A new approach to Poisson approximation and de-Poissonization - PowerPoint PPT Presentation

A new approach to Poisson approximation and de-Poissonization Hsien-Kuei Hwang Vytas Zacharovas Institute of Statistical Science Academia Sinica Taiwan 2008

Outline Combinatorial scheme Poisson approximation Improvements of Prokhorov’s results Depoissonization

Definition of combinatorial scheme Let { X n } n ≥ n 0 be a sequence of random variables. For a wide class of combinatorial problems the probability generating function ∞ � P ( X n = m ) w n P n ( w ) = m = 0 satisfies asymptotically P n ( z ) = e λ ( z − 1 ) z h ( g ( z ) + ε n ( z )) ( n → ∞ ) , where h is a fixed non-negative integer, – λ = λ ( n ) → ∞ with n ; – g is independent of n and is analytic for | z | ≤ η , where η > 1; g ( 1 ) = 1 and g ( 0 ) � = 0; – ε n ( z ) satisfies ε n ( z ) = o ( 1 ) , uniformly for | z | ≤ η .

Cauchy formula 1 � e λ ( z − 1 ) ( g ( z ) + ε n ( z )) dz P ( X n = m ) = z n + 1 2 π i | z | = r k ≈ e − λ λ m � a j C j ( λ, m ) (1) m ! j = 0 if g ( z ) ≈ a 0 + a 1 ( z − 1 ) + a 2 ( z − 1 ) 2 + · · · + ( z − 1 ) k

Charlier polynomials The Charlier polynomials C k ( λ, m ) are defined by formula λ m m ! C k ( λ, m ) = [ z m ]( z − 1 ) k e λ z , (2) or, equivalently ∞ λ m m ! C k ( λ, m ) z m = ( z − 1 ) k e λ z . � m = 0

Orthogonality relations Jordan in 1926 proved that Charlier polynomials are orthogonal with respect to Poisson measure e − λ λ m m ! , that is ∞ C k ( λ, m ) C l ( λ, m ) e − λ λ m k ! � m ! = δ k , l λ k , m = 0 Which means that if a sequence of complex numbers P 0 , P 1 , . . . satisfies condition ∞ | P j | 2 � < ∞ e − λ λ j j = 0 j ! then we can expand ∞ P m = e − λ λ m � a j C j ( λ, m ) . m ! j = 0

Suppose we have a generating function ∞ � P n z n P ( z ) = n = 0 then ∞ P m = e − λ λ m � a j C j ( λ, m ) . m ! j = 0 is equivalent to ∞ ∞ P n z n = e λ ( z − 1 ) � � a j ( z − 1 ) j n = 0 j = 0

P ( z ) = e λ ( z − 1 ) f ( z ) . e λ ( z − 1 ) is a generating function of Poisson distribution. Therefore if P ( z ) ≈ e λ ( z − 1 ) f ( 1 ) we can expect that P m ≈ f ( 1 ) e − λ λ m m ! .

Parseval identity for Charlier polynomials ∞ ∞ P m z m = e λ ( z − 1 ) f ( z ) = e λ ( z − 1 ) � � a n ( z − 1 ) n m = 0 n = 0 Theorem Suppose f ( z ) is analytic in the whole complex plain and | f ( z ) | ≪ e H | z − 1 | 2 as | z | → ∞ , then for any λ > 2 H we have 2 ∞ � � ∞ e − λ λ n P n n ! � � � � λ n | a n | 2 n ! = � � e − λ λ n � � � n ! � n = 0 n = 0

Application of the Parseval identity P ( z ) = e λ ( z − 1 ) g ( z ) Theorem Suppose g ( z ) is analytic in the whole complex plane and | g ( z ) | � Ae H | z − 1 | 2 , (3) for all z ∈ C with some positive constants A and H. Then uniformly for all N , n � 0 and λ � ( 2 + ǫ ) H with ǫ > 0 we have �   � � ( N + 1 ) / 2 N P n − e − λ λ n � � � ( 2 + ǫ ) H � � � a j C j ( λ, n ) � A � �   λ ( N + 2 ) / 2 n ! � � j = 0 � �

Theorem Under the conditions of the previous theorem � � � ( N + 1 ) / 2 ∞ N P n − e − λ λ n � � � ( 2 + ǫ ) H � � � � a j C j ( λ, n ) � A � � λ ( N + 1 ) / 2 n ! � � n = 0 j = 0 � � for all n , N � 0 .

