Bivariate Uniqueness in the Logistic Recursive Distributional Equation

Transcription

1 Bivariate Uniqueness in the Logistic Recursive Distributional Equation Antar Bandyopadhyay Technical Report # 629 University of California Department of Statistics 367 Evans Hall # 3860 Berkeley CA November 2002 Abstract In this ork e prove the bivariate uniqueness property of the Logistic fixed-point equation, hich arise in the study of the random assignment problem, as discussed by Aldous [4]. Using this and the general frameork of Aldous and Bandyopadhyay [2], e then conclude that the associated recursive tree process is endogenous, and hence the Logistic variables defined in Aldous ork [4] are measurable ith respect to the σ-field generated by the edge eights. The method involves construction of an explicit recursion to sho the uniqueness of the associated integral equation. Key ords and phrases. Bivariate uniqueness, distributional identity, fixed-point equation, Logistic distribution, measurability, non-linear integral equation, Poisson eighted infinite tree, random assignment problem.

2 Introduction Our result is much easier to state than to motivate, so e ill first state the main result Theorem of this ork. Theorem Let 0 < ξ < ξ 2 < be points of a Poisson point process of rate on 0,. Let X, Y, X j, Y j j be independent random variables ith some common distribution ν on R 2, hich are independent of ξ j j. Then X Y d = min j ξ j X j min j ξ j Y j, if and only if ν = µ, here µ is defined as the joint distribution of Z, Z on R 2, ith Z Logistic distribution, that is, the distribution function of Z is given by Hx := P Z x =, x R 2 + e Theorem is a concrete result falling ithin the frameork of recursive distributional equations surveyed in [2]. This particular problem arose in the study of a classical problem of combinatorial optimization, namely, the mean-field random assignment problem. The folloing section develops background and provides the details of our motivation for Theorem. In Section 3 e give a proof of Theorem using analytic techniques. Some technical results hich are not terribly important, but are needed for the proof, are given separately in Section 4. 2 Background and Motivation This section mainly provides background and motivation for Theorem. For a given n n matrix of costs C ij, consider the problem of assigning n jobs to n machines in the most cost effective ay. Thus the task is to find a permutation π of {, 2,..., n}, hich solves the folloing minimization problem A n := min π n C i,πi. 3 This problem has been extensively studied in literature for a fixed cost matrix, and there are various algorithms to find the optimal permutation π. A probabilistic model for the assignment problem can be obtained by assuming that the costs are independent random variables each ith distribution 2 i=

3 Uniform[0, ]. Although this model appears to be rather simple, careful investigations of it in the last fe decades have proven that, it has enormous richness in its structure. For a careful survey and other related orks see [8, 3]. Our interest in this problem is from another perspective. In 200 Aldous [4] shoed that lim E[A n] = ζ2 = π2 n 6, 4 confirming the earlier ork of Mézard and Parisi [6], here they computed the same limit using some non-rigorous arguments based on the replica method [7]. In an earlier ork Aldous [] shoed that the limit of E [A n ] as n exists for any cost distribution, and does not depend on the specifics of it, except only on the value of its density at 0, provided it exists and is strictly positive. So for calculation of the limiting constant one can assume that C ij s are independent and each has Exponential distribution ith mean n, and re-rite the objective function A n in the normalized form, A n = min π n n C i,πi. 5 Aldous [4] identified the limit constant ζ2 in terms of an optimal matching problem on a limit infinite tree ith random edge eights. This structure is called Poisson Weighted Infinite Tree, or, PWIT, it is described as follos see the survey of Aldous and Steele [3] for a more friendly account. Let T := V, E be the canonical infinite rooted tree ith vertex set V := m=0 Nm here N 0 := { }, and edge set E := {e = i, ij i V, j N}. We consider as the root of the tree, and ill rite j = j j N. For every vertex i V, let ξ ij j be points of independent Poisson process of rate on 0,. Define the eight of the edge e = i, ij as ξ ij. Aldous [4] shoed that on a PWIT one can construct random variables X i i V taking values in R, such that X i = min j ξ ij X ij, i V. { ξi X i is independent of j rite i = d if i N d. X i Logistic distribution. j i= } i < i, for all i V\{ }; here e 3

