Independence of some multiple Poisson stochastic integrals with variable-sign kernels

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1 Independence of some multiple Poisson stochastic integrals with variable-sign kernels Nicolas Privault Division of Mathematical Sciences School of Physical and Mathematical Sciences Nanyang Technological University SPMS-MAS, 21 Nanyang Link Singapore October 26, 216 Abstract It is known that the multiple Poisson stochastic integrals I n (f n ), I m (g m ) of two symmetric constant-sign functions are independent if and only if f n and g m satisfy a disjoint support condition. In this paper we present some examples in which this property extends to some variable-sign functions. Key words: Multiple stochastic integrals, Poisson process, independence. Mathematics Subject Classification: 6G57, 6G3, 6G55, 6H7. 1 Introduction The independence of two Gaussian random variables can be characterized by the vanishing of their covariance, and this property has been extended to multiple Wiener integrals in [9], [1], by stating that the multiple integrals I n (f n ), I m (g m ) with respect to Brownian motion are independent if and only if the contraction f n (x 1,..., x n 1, z)g m (y 1,..., y m 1, z)dz =, vanishes, x 1,..., x n 1, y 1,..., y m 1 IR, where L 2 (IR + ) n denotes the subspace of L 2 (IR + ) n = L 2 (IR n +) made of symmetric functions. 1

2 A proof of necessity has been provided in [2] using standard probabilistic arguments, while the original proof of [9], [1] relied on the Malliavin calculus. In addition, I n (f n ) is independent of I m (g m ) if and only if E[(I n (f n )) 2 (I m (g m )) 2 ] = E[(I n (f n )) 2 ] E[(I m (g m )) 2 ], (1.1) cf. Corollary 5.2 of [7]. Independence criteria for multiple stochastic integrals with respect to other Gaussian processes have been obtained in [1]. Coming back to the case of single integrals I 1 (f) with respect to Brownian motion, when X N (, s) and Y N (, t s) are two independent centered Gaussian random variables it is well known that X Y I 1 (f), with f = 1 [,s] 1 [s,t], is independent of X + Y I 1 (g), with g = 1 [,s] + 1 [s,t], since f, g = 1 [,s] 1 [s,t], 1 [,s] + 1 [s,t] =. The situation is completely different when X P(s) and Y P(t s) are independent centered Poisson random variables with means s and t s, in which case X Y is no longer independent of X + Y. Although we may also write X Y I 1 (f) = I 1 (1 [,s] 1 [s,t] ) and X + Y I 1 (g) = I 1 (1 [,s] + 1 [s,t] ) where I 1 (f) is the compensated Poisson integral of f, independence can occur only when fg =, as follows from Proposition 3.1 below. When I n (f n ) and I m (g m ) are multiple Poisson stochastic integrals on IR +, the condition f n 1 g m (x 1,..., x n, y 1,..., y n 1 ) := f n (x 1,..., x n )g m (y 1,..., y n 1, x n ) =, (1.2) 2

3 x 1,..., x n, y 1,..., y n 1 IR +, is sufficient for the independence of I n (f n ) and I m (g m ) on the Poisson space, due to the independence of increments of the Poisson process, since by symmetry of these functions, the supports of f n and g m can be respectively contained in two Borel sets of the form A n IR n + and B m IR m + with A B =, cf. [5]. On the other hand, it has been claimed in [4], [5] that (1.2) is also a necessary condition for the independence of I n (f n ) and I m (g m ) on the Poisson space. A similar result was stated independently in [8]. However that necessity claim was proved only when the functions f n and g m have constant signs. More precisely, the necessity condition therein is based on the relation (1.1) which, unlike on the Wiener space, turns out to be weaker than the independence of I n (f n ) and I m (g m ) when f n and g m have variable signs, as was shown in [7] by a counterexample for n = 2 and m = 1, cf. also Remark 3.4 below. In this note we present some examples in which (1.2) holds as a necessary and sufficient condition for the independence of I n (f n ), I m (g m ) when f n and g m are allowed to have variable signs. Namely we treat the following cases: - tensor powers: f n and g m take the form f n = f n and g m = g m, - integrals of first and multiple orders: f n L 2 (IR + ) n and g 1 has constant sign, - Integrals of 1 st and 2 nd orders: f 2 = f g, without sign restrictions on f 2 and g 1, for which we show that f n 1 g m = is necessary and sufficient for independence. The question whether (1.2) is a necessary condition for independence is still open in full generality. In Section 2 we recall the necessary conditions for independence obtained in [4], [5]. The examples of multiple integrals with variable-sign kernels for which the disjoint support condition is necessary and sufficient are given in Section 3. Finally in Section 4 we consider the expectation of multiple stochastic integrals. 3

