1) . As for the first sum, given q ∈Ω ≤ with positive components (the necessary modification when q has zero components is straightforward), let n = n(q) > k be such that q n < 1 4 q k and consider q's open neighborhood U = U (q) = p ∈Ω ≤ : p k > 2 3 q k , p n < 4 3 q n . In particular, for all p ∈ U , p n < 1 2 p k and thus, when j > i > n , p i + p j ≤ 2p n < p k , which means that M ij p and p coincide in their first k components, or that g i (p) = 0 for every i > n(q) and p ∈ U (q).
Finally, each g i is continuous because the series defining it converges uniformly. Indeed, for j > i and uniformly in p, which proves the uniform convergence.
Lemma 5. Let µ ∈ M 1 (Ω ≤ ) be K σ,βm,βs -invariant. Then Furthermore, if we set for n ≥ 1, then for all n ≥ 1, Proof. Let ε ∈ [0, 1], and consider the random variable X ε := i 1 ε<p i onΩ ≤ which counts the intervals longer than ε. We first prove (3). (The value ε = 0 is used in the subsequent proof of (5).) Assume that X ε is finite which is always the case for ε > 0 since onΩ ≤ , X ε ≤ 1/ε and is also true for ε = 0 if only finitely many p i are non zero. Then the expected (conditioned on p) increment ∆ ε of X ε after one step of the underlying Markov process is well-defined. It equals The right hand side of (6) converges as ε tends to 0 to Since µ is K σ,βm,βs -invariant we have ∆ ε dµ = 0 for all ε. Now (3) follows from (7) by dominated convergence since |∆ ε | ≤ 2.
Consequently, there are µ ∈ M 1 (Ω ≤ ) and a strictly increasing sequence (m n ) n of positive integers such thatν mn converges weakly towards µ as n → ∞. This limiting measure µ is invariant under K σ,βm,βs by the following standard argument.
Proof. Let k ≥ 1 be an integer and α ∈ {1, 2}. Given p, consider the expected increment ∆ α,k of i p k i after one step of the process driven by K σα,βm,αβs,α : Note that i p k i dµ is finite because of k ≥ 1. Therefore, by invariance, ∆ α,k dµ = 0, which implies Hence, for any k, Taking k → ∞ we conclude that γ = 1. This proves the second claim. In addition, we have for all k ≥ 1. Obviously, (9) also holds true for k = 0. Extend σ α to probability measures on [0, 1] which are supported on [0, 1/2]. It is enough for the proof of Fix such an f and choose a sequence of polynomials which converges uniformly on [0, 1] to f as n → ∞. Then π n (t) vanishes on the support of σ 1 and σ 2 we get for α = 1, 2, which is the same for α = 1 and α = 2 due to (9).

Support properties
Theorem 1 is a consequence of the following result.
Proof. We start by proving that 1/x dσ(x) < ∞ implies µ[S] = 1. Fix an arbitrary 0 < ϑ ≤ 1/2 and consider the random variables After one step of the process W n may increase, decrease or stay unchanged. If we merge two intervals then W n cannot decrease, but may increase by the mass of one or two intervals which are smaller than ϑ n but become part of an interval which is bigger than ϑ n . If we split an interval then W n cannot increase, but it decreases if the original interval was larger than ϑ n and at least one of its parts is smaller than ϑ n . Thus given p, the expected increment ∆ of W n after one step of the process is We bound ∆ + from below by and ∆ − from above by Since µ is invariant by assumption, 0 = ∆ dµ = ∆ + dµ− ∆ − dµ and therefore which is finite by assumption. Thus by Borel-Cantelli, W n ♯I n+1 is µ-a.s. eventually (for large n) less than 1/2. However, W n converges µ-a.s. to 1 as n tends to ∞. Thus even ♯I n+1 is µ-a.s. eventually less than 1/2, which means that I n+1 is µ-a.s. eventually empty, that is µ[S] = 1. Now we assume µ[S] > 0 in which case there exist some i ≥ 1 and ε > 0 such that δ := µ[p i > ε, p i+1 = 0] > 0. By i successive merges of the positive parts and µ's invariance we obtain Next, we assume µ[{p}] > 0 and note that K σ,βm,βs 1 S = 1 S and thus, definingμ := µ/µ[S], one obtains an invariant measure supported on S. The chain determined by K σ,βm,βs on S is δp-irreducible, and hasμ as invariant measure, withμ[{p}] > 0. Therefore, Kac's recurrence theorem [ where ξ has distribution σ. Furthermore, restricted to A and conditioned on ξ, H ≥ τ P p ξ -a.s., where in terms of the chain's sampling and merge/split interpretation, τ is the first time a marked part of size ξ is sampled, i.e. a geometric random variable with parameter 1 Corollary 9. If 1/x dσ(x) < ∞ then there exists a unique K σ,βm,βs -invariant probability measure µ ∈ M 1 (Ω 1 ).