Parseval identity for Charlier polynomials. Integral form. Theorem Suppose f ( z ) is analytic in the whole complex plain and | f ( z ) | ≪ e H | z − 1 | 2 as | z | → ∞ , then for any λ > 2 H we have 2 � ∞ ∞ � � e − λ λ n P n � � r /λ ) e − r dr , � � n ! = I ( � � e − λ λ n � � 0 � n ! � n = 0 where � π I ( r ) = 1 | f ( 1 + re it ) | 2 dt . 2 π − π

Consequences of the Parseval identity Suppose ∞ � P n z n . P ( z ) = n = 0 � π I ( P , λ ; r ) = 1 | P ( 1 + re it ) e − λ re it | 2 dt . 2 π − π Theorem �� ∞ ∞ � 1 / 2 r /λ ) e − r dr � � | P n | � I ( P , λ ; (4) 0 n = 0 and �� ∞ � 1 / 2 � 1 r /λ ) re − r dr � | P n | � √ I ( P , λ ; Z ( n ) , (5) λ 0 for all n � 0 and Z ( n ) � e − ( m − λ ) 2 2 ( m + λ )

Further inequalities Theorem If we additionally assume that P ( 1 ) = 0 , then �� ∞ ∞ √ � 1 / 2 r /λ ) r − 1 e − r dr � � | P 0 + P 1 + · · · + P n | � λ I ( P , λ ; , 0 n = 0 (6) and �� ∞ � 1 / 2 � r /λ ) e − r dr � | P 0 + P 1 + · · · + P n | � I ( P , λ ; Z ( n ) (7) 0 for all n � 0 .

Generalized binomial distribution Suppose S n = I 1 + I 2 + · · · + I n , (8) where the X j ’s are independent Bernoulli random variables with P ( I j = 1 ) = 1 − P ( I j = 0 ) = p j . Then P ( S n = m ) z m = � � ( 1 + p j ( z − 1 )) = e λ ( z − 1 ) g ( z ) . 0 ≤ m ≤ n 1 ≤ j ≤ n We will use notation λ = p 1 + p 2 + · · · + p n .

Generalized binomial distribution Suppose S n = I 1 + I 2 + · · · + I n , (8) where the X j ’s are independent Bernoulli random variables with P ( I j = 1 ) = 1 − P ( I j = 0 ) = p j . Then P ( S n = m ) z m = � � ( 1 + p j ( z − 1 )) = e λ ( z − 1 ) g ( z ) . 0 ≤ m ≤ n 1 ≤ j ≤ n We will use notation λ = p 1 + p 2 + · · · + p n .

Example of application to Poisson approximation θ := p 2 1 + p 2 2 + · · · + p 2 n , and λ := p 1 + p 2 + · · · + p n p 1 + p 2 + · · · + p n Theorem Suppose θ < 1 then the following inequalities hold 2 ∞ � � e − λ λ m θ 2 P ( S n = m ) m ! � e � � � − 1 ( 1 − θ ) 3 , � � e − λ λ m 2 � � � m ! � m = 0 √ e ∞ � � P ( S n = m ) − e − λ λ m � 1 θ � � � � � � � 2 m ! 2 3 / 2 ( 1 − θ ) 3 / 2 m = 0 Since √ e / 2 3 / 2 = 0 . 582 . . . the bound of the above theorem could be sharper than that of Barbour-Hall inequality if θ � 0 . 3 and λ is large enough.

