A functional equation whose unknown is P([0,1]) valued
aa r X i v : . [ m a t h . P R ] N ov A functional equation whose unknown is P ([0 , Giacomo Aletti
ADAMSS Center–Dipartimento di Matematica,Universit`a degli Studi di Milano, Milan, Italy [email protected]
Caterina May
Dipartimento SEMEQ,Universit`a del Piemonte Orientale, Novara, Italy [email protected]
Piercesare Secchi
MOX–Dipartimento di Matematica,Politecnico di Milano, Milan, Italy [email protected]
July 11, 2011
Keywords : Functional equation in unknown distribution functions; general-ized P´olya urn; reinforced urn process
AMS 2000 subject classifications : Primary: 62E10; secondary: 39B52; 62E20
Abstract
We study a functional equation whose unknown maps a Euclideanspace into the space of probability distributions on [0 , . We proveexistence and uniqueness of its solution under suitable regularity andboundary conditions, we show that it depends continuously on theboundary datum, and we characterize solutions that are diffuse on[0 , . A canonical solution is obtained by means of a Randomly Re-inforced Urn with different reinforcement distributions having equalmeans. The general solution to the functional equation defines a newparametric collection of distributions on [0 ,
1] generalizing the Betafamily. Introduction
The present work treats a particular functional equation whose unknownmaps a Euclidean space into the space P ([0 , , µ and ν on the interval [0 , β ],with β >
0, and assume that µ and ν have the same positive mean. Then,for ( x, y ) ranging over the subspace S = [0 , ∞ ) \ { (0 , } of R , define thefollowing equation with parameters µ and ν : x Z β ( G ( x, y ) −G ( x + k, y )) µ ( dk )+ y Z β ( G ( x, y ) −G ( x, y + k )) ν ( dk ) = 0; (1)the unknown is the function G : S → P ([0,1]) . Without additional constraints or requirements, equation (1) in its com-plete generality admits infinitely many solutions. For instance, any con-stant function G satisfies (1). We will show that under suitable regularityand boundary conditions, the problem described by (1) is well-posed in thesense that its solution exists, it is unique and depends continuously on theboundary datum. Moreover, we will also prove that the solution dependscontinuously on the parameters µ and ν and we will characterize a classof solutions G mapping the interior of S into the subspace of probabilitydistributions diffuse on [0 , . A particular instance of the problem considered in this paper has beenstudied in [1] where it is proved that, when the two parameters µ and ν areequal, there exists one and only one continuous solution to (1) that mapsthe x -axis and y -axis borders of S in the point mass at 1 or at 0, respectively,and that approaches the point mass at x/ ( x + y ) as x + y tends to infinity.We here extend this result to the case of different parameters µ and ν withthe same mean, and to more general boundary conditions. These will bedescribed by means of a continuous function ϕ : [0 , → P ([0 , boundary datum of the problem.From a probabilistic point of view, (1) is naturally connected to thedynamics of a two-color randomly reinforced urn with reinforcement distri-butions µ and ν. Indeed, for ( x, y ) ∈ S , consider an urn containing initially x black ballsand y white balls. The urn is sequentially sampled. At time n = 1 , , . . . aball is drawn from the urn and its color is observed: if the sampled ball is2lack, it is replaced in the urn together with a random number of black ballshaving distribution µ, if the sampled ball is white it is replaced in the urntogether with a random number of white balls having distribution ν. Theextra balls added every time the urn is sampled are called reinforcements.This urn scheme is called a two-color randomly reinforced urn (
RRU ); it hasbeen introduced in [5, 8, 4, 10, 11] and further studied in [3, 1, 2, 9, 12] asa general model for learning through reinforcement with direct applicationsin statistics as a device for adaptive sampling.If {G ( x, y ) , ( x, y ) ∈ S } is a family of distributions on [0 ,
1] parameterizedby the elements of S and γ is a distribution on S , we use G ( X, Y ) ∧ γ toindicate the distribution on [0 ,
1] obtained by mixing the distributions G according to γ. This can be thought of as the distribution of a random valuein [0 ,
1] generated through a two-step procedure: first sample (
X, Y ) ∈ S with distribution γ and then, given ( X, Y ) = ( x, y ) , sample a random valuein [0 ,
1] according to the distribution G ( x, y ) . Now let ( X , Y ) be the random number of black and white balls respec-tively present in a RRU after it has been sampled for the first time andindicate with γ ( x, y ) the distribution of ( X , Y ) on S , which depends on µ, ν and the initial urn composition ( x, y ) . Finally let A to be an operatoracting on the distributions G and defined as( A G )( x, y ) = G ( X, Y ) ∧ γ ( x, y ) . A solution G of equation (1) is a fixed point of A : A G = G . Indeedwe will also show that if G ( x, y ) is the distribution of the limit proportionof black balls of a RRU with initial composition ( x, y ) and reinforcementdistributions µ and ν, then G is a fixed point of A satisfying specific boundaryconditions.As a prototypical example, consider an RRU whose reinforcement dis-tributions µ and ν are both point masses at 1; this is a P´olya urn scheme.It is well known that if ( x, y ) ∈ S is the initial composition of a P´olya urn,then the limit proportion of black balls generated by this urn scheme hasdistribution Beta( x, y ) . In fact the family of Beta distributions is a fixedpoint of the operator A related to the P´olya urn. In this sense, the generalsolution of (1) defines a collection of distributions on [0 ,
1] parameterized byelements of S and generalizing the Beta.In the next section we set notation and terminology, we formally describethe functional equation problem and we state three results concerning itssolution; they will be proved in the rest of the paper. Section 3 deals withthe construction of the canonical solution to (1) for the special case when3he boundary datum ϕ ( t ) is the point mass at t, for all t ∈ [0 , RRU . Canonical solutions arethe building blocks for proving existence, uniqueness and regularity proper-ties of the solution to the functional equation problem with a general bound-ary datum; this will be shown in Section 4. Section 5 describes functionalequation problems whose solution maps the interior of S into the subspaceof P ([0 , In this section we set notation and terminology and we describe the func-tional equation problem in detail. We also state three main results concern-ing its solution; they will be proved in the rest of the paper.