Proof. In view of Theorem 1, for the study of invariant measures it is enough to restrict attention to the state space S, where the Markov chain (p(n)) n is δpirreducible, implying, see [11,Chapter 10], the uniqueness of the invariant measure.

Transience and recurrence
Proof of Theorem 2. The statement about positive recurrence is included in Theorem 8.
The idea for the proof of the transience statement is to show that under (1) the event that the size of the smallest positive part of the partition never increases has positive probability. By n 0 := 0 and n j+1 := inf{n > n j : p(n) = p(n − 1)} (j ≥ 0) we enumerate the times n j at which the value of the Markov chain changes. Denote by s n the (random) number of instants among the first n steps of the Markov chain in which some interval is split. Since j − s n j is the number of steps among the first n j steps in which two parts are merged and since this number can never exceed s n j if p(0) =p, we have that Pp-a.s., Let (τ l ) l≥1 denote the times at which some part is split. This part is split into two parts of sizes ℓ(l) and L(l) with 0 < ℓ(l) ≤ L(l). According to the model the random variables ξ l := ℓ(l)/(ℓ(l) + L(l)), l ≥ 1, are independent with common distribution σ. Further, for any deterministic sequence ξ = (ξ n ) n , let P ξ,p [ · ] denote the law of the process which evolves using the kernel K σ,βm,βs except that at the times τ l it uses the values ξ l as the splitting variables. Note that Now denote by q(n) := min{p i (n) : i ≥ 1, p i (n) > 0} (n ≥ 0) the size of the smallest positive part at time n. We prove that for N ≥ 0, (Here and in the sequel, we take ξ 1 ∧ . . . ∧ ξ s N = ∞ if s N = 0). Indeed, we need only consider the case s N > 0, in which case there exists a 1 ≤ t ≤ s N such that ξ t = ξ 1 ∧ . . . ∧ ξ s N , and τ t ≤ N . But clearly q(τ t ) ≤ ξ t , and then the condition q(1) ≥ · · · ≥ q(N ) and the fact that τ t ≤ N imply q(N ) ≤ q(τ t ) ≤ ξ t = ξ 1 ∧. . .∧ξ s N , as claimed.
Next, fix some ε ∈ (0, β 0 /2] where β 0 := min{β m , β s }. We will prove by induction over j ≥ 1 that For j = 1 the left hand side of (13) equals the probability that the unit interval is split in the first step with the smaller part being smaller than ε which equals β s 1 ξ 1 <ε . Assume that (13) has been proved up to j. Then, with F n j = σ(p(n), n ≤ n j ), Now choose k minimal such that p k (n j ) = q(n j ). One possibility to achieve q(n j ) ≥ q(n j+1 ) is not to merge the part p k (n j ) in the next step in which the Markov chain moves. The probability to do this is Therefore (14) is greater than or equal to By (12) this can be estimated from below by This is due to (11) greater than or equal to Along with the induction hypothesis this implies (13) for j + 1.
