Polynomial families and Boolean probability Michael Anshelevich - PowerPoint PPT Presentation

Polynomial families and Boolean probability Michael Anshelevich January 17, 2008

Derivative: ( x n ) ′ = nx n − 1 , x 0 = 1 . 1. Paul Appell 1880: Appell polynomials = “generalized powers” A n ( x ) ′ = nA n − 1 ( x ) , A 0 ( x ) = 1 and � A n ( x ) dµ ( x ) = 0 for some probability measure µ . Equivalently: X a random variable with distribution µ , denote A n by A X n , � A X � E n ( X ) = 0 . 1

Examples. 2 π e − x 2 / 2 dx , 1 Hermite polynomials, dµ = √ Bernoulli polynomials, dµ = 1 [0 , 1] dx . 2. Generating function: ∞ 1 n ! A n ( x ) z n = e xz − ℓ ( z ) , � n =0 where � e xz dµ ( x ) . ℓ ( z ) = log 2

3. Binomial property: if X, Y are independent random variables, then n � n A X + Y � A X k ( X ) A Y � ( X + Y ) = n − k ( Y ) n k k =0 (compare ( X + Y ) n ). 4. Martingale property: if { X t } is a L´ evy process, i.e. a random process with stationary independent increments, then � A X t � = A X s E n ( X t ) | ≤ s n ( X s ) . 3

C ONNECTION WITH FREE PROBABILITY . Start with the difference quotient ∂ ( f )( x, y ) = f ( x ) − f ( y ) . x − y n − 1 ∂ ( x n ) = x k y n − k − 1 . � k =0 So define the free Appell polynomials by n − 1 � ∂A n ( x, y ) = A k ( x ) A n − k − 1 ( y ) , A 0 ( x ) = 1 k =0 and � � A X � A n ( x ) dµ ( x ) = 0 or E n ( X ) = 0 . 4

� 4 − x 2 dx . Examples: Chebyshev polynomials, dµ = 1 2 π 2. Generating function: ∞ 1 A n ( x ) z n = � 1 − xz + zR ( z ) , n =0 where R ( z ) = R -transform of µ . 3. Binomial property: if X, Y are freely independent random variables, then A X + Y A X u (1) ( X ) A Y u (2) ( Y ) A X u (3) ( X ) A Y � ( X + Y ) = u (4) ( Y ) . . . n A Y v (1) ( Y ) A X v (2) ( X ) A Y v (3) ( Y ) A X � + v (4) ( X ) . . . , where u (1) + u (2) + . . . = v (1) + v (2) + . . . = n . 5

Example. A X + Y ( X + Y ) = A X 3 ( X ) + A X 2 ( X ) A Y 1 ( Y ) + A X 1 ( X ) A Y 1 ( Y ) A X 1 ( X ) 3 + A Y 1 ( Y ) A X 2 ( X ) + A X 1 ( X ) A Y 2 ( Y ) + A Y 1 ( Y ) A X 1 ( X ) A Y 1 ( Y ) + A Y 2 ( Y ) A X 1 ( X ) + A Y 3 ( Y ) . (again compare ( X + Y ) n ). 4. Martingale property: if { X t } is a free L´ evy process, i.e. a random process with stationary freely independent increments, then � A X t � = A X s n ( X t ) | ≤ s n ( X s ) . E 6

5. Polynomials with generating function ∞ 1 P n ( x ) z n = � 1 − xU ( z ) + U ( z ) R ( U ( z )) n =0 for some U ( z ) also martingales. Free Sheffer polynomials. 6. Free Meixner distributions = measures for which their orthogonal polyno- mials are free Sheffer (classical versions classified by Meixner 1934). In this case, U ( z ) = R ( z ) �− 1 � and R ( z ) = 1 + bR ( z ) + cR ( z ) 2 . z Examples. Semicircular, Marchenko-Pastur, limit of Jacobi / double Wishart, arcsine, Kesten measures, Bernoulli distributions. 7

In today’s talk: start with a very simple derivative operator Df ( x ) = f ( x ) − f (0) . x The q = 0 version of the q -derivative operator D q f ( x ) = f ( x ) − f ( qx ) . (1 − q ) x x 0 = 1 . D ( x n ) = x n − 1 , So define the (Boolean) Appell polynomials by DA n ( x ) = A n − 1 ( x ) , A 0 ( x ) = 1 and � � A X � A n ( x ) dµ ( x ) = 0 n ( X ) = 0 . or E 8

2. Generating function: ∞ A n ( x ) z n = 1 − η µ ( z ) � . 1 − xz n =0 What is η µ ( z ) ? � 1 − η (1 /z ) � 1 − η (1 /z ) 1 = dµ = z dµ = z (1 − η (1 /z )) G µ ( z ) . 1 − x/z z − x So 1 η (1 /z ) = 1 − zG ( z ) . This function appears in Boolean non-commutative probability theory. 9

A an algebra, ϕ a state on it. Non-unital subalgebras B 1 , B 2 , . . . , B k are Boolean independent if for b i ∈ B u ( i ) , u (1) � = u (2) � = . . . � = u ( n ) , ϕ [ b 1 b 2 . . . b n ] = ϕ [ b 1 ] ϕ [ b 2 ] . . . ϕ [ b n ] . Example. In C � x 1 , x 2 , . . . , x d � with the state � � = 0 , ϕ [1] = 1 , ϕ x u (1) x u (2) . . . x u ( n ) x 1 , . . . , x d are freely independent. In C � x 1 , x 2 , . . . , x d � with the state = e − n , � � ϕ x u (1) x u (2) . . . x u ( n ) x 1 , . . . , x d are Boolean independent. Combinatorics governed by the lattice of interval partitions, isomorphic to the Boolean lattice of subsets. 10