For any β ∈ (0 , ∞ ), we endow the set P ([0 , β ]) of probability distributionson the real interval [0 , β ] with the 1–Wasserstein metric d W which metrizesweak convergence. Recall that, for ξ , ξ ∈ P ([0 , β ]) ,d W ( ξ , ξ ) = Z β | F ξ ( t ) − F ξ ( t ) | dt = Z | q ξ ( t ) − q ξ ( t ) | dt, where F ξ and q ξ are the cumulative distribution function and the quantilefunction of ξ ∈ P ([0 , β ]), respectively (see [6] for more details). Moreover,by the Kantorovich-Rubinstein Theorem, d W ( ξ , ξ ) = inf { E ( | X − X | ) : X ∼ ξ , X ∼ ξ } (2)where the infimum is taken over all joint distributions for the vector ofrandom variables ( X , X ) with marginal distributions equal to ξ and ξ , respectively. The metric space ( P ([0 , β ]) , d W ) is complete and compact. P of parameters For 0 < m ≤ β < ∞ , endow the cartesian product P ([0 , β ]) × P ([0 , β ]) withthe Manhattan-distance d M (( µ , ν ) , ( µ , ν )) = d W ( µ , µ ) + d W ( ν , ν )4nd consider the subset P of couples ( µ, ν ) of probability distributions withsupport in [0 , β ] having means that are equal and that are both greater thanor equal to m , i.e. such that Z β kµ ( dk ) = Z β kν ( dk ) ≥ m . The elements of P will act as the parameters for the functional equation(1); note that P is a closed subset of the metric space P ([0 , β ]) × P ([0 , β ])and therefore it is compact. C ([0 , , P ([0 , of boundary data A boundary datum ϕ is defined as a continuous map from [0 ,
1] to P ([0 , C ([0 , , P ([0 , d ∞ ( ϕ , ϕ ) = sup t ∈ [0 , d W ( ϕ ( t ) , ϕ ( t ));then ( C ([0 , , P ([0 , , d ∞ ) is a complete metric space.From now on, δ will indicate the element of C ([0 , , P ([0 , δ ( t ) = δ t for t ∈ [0 , , where δ t denotes the point mass at t. C ( S , P ([0 , where solutions are to be found Let S = [0 , ∞ ) \ { (0 , } and C ( S , P ([0 , G : S → P ([0,1]).For n = 1 , , . . . let S n = { ( x, y ) ∈ S : x + y ≥ /n } and consider thedistance between elements G , G ∈ C ( S n , P ([0 , d n ( G , G ) = sup ( x,y ) ∈ S n d W ( G ( x, y ) , G ( x, y )) . We then define a new distance d by setting, for all G , G ∈ C ( S , P ([0 , ,d ( G , G ) = ∞ X n =1 n · d n ( G | S n , G | S n )1 + d n ( G | S n , G | S n )where G | S n indicates the restriction to S n of a G ∈ C ( S , P ([0 , . The distance d metrizes the uniform weak convergence in any closedsubset of S S { (0 , } which does not contain the origin. Note that conver-gence with respect to d is equivalent to convergence with respect to all d n of the corresponding restrictions. The set ( C ( S , P ([0 , , d ) is a completemetric space; we will look for elements of this space that are solutions of thefunctional equation (1). 5 .5 The functional equation problem The Problem object of this paper is now easily stated: given ( µ, ν ) ∈ P and the boundary datum ϕ ∈ C ([0 , , P ([0 , , find G ∈ C ( S , P ([0 , x, y ) ∈ S , x Z β ( G ( x, y ) − G ( x + k, y )) µ ( dk )+ y Z β ( G ( x, y ) − G ( x, y + k )) ν ( dk ) = 0 , (3b) G (0 , y ) = ϕ (0) , (3c) G ( x,
0) = ϕ (1) , (3d) d W (cid:16) G ( x, y ) , ϕ (cid:0) xx + y (cid:1)(cid:17) −→ x + y →∞ . (3e) Our first result states that Problem (3) is well-posed in the sense of Hadamard.
Theorem 2.1.
A solution to Problem (3) exists, it is unique, and it dependscontinuously on the boundary datum.
In the rest of the paper, we denote with G ϕ ( µ,ν ) the unique solution toProblem (3). Theorem 2.1 will be proved first in the special case when theboundary datum is the map δ. Indeed G δ ( µ,ν ) is a canonical solution for theproblem since, for any other boundary datum ϕ ∈ C ([0 , , P ([0 , G ϕ ( µ,ν ) = Ψ ϕ ( G δ ( µ,ν ) ) , (4)where Ψ ϕ : C ( S , P ([0 , → C ( S , P ([0 , G ∈ C ( S , P ([0 , ϕ ( G )( x, y ) = Z ϕ ( t ) G ( x, y )( dt ) (5)with ( x, y ) ranging over S . G ϕ the set of solutions to Problem (3) obtained by holding fixed the boundarydatum ϕ ∈ C ([0 , , P ([0 , µ, ν ) range over P : G ϕ = n G ϕ ( µ,ν ) : ( µ, ν ) ∈ P o . Theorem 2.2.
For any given boundary datum ϕ ∈ C ([0 , , P ([0 , , themap ( µ, ν )
7→ G ϕ ( µ,ν ) , from P to G ϕ , is uniformly continuous and G ϕ is compact. To prove Theorem 2.2 we will first show that it holds for canonical solu-tions, i.e. for G δ , and then we will prove that the map Ψ ϕ is continuous.The third result regards a different regularity property of the solutionto Problem (3), which depends on the boundary datum ϕ but not on theparameters ( µ, ν ) . Indeed we characterize solutions G ϕ ( µ,ν ) mapping the inte-rior of S into the class of probability distributions on [0 ,
1] having no pointmasses; such solutions will be called diffuse.
A boundary datum ϕ ∈ C ([0 , , P ([0 , monotonic if, forall s, t ∈ [0 , , s ≤ t, ϕ ( s ) ≤ st ϕ ( t ) . For a given ϕ ∈ C ([0 , , P ([0 , , indicate with Φ the probability distribu-tion on [0 ,
1] obtained as the convex combination with uniform weights ofthe members of the family { ϕ ( t ) : t ∈ [0 , } ; i.e. Φ = R ϕ ( t ) dt. Theorem 2.3.
Assume that the boundary datum ϕ is monotonic and let G ϕ ( µ,ν ) be the unique solution to Problem (3) . Then:1. If there is ( x , y ) in the interior of S such that G ϕ ( µ,ν ) ( x , y ) has nopoint masses in [0 , , then Φ = R ϕ ( t ) dt has no point masses in [0 , .
2. If
Φ = R ϕ ( t ) dt has no point masses in [0 , , then G ϕ ( µ,ν ) ( x, y ) has nopoint masses in [0 , for all ( x, y ) in the interior of S . Once again, in Section 5, we will first prove Theorem 2.3 for canonicalsolutions and then for the general solution G ϕ ( µ,ν ) . Existence of canonical solutions: a RandomlyReinforced Urn
In this section we assume that the boundary datum of Problem (3) is themap δ ∈ C ([0 , , P ([0 , RRU scheme.This solution will be called canonical since the solution to Problem (3) fora general boundary datum will be obtained by transforming the canonicalsolution through a suitable map. While constructing canonical solutions, wewill also provide two novel results concerning the continuity of the distribu-tion of the limit proportion of black balls generated by an
RRU by provingits continuity with respect to the initial urn composition (Lemma 3.3) aswell as with respect to the reinforcement distributions (Proposition 3.2);the continuity is uniform in any closed subset of S which does not containsthe origin.On a rich enough probability space (Ω , A , P ) , define two independentinfinite sequences of random elements, { U n } and { ( V n , W n ) } ; { U n } is asequence of i.i.d. random variables uniformly distributed on [0 , , while { ( V n , W n ) } is a sequence of i.i.d. bivariate random vectors with componentsuniformly distributed on [0 , . Then, define an infinite sequence { ( R X ( n ) , R Y ( n )) } of bivariate random vectors by setting, for all n,R X ( n ) = q µ ( V n ) and R Y ( n ) = q ν ( W n ) , where q µ and q ν are the quantile functions of two distributions µ and ν having support in [0 , β ] , with β > . Let x and y be two non-negative realnumbers such that x + y > . Set X = x , Y = y , and, for n = 0 , , , ... , let (cid:26) X n +1 = X n + R X ( n + 1) I ( n + 1) ,Y n +1 = Y n + R Y ( n + 1)(1 − I ( n + 1)) , (6)where the variable I ( n + 1) is the indicator of the event { U n +1 ≤ X n ( X n + Y n ) − } . The law of { ( X n , Y n ) } is that of the stochastic process counting,along the sampling sequence, the number of black and white balls present ina RRU with initial composition ( x, y ) and reinforcement distributions equalto µ and ν, respectively.For n = 0 , , , . . . let D n = X n + Y n be the total number of balls presentin the urn at time n and set Z n ( x, y ) = X n /D n ;8 n ( x, y ) represents the proportion of black balls in a RRU with initial com-position ( x, y ) , before the ( n + 1)-th ball is sampled from it. In [11] it isproved that { Z n ( x, y ) } is eventually a bounded sub- or super-martingale,and it thus converges almost surely, and in L p , for 1 ≤ p ≤ ∞ , to a randomvariable Z ∞ ( x, y ) ∈ [0 , µ and ν have different means, Z ∞ ( x, y ) is the point mass concentrated in 1 or 0, according to whether themean of µ is greater or smaller than that of ν. However, when the means of µ and ν are the same, the distribution of Z ∞ ( x, y ) is unknown, apart froma few special cases, see [1] and [9].For a given couple ( µ, ν ) ∈ P , let L ( µ,ν ) : S → P ([0 , x, y ) ∈ S the distribution of the limitproportion Z ∞ ( x, y ) of a RRU with initial composition ( x, y ) and reinforce-ment distributions µ and ν . In the special case where µ = ν , the map L ( µ,µ ) has been characterized in [1] as the unique solution to Problem (3) whenthe boundary datum is δ. We now extend this result to the general case( µ, ν ) ∈ P . Proposition 3.1. L ( µ,ν ) is a solution to Problem (3) when its boundarydatum is equal to δ . In order to prove Proposition 3.1 we need some auxiliary results; whenthey do not depend on the parameters ( µ, ν ) ∈ P , and there is no place formisunderstanding, we write L for L ( µ,ν ) . Some technicalities connected withthe Doob’s decomposition of the process { Z n } have been postponed to theAppendix.The distance between L , evaluated at ( x, y ) , and the boundary datum,evaluated at x/ ( x + y ) , is controlled in the following lemma; this distance isuniformly bounded, provided that the size of the urn initial composition issufficiently large. Lemma 3.1. If x + y ≥ β , d W (cid:0) L ( x, y ) , δ xx + y (cid:1) < s βx + y . Proof.