4 As defined in Theorem, e say a random variable taking real values has Logistic distribution if it has distribution function given by 2. The key calculation ingredient of the construction above is the folloing recursive distributional equation RDE, hich e ill refer as the Logistic RDE. X = d min ξ j X j, 6 j here ξ j j are points of a Poisson point process of rate on 0,, and are independent of X j j, hich are independent and identically distributed ith same la as of X. It is easy to sho [4] that 6 has unique solution as Logistic distribution. The existence of X i s then follos from Kolmogorov s consistency. The construction of the X i from the Logistic RDE is abstracted in [2]; to any RDE one can associate a recursive tree process RTP. In [4] Aldous has heuristic interpretation of X i s through the edge eights. Thus it is natural to ask if they are actually measurable ith respect to the sigma-field generated by the edge eights. We note that the abstract construction of the X i s using Kolmogorov s consistency does not thro any light toards measurability issues. Notationally, if } G := σ {ξ ij j i V, 7 then the question above can be restated as is X measurable ith respect to the σ-field G? 8 Our ork is motivated to anser this question of Aldous see remarks 4.2.d and 4.2.e in [4]. Remarkably enough, one can sho in the abstract setting of RTP that the question 8 has an affirmative anser if and only if the bivariate uniqueness property of Theorem holds. This has been extensively discussed in the ork of Aldous and Bandyopadhyay [2], hose Theorem specializes to our current setting as follos. Theorem 2 Suppose S is the set of all probabilities on R 2 and e define Λ : S S as d minj ξ Λν = j X j, 9 min j ξ j Y j here ξ j j are points of a Poisson process ith mean intensity on 0,, and are independent of X j, Y j j, hich are i.i.d ith distribution ν on R 2. Suppose further that Λ is continuous ith respect to the eak convergence topology hen restricted to the subspace S of S defined as { } S := ν S both the marginals of ν are Logistic distribution. 0 4

5 If the fixed-point equation Λν = ν has unique solution as µ as defined in Theorem then X as defined above is measurable ith respect to the σ-field G. Notice that Theorem basically states that Λν = ν has unique solution as µ. Further it is easy to see that the operator Λ is continuous ith respect to the eak convergence topology hen restricted to the subspace S see Proposition 9 of Section 4 for a proof. Thus using Theorem 2 e get the folloing immediate corollary of Theorem hich ansers the question 8 affirmatively, and hence proving that the RTP associated ith Logistic RDE 6 is endogenous. Corollary 3 X as defined above is measurable ith respect to the σ-field G, generated by the edge eights of the PWIT. 3 Proof of Theorem First observe that if the equation has a solution then, the marginal distributions of X and Y solve the Logistic RDE, and hence they are both Logistic. Further by inspection µ is a solution of. So it is enough to prove that µ is the only solution of. Let ν be a solution of. Notice that the points {ξ j ; X j, Y j j } form a Poisson point process, say P, on 0, R 2, ith mean intensity ρt; x, y dt dx, y := dt νdx, y. Thus if Gx, y := P X > x, Y > y, for x, y R, then Gx, y = P min ξ j X j > x, and, min ξ j Y j > y j j { } = P No points of P are in t; u, v t u x, or, t v y = exp ρt; u, v dt du, v = exp t u x, or, t v y 0 [ Ht x + Ht y Gt x, t y ] dt = Hx Hy exp Gt x, t y dt. 0 The last equality follos from properties of the Logistic distribution see Fact of appendix. For notational convenience in this paper e ill rite F := F, for any distribution function F. 5

6 The folloing simple observation reduces the bivariate problem to a univariate problem. Lemma 4 For any to random variables U and V, U = V a.s. if and only if U d = V d = U V. Proof : First of all if U = V a.s. then U V = U a.s. Conversely suppose that U = d V = d U V. Fix a rational q, then under our assumption, P U q < V = P V > q P U > q, V > q = P V > q P U V > q = 0 A similar calculation ill sho that P V q < U = 0. These are true for any rational q, thus P U V = 0. Thus if e can sho that X Y also has Logistic distribution, then from the lemma above e ill be able to conclude that X = Y a.s., and hence the proof ill be complete. Put g := P X Y >, e ill sho g = H. No, for every fixed x R, gx = Gx, x by definition. So using e get gx = H 2 x exp gs ds, x R. 2 Notice that from A see Fact 3 of appendix g = H is a solution of this non-linear integral equation 2, hich corresponds to the solution ν = µ of the original equation. To complete the proof of Theorem e need to sho that this is the only solution. For that e ill prove that the operator associated ith 2 defined on an appropriate space is monotone and has unique fixed-point as H. The techniques e ill use here are similar to Eulerian recursion, and are heavily based on analytic arguments. Let F be the set of all functions f : R [0, ] such that H 2 x fx Hx, x R, f is a tail of a distribution, that is, random variable say W such that fx = P W > x, x R. 6