Taking expectations with respect to ξ in (13) yields By independence of ξ 1 from ξ i , i ≥ 2, the right hand side of (15) equals Observe that (1) implies P [ξ 1 < ε] = σ[(0, ε)] > 0. By Jensen's inequality and monotone convergence, (16) can be estimated from below by where we used that due to independence Due to assumption (1), (17) and therefore also the left hand side of (15) are positive. This implies transience ofp.

Poisson-Dirichlet invariant probability measures
Throughout this section, the splitting measure is the uniform measure on (0, 1/2]. To emphasize this, we use K βm,βs instead of K σ,βm,βs throughout. Recall that θ = β s /β m . It will be convenient to equip Ω (but notΩ ≤ ) with the ℓ 1 topology (noting that the Borel σ-algebra is not affected by this change of topology), and to replace the kernel K βm,βs by Both kernels coincide on Ω 1 (not on Ω ≤ ). However, K H βm,βs has the advantage that it is well defined on all of Ω and is homogeneous (hence the superscript H) in the sense of the first of the following two lemmas, whose proof is straightforward and in which by a slight abuse of notation K H βm,βs will denote both the kernel in Ω 1 and in Ω and also the operators induced by these kernels, the distinction being clear from the context. Lemma 10. For all p ∈ Ω, K H βm,βs (πp, ·) = K H βm,βs (p, ·) • π −1 . More generally, denoting (Πf )(p) = f (π(p)), we have K H βm,βs Π = ΠK H βm,βs .
Proof of Lemma 11. Note that we work with the ℓ 1 topology, and hence have to modify the proof in Lemma 4. The ℓ 1 topology makes the mapping p → p continuous (when Ω 1 is equipped with the induced ℓ 1 topology). Fix F ∈ C b (Ω). By (18) we have Note that for l = 1, 2 , K l (p) is of the form T l (p) p, p , with T l (·) ∈ C (Ω; L(ℓ 1 , ℓ ∞ )), and ·, · denoting the standard duality pairing. In stating this we have used the facts that F is continuous and bounded, and that all the mappings M ij and S u i are contractive.
The continuity of K l , l = 1, 2 , then follows from after observing that | q| and T l (q) remain bounded in any ℓ 1 neighborhoods of p.
The continuity of K 3 is obvious being the product of two continuous functions of p. It has thus been shown that K H βm,βs F ∈ C(Ω).
Theorem 12. The Poisson-Dirichlet measure µ θ ∈ M 1 (Ω 1 ) is reversing for K σ,βm,βs with σ = U (0, 1/2]. Proof. By Lemma 10 it suffices to verify that µ θ ∈ M 1 (Ω) is reversing for the kernel K H βm,βs , which for simplicity will be denoted by K for the rest of this proof. We thus need to show that Because µ θ • M −1 ij and µ θ • (S u i ) −1 are absolutely continuous with respect to µ θ , it follows from (19) that if F, {F n } n are uniformly bounded functions such that |F n −F |µ θ (dp) → n→∞ 0, then |KF n −KF |µ θ (dp) → n→∞ 0. Thus, by standard density arguments we may and shall assume F and G to be continuous.
Define for each ε > 0 the truncated intensity measure ν ε θ ≡ 1 (ε,∞) ν θ , and the corresponding Poisson measure µ ε θ , with expectation operator E ε θ . Alternatively, if X is distributed in Ω according the µ θ , then µ ε θ is the distribution of T ε X : implying that the measures µ ε θ converge weakly to µ θ as ε → 0. To prove (20) we first write and conclude that the first and third terms in (21) converge to 0 as ε → 0 by virtue of the weak convergence of µ ε θ to µ θ and K's Feller property, established in Lemma 11. It thus remains to be shown that, for all F, G ∈ B(Ω) and ε > 0, The truncated intensity ν ε θ has finite mass V ε θ = θ ∞ ε x −1 e −x dx, and thus N (R + ) < ∞, µ ε θ -a.s. In particular each F ∈ B(Ω) can be naturally represented as a sequence (F n ) ∞ n=0 of symmetric F n 's ∈ B R n + , with F n ∞ ≤ F ∞ for each n. As a result, and in terms of the expectation operators E ε θ,n of µ ε θ conditioned on {N (R + ) = n}, we may write while by the definition (18) of K H βm,βs and the properties stated above of the Poisson random measure conditioned on {N (R + ) = n}, where x = (x 1 , . . . , x n ). Our goal is to prove that this expression, after summing in n, is roughly symmetric in F and G (as stated precisely in (22)). Obviously I n (F, G) (with an error appropriately small as ε → 0). This will be achieved by a simple change of variables, including the splitting coordinate u in I (2) .
