MOMENTS AND CUMULANTS OF THE SUBORDINATED LEVY PROCESSES

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1 MOMENTS AND CUMULANTS OF THE SUBORDINATED LEVY PROCESSES PENKA MAYSTER ISET de Rades UNIVERSITY OF TUNIS 1

2 The formula of Faa di Bruno represents the n-th derivative of the composition of two functions f(g(t)), involving the Bell polynomials. Various identities relating moments and cumulants of random variables provide applications of Bell polynomials. As the Laplace exponent of the subordinated Lévy processes is exactly the composition of two Laplace exponents, 2

3 we utilize the Faa di Bruno formula to describe the moments and cumulants of the subordinated Lévy processes. 3

4 Subordinated Lévy processes Lévy process is an additive process X = (X(t), t 0) on a probability space (Ω X, B X, P X ), which is a right continuous process having left limits and X(0) = 0. Additive process, it means: independent and stationary increments. If (F t, t 0) denotes the natural filtration generated by X, then the increment (X t+s X t ) is independent of F t and has the same law as X s for every s, t 0. The distribution of X is determined by its transition probability p X (t, dx) measure and thus by its characteristic function. Let E(exp(iλX(t))) = exp(tf X (λ)), λ R, t 0. 4

5 We have the Lévy-Khinchine formula f X (λ) = iaλ 1 2 σ2 λ 2 + R (eiλx 1 iλx1 { x <1} )Π X (dx). Obviously a and σ are constant, i = 1 and (1 0+ x2 )Π X (dx) <. The Lévy measure Π X (dx) represent the mean of the numbers of increments with altitude x. 5

6 They say, that the Lévy process is defined by its triplet: (a, σ 2, Π X (dx)). It is well known that a Lévy process is of unbounded variation if σ 2 > 0 or x Π X(dx) =. 6

7 Let T = (T (t), t 0) be a Subordinator i.e, the Lévy process with non decreasing sample paths. Equivalently, this means that the Gaussian coefficient σ 2 is equal to zero and the Lévy measure Π T does not charge the interval ], 0] and fulfills 0+ (1 x)π T (dx) <. 7

8 All the subordinators are of bounded variation, positive drifts and jump measure concentrated on the interval (0, ). triplet of the subordinator is The (b, 0, Π T ) 8

9 To describe the subordinators it is more convenient to use Laplace transform rather than characteristic function. We denote the transition probability of T by p T (t, dx), x 0, t 0, and its Laplace transform by E(exp( λt (t))) = exp( tψ T (λ)), λ > 0, t > 0. where ψ : [0, + [ [0, + [ denotes the so-called Bernstein function ψ T (λ) = bλ + where b > 0 is the constant drift. 0 +(1 e λx )Π T (dx), 9

10 Natural examples of subordinators are: Gamma process, one-side stable, the quadratic variation of any Lévy process and so on. The compound Poisson process includes many explicitly known Lévy processes as special cassis, especially integer-valued Lévy processes. 10

11 Let X and T be independent Lévy processes, supposed T is a subordinator. The subordinated process Y = (Y (t), t 0) is defined by Y (t) = X(T (t)) on the probability space (Ω Y, B Y, P Y ), where Ω Y is the cartesian product Ω X Ω T and B Y = B X B T, P Y = P X P T 11

12 and the transition probability is given by p Y (t, dy) = 0 p X(s, dy)p T (t, ds). The subordinated process Y has stationary independent increments and characteristic exponent given by f Y (λ) = ψ T ( f X (λ), λ R, t 0. The Lévy measure of the subordinated process is Π Y (dy) = d T Π X (dy) + 0 p X(s, dy)π T (ds) where d T is the constant drift of the subordinator T. 12

13 The drift of the subordinated process is: d Y = d X d T + 0 ( x <1 xp X(s, dx))π T (ds). The continuous part of the subordinated process has the coefficient σ 2 Y = d T σ 2. The triplet of the subordinated process is {d Y, d T σ 2, Π Y (dx)}. 13

14 Moments and cumulants In general, the n-th moments and cumulants can be calculated as the derivatives of the characteristic functions. The cumulants are the derivatives of the Bernstein function at zero, or the moments of the Lévy measure. The relations between them can been expressed via the Faa di Bruno formulas and the Bell polynomials B n and B n,k, Charalambides,Ch.A.: Enumerative Combinatorics,Chapman, Hall,(2002 Roman, S.M.: The formula of Faa di Bruno, AMS Monthly, 87 (1980)