7 Observe that by definition H F. Further, from 2 it follos that gx H 2 x, x R, as ell as, gx = P X Y > x P X > x = Hx, x R. So it is appropriate to search for solutions of 2 in F. Let T : F F be defined as T fx := H 2 x exp fs ds, x R. 3 Proposition 0 of Section 4 shos that T does indeed map F into itself. Observe that the equation 2 is nothing but the fixed-point equation associated ith the operator T, that is, g = T g on F. 4 We here note that using A see Fact 3 of appendix T can also be ritten as T fx := Hx exp Hs fs ds, x R, 5 hich ill be used in the subsequent discussion. Define a partial order on F as, f f 2 in F if f x f 2 x, x R, then the folloing result holds. Lemma 5 T is a monotone operator on the partially ordered set F,. Proof : Let f f 2 be to elements of F, so from definition f x f 2 x, x R. Hence f s ds f 2 s ds, x R T f x T f 2 x, x R T f T f 2. Put f 0 = H 2, and for n N, define f n F recursively as, f n = T f n. No from Lemma 5 e get that if g is a fixed-point of T in F then, f n g, n 0. 6 If e can sho f n H pointise, then using 6 e ill get H g, so from definition of F it ill follo that g = H, and our proof ill be complete. For that, the folloing lemma gives an explicit recursion for the functions {f n } n 0. 7

8 Lemma 6 Let β 0 s = s, 0 s. Define recursively Then for n, β n s := s e β n d, 0 < s. 7 f n x = Hx exp β n Hx, x R. 8 Proof : We ill prove this by induction on n. Fix x R, for n = e get f x = T f 0 x = Hx exp = Hx exp = Hx exp = Hx exp = Hx exp Hx = Hx exp β 0 Hx Hs H 2 s ds [ using 5 ] Hs Hs ds Hs Hs ds H s ds [ using Fact of appendix ] No, assume that the assertion of the Lemma is true for n {, 2,..., k}, for some k, then from definition e have f k+ x = T f k x = Hx exp = Hx exp = Hx exp Hx Hs fk s ds Hs [ using 5 ] e β k Hs ds e β k d 9 The last equality follos by substituting = Hs and thus from Fact and Fact 2 of the appendix e get that d = Hs ds and H = Hx. Finally by definition of β n s and using 9 e get f k+ = T f k. 8

9 To complete the proof it is no enough to sho that β n 0 pointise, hich ill imply by Lemma 6 that f n H pointise, as n. Using Proposition see Section 4 e get the folloing characterization of the pointise limit of these β n s. Lemma 7 There exists a function L : [0, ] [0, ] ith L = 0, such that Ls = e L d, s [0,, 20 and Ls = lim n β ns, 0 s. s Proof : From part b of Proposition e kno that for any s [0, ] the sequence {β n s} is decreasing, and hence a function L : [0, ] [0, ] such that Ls = lim β ns. No observe that β n β 0 = n, 0, and hence 0 e β n β n, 0. Thus by taking limit as n in 7 and using the dominated convergence theorem along ith part a of Proposition e get that Ls = s e L d, 0 s <. The above lemma basically translates the non-linear integral equation 2 to the non-linear integral equation 20, here the solution g = H of 2 is given by the solution L 0 of 20. So at first sight this may not lead us to the conclusion. But fortunately, something nice happens for equation 20, and e have the folloing result hich is enough to complete the proof of Theorem. Lemma 8 If L : [0, ] [0, ] is a function hich satisfies the non-linear integral equation 20, namely, Ls = s and if L = 0, then L 0. e L d, 0 s <, 9

