Convolution semigroups with linear Jacobi parameters

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1 Convolution semigroups with linear Jacobi parameters Michael Anshelevich; Wojciech Młotkowski Texas A&M University; University of Wrocław February 14, 2011

2 Jacobi parameters. µ = measure with finite moments, m i = x i dµ(x) <. Jacobi parameters Classical result. µ J(µ) = and all {β i }, {γ i 0} occur. ( β0, β 1, β 2, )... γ 0, γ 1, γ 2,... µ positive all γ i 0,

3 Jacobi parameters. Definition 1. Tridiagonal matrix. β 0 γ β 1 γ m n J =. 0 1 β 2 γ.. 2 ; J n = β

4 Jacobi parameters. Definition 2. Cauchy transform 1 G µ (z) = z x dµ(x) = 1 z + m 1 z 2 + m 2 z 3 + m 3 z Then G µ (z) = R z β 0 z β 1 1 γ 0 z β 2 γ 1 γ 2. z β 3 γ 3 z...

5 Jacobi parameters. Definition 3. {P n } = monic orthogonal polynomials with respect to µ. Then xp n (x) = P n+1 (x) + β n P n (x) + γ n 1 P n 1 (x).

6 Convolution semigroups. Measures frequently come in convolution semigroups {µ t }, µ t µ s = µ t+s, (ν τ)(f) = f(x + y) dν(x) dτ(y). Processes with independent increments. ( ) β0 (t), β J(µ t ) = 1 (t), β 2 (t),... γ 0 (t), γ 1 (t), γ 2 (t),... In general, rational functions of t. But:

7 Examples. Gaussian semigroup µ t (x) = 1 2πt e x2 /2t. Jacobi parameters β n (t) = 0, γ n (t) = (n + 1)t. Poisson semigroup µ t (x) = e t k=0 1 k! tk δ k (x) Jacobi parameters β n (t) = t + n, γ n (t) = (n + 1)t. Linear in t!

8 Meixner class. More generally: µ in Meixner class if β n = a + nb, γ n = (n + 1)[c + nd]. Then (not obvious!) for µ t, β n (t) = at + nb, γ n (t) = (n + 1)[ct + nd].

9 Main result. Question. What are all convolution semigroups with Jacobi parameters polynomial in t? Theorem. Only Meixner. In particular, all linear in t.

10 Ingredients of the proof. Recall. β 0 γ β 1 γ J n =. 0 1 β 2 γ β n m n =. Formula for m n in terms of β i, γ i. Viennot-Flajolet: lattice paths. Accardi-Bożejko: non-crossing partitions.

11 Moment-Jacobi formula. m n = π NC 1,2 (n) V π V =1 β d(v,π) V π V =2 γ d(v,π). NC 1,2 = non-crossing partitions into pairs and singletons. d(v, π) = depth of V in π. m 1 = β 0, m 2 = β0 2 + γ 0, m 3 = β β 0 γ 0 + β 1 γ 0, m 4 = β β0γ β 0 β 1 γ 0 + β1γ γ0 2 + γ 0 γ 1.

12 Moment-cumulant formula. m n (µ t ) = complicated. Use cumulants instead. r 1, r 2, r 3,... r n (t) = t r n (linearize convolution). m n = r V. π P(n) V π No simple relation between {r n } and {β n, γ n }.

13 Indirect proof. β 0 (t)= at, β 1 (t)= at + b, γ 0 (t)= ct, γ 1 (t)= 2ct + d. To show: β 2, γ 2,... the same as for Meixner(a, b, c, d). By induction. m 2n+1 (µ t ) = m 2n+1 (µ M t ) + [β n (µ t ) β n (µ M t )] γ n 1 (t)... γ 1 (t)γ 0 (t). So m k = r k + Q k (r 1, r 2,..., r k 1 ). r 2n+1 (µ t ) = r 2n+1 (µ M t ) + [β n (µ t ) β n (µ M t )] γ n 1 (t)... γ 1 (t)γ 0 (t). Know γ 0 (t) = ct and γ 1 (t) depends on t, the rest of γ i (t) polynomial.

14 Indirect proof. By induction. m 2n+1 (µ t ) = m 2n+1 (µ M t ) + [β n (µ t ) β n (µ M t )] γ n 1 (t)... γ 1 (t)γ 0 (t). So m k = r k + Q k (r 1, r 2,..., r k 1 ). r 2n+1 (µ t ) = r 2n+1 (µ M t ) + [β n (µ t ) β n (µ M t )] γ n 1 (t)... γ 1 (t)γ 0 (t). Know γ 0 (t) = ct and γ 1 (t) depends on t, the rest of γ i (t) polynomial. β n (µ t ) = β n (µ M t ). Use m 2n+2 to show γ n (µ t ) = γ n (µ M t ).