Note that, by (2), d W (cid:0) L ( x, y ) , δ xx + y (cid:1) = E ( | Z ∞ ( x, y ) − xx + y | ). More-over, E ( | Z ∞ ( x, y ) − xx + y | ) = E ( | A ∞ ( x, y ) + M ∞ ( x, y ) | ), where ( A n ) n and( M n ) n are the Doob’s decomposition processes (see Appendix) of ( Z n ) n .9ote that E ( | M ∞ | ) ≤ p E ( h M i ∞ ) by Jensen inequality and thus, fromtriangular inequality, Lemma A.2, Lemma A.3 with n = 0, and because x + y ≥ β implies q βx + y <
1, we get d W (cid:0) L ( x, y ) , δ xx + y (cid:1) ≤ βx + y + s βx + y < s βx + y + s βx + y . The Markov inequality together with Lemma 3.1 imply the followingcorollary.
Corollary 3.1. If x + y ≥ β, P (cid:16)(cid:12)(cid:12) Z ∞ ( x, y ) − xx + y (cid:12)(cid:12) > h (cid:17) ≤ h s βx + y for every h > . Lemma 3.2.
For all n ≥ and ǫ > , there is N = N ( ǫ, n ) such that, E (cid:0) | Z n ( x, y ) − Z ∞ ( x, y ) | (cid:1) ≤ ǫ, if n ≥ N and x + y ≥ /n . Proof.
Equation (A.28) yields, for all t > P ( D n < t ) = P (cid:16) D n > t (cid:17) ≤ t E D n < t n ( β − m ) m ( n −
1) + β (7)where m is given in Section 2.2. Set t = max { β/ǫ , β } (8)and N ≥ tǫ (1 + n ( β − m )) − βm + 1 . (9)From (7) and (9), we get P ( D N < t ) < ǫ . (10)10oreover, since the process { ( X n , Y n ) } is Markov, it follows from Lemma 3.1and (8) that, for n ≥ N and ω ∈ { D N ≥ t } , E (cid:16) | Z ∞ − Z n | (cid:12)(cid:12)(cid:12) ( X n , Y n ) (cid:17) ( ω ) = d W (cid:16) L ( X n ( ω ) , Y n ( ω )) , δ Xn ( ω ) Xn ( ω )+ Yn ( ω ) (cid:17) ≤ ǫ . (11)Since { D n +1 < t } ⊆ { D n < t } for all n, (10) and (11) imply that E [ | Z ∞ − Z n | ] = E [ | Z ∞ − Z n | ; { D n ≥ t } ] + E [ | Z ∞ − Z n | ; { D n < t } ] ≤ E (cid:0) E ( | Z ∞ − Z n | { D n ≥ t } (cid:12)(cid:12) ( X n , Y n )) (cid:1) + P ( D N < t ) ≤ ǫ for n ≥ N. The next result can be read as a bound on the modulus of continuity of L when evaluated at the inner points of S . Lemma 3.3.
For all n ≥ and ǫ > , there is η = η ( ǫ, n ) , increasingwith ǫ and /n , such that d W ( L ( x, y ) , L (¯ x, ¯ y )) < ǫ, if | x − ¯ x | + | y − ¯ y | < η and min { x + y, ¯ x + ¯ y } ≥ /n . Proof.
Let N = N ( ǫ/ , n ) be given by Lemma 3.2. Then: d W ( L ( x, y ) , L (¯ x, ¯ y )) ≤ E [ | Z ∞ ( x, y ) − Z N ( x, y ) | ] + E [ | Z ∞ (¯ x, ¯ y ) − Z N (¯ x, ¯ y ) | ]+ E [ | Z N ( x, y ) − Z N (¯ x, ¯ y ) | ] ≤ ǫ E [ | Z N ( x, y ) − Z N (¯ x, ¯ y ) | ] . For controlling the last term, we adopt a coupling argument as in [1].Consider two different randomly reinforced urns, the first one with initialcomposition ( x, y ) and second one with (¯ x, ¯ y ). The two urns are coupledin the sense that the same processes { U n } and { ( V n , W n ) } generate both { ( X n ( x, y ) , Y n ( x, y )) } and { ( X n (¯ x, ¯ y ) , Y n (¯ x, ¯ y )) according to the dynamicsdescribed in (6). With the same arguments as in [1, pages 701-702], onemay show that E [ | Z N ( x, y ) − Z N (¯ x, ¯ y ) | ] ≤ (1 + N ) | x − ¯ x | + | y − ¯ y | min { x + y, ¯ x + ¯ y } ;11herefore, if η ≤ ǫ N ) n , E [ | Z N ( x, y ) − Z N (¯ x, ¯ y ) | ] ≤ ǫ . Proof of Proposition 3.1.
By considering the conditional distribution of Z ∞ ( x, y ) , given I (1) , R X (1) and R Y (1), and taking the expected values, one immedi-ately verifies that L ( µ,ν ) satisfies equation (3b) for all ( x, y ) ∈ S . Conditions(3c) and (3d) are also easily verified when ϕ = δ. Finally, (3a) and (3e) areconsequences of Lemma 3.3 and of Lemma 3.1, respectively.The next result proves a further regularity property of L ( µ,ν ) . Proposition 3.2.