4 The results of this paper are stated for a standard Poisson process on the half line IR +, however they can be extended without difficulty to Poisson measures on metric spaces. 2 Necessary condition for independence We start by recalling some results of [4]. Given f n L 2 (IR) n a symmetric squareintegrable function of n variables on IR n +, the multiple stochastic integral of f n with respect to the standard Poisson process (N t ) t IR+ I n (f n ) = n! t n We will need the multiplication formula is defined as t 2 f n (t 1,..., t n )d(n t1 t 1 ) d(n tn t n ). (2.1) I n (f n )I m (g m ) = 2(n m) cf. e.g. Proposition of [6], provided ( n h n,m,s := k! k s 2k 2(s n m) s= )( m k I n+m s (h n,m,s ), (2.2) )( k s k ) f n s k k is in L 2 (IR + ) n+m s, s 2(n m) where f n l k g m, l k, is the symmetrization in n + m k l variables of the function f n l k g m (x l+1,..., x n, y k+1,..., y m ) := f n (x 1,..., x n )g m (x 1,..., x k, y k+1,..., y m )dx 1 dx l. for f n L 2 (IR + ) n and g m L 2 (IR + ) m such that f n l k g m L 2 (IR + ) n+m k l, l k n m. g m The following proposition shows that f n 1 g m = is a necessary condition for the independence of I n (f n ) and I m (g m ). 4

5 Proposition 2.1 Let f n L 2 (IR + ) n and g m L 2 (IR + ) m such that E[(I n (f n )) 2 (I m (g m )) 2 ] = E[(I n (f n )) 2 ] E[(I m (g m )) 2 ]. Then we have f n 1 g m =. Proof. ([4]) If I n (f n ) and I m (g m ) are independent, then I n (f n )I m (g m ) L 2 (Ω) and using the isometry formula E [I n (f n )I m (g m )] = n! f n, g m L 2 (IR + ) n1 {n=m}, (2.3) for f n L 2 (IR + ) n, g m L 2 (IR + ) m, we get (n + m)! f n g m 2 L 2 (IR + ) n+m n!m! f n g m 2 L 2 (IR + ) n+m = n!m! f n 2 L 2 (IR + ) n g m 2 L 2 (IR + ) m = E [ I n (f n ) 2] E [ I m (g m ) 2] = E [ (I n (f n )I m (g m )) 2] = 2(n m) r= (n + m r)! h n,m,r 2 L 2 (IR + ) n+m r (n + m)! h n,m, 2 L 2 (IR + ) n+m +(n + m 1)! h n,m,1 2 L 2 (IR + ) n+m 1 (n + m)! f n g m 2 L 2 (IR + ) n+m +nm(n + m 1)! f n 1 g m 2 L 2 (IR + ) n+m 1 from (2.2), which implies f n 1 g m =. It follows from Proposition 2.1 that if I n (f n ) and I m (g m ) are independent then we have f n 1 g m =. However, f n 1 g m = is in general only a necessary and not sufficient condition for f n 1 g m to vanish, since f n 1 g m, f n 1 g m f n 1 g m, f n 1 g m, which follows from the fact that f n 1 g m is the symmetrization of f n 1 g m in n+m 1 variables. A counterexample for which we have f n 1 g m = f m 1 1 g m = and f n 1 g m can be found in [7], Example 5.3, as f 2 (s, t) = (2.4) 5

6 1 {st<} (1 [ 1, 1/2] (s) 1 [ 1/2,1/2] (s) + 1 [1/2,1] (s))(1 [ 1, 1/2] (s) 1 [ 1/2,1/2] (s) + 1 [1/2,1] (s)) and g 1 (t) = 1 [ 1,] (t) + 1 (,1] (t). In this example the integrals I 2 (f 2 ) and I 1 (g 1 ) are not independent, however (1.1) holds and therefore (1.1) and (1.2) cannot characterize the independence of I n (f n ) and I m (g m ). In the case n = 2 and m = 1 we are able to show that (1.1) entails f g 1 = in addition to f 2 1 g 1 =. Proposition 2.2 Let f 2 L 2 (IR + ) 2 and g 1 L 2 (IR + ) such that E[(I 2 (f 2 )) 2 (I 1 (g 1 )) 2 ] = E[(I 2 (f 2 )) 2 ] E[(I 1 (g 1 )) 2 ]. Then we have f 2 1 g 1 =, and f g 1 =. Proof. By the same argument as in the proof of Proposition 2.1 we find h 2,1,1 = f 2 1 g 1 =, and h 2,1,2 := 2f g 1 =. When f 2 = f g and g 1 = h, Proposition 2.2 yields (f g) 1 h = 1 (f (gh) + (gh) f + g (fh) + (fh) g) =, (2.5) 4 and hence (f g) 1 1 h = f g, h + g f, h =, (2.6) g, h = f, h =, (2.7) provided f g. 6