15 Denote by a n, b n, c n the cumulants. Let x n (s), t n (t), y n (t) be the moments of the processes X(s), T (t), Y (t) respectively and x n = x n (1), t n = t n (1), y n = y n (1) be the moments of the representative random variables X(1), T (1), Y (1). 15

16 For the convenience, we shall denote the sequences by and so on. a = (a 1, a 2,...), b = (b 1, b 2,...) The sequences of the powers in particular: a = (a, a 2,...), 1 = (1, 1,...). 16

17 = (1, 2, 3...)., ( 1) = (0, 1, 2,...). J. PITMAN, Combinatorial Stochastic Processes, LECTURE NOTES IN MATHEMATICS v. 1875, 2002, 17

18 The (n, k) th partial Bell polynomial is defined as a sum of products as follows: or equivalently: B n,k (a ) = n! k! k (k 1,k 2,...k n ) i=1 a ki k i!, B n,k (a ) = (k 1,k 2,...k n ) n!a k ak n n k 1!(1!) k 1...k n!(n!) k, n where the sum is over all partitions of n into k parts, that is over all nonnegative integer solutions (k 1, k 2,..., k n ) of the equations: k 1 + 2k nk n = n, k 1 + k k n = k. 18

19 The (n) th partition polynomial of two variables b, a is defined as a sum of the partial Bell polynomials as follows: or equivalently: B n (b, a ) = n b k B n,k (a ), B n (b, a ) = (k 1,k 2,...k n ) n!b k a k ak n n k 1!(1!) k 1...k n!(n!) k, n where the sum is over all nonnegative integer solutions (k 1, k 2,..., k n ) of the equation: k 1 + 2k nk n = n. 19

20 For the sequence of: (a bx ) = (abx 1, a 2 bx 2,..., a n bx n,...) The definitions imply the following relations B n,k (a bx ) = a n b k B n,k (x ) 20

21 The Faa di Bruno formula is manifested by the following x n (t) = B n (t, a ), c n = B n (b, a ) If X is a Poisson process, then (a ) = (a1 ), and x n (s) will be expressed by the Stirling numbers of the second kind. 21

22 If X is a Gamma process, then (a ) = (a ( 1)!). obtain, as the auxiliary result We shall B n (1, ( 1)!) = n! 22

23 The relation between moments and cumulants is given by the following combination: x n = n 1 ( n 1 k=0 k )x k a n k, a n = n ( 1) k 1 (k 1)!B n,k (x ). 23

24 x 1 (s) = a 1 s x 2 (s) = a 2 s + (a 1 s) 2 x 3 (s) = a 3 s + 3a 1 a 2 s 2 + (a 1 s) 3 x 4 (s) = a 4 s + (4a 1 a 3 + 3a 2 2 )s2 + 6a 2 1 a 2s 3 + (a 1 s) 4 24

25 x 5 (s) = a 5 s+(5a 1 a 4 +10a 2 a 3 )s 2 +(10a 1 a 3 +15a 1 a 2 2 )s3 +10a 3 1 a 2s 4 +(a 1 s) 5 x 6 = a 6 s+(15a 2 a 4 +6a 1 a 5 +10a 2 3 )s3 +(60a 1 a 2 a 3 +15a 2 1 a 4+15a 3 2 )s3 + 20a 3 1 a 3s a 4 1 a 2s 5 + (a 1 s) 6 and in particular B 6,3 (c ) = 60c 1 c 2 c c 2 1 c c 2 2.

26 The exponential Bell polynomials are given by: B n (c, a ) and B n (1, a ) 25

27 The Stirling numbers of the second kind we denote by S n,k := B n,k (1 ) = 1 k! k j=0 ( 1) k j ( k j )jn The unsigned Stirling numbers of the first kind. are given by s n,k := B n,k (( 1)!) 26

28 For the sequence of factorials,! = (1, 2!, 3!,...) we obtain the Lah numbers: B n,k (!) = ( n 1 k 1 )n! k!. the idempotent numbers are equal to: B n,k ( ) = ( n k )kn k. 27

29 For the sequence we have (a 1 (!)) = (1a 0, a2!, a 2 3!,...a n 1 n!,...), B n,k (a 1 (!)) = a n k ( n 1 k 1 )n! k!. 28