The map ( µ, ν )
7→ L ( µ,ν ) , from ( P , d M ) to ( C ( S , P ([0 , , d ) , is uniformly continuous.Proof. Let A : C ( S , P ([0 , × P → C ( S , P ([0 , H ∈ C ( S , P ([0 , µ, ν ) ∈ P ,A ( H , ( µ, ν ))( x, y ) = xx + y Z β H ( x + k, y ) µ ( dk ) + yx + y Z β H ( x, y + k ) ν ( dk )= xx + y Z H ( x + q µ ( t ) , y ) dt + yx + y Z H ( x, y + q ν ( t )) dt, where ( x, y ) ranges over S . Let n ≥
1. Then d n ( A ( H , ( µ, ν )) | S n , A ( H , ( µ, ν )) | S n ) ≤ d n ( H | S n , H | S n ) , (12)for every H , H ∈ C ( S , P ([0 , µ, ν ) ∈ P . Indeed, for every ( x, y ) ∈ S n , d W ( A ( H , ( µ, ν ))( x, y ) , A ( H , ( µ, ν ))( x, y )) ≤ xx + y Z β d W ( H ( x + k, y ) , H ( x + k, y )) µ ( dk )+ yx + y Z β d W ( H ( x, y + k ) , H ( x, y + k )) ν ( dk ) ≤ sup ( x ′ ,y ′ ) ∈ S n d W ( H ( x ′ , y ′ ) , H ( x ′ , y ′ )) . H ∈ C ( S , P ([0 , S n with Lipschitz constant K n , then, for every ( µ , ν ) , ( µ , ν ) ∈ P ,d n ( A ( H , ( µ , ν )) | S n , A ( H , ( µ , ν )) | S n ) ≤ K n d M (( µ , ν ) , ( µ , ν )) , (13)since, for every ( x, y ) ∈ S n ,d W ( A ( H , ( µ , ν ))( x, y ) , A ( H , ( µ , ν ))( x, y )) ≤ xx + y Z d W ( H ( x + q µ ( t ) , y ) , H ( x + q µ ( t ) , y )) dt + yx + y Z d W ( H ( x, y + q ν ( t )) , H ( x, y + q ν ( t ))) dt ≤ K n ( xx + y d W ( µ , µ ) + yx + y d W ( ν , ν )) ≤ K n d M (( µ , ν ) , ( µ , ν )) . (14)Now, for every H ∈ C ( S , P ([0 , µ, ν ) ∈ P , set A ( H , ( µ, ν )) = H and, for N = 1 , , ... define iteratively A N ( H , ( µ, ν )) = A ( A N − ( H , ( µ, ν )) , ( µ, ν )) . Consider H ∈ C ( S , P ([0 , H ( x, y ) = δ ( xx + y ) for every( x, y ) ∈ S ; then Z ( µ,ν )0 ( x, y ) has distribution H ( x, y ) , while, for N = 1 , , ...,Z ( µ,ν ) N ( x, y ) has distribution A N ( H , ( µ, ν ))( x, y ) , where, for clarity of ex-position, the exponent of the Z variables is evidence for the reinforcementdistributions of the RRU under consideration. Note that, for n ≥ H isa Lipschitz map from S n to P ([0 , n . Moreover,it is not difficult to show, with computations analogous to those appear-ing in (14), that the operator A preserves the Lipschitz property with thesame constant; hence, for ( µ, ν ) ∈ P and N = 1 , , ..., A N ( H , ( µ, ν )) is aLipschitz map from S n to P ([0 , n .Let ( µ , ν ) , ( µ , ν ) ∈ P , n, N ≥ H i = A N − ( H , ( µ i , ν i )) , for i = 1 ,
2; then d n ( A N ( H , ( µ , ν )) | S n ,A N ( H , ( µ , ν )) | S n )= d n ( A ( H , ( µ , ν )) | S n , A ( H , ( µ , ν )) | S n ) ≤ d n ( A ( H , ( µ , ν )) | S n , A ( H , ( µ , ν ) | S n ))+ d n ( A ( H , ( µ , ν )) | S n , A ( H , ( µ , ν )) | S n ) ≤ d n ( H | S n , H | S n ) + nd M (( µ , ν ) , ( µ , ν )) , (15)13he last inequality being a consequence of (12) and (13). By iterativelyapplying (15), it follows that d n ( A N ( H , ( µ , ν )) , A N ( H , ( µ , ν ))) ≤ nN d M (( µ , ν ) , ( µ , ν )) . Therefore, for every n ≥ ǫ >
0, if N = N ( ǫ, n ) is chosen according toLemma 3.2, we obtain d n ( L ( µ ,ν ) | S n , L ( µ ,ν ) | S n ) ≤ d n ( L ( µ ,ν ) | S n , A N ( H , ( µ , ν )) | S n )+ d n ( A N ( H , ( µ , ν )) | S n , A N ( H , ( µ , ν )) | S n )+ d n ( A N ( H , ( µ , ν )) | S n , L ( µ ,ν ) | S n ) ≤ nN d M (( µ , ν ) , ( µ , ν ))+ sup ( x,y ) ∈ S n h E (cid:0) | Z ( µ ,ν ) ∞ ( x, y ) − Z ( µ ,ν ) N ( x, y ) | (cid:1) + E (cid:0) | Z ( µ ,ν ) N ( x, y ) − Z ( µ ,ν ) ∞ ( x, y ) | (cid:1)i ≤ nN d M (( µ , ν ) , ( µ , ν )) + 2 ǫ. This shows that the map ( µ, ν )
7→ L ( µ,ν ) | S n , from ( P , d M ) to ( C ( S n , P ([0 , , d n ) , is continuous for every n. Hence the map ( µ, ν )
7→ L ( µ,ν ) from ( P , d M ) to ( C ( S , P ([0 , , d ) is con-tinuous; since P is compact, it is also uniformly continuous. In this section we prove Theorem 2.1 and Theorem 2.2. In particular we showexistence and uniqueness of the solution to Problem (3) when the boundarydatum is a generic element of C ([0 , , P ([0 , . Existence is shown by meansof a constructive proof based on the canonical solution described in Section3. Uniqueness is proved through a fixed point argument.Given ϕ ∈ C ([0 , , P ([0 , , define the map Γ ϕ : P ([0 , → P ([0 , ξ ∈ P ([0 , , Γ ϕ ( ξ )( B ) = Z ϕ ( t )( B ) ξ ( dt ) , where B ranges over the Borel sets in [0 , . Lemma 4.1.
For any given ϕ ∈ C ([0 , , P ([0 , , the map Γ ϕ is uniformlycontinuous. roof. Since ϕ ∈ C ([0 , , P ([0 , ϕ is uniformly continuous and bounded:i.e. for any ǫ >
0, there is an η = η ( ǫ, ϕ ) such that d W ( ϕ ( t ) , ϕ ( t )) ≤ ǫ, if | t − t | ≤ η, (16)while d W ( ϕ ( t ) , ϕ ( t )) ≤ , for all t , t ∈ [0 , . (17)Now, take ξ , ξ ∈ P ([0 , d W ( ξ , ξ ) < ǫη . We are going to provethat d W (Γ ϕ ( ξ ) , Γ ϕ ( ξ )) ≤ ǫ .Because of (2), there is a probability space ( e Ω , e A , e P ) carrying a coupleof random variables X , X such that X ∼ ξ , X ∼ ξ , and d W ( ξ , ξ ) = E ( | X − X | ) ≤ ǫη ; by Markov inequality, e P ( | X − X | > η ) ≤ ǫ. (18)On the product probability space ( e Ω × [0 , , e A ⊗ B ([0 , , e P ⊗ λ [0 , ) definethe random variables ζ ( ω, t ) = inf n z : Z [0 ,z ] ϕ ( η ( ω ))( ds ) ≥ t o = q ϕ ( η ( ω )) ( t )and ζ ( ω, t ) = inf n z : Z [0 ,z ] ϕ ( η ( ω ))( ds ) ≥ t o = q ϕ ( η ( ω )) ( t ) , where q ξ indicates the quantile function relative to the probability distribu-tion ξ ∈ P ([0 , . For i = 1 , , note that ϕ ( X i ) is the conditional distribution of ζ i , given X i ; thus, ζ i ∼ Γ ϕ ( ξ i ) . Moreover, for all ω ∈ e Ω, d W ( ϕ ( X ( ω )) , ϕ ( X ( ω ))) = Z | q ϕ ( X ( ω )) ( t ) − q ϕ ( X ( ω )) ( t ) | dt = Z | ζ ( ω, t ) − ζ ( ω, t ) | dt. Hence, d W (Γ ϕ ( ξ ) , Γ ϕ ( ξ )) ≤ E ( | ζ − ζ | )= E (cid:0) E (cid:0) | ζ − ζ | (cid:12)(cid:12) X , X (cid:1)(cid:1) = E (cid:0) d W ( ϕ ( X ) , ϕ ( X )) (cid:1) . (19)Now, let F = {| X − X | > η } . From (16), (17) and (18) one obtains: E (cid:0) d W ( ϕ ( X ) , ϕ ( X )) (cid:1) = E (cid:0) d W ( ϕ ( X ) , ϕ ( X )); F (cid:1) + E (cid:0) d W ( ϕ ( X ) , ϕ ( X )); F c (cid:1) ≤ e P ( F ) + ǫ e P ( F c ) ≤ ǫ. The last inequality, together with (19), concludes the proof.15 roof of Theorem 2.1. (i)
Existence.
