On some distributions related to digital expansions Ligia-Loretta - PowerPoint PPT Presentation

Journ´ ees de Num´ eration, Graz 2007 On some distributions related to digital expansions Ligia-Loretta Cristea Institut f¨ ur Finanzmathematik, Johannes Kepler Universit¨ at Linz FWF Project S9609 joint work with Helmut Prodinger () April 24, 2007 1 / 31

Overview The binomial distribution and its moments The Gray code distribution and its moments () April 24, 2007 2 / 31

The binomial distribution and its moments () April 24, 2007 3 / 31

The binomial measure. Definitions and notations Definition (Okada, Sekiguchi, Shiota, 1995) Let 0 < r < 1 and I = I 0 , 0 = [0 , 1], � j � � 2 n − 1 � 2 n , j + 1 , for j = 0 , 1 , . . . , 2 n − 2 , I n , j = I n , 2 n − 1 = , 1 , 2 n 2 n for n = 1 , 2 , 3 , . . . . The binomial measure µ r is a probability measure on I uniquely determined by the conditions µ r ( I n +1 , 2 j ) = r µ r ( I n , j ) , µ r ( I n +1 , 2 j +1 ) = (1 − r ) µ r ( I n , j ) , for n = 0 , 1 , 2 , . . . and j = 0 , 1 , . . . , 2 n − 1 . I n , j = elementary intervals of level n , n ∈ { 1 , 2 , 3 , . . . } , j ∈ { 0 , . . . , 2 n − 1 } () April 24, 2007 4 / 31

The binomial measure. Definitions and notations Notations W = the set of all infinite words over the alphabet D = { 0 , 1 } W m = the set of all words of length m ( m ≥ 1) over the alphabet D . For every word ω ∈ W , ω = ω 1 ω 2 . . . ω n . . . we define its value � ω i · 2 − i . val ( ω ) = i ≥ 1 Thus we assign to every infinite word ω = ω 1 ω 2 . . . the binary fraction 0 .ω 1 ω 2 . . . . Analogously we define the value of any word of W m () April 24, 2007 5 / 31

Remark. In the case of choosing in a random way (with respect to µ r ) a word ω ∈ W , we have P ( ω k = 0) = P ( ω k = 1) = 1 2 , for k = 1 , 2 , . . . , These probabilities depend neither on the parameter r nor on k . () April 24, 2007 6 / 31

The moments of the binomial distribution We study the moments of the function val with respect to the distribution defined by µ r . M n the moment of order n � µ r ( ω ) · ( val ( ω )) n . M n = ω ∈W Let � M nm = µ r ( ω ) · ( val ( ω )) n . ω ∈W m We have M n =lim m →∞ M nm . It is easy to verify that val ( d ω ) = d · 2 − 1 + 2 − 1 · val ( ω ) . () April 24, 2007 7 / 31

The moments of the binomial distribution Notation: W k m = the set of words of W m containing exactly k times the character 0. We have m � � M m r k (1 − r ) m − k ( val ( ω )) n . n = k =0 ω ∈W mk By analysing the first character of the words occurring in the last sum we get m − 1 � � n = r · 1 M m r k (1 − r ) m − 1 − k ( val ( ω )) n 2 n k =0 ω ∈W k m − 1 m − 1 � � + (1 − r ) · 1 r k (1 − r ) m − 1 − k (1 + val ( ω )) n 2 n k =0 ω ∈W k m − 1 � n � n − 1 � = 1 + (1 − r ) · 1 2 n M m − 1 M m − 1 . n 2 n j j j =0 () April 24, 2007 8 / 31

Theorem The moments of the binomial distribution µ r satisfy the relations: M 0 = 1 , � n � n � M n = r 2 n M n + 1 − r M j , for all integers n ≥ 1 . 2 n j j =0 Remarks One can use this recursion in order to compute a list of the first moments M 1 , M 2 , M 3 , . . . . From above one can express M n with the help of the previous moments: � n � n − 1 � M n = 1 − r M j . 2 n − 1 j j =0 () April 24, 2007 9 / 31

The asymptotics of the moments M n We define the exponential generating function � z n M ( z ) = M n n ! . n ≥ 0 We obtain z M ( z ) = r · M ( z 2 ) + (1 − r ) · M ( z 2 ) · e 2 . (The above functional equation could also have been derived by using the self-similar properties of µ r .) M ( z ) = M ( z ) · e − z satisfies � The Poisson transformed function 2 ) · e − z M ( z ) = r · � � M ( z 2 + (1 − r ) · � M ( z 2 ) . () April 24, 2007 10 / 31

The asymptotics of the moments M n Herefrom, by iteration: � � � r · e − z 2 k + (1 − r ) � M ( z ) = . k ≥ 1 As we are looking for the asymptotics of the moments M n we are going to study the behaviour of � M ( z ) for z → ∞ . This is based on the fact that M n ∼ � M ( n ). Justification: by using depoissonisation . The basic idea: extract the coefficients M n from M ( z ) using Cauchy’s integral formula and the saddle point method. () April 24, 2007 11 / 31

The asymptotics of the moments M n This leads in our applications to an approximation � � 1 + O (1 M n = � M ( n ) n ) , with more terms being available in principle. We rewrite 2 ) · e − z M ( z ) = r · � � 2 + (1 − r ) · � M ( z M ( z 2 ) . as M ( z ) = (1 − r ) · � � M ( z 2 ) + R ( z ) , 2 ) · e − z where R ( z ) = r · � M ( z 2 is considered to be an auxiliary function which we treat as a known function. () April 24, 2007 12 / 31