30 For the sequence (( a) 1 ) = ((1a) 0, (2a) 1, (3a) 2,..., (na) n 1 ), B n,k (( a) 1 ) = ( n 1 k 1 )(an)n k. B n (1, a x ) = a n B n (x ) S.Bouroubi, M.Abbas, New identities for Bell s polynomials. New approaches, Rostock. Math. Kolloq. v.61, 49-55,(2006) 29

31 The central moments can be expressed by the cumulants as follows: E(X x 1 ) = 0, E(X x 1 ) 2 = V arx = a 2, E(X(s) x 1 (s)) 2 = V arx(s) = a 2 s E(X x 1 ) 3 = a 3, E(X(s) x 1 (s)) 3 = a 3 s 30

32 Obviously, the first moment coincides with the first cumulant, the second cumulant coincides with the variance and the third cumulant is equal to the third central moment. 31

33 E(X x 1 ) 4 = a 4 + 3a 2 2, E(X(s) x 1(s)) 4 = a 4 s + 3a 2 2 s2 E(X x 1 ) 5 = a a 2 a 3, E(X(s) x 1 (s)) 5 = a 5 s + 10a 2 a 3 s 2 E(X x 1 ) 6 = a 6 + (15a 2 a a 2 3 ) + 15a3 2, E(X(s) x 1 (s)) 6 = a 6 s + (15a 2 a a 2 3 )s2 + 15a 3 2 s3 32

34 Suppose T (t) is any subordinator and Let X(s) be any Lévy process with finite moments t n, x n and cumulants b n, a n. cumulants of the subordinated process: Then the c n = B n (b, a ) and moments are: y n (t) = n t k B n,k (c ) = n t k (t)b n,k (a ), 33

35 We have, in general: y n (t) = Consequently: y n (t) = 0 xn p Y (t, dx) = 0 xn 0 p X(s, dx)p T (t, ds). x n(s)p T (t, ds) = B n(s, a )p T (t, ds). 0 0 Matrix representation of composition,(jabtinsky) and Change of basis ( Kolchin, Moskow) 34

36 B n (t, c ) = n B k (t, b )B n,k (a ). Remember, that Finally, we have: B n (s, a ) = n s k B n,k (a ). y n (t) = B n (t (t), a ). 35

37 Suppose T (t) is any subordinator and Let X(s) be a Poisson process with parameter a with finite moments t n, x n and cumulants b n, a n. Then Y (t) is a compound Poisson process with the following moments and cumulants: y n (t) = n t k (t)a k S n,k, c n = n b k a k S n,k. 36

38 Suppose T (t) is any subordinator and Let X(s) be a Gamma process with parameter a with finite moments t n, x n and cumulants b n, a n. Then Y (t) is a Lévy process with the cumulants: c n = a n n b k B n,k (( 1)!). y n (t) = a n n t k (t)b n,k (( 1)!), 37

39 Suppose T (t) is any subordinator and Let X(s) be a Brownian motion with finite moments and cumulants, then the subordinated process Y (t) has the following moments and cumulants: c n = (2k 1)!!σ 2k b k, y n = (2k 1)!!σ 2k t k, n = 2k. 38

40 The transition probability density of Brownian motion is : P (X(s) = x) = exp( x2 2sσ 2) 2sπσ 2. The moments and cumulants of the Browian motion are a 2 = σ 2 and a n = 0 for n 2. The sequence a = (0, σ 2, 0,...), and x n (s) = s k (2k 1)!!(σ 2 ) k, n = 2k. 39

41 By the main definition of the partial Bell polynomials B n,k we see that non-null terms are only those having k 2 0, k 1 = 0, k 3 = 0,...k n = 0. Consequently: k = k 2, 2k = n and For example, B n,k (a ) = n!(σ2 ) k k!2 k. B 2,1 (a ) = σ 2, B 4,2 (a ) = 3σ 4, B 6,3 (a ) = 15σ 6 40

42 (Brow, T),(Brow,Po) If X is Brownian and T is any subordinator, or Poisson subordinator, then the cumulants are given by (X, T ) T : b n T is Po: b n = b X : a n c n = B n (b, a ) c n = bb n (1, a ) Bro : a 2 = a, c n = (2k 1)!!σ 2k b k, n = 2k c n = (2k 1)!!σ 2k b 41

43 (X,Po) suppose T (t) is a Poisson process then c = bx t k (t) = B k (t, b ) = k j=1 (tb) j S k,j. (X, T ) T : b n T is Po: b n = b X : a n c n = B n (b, a ) c n = bb n (1, a ) y n (t) = B n (t (t), a ) = n k j=1 (tb) j S k,j B n,k (a ). 42