When the boundary datum ϕ = δ, theexistence of a solution G δ ( µ,ν ) is guaranteed by Proposition 3.1, and this is F ( µ,ν ) .Now let ϕ ∈ C ([0 , , P ([0 , x, y ) ∈ S , G ϕ ( µ,ν ) ( x, y ) = Γ ϕ ( G δ ( µ,ν ) ( x, y ));we are going to show that G ϕ ( µ,ν ) is indeed a solution to Problem (3) whenthe bondary datum is ϕ. In other words, the composition( x, y )
7→ G δ ( µ,ν ) ( x, y ) Γ ϕ ( G δ ( µ,ν ) ( x, y ))gives a solution to Problem (3), i.e., (4) holds if the map Ψ ϕ is defined bysetting, for all G ∈ C ( S , P ([0 , , Ψ ϕ ( G )( x, y ) = Γ ϕ ( G ( x, y )) = Z ϕ ( t ) G ( x, y )( dt )with ( x, y ) ranging over S . Because of Proposition 3.1, G δ ( µ,ν ) satisfies (3b) when the border datumis δ. Since Γ ϕ is linear, this implies that, for all ( x, y ) ∈ S , Ψ ϕ ( G δ ( µ,ν ) )( x, y ) = Γ ϕ ( G δ ( µ,ν ) ( x, y ))= Γ ϕ (cid:16) xx + y Z G δ ( µ,ν ) ( x + k, y ) µ ( dk ) + yx + y Z G δ ( µ,ν ) ( x, y + k ) ν ( dk ) (cid:17) = xx + y Z Ψ ϕ ( G δ ( µ,ν ) )( x + k, y ) µ ( dk ) + yx + y Z Ψ ϕ ( G δ ( µ,ν ) )( x, y + k ) ν ( dk );hence Ψ ϕ ( G δ ( µ,ν ) ) satisfies (3b) when the border datum is ϕ. Now, by Lemma 4.1,Ψ ϕ ( G δ ( µ,ν ) ) is a continuous map from S to P ([0 , ϕ and G δ ( µ,ν ) ; hence (3a), (3c), (3d) and (3e) alsohold true.(ii) Uniqueness. Sketch of the argument.
Our argument in [1, Section 5]can be easily extended to this more general situation.Condition (3e) requires G to be continuous at the projective infinitepoints. It is therefore convenient to transform the space S along the projec-tive automorphism τ of P so defined:( x : y : u ) τ ( u : x : x + y ) . The automorphism τ has the following properties:16 the space S is mapped into the affine space S ∗ = [0 , ∞ ) × [0 , x –axis is mapped into itself by ( x, → (1 /x, y –axis is mapped into the semiline { y ∗ = 1 , x ∗ > } by(0 , y ) (1 /y, xx + y = k is mappedin the point (0 , k );- the origin is mapped in the projective infinite point (1 : 0 : 0) . The inverse map of τ is ( x ∗ : y ∗ : u ∗ ) τ − ( y ∗ : u ∗ − y ∗ : x ∗ ) . Problem (3) canbe equivalently formulated on S ∗ as follows: given ( µ, ν ) ∈ P and the boundary datum ϕ ∈ C ([0 , , P ([0 , , find G ∗ ∈ C ( S ∗ , P ([0 , x, y ) ∈ S ∗ , G ∗ ( x ∗ , y ∗ ) = y ∗ Z G ∗ (cid:16) x ∗ kx ∗ , y ∗ + kx ∗ kx ∗ (cid:17) µ ( dk )++ (1 − y ∗ ) Z G ∗ (cid:16) x ∗ kx ∗ , y ∗ kx ∗ (cid:17) ν ( dk ) , (20b) G ∗ ( x ∗ ,
0) = ϕ (0) , (20c) G ∗ ( x ∗ ,
1) = ϕ (1) , (20d) G ∗ (0 , y ∗ ) = ϕ ( y ∗ ) . (20e)In fact, (20b) is just (3b) in the new coordinates. Indeed, the transforma-tions G ( x, y ) = G ∗ ( τ ( x, y )) G ∗ ( x ∗ , y ∗ ) = ( G ( τ − ( x ∗ , y ∗ )) if ( x ∗ , y ∗ ) ∈ (0 , ∞ ) × [0 , s ∗ → G ( τ − ( s ∗ , y ∗ )) if x ∗ = 0 , y ∗ ∈ [0 , , show the equivalence of Problem (3) and Problem (20).Now, let C ∗ ϕ ( S ∗ ) be the space of continuous function H ∗ : S ∗ → P ([0 , x ∗ , y ∗ ) ∈ S ∗ , H ∗ ( x ∗ ,
0) = ϕ (0) , H ∗ ( x ∗ ,
1) = ϕ (1) and H ∗ (0 , y ∗ ) = ϕ ( y ∗ ) . A ∗ mapping C ∗ ϕ ( S ∗ ) into C ∗ ϕ ( S ∗ ): A ∗ ( H ∗ )( x ∗ , y ∗ ) = y ∗ Z H ∗ (cid:16) x ∗ kx ∗ , y ∗ + kx ∗ kx ∗ (cid:17) µ ( dk )++ (1 − y ∗ ) Z H ∗ (cid:16) x ∗ kx ∗ , y ∗ kx ∗ (cid:17) ν ( dk )with ( x ∗ , y ∗ ) ranging over S ∗ . With the same argument used in [1, Theorem 5.2], one can prove that A ∗ has at most one fixed point; hence Problem (20) has at most one solution.(iii) Continuity with respect to the boundary datum.
We prove this lastpart by showing that d ( G ϕ ( µ,ν ) , G ϕ ( µ,ν ) ) ≤ d ∞ ( ϕ , ϕ ) (21)for all ϕ , ϕ ∈ C ([0 , , P ([0 , d W ( η , η ) = sup n(cid:12)(cid:12)(cid:12) Z h ( t ) η ( dt ) − Z h ( t ) η ( dt ) (cid:12)(cid:12)(cid:12) : k h k L ≤ o (22)where k h k L is the Lipschitz norm. Then, for h such that k h k L ≤ x, y ) ∈ S , we get (cid:12)(cid:12)(cid:12) Z h ( s ) G ϕ ( µ,ν ) ( x, y )( ds ) − Z h ( s ) G ϕ ( µ,ν ) ( x, y )( ds ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z h ( s ) Z ϕ ( t )( ds ) G δ ( µ,ν ) ( x, y )( dt ) − Z h ( s ) Z ϕ ( t )( ds ) G δ ( µ,ν ) ( x, y )( dt ) (cid:12)(cid:12)(cid:12) ≤ Z (cid:12)(cid:12)(cid:12) Z h ( s ) ϕ ( t )( ds ) − Z h ( s ) ϕ ( t )( ds ) (cid:12)(cid:12)(cid:12) G δ ( µ,ν ) ( x, y )( dt ) ≤ Z d W ( ϕ ( t ) , ϕ ( t )) G δ ( µ,ν ) ( x, y )( dt ) ≤ d ∞ ( ϕ , ϕ );hence d W ( G ϕ ( µ,ν ) ( x, y ) , G ϕ ( µ,ν ) ( x, y )) ≤ d ∞ ( ϕ , ϕ ), again by (22). Inequality(21) follows easily. Remark 4.1.
Given ( µ, ν ) ∈ P , the inequality (21) can be completed asfollows: for all ϕ , ϕ ∈ C ([0 , , P ([0 , ,d ( G ϕ ( µ,ν ) , G ϕ ( µ,ν ) ) ≤ d ∞ ( ϕ , ϕ ) ≤ d ( G ϕ ( µ,ν ) , G ϕ ( µ,ν ) ) . (23)18 ence, for any given ( µ, ν ) ∈ P , we have an embedding C ([0 , , P ([0 , Ψ ϕ ֒ → C ( S , P ([0 , . Indeed, for n = 1 , , ..., (3e) implies that d ∞ ( ϕ , ϕ ) ≤ d n ( G ϕ ( µ,ν ) , G ϕ ( µ,ν ) ); since d n ≤ , and thus d n ≤ d n d n , this implies the right inequality in (23) . Remark 4.2.
Let m be the common mean of ( µ, ν ) ∈ P . For p ∈ [ m /m, ,set ( µ ′ , ν ′ ) = ( pµ + (1 − p ) δ , pν + (1 − p ) δ ) . Then ( µ ′ , ν ′ ) ∈ P and G ϕ ( µ ′ ,ν ′ ) = G ϕ ( µ,ν ) . Remark 4.3.