In the integral of the i-th term in I (2) n−1 (G, F ) perform the change of variables (u, x 1 , . . . , x n−1 ) → (y 1 , . . . , y n ) given by y = S u i x (or (u, x) = ( y i y 0 +y i , M i n y)).
More precisely, u)x i for which |y| = |x| and dy 1 . . . dy n = x i du dx 1 . . . dx n−1 , so that G n (y)F n−1 (M i n y) e −|y| dy 1 . . . dy n |y| 2 y 1 . . .y i . . . y n−1 + C ε n (C ε n is as the term preceding it but with the dy i and dy n integrals taken in [0, ε], and the notationy i means that the variable y i has been eliminated from the denominator) (by F n−1 's symmetry, the sum's (n − 1) terms are equal, hence the last equality).
On the other hand, and for the same reason of symmetry, the n(n − 1) terms in I (1) n (F, G) are all equal so that Comparing (25) with (26), and observing that by definition β m θ = β s , we conclude that there exists a C > 0 such that, for n ≥ 2, Applying (27) via (24) in (23) twice, once as written and once reversing the roles of F and G, and noting that I The Poisson-Dirichlet law µ = µ θ is reversing by Theorem 12, and hence invariant. We now show that it belongs to A. Note first that µ k is absolutely continuous with respect to leb k : for any D ⊂ Ω k < with leb k (D) = 0, it holds that where we used the fact that under µ θ , π(X) = X/|X| and |X| are independent, with |X| being distributed according to the Gamma law γ θ (dx) of density 1 x≥0 x θ−1 e −x /Γ(θ) (see [9]). It thus suffices to compute the limit where all x i are pairwise distinct and nonzero, to have For such x 1 , . . . , x k , set By the memoryless property of the Poisson process, for any Borel subset A ⊂ R, where (28), as above, is due to [9]. Further, recall that N and ( X i ) i are independent. Recall that the density of the Poisson process at (y 1 , . . . , y k ) is Performing the change of variables y i /(z + |y|) = x i , one finds that the Jacobian of this change of coordinate is (z + |y|) k /(1 − |x|) (in computing this Jacobian, it is useful to first make the change of coordinates (y 1 , . . . , y k−1 , |y|) → (x 1 , . . . ,x k−1 , |x|) where |y|, |x| are considered as independent coordinates, and note the block-diagonal structure of the Jacobian). It follows that which is real analytic on {x ∈ R k : |x| < 1}. Thus, µ θ ∈ A. In passing, we note that m k (·) = 1 onΩ k < when θ = 1. (b) 1) First we show that the family of functions (m k ) k≥1 associated with µ, determines µ. To this end, define for j ∈ N k (k ∈ N) functions g j ,ĝ j : Ω 1 → [0, 1] by Note that any functionĝ j with j ∈ N k can be written after expansion of the product as a (finite) linear combination of functions g h with h ∈ N n , n ≥ 1. Since we have by the definition of µ k that the family (m k ) k≥1 therefore determines the expectations ĝ j dµ (j ∈ N k , k ≥ 1). Consequently, (m k ) k≥1 determines also the joint laws of the random variables (Z 1 , . . . , Z k ), k ≥ 1, under µ. We claim that these laws characterize µ. Indeed, let µ be the distribution of the random variable π := (Z n ) n≥0 : Ω 1 → [0, 1] N under µ. Since π is injective it suffices