The asymptotics of the moments M n We compute the Mellin transform � M ∗ ( s ) of the function � M ( z ) . We get R ∗ ( s ) M ∗ ( s ) = (1 − r ) · 2 s · � � M ∗ ( s ) + R ∗ ( s ) = 1 − (1 − r ) · 2 s . Now the function � M ( z ) can be obtained by applying the Mellin inversion formula, namely � c + i ∞ � c + i ∞ R ∗ ( s ) 1 1 � � M ∗ ( s ) · z − s ds = 1 − (1 − r ) · 2 s · z − s ds , M ( z ) = 2 π i 2 π i c − i ∞ c − i ∞ 1 where 0 < c < log 2 1 − r . () April 24, 2007 13 / 31

The asymptotics of the moments M n We shift the integral to the right and take the residues with negative sign into account in order to estimate � M ( z ). The function under the integral has simple poles at 1 − r + 2 k π i 1 s k = log 2 log 2 , k ∈ Z . For these the residues with negative sign are log 2 R ∗ � � 1 1 − r + 2 k π i 1 1 − r − 2 k π i 1 z − log 2 log 2 , log 2 log 2 � ∞ 2 ) · e − z 0 r � M ( z 2 · z s − 1 dz . with R ∗ ( s ) = () April 24, 2007 14 / 31

For k = 0 the residue with negative sign is, using the definition of R ( z ) , � ∞ 1 M ( z 1 2) · e − z log 2 · z log 2 (1 − r ) 2 · z log 2 1 − r − 1 dz . r � 0 This term plays an important role in the asymptotic behaviour of the n th moment M n of the binomial distribution. In order to get this one collects all mentioned residues into a periodic function. () April 24, 2007 15 / 31

Theorem The nth moment M n of the binomial distribution µ r admits the asymptotic estimate M n =Φ( − log 2 n ) · n log 2 (1 − r ) � �� � 1 1 + O , n for n → ∞ , where Φ( x ) is a periodic function having period 1 and known Fourier coefficients. The mean (zeroth Fourier coefficient) of Φ is given by the expression � ∞ 1 M ( z 1 2) · e − z r � 2 · z log 2 1 − r − 1 dz . log 2 0 () April 24, 2007 16 / 31

Remark. One can compute this integral numerically by taking for � M ( z 2 ) the first few terms of its Taylor expansion. These can be found from the recurrence for the numbers M n . The integral in the expresion of the zeroth Fourier coefficient can be written as � ∞ � ∞ e − z � z k M ( z 1 1 2) · e − z 2 · z log 2 1 − r − 1 dz = r 2 k k ! z log 2 1 − r − 1 dz r � M k 0 0 k ≥ 0 � � � M k 1 = r 2 k k ! · Γ k + log 2 . 1 − r k ≥ 0 This series is well suited for numerical computations. For example, let r = 0 . 6 , then M 100 = 0 . 002453 . . . and the value predicted in the Theorem (without the oscillation) is 0 . 002491 . . . . () April 24, 2007 17 / 31

Generalisation: the multinomial distribution () April 24, 2007 18 / 31

The Gray code distribution and its moments () April 24, 2007 19 / 31

The Gray code distribution. Definition Definition (Kobayashi) Let I = I 0 , 0 = [0 , 1] and � j � � 2 n − 1 � 2 n , j + 1 , for j = 0 , 1 , . . . , 2 n − 2 , I n , j = I n , 2 n − 1 = , 1 , 2 n 2 n for n = 1 , 2 , 3 , . . . . For each 0 < r < 1 there exists a unique probability measure ˜ µ r on I such that, for j = 0 , 1 , . . . , 2 n − 1 and n = 0 , 1 , 2 , . . . , � r ˜ µ r ( I n , j ) j : even, µ r ( I n +1 , 2 j ) = ˜ (1 − r )˜ µ r ( I n , j ) j : odd, � (1 − r )˜ µ r ( I n , j ) j : even, µ r ( I n +1 , 2 j +1 ) = ˜ r ˜ µ r ( I n , j ) j : odd. We call ˜ µ r the Gray code measure . () April 24, 2007 20 / 31

The moments of the Gray code distribution We have � µ r ( ω ) · ( val ( ω )) n M n = ˜ ω ∈W and m →∞ M m M n = lim n , where � M m µ r ( ω ) · ( val ( ω )) n . n = ˜ ω ∈W m () April 24, 2007 21 / 31

We are looking for a recurrence relation between the moments of different orders. Idea: study first the recursive behaviour of the moments of finite words M m n . Reading a word ω ∈ W m , ω = ω 1 ω 2 . . . ω m If ω 1 = 0 then val ( ω ) lies in the left elementary interval of level 1. If ω 1 = 1 then val ( ω ) lies in the right elementary interval of level 1. Reading ω 2 indicates for val ( ω ) the interval of level 2 inside the interval of level 1 indicated by ω 1 . ω 2 = 0 − → left interval, ω 2 = 1 − → right interval. ω k indicates the position of the interval of level k ( k ≤ m ) that contains val ( ω ). () April 24, 2007 22 / 31