Let h : [0 , → [0 , be a continuous function and ϕ ∈ C ([0 , , P ([0 , a boundary datum. For ξ ∈ P ([0 , , denote with h ◦ ξ thedistribution of the random variable h ( W ) , where W is a random variablewith probability distribution ξ. Then G h ◦ ϕ ( µ,ν ) ( x, y ) = h ◦ G ϕ ( µ,ν ) ( x, y ) , for all ( x, y ) ∈ S . Remark 4.4.
One may notice that the boundary conditions (3c) and (3d) are redundant. Indeed, if (3b) and (3e) are true for a G : S → P ([0 , , then G satisfies (3c) and (3d) . To see this, let x > and consider G ( x, . By iteratively applying (3b) , one obtains G ( x,
0) = Z β G ( x + k , µ ( dk )= Z β Z β G ( x + k + k , µ ( dk ) µ ( dk ) · · · = Z β · · · Z β G ( x + k + · · · + k n , µ ( dk ) · · · µ ( dk n ) for all n = 1 , , ... . However, because of (3e) , if P ni =1 k i → ∞ as n → ∞ ,then lim n →∞ d W ( G ( x + n X i =1 k i , , ϕ (1)) = 0 . Hence, the Law of Large Numbers and the Dominated Convergence Theoremimply that G ( x,
0) = ϕ (1) . The argument for proving that G (0 , y ) = ϕ (0) , if y > , is analogous. emark 4.5. For n = 1 , , ... let γ n ( x, y ) be the distribution of ( X n , Y n ) , thenumber of black and white balls respectively present in a RRU , with initialcomposition ( x, y ) ∈ S and reinforcement distribution µ and ν, after is hasbeen sampled for the n -th time.Since { ( X n , Y n ) } is Markov, if G is a solution to Problem (3) , it is easyto prove, by induction on n, that G ( X, Y ) ∧ γ n ( x, y ) = G ( x, y ) for all n and ( x, y ) ∈ S . However one can show that (3e) implies that d W ( G ( X, Y ) ∧ γ n ( x, y ) , ϕ (cid:16) XX + Y (cid:17) ∧ γ n ( x, y )) −→ n →∞ . Hence ϕ (cid:16) XX + Y (cid:17) ∧ γ n ( x, y ) d W −→ n →∞ G ( x, y ) and this shows the uniqueness of the solution G . This interesting remark hasbeen pointed out by a referee.
We are finally in the position to prove Theorem 2.2.
Proof of Theorem 2.2.
The theorem is true when the boundary datum is δ ,as follows from Proposition 3.2 and the fact that P is compact. For a generalboundary datum ϕ ∈ C ([0 , , P ([0 , ϕ is continuous; but this is a consequence of Lemma 4.1. As an immediate consequence of Proposition 3.1 and [2, Theorem 3.2] wehave the following result, which is a particular instance of Theorem 2.3 andrepresents a tool for proving it.
Proposition 5.1.
For all ( x, y ) in the interior of S and z ∈ [0 , , G δ ( µ,ν ) ( x, y )( { z } ) = 0 . Proof of Theorem 2.3.
Given z ∈ [0 , , note that ϕ z ( t ) = ϕ ( t )( { z } ) is ameasurable function of t , since it is the monotone limit of the sequence ofcontinuous functions k ( n ) z defined by setting, for n = 1 , , ... and t ∈ [0 , ,k ( n ) z ( t ) = Z h ( n ) z ( s ) ϕ ( t )( ds ) , h ( n ) z ( s ) = ns − nz + 1 , if z − /n ≤ s ≤ z,nz − ns + 1 , if z < s ≤ z + 1 /n, , otherwise . The two functions ϕ − z ( t ) = sup w Let ( x , y ) be a point in the interior of S such that G ϕ ( µ,ν ) ( x , y ) has no point masses in [0 , . By way of contradiction, sup-pose there is a z ∈ [0 , 1] such that Φ( { z } ) > . Since Φ( { z } ) > , there are ǫ > , a > z ∗ ∈ [0 , 1] such that ϕ z ( t ) > ǫ, for all t ∈ I ∗ = [ z ∗ − a, z ∗ + a ] ∩ [0 , R = n ( x, y ) ∈ S : x ≥ max(2 β, βa ) ,y = (cid:0) z − (cid:1) x (cid:0) z = xx + y (cid:1) ,z ∈ (cid:2) z ∗ − a , z ∗ + a (cid:3) ∩ [0 , o . For all ( x, y ) ∈ R, Corollary 3.1 with h = a implies that P (cid:16) Z ∞ ( x, y ) I ∗ (cid:17) ≤ P (cid:16) | Z ∞ ( x, y ) − z | > a (cid:17) ≤ a r βx ≤ G δ ( µ,ν ) ( x, y )( I ∗ ) ≥ . Then, because of (24), for all ( x, y ) ∈ R , G ϕ ( µ,ν ) ( x, y )( { z } ) = Z ϕ z ( t ) G δ ( µ,ν ) ( x, y )( dt ) ≥ ǫ Z I ∗ G δ ( µ,ν ) ( x, y )( dt ) ≥ ǫ . Now, set τ = inf { n ≥ X n ( x , y ) , Y n ( x , y )) ∈ R } (inf ∅ = ∞ );21t is not difficult to show that P ( τ < ∞ ) = p > 0. Thus the strong Markovproperty implies that G δ ( µ,ν ) ( x , y )( I ∗ ) ≥ p ; therefore G ϕ ( µ,ν ) ( x , y )( { z } ) ≥ p ǫ contradicting the assumption that G ϕ ( µ,ν ) ( x , y ) has no point masses in[0 , . This concludes the proof of part Let now z ∈ [0 , { z } ) = 0. Since ϕ z has at most a countable number of points of discontinuity, ϕ z ≥ R ϕ z ( t ) dt = 0 , there is a sequence t , t , ... such that ϕ z ( t ) = 0 for all t ∈ F = ( ∪ i { t i } ) c . Then, given any ( x, y ) in the interior of S , G ϕ ( µ,ν ) ( x, y )( { z } ) = Z [0 , ϕ z ( t ) G δ ( µ,ν ) ( x, y )( dt )= Z [0 , ∩ F ϕ z ( t ) G δ ( µ,ν ) ( x, y )( dt ) + X t i ϕ z ( t i ) G δ ( µ,ν ) ( x, y )( { t i } )= 0the last term being zero because of Proposition 5.1. In this section we give explicit descriptions of the solution G ϕ ( µ,ν ) for somespecific choices of the reinforcement distributions ( µ, ν ) ∈ P and of theboundary datum ϕ ∈ C ([0 , , P ([0 , . The first example is prototypicalsince it considers the P´olya urn scheme and the family of Beta distributions,whose properties had a central role in originating most of the problemstackled in this paper. We indicate with Beta( x, y ) the beta distribution on [0 , 1] with parameters( x, y ) ∈ S . If ( x, y ) is a point in the interior of S , Beta( x, y ) has a densitygiven by f Beta( x,y ) ( t ) = Γ( x + y )Γ( x )Γ( y ) t x − (1 − t ) y − , for t ∈ [0 , , y ) and Beta( x, 0) thepoint mass at 0 or at 1, respectively.The random limit composition of a P´olya urn with initial composition( x, y ) ∈ S and constant reinforcement equal to 1 has distribution Beta( x, y );22ndeed a P´olya urn is a special RRU with reinforcements µ = ν = δ . Thesame result holds when µ = ν = Bernoulli( p ) , for p > 0; see [4] and thereferences therein. Hence, for p > , G δ ( δ ,δ ) ( x, y ) = G δ (Bernoulli( p ) , Bernoulli( p )) ( x, y ) = Beta( x, y ) , for all ( x, y ) ∈ S . Different families of distributions related to the Beta can be generatedthrough the P´olya urn