44 (Po,Po) If X is Poisson process and T is a Poisson process (X, T ) T : b n T isp o : b n = b X : a n c n = B n (b, a ) c n = bb n (1, a ) P o : a n = a c n = n b k a k S n,k c n = b n a k S n,k 43

45 Poisson subordinated by Poisson Neyman process Let X(t) and T (t) be two Poisson processes with intensity a and b respectively. Then the subordinated process Y (t) is a compound Poisson process with intensity equal to the total mass of the Lévy measure ψ Y ( ) = b(1 e a ). The intensity of Y (t) is less than the intensity of T (t). The Lévy measure is a zero truncated Poisson probability measure: Π Y ({k}) = 0 p X(s, {k})bδ(s 1) = b ak k! e a, k = 1, 2,... 44

46 Obviously, a n = a and b n = b and the moments Then, x n = B n (1, a ) = n a k S n,k. c n = b n B n,k (a1 ) = b n a k S n,k. y n (t) = B n (t (t), a ) = n k j=1 (tb) j S k,j a k S n,k. 45

47 The transition probability of the Neyman process is : p Y (t, {k}) = ak e bt k! j=0 j k (bte a ) j, k = 0, 1, 2,... j! This probability distribution had been introduced by Neyman in

48 (Ga,Po) If X is Gamma process and T is a Poisson process (X, T ) T : b n T is P o : b n = b X : a n c n = B n (b, a ) c n = bb n (1, a ) Ga : a n = a n (n 1)! c n = a n n b k B n,k (( 1)!) c n = ba n n! 47

49 Let X(s) be a Gamma process with parameter a and T (t) be a Poisson process with parameter b, then the subordinated process Y (t) is a compound Poisson process with the following moments y n (t) = a n n k j=0 (bt) j S k,j B n,k (( 1)!), y n (t) = a n n! n (bt) k ( n 1 k 1 )n! k!. 48

50 Gamma process X(s) subordinated by Poisson is a compound Poisson, but not integer-valued, with transition distribution characterized by the Bessel function of the first kind The jumps times of the subordinated process are the same as the jumps times of the subordinator. 49

51 Namely, the Bernstein function the Lévy measure is equal to ψ Y (λ) = b( aλ ) = baλ 1 + aλ, Π Y (dx) = b e x/a a dx and has the total mass b, and it is easy to calculate the cumulants. c n = b n B n,k (a k (k 1)!) = ba n n!. 50

52 The transition probability: p Y (t, dx) = e (bt+x a )tb a ( xbt 1 a )k 1 k!γ(k) dx. Bessel function of the first kind with parameter α denoted by J α (x) is a solution of the equation: x 2 y + xy + (x 2 α 2 )y = 0 and has the following series expression J α (x) = k=0 ( 1) k k!γ(k + α + 1) (x 2 )2k+1. In our example α = 1,e.i. is a negative integer, the first term of the series is vanished and J 1 (x) = ( 1) 1 J 1 (x). 51

53 Obviously, it is more convenient to calculate the moments via the Bell polynomials and cumulants via the Lévy measure. y n (t) = n B n,k (a ) k j=0 For the completeness we give the calculus: (bt) j S k,j, y n (t) = 0 xn e (bt+x a )tb a ( xbt 1 a )k 1 k!γ(k) dx y n (t) = a n n (bt) k B n,k (!). 52

54 (X, T ) T : b n T is P o : b n = b X : a n c n = B n (b, a ) c n = bb n (1, a ) a n = a c n = n b k a k S n,k c n = b n a k S n,k a n = a n (n 1)! c n = a n n b k B n,k (( 1)!) c n = ba n n! Bro : a 2 = a, c n = (2k 1)!!σ 2k b k b(2k 1)!!σ 2k 53

55 suppose T (t) is a Gamma process -cumulants (X, T ) T : b n Ga : b n = b n (n 1)! X : a n B n (b, a ) n b k (k 1)!B n,k (a ) P o : a n = a n b k a k S n,k n (ab) k (k 1)!S n,k a n = a n (n 1)! a n n b k B n,k (( 1)!) a n n b k (k 1)!B n,k (( 1 Bro : a 2 = a, (2k 1)!!σ 2k b k (2k 1)!!σ 2k b k (k 1)! 54