10 Proof : First note that L 0 is a solution. No let L be any solution of 20, then L is infinitely differentiable on the open interval 0,, by repetitive application of Fundamental Theorem of Calculus. Consider, η := e L + e L, [0, ]. 2 Observe that η0 = η = 0 as L = 0. No, from 20 e get that L = e L, 0,. 22 Thus differentiating [ the function η e get η = e L 2 e L + e L ] 0, 0,. 23 So the function η is decreasing in 0, and is continuous in [0, ] ith boundary values as 0, hence η 0 L 0. 4 Some Technical Details In this section e prove some results hich ere used in Sections 2 and 3 for proving Theorems and Corollary 3. These results are mainly technical details and hence have been omitted in the previous sections. Proposition 9 The operator Λ defined in Theorem 2 is eakly continuous hen restricted to the subspace S as defined in 0. Proof : Let {ν n } n= S d and suppose that ν n ν S. We ill d sho that Λν n Λν. Let Ω, F, P be a probability space such that, {X n, Y n } n= and X, Y random vectors taking values in R 2, ith X n, Y n ν n, n, and X, Y ν. Notice that by definition X n d = Yn d = X d = Y, and each has Logistic distribution. Fix x, y R, then using similar calculations as in e get G n x, y := Λν n x, y, = HxHy exp P X n > t x, Y n > t y dt 0 = HxHy exp P X n + x Y n + y > t dt 0 = HxHy exp E [ X n + x + Y n + y +], 24 0

11 and a similar calculation ill also give that Gx, y := Λν x, y, = HxHy exp E [ X + x + Y + y +]. 25 No to complete the proof all e need is to sho E [ X n + x + Y n + y +] E [ X + x + Y + y +]. Since e assumed that X n, Y n d X, Y thus X n + x + Y n + y + d X + x + Y + y +, x, y R. 26 Fix x, y R, define Zn x,y Y + y +. Observe that := X n + x + Y n + y +, and Z x,y := X + x + 0 Z x,y n X n + x + X n + x, n. 27 d But, X n + x = X + x, n. So clearly {Zn x,y } n= is uniformly integrable. Hence e conclude using Theorem 25.2 of Billingsley [5] that E [Zn x,y ] E [Z x,y ]. This completes the proof. Proposition 0 The operator T maps F into F. Proof : First note that if f F, then by definition T fx H 2 x, x R. Next by definition of F e get that f F f H, thus fs ds Hs ds, x R T fx H 2 x exp Hs ds = Hx, x R The last equality follos from A see Fact 3 of appendix. So, H 2 x T fx Hx, x R. 28

12 No e need to sho that for f F, T f is a tail of a distribution. From the definition T f is continuous in fact infinitely differentiable. Further using 28 and the fact that H is a tail of a distribution e get that lim T fx = 0, and lim T fx =. 29 x x Finally let x y be to real numbers, then Hs fs ds Hs fs ds, y because f H. Also Hx Hy, thus using 5 e get T fx T fy 30 So using 28, 29, 30 e conclude that T f F if f F. Proposition The folloing are true for the sequence of functions {β n } n 0 as defined in 7. a For every n, lim s 0+ β ns exists, and is given by 0 e ill rite this as β n 0. e β n d, b For every fixed s [0, ], the sequence {β n s} is decreasing. Proof : a Note that for n =, β s = Thus lim s 0+ β s exists and is given by 0 s e d, s 0, ], e β 0 d. No e assume that the assertion is true for n {, 2,..., k} for some k, e ill sho that it is true for n = k +. For that note β k+ s = s e β k d, s 0, ]. 2

13 But, lim e β k 0+ e β k = lim 0+ β k = lim 0+ = e β k 0 v β k e β k v dv The last equality follos from mean-value theorem and the induction hypothesis. The rest follos from the definition. b Notice that β 0 s = s for s [0, ], thus β s = s e d < s = β 0 s, s [0, ]. No assume that for some n e have β n s < β n s < < β 0 s, s [0, ], if e sho that β n+ s < β n s, s [0, ] then by induction the proof ill be complete. For that, fix s [0, ] then β n+ s = s < s = β n s Hence the proof of the proposition. e β n d e β n d 5 Final Remarks 5. Comments on the proof of Theorem a Intuitively, a natural approach to sho that the fixed-point equation Λν = ν on S has unique solution, ould be to specify a metric ρ on S such that the operator Λ becomes a contraction ith respect to it. Unfortunately, this approach seems rather hard or may even be impossible. For this reason e have taken a complicated route of proving the bivariate uniqueness using analytic techniques similar to Eulerian recursion. 3