scheme, where both reinforcement distributions areequal to the same point mass, by modifying the boundary datum and solvingProblem (3).For instance, let λ > , define ϕ λ : [0 , → P ([0 , ϕ λ ( t ) = δ ( t /λ ) , for t ∈ [0 , , and consider Problem (3) with µ = ν = δ and bound-ary datum equal to ϕ λ . Note that ϕ λ is monotone and thus the uniquesolution to the problem is diffuse. Indeed for ( x, y ) in the interior S , thedistribution G ϕ λ ( δ ,δ ) ( x, y ) has density f G ϕλ ( δ ,δ ( x,y ) ( t ) = λ Γ( x + y )Γ( x )Γ( y ) t λx − (1 − t λ ) y − , for t ∈ [0 , . For x = 1 and y > , the solution G ϕ λ ( δ ,δ ) (1 , y ) is called theKumaraswamy distribution with shape parameters λ and y ; see [7].A captivating family of distributions is generated along these lines bysetting the boundary datum ϕ to be the function mapping t ∈ [0 , 1] to theexponential distribution with parameter t truncated to [0 , 1] : hence, for t ∈ (0 , , the density of ϕ ( t ) is f ϕ ( t ) ( z ) = t exp( − zt )1 − exp( − t ) 1 [0 , ( z ) , while f ϕ (0) ( z ) = 1 [0 , ( z ).Then (4) and (5) imply that, for ( x, y ) in the interior S , the distribution G ϕ ( δ ,δ ) ( x, y ) has density f G ϕ ( δ ,δ ( x,y ) ( z ) = Z t exp( − zt )1 − exp( − t ) Γ( x + y )Γ( x )Γ( y ) t x − (1 − t ) y − dt. This density admits an interesting representation in terms of the Hurwitzzeta function.Indeed, since (1 − exp( − t )) − = P n ≥ (exp( − t )) n , G ϕ ( δ ,δ ( x,y ) ( z ) = Γ( x + y )Γ( x )Γ( y ) ∞ X n =0 Z te − ( z + n ) t t x − (1 − t ) y − dt = Γ( x + y )Γ( x )Γ( y ) Γ( x + 1)Γ( y )Γ( x + y + 1) X n ≥ Γ( x + y + 1)Γ( x + 1)Γ( y ) Z e − ( z + n ) t t x (1 − t ) y − dt = xx + y X n ≥ M ( x + 1 , x + y + 1 , − ( z + n )) , where M ( a, b, z ) is the Kummer’s (confluent hypergeometric) function. Whenthe real parts ℜ ( a ) and ℜ ( b ) of a and b are such that ℜ ( b ) > ℜ ( a ) > , thefunction M can be represented as a Barnes integral: M ( a, b, − z ) = 12 πi Γ( b )Γ( a ) Z i ∞− i ∞ Γ( − s )Γ( a + s )Γ( b + s ) z s ds where the contour of integration separates the poles of Γ( a + s ), whichare {− a, − a + 1 , − a + 2 , . . . } , from those of Γ( − s ), which are { , , , . . . } .Therefore this contour may be taken in the halfplane ℜ ( s ) < − ℜ ( a ) > 1. Moreover, for z ∈ (0 , 1) and ℜ ( s ) < − X n ≥ ( z + n ) s = ζ ( − s, z )where ζ is the Hurwitz zeta function. Therefore f G ϕ ( δ ,δ ( x,y ) ( z ) = xx + y X n ≥ πi Γ( x + y + 1)Γ( x + 1) Z i ∞− i ∞ Γ( − s )Γ( x + 1 + s )Γ( x + y + 1 + s ) ( z + n ) s ds = xx + y πi Γ( x + y + 1)Γ( x + 1) Z i ∞− i ∞ Γ( − s )Γ( x + 1 + s )Γ( x + y + 1 + s ) ζ ( − s, z ) ds = 12 πi Γ( x + y )Γ( x ) Z i ∞− i ∞ Γ( − s )Γ( x + 1 + s )Γ( x + y + 1 + s ) ζ ( − s, z ) ds where the contour of integration may be chosen in the halfplane ℜ ( s ) < − ℜ ( x + 1) > A more intriguing extension which goes well beyond the P´olya urn schemeis obtained by considering reinforcement distributions ( µ, ν ) ∈ P different24rom equal point masses; we here treat the case where µ and ν are scaledBernoulli distributions with the same mean. Let m ≥ k µ ≥ k ν > , andassume that µ and ν are the distributions of two random variables, say R X and R Y , such that R X /k µ has distribution Bernoulli( m/k µ ) while R Y /k ν has distribution Bernoulli( m/k ν ).Equation (1), with ( µ, ν ) as above, reads xk µ (cid:16) G ( x, y ) − G ( x + k µ , y (cid:1)(cid:17) + yk ν (cid:16) G ( x, y ) − G ( x, y + k ν (cid:1)(cid:17) = 0 , which does not depend on m. One easily verifies that the equation is satisfiedby the continuous map G : S → P ([0 , x, y ) ∈ S , G ( x, y ) = Beta( xk µ , yk ν ) . Moreover, note that, d W (cid:0) Beta( xk µ , yk ν ) , δ ( xk ν xk ν + yk µ ) (cid:1) −→ x + y →∞ . Hence, if h : [0 , → [0 , 1] is defined by setting h ( t ) = tk ν tk ν + (1 − t ) k µ for all t ∈ [0 , , then G h ◦ δ ( µ,ν ) ( x, y ) = Beta( xk µ , yk ν ) , for ( x, y ) ∈ S , is the unique solution to Problem (3) when µ and ν are thescaled Bernoulli distributions defined above and the boundary datum is thecontinuous map h ◦ δ : [0 , → P ([0 , h ◦ δ ( t ) = δ ( tk ν tk ν + (1 − t ) k µ )for all t ∈ [0 , . We now want to find the distribution of the limit composition of a RRU whose reinforcements are distributed according to the scaled Bernoulli dis-tributions µ and ν. Note that h is continuous, monotonically increasing andits inverse is h − ( u ) = uk µ uk µ + (1 − u ) k ν u ∈ [0 , . Then it follows from Remark 4.3 that, G δ ( µ,ν ) ( x, y ) = G h − ◦ h ◦ δ ( µ,ν ) ( x, y ) = h − ◦ Beta( xk µ , yk ν ) , for all ( x, y ) ∈ S . For ( x, y ) in the interior of S , G δ ( µ,ν ) ( x, y ) has a density andthis is f G δ ( µ,ν ) ( x,y ) ( t ) = k ykν µ k xkµ ν Γ( x/k µ + y/k ν )Γ( x/k µ )Γ( y/k ν ) t xkµ − (1 − t ) ykν − [ tk ν + (1 − t ) k µ ] xkµ + ykν for t ∈ [0 , . Apart from the Polya urn scheme, to the best of our knowledgethis is the first example of an RRU where the analytical expression of thedensity of the urn limit composition is known; notably it has been found bysolving Problem (3). A Doob decomposition of the RRU process This appendix provides a series of auxiliary results necessary to prove Propo-sitions 3.1 and 3.2. We will refer to the notations introduced in Section 3.For n = 1 , , ... let A n = σ ( δ , R X (1) , R Y (1) , . . . , δ n , R X ( n ) , R Y ( n )) and con-sider the filtration {A n } ; then, given the initial urn composition ( x, y ) ∈ S , the Doob’s semi-martingale decomposition of Z n ( x, y ) is Z n ( x, y ) = Z ( x, y ) + M n ( x, y ) + A n ( x, y )where { M n } is a zero mean martingale and the previsible process { A n } iseventually increasing (decreasing), again by [11, Theorem 2]. We also denoteby {h M i n } the bracket process associated to { M n } , i.e. the previsible processobtained by the