3. Binomial property: if X, Y are Boolean independent random variables, then n − 1 A X + Y ( X + Y ) = A X ( X + Y ) k − 1 Y A X � n ( X ) + n − k ( X ) n k =1 n − 1 ( X + Y ) k − 1 XA Y + A Y � n ( Y ) + n − k ( Y ) . k =1 Example. ( X + Y ) 3 = X 3 + Y X 2 + ( X + Y ) Y X + ( X + Y ) 2 Y + Y 3 + XY 2 + ( X + Y ) XY + ( X + Y ) 2 X. 11

4. Martingale property: if { X t } is a Boolean L´ evy process, i.e. a random process with stationary Boolean independent increments, then E [ A n ( X t ) | ≤ s ] = A n ( X s ) . Boolean states typically not tracial, so this does not immediately imply the Markov property; known due to Franz 2003. 12

5. Boolean Sheffer polynomials ∞ P n ( x ) z n = 1 − η ( V ( z )) � 1 − xV ( z ) . n =0 Proposition. These are the same as free: 1 − xU ( z ) + U ( z ) R ( U ( z )) = 1 − η ( V ( z )) 1 1 − xV ( z ) , where � − 1 � V ( z ) = 1 + U ( z ) R ( U ( z )) U ( z ) . Remark. Everything works in the multivariate situation. Start with “left” partial derivatives D 1 , D 2 , . . . , D d , D i ( x j x u (1) x u (2) . . . x u ( k ) ) = δ ij x u (1) x u (2) . . . x u ( k ) 13

6. Corollary. Boolean Meixner distributions = free Meixner distributions. Moreover, V ( z ) = ( Dη ( z )) �− 1 � and D 2 η ( z ) = 1 + bDη ( z ) + (1 + c )( Dη ( z )) 2 . Recall D 2 ( zR ( z )) = 1 + bD ( zR ( z )) + c ( D ( zR ( z ))) 2 and ℓ ( z ) ′′ = 1 + βℓ ( z ) ′ + γ ( ℓ ( z ) ′ ) 2 . Bercovici, Pata: there are bijections between infinitely divisible, freely infinitely divisible, Boolean infinitely divisible distributions. 14

ℓ µ ( z ) = zR ν ( z ) = η ζ ( z ) , µ ↔ ν ↔ ζ, Gaussian ↔ Semicircular ↔ Symmetric Bernoulli , Poisson ↔ Marchenko-Pastur ↔ Asymmetric Bernoulli . Does not take classical Meixner to free Meixner. Takes free Meixner to Boolean Meixner: µ b,c �→ µ b, 1+ c . More general results on the behavior under the Belinschi-Nica transformation. Again, this is all true in the multi-variable case. 15

If µ is a Meixner distribution, the orthogonal polynomials for µ ∗ t satisfy recur- sion relations xP n ( x ) = P n +1 ( x ) + ( tβ 0 + nb ) P n ( x ) + n ( tγ 1 + ( n − 1) c ) P n − 1 . If µ is a free / Boolean Meixner distribution, the orthogonal polynomials for µ ⊞ t satisfy recursion relations xP 0 ( x ) = P 1 ( x ) + tβ 0 P 0 ( x ) , xP 1 ( x ) = P 2 ( x ) + ( tβ 0 + b ) P 1 ( x ) + tγ 1 P 0 , xP n ( x ) = P n +1 ( x ) + ( tβ 0 + b ) P n ( x ) + ( tγ 1 + c ) P n − 1 . In contrast, if µ is any distribution, the orthogonal polynomials for µ ⊎ t satisfy recursion relations xP 0 ( x ) = P 1 ( x ) + tβ 0 P 0 ( x ) , xP 1 ( x ) = P 2 ( x ) + β 1 P 1 ( x ) + tγ 1 P 0 , xP n ( x ) = P n +1 ( x ) + β n P n ( x ) + γ n P n − 1 . Proof using (multivariate) continued fractions. 16

ϕ = state (with monic orthogonal polynomials). � � � � 1 + M ( z ) = 1 + ϕ [ x i ] z i + ϕ x i x j z i z j + . . . i i,j its moment generating function. Stieltjes continued fraction: one-variable case. z 2 + . . . � x 2 � 1 + M ( z ) = 1 + ϕ [ x ] z + ϕ 1 = . ω 1 z 2 1 − α 0 z − ω 2 z 2 1 − α 1 z − 1 − α 2 z − ω 3 z 2 1 − . . . 17

For k = 1 , 2 , . . . , there are diagonal non-negative d k × d k Proposition. matrices C ( k ) and Hermitian matrices T ( k ) , such that i 1 + M ( z ) = 1 j 1 z j 1 E j 1 C (1) | � k 1 E k 1 z k 1 � i 0 z i 0 T (0) 1 − � − i 0 j 2 z j 2 E j 2 C (2) | � � k 2 E k 2 z k 2 i 1 z i 1 T (1) 1 − � − i 1 j 3 z j 3 E j 3 C (3) | � k 3 E k 3 � i 2 z i 2 T (2) 1 − � − i 2 1 − . . . M d 2 × d 2 = M d × d ⊗ M d × d . 18

L AHA -L UKACS PROPERTY . Proposition. Suppose X, Y are (appropriately) independent, self-adjoint, non-degenerate and there are numbers α, α 0 , C, a, b ∈ R such that ϕ [ X | X + Y ] = α ( X + Y ) + α 0 and � � 1 + a ( X + Y ) + b ( X + Y ) 2 Var [ X | X + Y ] = C . X, Y independent ⇒ Meixner (Laha, Lukacs). X, Y freely independent ⇒ free Meixner (Bo˙ zejko, Bryc). X, Y Boolean independent ⇒ Bernoulli. 19