14 b Although at first glance it seems that the operator T as defined in 3 is just an analytic tool to solve the equation 2, it has a nice interpretation through Logistic RDE 6. Suppose A is the operator associated ith Logistic RDE, that is, Aµ d = min j ξ j X j, 3 here ξ j j are points of a Poisson point process of mean intensity on 0,, and are independent of X j j, hich are i.i.d ith distribution µ on R. It is easy to check that the domain of definition of A is the space A := { F F is a distribution function on R and 0 } F s ds <. 32 Note that in probabilistic terminology the condition 0 F s ds < means E F [X + ] <. Notice that from definition F A, and T can be naturally extended to the hole of A. In that case the folloing identity holds T µ H Aµ H =, µ A. 33 This at least explains the monotonicity of T through anti-monotonicity property easy to check of the Logistic operator A. The identity 33 seems rather interesting and might be useful for deeper understanding of the bivariate uniqueness property of the Logistic fixed-point equation. 5.2 Domain of attraction As discussed in Aldous and Bandyopadhyay [2], related to any fixed-point equation there is alays the question of its domain of attraction. From the recursion proof e can clearly see that the equation 2 has the hole of F ithin its domain of attraction. Thus it is natural to believe that one might be able to derive the uniqueness by a contraction argument. It still remains an open problem to determine the exact domain of attraction of Logistic RDE. Unfortunately, the identity 33 does not seem to be useful in that regard. 5.3 Everyhere discontinuity of the operators Λ and A From Proposition 9 e get that the operator Λ is continuous ith respect to the eak convergence topology hen restricted to the subspace S of its 4

15 domain of definition, and e sa that this is enough regularity to conclude nice result like Corollary 3. It is still interesting to see if Λ is continuous on hole of its domain of definition. Unfortunately, it is just the opposite. Λ is discontinuous everyhere on its domain of definition. In fact, even the operator A associated ith the Logistic RDE as defined in 3, is discontinuous everyhere on its domain of definition A. To see this e note that if d µ A, and µ n µ, here {µ n } n= A, then Aµ d n Aµ provided E µn [X n + ] E µ [X + ]. Thus clearly A is discontinuous everyhere on A and hence so is Λ. Appendix Here e provide some knon basic facts about the Logistic distribution hich are used in various places in the proofs. First recall that e say a real valued random variable Z has Logistic distribution if its distribution function is given by 2, namely, Hx = P Z x = The folloing facts hold for the function H., x R. + e Fact H is infinitely differentiable, and H = H H, here H = H. Proof : From the definition it follos that H is infinitely differentiable on R. Further, H x = e + e + e = Hx Hx x R Fact 2 H is symmetric around 0, that is, H = Hx x R. Proof : From the definition e get that for any x R, H = + e x = e = Hx. + e 5

16 Fact 3 H is the unique solution of the non-linear integral equation Hx = exp Hs ds, x R. A Proof : Notice that the equation A is nothing but Logistic RDE, since AHx = exp Hs ds, x R see proof of Lemma 5 in Aldous [4]. Thus from the fact that H is the unique solution of Logistic RDE Lemma 5 of Aldous [4] e conclude that H is unique solution of equation A. Acknoledgments This ork is a part of my doctoral dissertation, ritten under the guidance of Professor David J. Aldous, hom I ould like to thank for suggesting the problem and for many illuminating discussions. My sincere thanks to him also for his continuous support and help. References [] David Aldous. Asymptotics in the random assignment problem. Probab. Theory Related Fields, 934: , 992. [2] David Aldous and Antar Bandyopadhyay. Probability Distributions Arising as Fixed-points of Max-type Operators. in preparation, [3] David Aldous and J. Michael Steel. The Objective Method : Probabilistic Combinatoria Optimization and Local Weak Convergence. in preparation, [4] David J. Aldous. The ζ2 Limit in the Random Assignment Problem. Random Structures Algorithms, 84:38 48, 200. [5] Patrick Billingsley. Probability and measure. John Wiley & Sons Inc., Ne York, third edition, 995. A Wiley-Interscience Publication. [6] M. Mézard and G. Parisi. On the solution of the random link matching problem. J. Physique, 48:45 459,

17 [7] M. Mézard, G. Parisi, and M. A. Virasoro. Spin Glass Theory and Beyond. World Scientific, Singapore, 987. [8] J. Michael Steele. Probability theory and combinatorial optimization. Society for Industrial and Applied Mathematics SIAM, Philadelphia, PA,