56 (X,Ga) Let X(s) be any Levy process with finite moments x n and cumulants a n and suppose T (t) is a Gamma process with finite moments t n (t) = b n Γ(t+n) Γ(t), and cumulants b n = b n (n 1)!. Then Y (t) is a Levy process with the following moments y n (t) = n b kγ(t + k) B n,k (a ) Γ(t) and cumulants: c n = n b k (k 1)!B n,k (a ). 55

57 (Po,Ga) Let X(s) be a Poisson process with parameter a and suppose T (t) is a Gamma process with parameter b then the subordinated process Y (t) is a Negative-Binomial process with the following moments and cumulants: y n = n (ab) k k!s n,k, c n = n (ab) k (k 1)!S n,k. 56

58 The transition probability density of Gamma process T (t) with parameter b is given by: p T (t, dx) = 1 dx Γ(t) b (x b )t 1 e x/b. The Bernstein function is equal to: ψ T (λ) = log(1 + bλ) and the Lévy measure Π T (dx) = x 1 e x/b dx, x > 0. It easy to calculate the moments and cumulants: t k = b k k!, b k = b k (k 1)!. 57

59 Namely: t k (t) = b kγ(t + k), Γ(t) b k = Then x=0 xk Π T (dx) = x=0 xk 1 e x/b dx = b k Γ(k) = b k (k 1)!. y n = B n (t, a ) = In the same way: n t k B n,k (a ) = n b k k!a k S n,k. c n = B n (b, a ) = n b k B n,k (a ) = n b k (k 1)!a k S n,k.

60 (Ga,Ga) Gamma processx(s) with parameter a subordinated by Gamma process with parameter b has the following moments: and cumulants: y n (t) = a n n b kγ(t + k) B n,k (( 1)!), Γ(t) c n = a n n b k (k 1)!B n,k (( 1)!). 58

61 Cumulants of the subordinated processes Y (t) = X(T (t)): (X, T ) b n Po:b n = b Ga:b n = b n (n 1)! a n bb n (1, a ) n b k (k 1)!B n,k (a ) Po b n a k S n,k n (ab) k (k 1)!S n,k Ga ba n n! a n n b k (k 1)!B n,k (( 1)!) Bro (2k 1)!!σ 2k b (2k 1)!!σ 2k b k (k 1)! 59

62 The moments of the subordinated Levy processes are (X, T ) T P o Ga X y n (t) = B n (t, c ) n (bt) k B n,k (x ) Po n t k (t)a k S n,k n kj=1 (tb) j S k,j a k S n,k Ga a n n t k (t)b n,k (( 1)!) a n n (bt) k B n,k (!) Bro (2k 1)!!σ 2k t k (t) (2k 1)!!σ 2k k j=1 (tb) j S k,j 60

63 The moments of the subordinated Levy processes are 61

64 (X, T ) T P o Ga X n (bt) k B n,k (x ) n b k Γ(t+k) Γ(t) B n,k(a ) Po n kj=1 (tb) j S k,j a k S n,k n (ab) k k!s n,k Ga a n n (bt) k B n,k (!) a n n b k k!b n,k (( 1)!) Bro (2k 1)!!σ 2k k j=1 (tb) j S k,j (2k 1)!!σ 2k b k k!

65 The symmetric Variance-Gamma process (Brow,Ga) The Variance-Gamma process may also be expressed as the difference of two independent Gamma processes. c n = b k (k 1)!(2k 1)!!σ 2k, y n (t) = (2k 1)!!σ 2k b kγ(t + k). Γ(t) n = 2k 62

66 The symmetric Meixner process Biane, P., Pitman J. and Yor, M.: Probability laws related to the Jacobi theta and Riemann zeta functions, and Brownian excursions, Bulletin of the American Mathematical Society 38 (2001) The Meixner(α, β, δ) density is given by: where p(x) = (2 cos β 2 )2δ 2απΓ(2α) exp(βx α ) Γ(δ + ix α ) 2, x (, ), α > 0, δ > 0, β < π. 63

67 The characteristic function is cos β E(exp(iλY (t))) = ( 2 ) cosh( αλ iβ 2 ) )2δt, λ R, t 0. Let X(t) be a Brownian motion with Laplace exponent ψ X = λ2 σ 2 2 and let the subordinator T (t) = 2 π 2 n=1 G n (t), where G n (t) is a Gamma process with mean β n t 64

68 and Lévy measure Π n (dx) = 1 x exp( x β n ), where the constant β n = b (n 1 2 )2 then the subordinated process Y (t) = X(T (t)) is a Meixner(α, 0, δ) process with parameters α = 2σ b, δ =

69 THANKS FOR ATTENTION 66