Doob’s decomposition of M n .We first provide some auxiliary inequalities. As a consequence of [2,Lemma 4.1], we can bound the increments ∆ A n of the Z n -compensatorprocess and the increments ∆ h M i n of the bracket process associated to { M n } . In fact, an easy computation gives∆ A n +1 = E (∆ Z n +1 |A n ) = Z n (1 − Z n ) A ∗ n +1 and E ((∆ Z n +1 ) |A n ) = Z n (1 − Z n ) Z ∗ n +1 . where A ∗ n +1 = E (cid:16) R X ( n +1) D n R X ( n +1) D n − R Y ( n +1) D n R Y ( n +1) D n (cid:12)(cid:12)(cid:12) A n (cid:17) , Z ∗ n +1 = E (cid:16) (1 − Z n ) (cid:16) R X ( n +1) D n R X ( n +1) D n (cid:17) + Z n (cid:16) R Y ( n +1) D n R Y ( n +1) D n (cid:17) (cid:12)(cid:12)(cid:12) A n (cid:17) . Now, [2, Lemma 4.2] with m = R β kµ ( dk ) = R β kν ( dk ) gives | A ∗ n +1 | ≤ mm + D n − mβ + D n . (A.25)By applying [2, Lemma 4.1] with h ( x, t ) = ( xx + t ) , B D = [2 β, ∞ ), D = D n , R = R X ( n + 1) or R = R Y ( n + 1) and A = A n , one obtains: Z n (1 − Z n ) (cid:16) mm + D n (cid:17) ≤ E ((∆ Z n +1 ) |A n ) ≤ Z n (1 − Z n ) mβ ( β + D n ) , (A.26)on the set { D n ≥ β } . Since E ((∆ Z n +1 ) |A n ) = E ((∆ A n +1 + ∆ M n +1 ) |A n ) = (∆ A n +1 ) + ∆ h M i n +1 , if D ≥ β, and thus β + D n ≥ β , (A.25) together with (A.26) yields∆ h M i n +1 ≥ Z n (1 − Z n ) (cid:16) mm + D n (cid:17) (cid:16) − (cid:16) β − mβ + D n (cid:17) (cid:17) ≥ Z n (1 − Z n ) (cid:16) mm + D n (cid:17) , ∆ h M i n +1 ≤ Z n (1 − Z n ) mβ ( β + D n ) . (A.27) Lemma A.1. For all k = 1 , , ... , E ( 1 D k ) ≤ β − m ) /D D + m ( k − 1) + β . (A.28) If, in addition, D ≥ β then, for all k, n = 1 , , ..., (cid:12)(cid:12)(cid:12) E (cid:16) c + D k + n − d + D k + n (cid:12)(cid:12)(cid:12) A n (cid:17)(cid:12)(cid:12)(cid:12) ≤ β − m + dm (cid:16) b k − b k +1 (cid:17) , (A.29) when d ≥ c ≥ and b k = c + D n − β + mk .Proof. Let η ∗ be a random variable independent of A ∞ and let η be arandom variable independent of σ ( A ∞ , η ∗ ) and such that η /β has distri-bution Binomial(1 , m/β ). Define A ∗ k + n − = σ ( η ∗ , A k + n − , I ( k + n )); by [2,27emma 4.1], if D > A ∗ k + n − -measurable and 0 ≤ R ≤ β with E ( R ) = m is independent of A ∗ k + n − , one obtains E (cid:16) D + R (cid:12)(cid:12)(cid:12) A ∗ k + n − (cid:17) ≤ mβ D + β + β − mβ D = E (cid:16) D + η (cid:12)(cid:12)(cid:12) A ∗ k + n − (cid:17) , and thus E (cid:16) D k + n + η ∗ (cid:12)(cid:12)(cid:12) A ∗ k + n − (cid:17) = I ( k + n ) E (cid:16) D k + n − + η ∗ + R X ( k + n ) (cid:12)(cid:12)(cid:12) A ∗ k + n − (cid:17) + (1 − I ( k + n )) E (cid:16) D k + n − + η ∗ + R Y ( k + n ) (cid:12)(cid:12)(cid:12) A ∗ k + n − (cid:17) ≤ I ( k + n ) E (cid:16) D k + n − + η ∗ + η (cid:12)(cid:12)(cid:12) A ∗ k + n − (cid:17) + (1 − I ( k + n )) E (cid:16) D k + n − + η ∗ + η (cid:12)(cid:12)(cid:12) A ∗ k + n − (cid:17) = E (cid:16) D k + n − + η ∗ + η (cid:12)(cid:12)(cid:12) A ∗ k + n − (cid:17) . (A.30)Therefore, for c ≥ 0, by applying (A.30) k -times, we get E (cid:16) D k + n + c (cid:12)(cid:12)(cid:12) A n (cid:17) ≤ E (cid:16) D n + c + η k (cid:12)(cid:12)(cid:12) A n (cid:17) (A.31)where η k is independent of σ ( A ∞ ) and η k /β has distribution Binomial( k, m/β ).Equation (A.28) is now a consequence of [13, Eq. (21)]: if ˜ η k ∼ Binomial( k, r )and l > E (cid:16) l + ˜ η k (cid:17) ≤ (cid:16) − rl (cid:17) l + kr + (1 − r ) . Apply this to (A.31) with n = 0, ˜ η k = η k /β , l = D /β and r = m/β toobtain (A.28).Equation (A.29) is a consequence of [13, Eq. (25)]: if ˜ η k ∼ Binomial( k, r )and l > E (cid:16) l + ˜ η k (cid:17) ≤ l + kr − (1 − r ) . Apply this to (A.31) with ˜ η k = η k /β , l = D n + c/β (which is greater than2) and r = m/β to obtain E (cid:16) c + D n + k (cid:12)(cid:12)(cid:12) A n (cid:17) ≤ c + D n + m ( k + 1) − β . E (( d + D n + k ) − |A n ) ≥ ( d + D n + mk ) − , and thus (cid:12)(cid:12)(cid:12) E (cid:16) c + D n + k − d + D n + k (cid:12)(cid:12)(cid:12) A n (cid:17)(cid:12)(cid:12)(cid:12) ≤ β − m + d − c ( c + D n + m ( k + 1) − β )( d + D n + mk ) . Since 1 b k − b k +1 = m ( c + D n − β + mk )( c + D n − β + m ( k + 1))we get (A.29): (cid:12)(cid:12)(cid:12) E (cid:16) c + D n + k − d + D n + k (cid:12)(cid:12)(cid:12) A n (cid:17)(cid:12)(cid:12)(cid:12) b k − b k +1 ≤ β − m + d − cm c + D n − β + mkd + D n + mk ≤ β − m + d − cm . The following Lemma A.2 and Lemma A.3 provide inequalities whichcontrol the previsible and the martingale part of the process Z n respectively;they require that the initial composition of the urn is sufficiently large. Lemma A.2. If D ≥ β , then E (sup r | A r | ) ≤ βD . Proof. Apply (A.29) with n = 0, c = m , d = β . Equation (A.25) then reads E ( | A ∗ k +1 | ) ≤ (2 β − m ) (cid:16) b k − b k +1 (cid:17) , if b k = km + D − ( β − m ). Since A = 0, E (sup r | A r | ) ≤ E (cid:16) X k | ∆ A k +1 | (cid:17) ≤ X k E ( | A ∗ k +1 | ) ≤ β − m X k (cid:16) b k − b k +1 (cid:17) = 2 β − m D − ( β − m ) ≤ βD , where the last inequality is true because β − m ≤ β ≤ D / emma A.3. Let D ≥ β . For all n ≥ , E ( h M i ∞ − h M i n |A n ) ≤ βD . Proof. Since Z n + k (1 − Z n + k ) ≤ / 4, by (A.27), one gets∆ h M i n + k +1 ≤ mβ β + D n + k ) ≤ m (cid:16) D n + k − β + D n + k (cid:17) . Apply (A.29) with c = 0 and d = β , obtaining E (∆ h M i n + k +1 |A n ) ≤ m β − mm (cid:16) b k − b k +1 (cid:17) , if b k = km + D n − β . Thus E ( h M i ∞ − h M i n |A n ) = E (cid:16) X k ≥ ∆ h M i k + n +1 (cid:12)(cid:12)(cid:12) A n (cid:17) ≤ β − m X k ≥ (cid:16) b k − b k +1 (cid:17) ≤ β D n − β ) ≤ βD , since 2( D n − β ) ≥ D − β ) ≥ D . Acknowledgement. The authors thank two anonymous referees whose usefulcomments helped to clarify some key points of the paper. References [1] G. Aletti, C. May, and P. Secchi. 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