to show that the distributions of (Z 1 , . . . , Z k ), k ≥ 1, under µ determineμ. But, since any continuous test function F on the compact space [0, 1] N can be uniformly approximated by the local function F k ((x n ) n≥1 ) := F (x 1 , . . . , x k , 0, . . . ), this is true due to 2) For µ ∈ A, the set of numbers with k ≥ 1 and n 1 ≥ n 2 ≥ . . . ≥ n k ≥ 0 yare enough to characterize (m k ) k , and hence by the first part of the proof of b), to characterize µ. It is thus enough to prove that K βm,βs uniquely determines these numbers. Toward this end, first note that To simplify notations, we define m 0 ≡ 1 and extend m k to a function on [0, 1] k by setting it 0 on the complement ofΩ k < . For k ≥ 1 we have Indeed, for k = 1 this is (31) while for k ≥ 2, and arbitrary B ∈ BΩk−1 which implies (32). Now we fix k ≥ 1, apply K βm,βs to the test function #{j ∈ N k = : p j i ∈ (x i , x i + δ) , i = 1, . . . , k}δ −k , with (x 1 , . . . , x k ) ∈Ω k < having pairwise distinct coordinates, and take δ ց 0, which yields the basic relation Here the left hand side represents mergings and splittings that produce a new part roughly at one of the x i -s; the right hand side represents parts near one of the x i -s that merge or split. After multiplying by x 1 · · · x k , rearranging and using (32) to get rid of the integral with upper limit 1, we obtain the equality We now proceed to show how (33) -(37) yield all the required information. As starting point for a recursion, we show how to compute m k (0, . . . , 0) for all k ≥ 1. . Fix n such that n 1 ≥ n 2 ≥ . . . ≥ n k , with n 1 ≥ 2. Our analysis rests upon differentiating (33) -(37) n 1 times with respect to x 1 ; to make this differentiation easy, call a term a G term of degree ℓ if it is a linear combination of terms of the form k+1 (x 1 , 0, x 2 , . . . , x k ) and x 1 m k+1 (x 1 , 0, x 2 , . . . , x k ) . Note that (33) -(37) contains one G term of degree −1 in (33) and that differentiating a G term of degree ℓ once yields a G term of degree ℓ + 1. Thus, differentiating the G term in (33) n 1 ≥ 2 times and substituting x 1 = 0, we recover a constant multiple of m  Observe, that differentiating an H term of degree ℓ produces an H term of degree ℓ+1. If we differentiate twice the term x 1 x 1 0 m k (z, x 2 , . . . , x k )dz in (34) we get an H term of degree 0. Therefore differentiating this term n 1 ≥ 2 times results in an H term of degree n 1 − 2. Since the term x 2 1 m k (x 1 , . . . , x k ) in (37) is an H term of degree -2, differentiating this term n 1 times produces also an H term of degree n 1 − 2. Thus both terms produce after n 1 -fold differentiation and evaluation at x 1 = 0 a constant multiple of m (n 1 −2) k (0, x 2 , . . . , x k ). The H term x 1 m k (x 1 , . . . , x k ) in (37) is treated more carefully. It is easy to see by induction that its n 1 -th derivative equals n 1 m k (x 1 , . . . , x k ). Evaluating it at x 1 = 0 gives n 1 m (n 1 −1) k (0, x 2 , . . . , x k ). Moreover, the terms in (35) and (36) for i = 1 vanish when differentiated twice with respect to x 1 .