15 Other convolutions. Free cumulants and convolution. m n = r V. π NC (n) V π In particular rn (µ t ) = t rn (µ). Boolean cumulants and convolution. m n = r V. π Int(n) V π r n(µ t ) = t r n(µ).

16 Generalization. Same proof works in the abstract (Hasebe, Saigo) setting if 1 r n (µ t ) = t r n (µ). 2 m n = r n + Q(r 1,..., r n 1 ). 3 γ 1 (t) depends on t. Theorem: if have a 4-parameter Meixner-type family, have no others.

17 Free case. γ 1 (t) = ct + d. Answer: free Meixner class. If for µ, J(µ) = ( a, a + b, a + b, )... c, c + d, c + d,... then for its free convolution power (not obvious!) ( ) J(µ t at, at + b, at + b,... ) =. ct, ct + d, ct + d,... 4-parameter. The only free convolution semigroups with Jacobi parameters polynomial in t.

18 Boolean case. γ 1 (t) = d independent of t. General theorem does not apply! In fact, for any µ. J(µ t ) = ( β0 t, β 1, ) β 2,... γ 0 t, γ 1, γ 2,...

19 Better proof for the free case. Recall m n = π NC 1,2 (n) V π V =1 m n = β d(v,π) π NC (n) V π V π V =2 r V. Both non-crossing. Młotkowksi: direct relation. π NC 1,2 (n). u a labeling of π: Some labelings connected. r n = connected labelings u(v ) {0, 1, 2,..., d(v, π)}. γ d(v,π). (β u(v ) β u(v ) 1 ) (γ u(v ) γ u(v ) 1 ). V π V π V =1 V =2

20 Free cumulant-jacobi formula. r 1 = β 0, r 2 = γ 0, r 3 = γ 0 [ (β1 β 0 ) ], r 4 = γ 0 [ (β1 β 0 ) 2 + (γ 1 γ 0 ) ], r 5 = γ 0 [ (β1 β 0 ) 3 + 3(γ 1 γ 0 )(β 1 β 0 ) + (γ 1 γ 0 )(β 2 β 1 ) + γ 0 (β 2 β 1 ) ], r 6 = γ 0 [ (β1 β 0 ) 4 + 6(γ 1 γ 0 )(β 1 β 0 ) 2 + 4γ 1 (β 2 β 1 )(β 1 β 0 ) 2 + γ 1 (β 2 β 1 ) 2 + 2(γ 1 γ 0 ) 2 + γ 1 (γ 2 γ 1 ) ]. β 0 (t) = at, γ 0 (t) = ct, β 1 (t) β 0 (t) = b, γ 1 (t) γ 0 (t) = d, β 2 (t) β 1 (t) = 0, γ 2 (t) γ 1 (t) = 0, etc.

21 Two-state convolution. Two measures ( µ, µ). r n (µ) = free cumulants of µ. Two-state free cumulants R n ( µ, µ) defined via m n ( µ) = R V ( µ, µ) r U (µ). π NC (n) V Outer(π) U Inner(π) Convolution powers: r n (µ t ) = t r n (µ) and R n ( µ t, µ t ) = t R n ( µ, µ). Generalizes both free and Boolean convolution.

22 Two-state convolution. Theorem. If neither µ nor µ are point masses, Jacobi parameters of µ t polynomial in t if and only if ( ãt, at + J( µ t ) = b, ) at + b, at + b,... ct, ct + d, ct + d, ct + d,... and J(µ t ) = ( ) at, at + b, at + b,.... ct, ct + d, ct + d,... Note: polynomial dependence of the Jacobi parameters of µ t automatic.

23 Extended Boolean case. Proposition. µ and u arbitrary. ( µ, δ u ) can be extended to a c convolution semigroup ( µ t, δ tu ) with linear Jacobi parameters.

24 Properties of µ. 1 Satisfy quadratic d µ(x) = + at most 3 atoms. cubic 2 Appear in the two-state free Central and Poisson limit theorems. 3 Appear in the q = 0 case of the Bryc, Matysiak, Wesołowski (2007) quadratic harnesses. 4 Are precisely the measures found by Bożejko and Bryc (2009) to have a two-state free Laha-Lukacs characterization. 5 Satisfy R n+2 ( µ, µ) = b n R n+1 ( µ, µ) + c R k ( µ, µ) r n k (µ). k=2

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