7 3 Variable-sign kernels In this section we consider examples of kernels f n L 2 (IR + ) n, g m L 2 (IR + ) m for which f n 1 g m = = f n 1 g m =, in which case the condition f n 1 g m = f n (x 1,..., x n 1, z)g m (y 1,..., y m 1, z) =, x 1,..., x n 1, y 1,..., y m 1, z IR, becomes both necessary and sufficient for the independence of I n (f n ) and I m (g m ), and is equivalent to (1.1), which also becomes necessary and sufficient. We note that f n 1 g m, f n 1 g m 1 = (n + m 1)! σ Σ n+m 1 (3.1) = IR n+m 1 + (f n 1 g m )(x 1,..., x n+m 1 )(f n 1 g m )(x σ1,..., x σn+m 1 )dx 1 dx n+m 1 (n 1)!(m 1)! f n 1 g m, f n 1 g m (n + m 1)! 1 + (n + m 1)! σ Θ n,m f n 1 g m (x 1,..., x n+m 1 )f n 1 g m (x σ1,..., x σn+m 1 )dx 1 dx n+m 1, IR n+m 1 + where Θ n,m = {σ Σ n+m 1 : σ({1,..., n 1}) {1,..., n 1} or σ({n + 1,..., n + m 1}) {n + 1,..., n + m 1}}, and our method of proof will rely on this decomposition to show that f n 1 g m =. 7

8 Tensor powers In the next proposition we obtain a necessary and sufficient condition for the independence of integrals of (symmetric) tensor powers. Proposition 3.1 Let f, g L 2 (IR). Then I n (f n ) is independent of I m (g m ) if and only if f n 1 g m =, which is equivalent to fg =. Proof. We have f n 1 g n = (fg) f (n 1) g (m 1), hence f n 1 g n = is equivalent to fg = and to f n 1 g n = f (n 1) (fg) g (m 1) =. In particular, I 1 (f) is independent of I 1 (g) if and only if fg =, for all f, g L 2 (IR + ). Integrals of first and multiple orders Here we consider the case of I n (f n ) and I 1 (g 1 ) when only g 1 has constant sign. Proposition 3.2 Let f n L 2 (IR + ) n, and assume that g 1 L 2 (IR + ) only has constant sign. Then I n (f n ) is independent of I 1 (g 1 ) if and only if f n 1 g 1 =, i.e. Proof. We have and, as in (3.1), f n 1 g 1, f n 1 g 1 = 1 n 2 = 1 n f n (x 1,..., x n )g 1 (x 1 ) =, x 1,..., x n IR +. f n 1 g 1 (x 1,..., x n ) = 1 n f n(x 1,..., x n ) + n 1 n f 2 n(x 1,..., x n ) n i=1 n g 1 (x k ), k=1 n g 1 (x i )g 1 (x j )dx 1 dx n j=1 f 2 n(x 1,..., x n ) g 1 (x i ) 2 dx 1 dx n f 2 n(x 1,..., x n )g 1 (x 1 )g 1 (x 2 )dx 1 dx n = 1 n f n 1 g 1, f n 1 g 1 + n 1 n f n n n 2 f n, g 1 g 1, (3.2) hence f n 1 g 1 = implies f n 1 g 1 = since f n n n 2 f n. 8