Summarizing the above, we conclude by differentiating (33) -(37) n 1 ≥ 2 times with respect to x 1 and subsequent evaluation at x 1 = 0 that there are some constants C i (n 1 ), such that Further differentiating with respect to x 2 , . . . , x k , one concludes that there exist constants D i n,n ′ such that β m n 1 m (n 1 −1,n 2 ,... ,n k ) k = n ′ :|n ′ |≤|n|−2,n ′ i ≤n i It is tempting to use the technique leading to (3) in order to prove the conjecture by characterizing the expectations with respect to µ of suitable test functions. One possible way to do that is to consider a family of polynomials defined as follows. Let n = (n 2 , n 3 , . . . , n d ) be a finite sequence of nonnegative integers, with n d ≥ 1. We set |n| = d j=2 jn j , i.e. we consider n as representing a partition of |n| having n j parts of size j, and no parts of size 1. Recall next Z j = Z j (p) = i p j i and the n-polynomial |n| is the degree of P n , and, with n and d as above, d is the maximal monomial degree of P n . Because we do not allow partitions with parts of size 1, it holds that P n = P n ′ if n = n ′ (i.e, there exists a point p ∈ Ω 1 such that P n (p) = P n ′ (p)). It is easy to check that the family of polynomials {P n } is separating for M 1 (Ω). Letting ∆ n denote the expected increment (conditioned on p) of P n after one step of the process, we have that ∆ n is uniformly bounded. Hence, by invariance of µ, ∆ n dµ = 0. Expanding this equality, we get that Note that all terms in (40) are positive. Note also that the right hand side of (40) is a polynomial of degree |n| + 2, with maximal monomial degree d + 2, whereas the left hand side is a polynomial of degree at most |n| + 2 and maximal monomial degree at most d. Let π(k) denote the number of integer partitions of k which do not have parts of size 1. Then, there are π(k) distinct polynomials of degree k, whereas (40) provides at most π(k − 2) relations between their expected values (involving possibly the expected value of lower order polynomials). Since always π(k) > π(k − 2), it does not seem possible to characterize an invariant probability measure µ ∈ M 1 (Ω 1 ) using only these algebraic relations.
3) We have not been able to resolve whether the statep = (1, 0, 0, . . . ) is transient or null-recurrent for K σ,1,1 with σ = U (0, 1/2]. 4) There is much literature concerning coagulation-fragmentation processes. Most of the recent probabilistic literature deals with processes which exhibit either pure fragmentation or pure coagulation. For an extensive review, see [1], and a sample of more recent references is [2], [4] and [6]. Some recent results on coagulationfragmentation processes are contained in [8]. However, the starting point for this and previous studies are the coagulation-fragmentation equations, and it is not clear how to relate those to our model. The functions m k introduced in the context of Theorem 3 are related to these equations. 5) A characterization of the Poisson-Dirichlet process as the unique measure coming from an i.i.d. residual allocation model which is invariant under a split and merge transformation is given in [7]. J. Pitman has pointed out to us that a slight modification of this transformation, preceded by a size biased permutation and followed by ranking, is equivalent to our Markov transition K σ,βm,βs . Pitman [13] then used this observation to give an alternative proof of part (a) of Theorem 3. 6) Yet another proof of part a) of Theorem 3 which avoids the Poisson representation and Theorem 12 can be obtained by computing the expectation of the polynomials P n (p), defined in remark 1) above, under the Poisson-Dirichlet law. We prefer the current proof as it yields more information and is more transparent. 7) A natural extension of Poisson-Dirichlet measures are the two parameter Poisson-Dirichlet measures, see e.g. [14]. Pitman raised the question, which we have not addressed, of whether there are splitting measures σ which would lead to invariant measures from this family. 8) While according to Theorem 3 there is a reversing probability measure for σ = U (0, 1/2] this does not hold for general σ ∈ M 1 ((0, 1/2]). For instance, let us assume that the support of σ is finite. Then there exist 0 < a < b ≤ 1/2 such that σ[(a, b)] = 0. To show that any invariant measure µ is not reversing it suffices to find s, t ∈ Ω 1 such that the detailed balance equation fails. Due to Theorem 8, µ[{p}] > 0. Now we first refine the partitionp by successive splits until we reach a state p ∈ Ω 1 with p 1 < ε, where ε > 0 is a small number. Since µ has finite support, µ[{p}] > 0. Then we create from p by successive mergings some s ∈ Ω 1 with a < s 2 /s 1 < b, which is possible if ε was chosen small enough. Again, µ[{s}] > 0. If we call now t the state which one gets from s by merging s 1 and s 2 , then the left hand side of (41) is positive. On the other hand, the right hand side of (41) is zero because of K(t, {s}) = 0 due to the choice of a and b.