Example of application to Poisson approximation θ := p 2 1 + p 2 2 + · · · + p 2 n , and λ := p 1 + p 2 + · · · + p n p 1 + p 2 + · · · + p n Theorem Suppose θ < 1 then the following inequalities hold 2 ∞ � � e − λ λ m θ 2 P ( S n = m ) m ! � e � � � − 1 ( 1 − θ ) 3 , � � e − λ λ m 2 � � � m ! � m = 0 √ e ∞ � � P ( S n = m ) − e − λ λ m � 1 θ � � � � � � � 2 m ! 2 3 / 2 ( 1 − θ ) 3 / 2 m = 0 Since √ e / 2 3 / 2 = 0 . 582 . . . the bound of the above theorem could be sharper than that of Barbour-Hall inequality if θ � 0 . 3 and λ is large enough.

Example of application to Poisson approximation θ := p 2 1 + p 2 2 + · · · + p 2 n , and λ := p 1 + p 2 + · · · + p n p 1 + p 2 + · · · + p n Theorem Suppose θ < 1 then the following inequalities hold 2 ∞ � � e − λ λ m θ 2 P ( S n = m ) m ! � e � � � − 1 ( 1 − θ ) 3 , � � e − λ λ m 2 � � � m ! � m = 0 √ e ∞ � � P ( S n = m ) − e − λ λ m � 1 θ � � � � � � � 2 m ! 2 3 / 2 ( 1 − θ ) 3 / 2 m = 0 Since √ e / 2 3 / 2 = 0 . 582 . . . the bound of the above theorem could be sharper than that of Barbour-Hall inequality if θ � 0 . 3 and λ is large enough.

Kolmogorov distance θ := p 2 1 + p 2 2 + · · · + p 2 n , and λ := p 1 + p 2 + · · · + p n p 1 + p 2 + · · · + p n Theorem Whenever θ < 1 we have √ e � � e − λ λ m � � θ � � � � P ( S n � j ) − Z ( j ) , � � � 2 1 / 2 ( 1 − θ ) 3 / 2 m ! � � m � j � � where   e − λ λ m e − λ λ m  � e − ( m − λ ) 2   � � Z ( n ) = min m ! , 2 ( m + λ ) m !  j � n j > n

Compound poisson distribution λ 3 := p 3 1 + p 3 2 + · · · + p 3 n Theorem Suppose θ < 1 / 3 then ∞ � � � λ 3 2 e 1 e λ ( z − 1 ) − λ 2 � � 2 ( z − 1 ) 2 �� � � P ( S n = m ) − [ z m ] ( 1 − 3 θ ) 2 , � � λ 3 / 2 3 m = 0 � � � � λ 3 8 e Z ( m ) e λ ( z − 1 ) − λ 2 � � 2 ( z − 1 ) 2 �� � P ( S n = m ) − [ z m ] ( 1 − 3 θ ) 5 / 2 . � � λ 2 3

Generalized binomial distribution in combinatorics Can be used if the discrete random variable X n is Bernoulli decomposable X n = I 1 + I 2 + · · · + I n This happens if a probability generating function F n ( z ) of a discrete random variable X n is a polynomial whose root are real and negative Example ◮ Hypergeometric distribution. ◮ Number of cycles in a random permutation

Generalized binomial distribution in combinatorics Can be used if the discrete random variable X n is Bernoulli decomposable X n = I 1 + I 2 + · · · + I n This happens if a probability generating function F n ( z ) of a discrete random variable X n is a polynomial whose root are real and negative Example ◮ Hypergeometric distribution. ◮ Number of cycles in a random permutation

Generalized binomial distribution in combinatorics Can be used if the discrete random variable X n is Bernoulli decomposable X n = I 1 + I 2 + · · · + I n This happens if a probability generating function F n ( z ) of a discrete random variable X n is a polynomial whose root are real and negative Example ◮ Hypergeometric distribution. ◮ Number of cycles in a random permutation