9 Integrals of first and second orders In the next result the necessary and sufficient condition for independence of multiple stochastic integrals is obtained without any sign assumption on the integrands in the first and second order case. In this section we work in the tensor case f 2 = f g, which does not include Example 2.4, see (3.5) below for an example in the tensor case. Proposition 3.3 Let f, g, h L 2 (IR + ). The double Poisson stochastic integral I 2 (f g) = 2 is independent of the single integral I 1 (h) = which is equivalent to fh = gh =. Proof. t f(t) g(s)d(n s s)d(n t t) (f g) 1 h =, h(t)d(n t t) if and only if We assume for simplicity that f, f = g, g = 1. If I 2 (f g) is independent of I 1 (h), Proposition 2.1 shows that (f g) 1 h = and the conclusion follows from Lemma 3.5 in case f 2, h g 2, h. Next, if Lemma 3.6 below shows that f 2, h g 2, h <, h = λ1 {f } λ1 {g }, (3.3) where λ = f 2, h 2 1/2 = g 2, h 2 1/2. (3.4) In addition we have fg = by Lemma 3.6, and by Proposition 2.2, (2.6) and (3.3), we get hence, letting α = f(x)dx = g(x)dx =, 1 {g(x) } dx <, and α + = 9 1 {f(x) } dx <,

10 which are both finite since h L 2 (IR + ), we obtain E[(I 2 (f g)) 2 1 {I1 (h)=λn}] = E[I 1 (f) 2 I 1 (g) 2 1 {I1 (h)=λn}] ( k ) 2 = e α α + 1 f(x i ) dx 1 dx k (2k + n)! k= ( n) = e α α + k= ( n) = e α α + k= ( n) IR k + 1 (2k + n)! IR k + k(k + n) (2k + n)!. i=1 k f 2 (x i )dx 1 dx k i=1 IR k+n + IR k+n + ( n+k 2 g(y i )) dy 1 dy k+n i=1 n+k g 2 (y i )dy 1 dy k+n On the other hand, being the difference of two Poisson random variables, λ 1 I 1 (h) has the Skellam distribution. n Z, where P (I 1 (h) = λn) = e α α + I n (x) = k= ( n) α n+k + α k k!(n + k)! i=1 ( ) n/2 = e α α+ α + I n (2 α + α ), α k= (x/2) n+2k k!(n + k)!, x >, is the modified Bessel function of the first kind with parameter n. It follows that ( ) n/2 k(k+n) E[(I 2 (f g)) 2 α+ k= (2k+n)! I 1 (h) = λn] = I n (2 α + α ), which cannot be constant in n because the Bessel function I n (x) is not separable in its variables x and n. Therefore the independence of I 2 (f g) with I 1 (h) imposes λ = which concludes the proof by contradiction with (3.4). From the above proof we note again that, although independence of I 2 (f g) and I 1 (h) is equivalent to (f g) 1 h =, in general the statement (f g) 1 h = does not imply (f g) 1 h =, as shown by the following example: A, B IR +, with A B =, where we have α f = 1 A, g = 1 B, h = 1 A 1 B, (3.5) (f g) 1 h = (1 A 1 B ) 1 (1 A 1 B ) = 1 A 1 B + 1 B 1 A, 1

11 and (f g) 1 h =, while (f g) 1 h, i.e. (1.2) does not hold. In this case the independence of I 2 (f g) and I 1 (h) does not hold, as in Example (2.4) above. Remark 3.4 Unlike in the Wiener case, Condition (1.1) is in general not sufficient for the independence of I n (f n ) and I m (g m ), as shown by the example (2.4). Therefore the search for a necessary and sufficient condition for independence in the general case has to involve more complex criteria than (1.1). The next lemma has been used in the proof of Proposition 3.3. Lemma 3.5 Let f, g, h L 2 (IR + ) such that (f g) 1 h = and f 2, h g 2, h. Then we have (f g) 1 h =, and fh = gh =. Proof. For simplicity and without loss of generality, we assume that f, f = g, g = 1. By Proposition 2.2 we have (f g) 1 h =, and from (3.2) or (2.5) we find = (f g) 1 h, (f g) 1 h = 1 2 (f g) 1 h, (f g) 1 h (f g) 2 (f g), h h = 1 2 (f g) 1 h, (f g) 1 h fg, h f 2, h g 2, h. This shows that (f g) 1 h =, provided f 2, h g 2, h, and we conclude by Lemma 3.7 below. The next lemma has been used in the proof of Proposition 3.3. Lemma 3.6 Let f, g, h L 2 (IR + ) such that (f g) 1 h = and g 2, h f 2, h. (3.6) Then we have fg = and h = λ1 {f } λ1 {g }, 11

12 for some λ such that In particular it holds that λ 2 = f 2, h 2 f, f = g2, h 2 g, g. g 2, h f 2, h <. Proof. For simplicity of exposition we assume that f, f = g, g = 1. The condition (f g) 1 h = implies (f g) 1 h, f f = f, gh + f 2, h f, g =, (f g) 1 h, g g = f, gh + g 2, h f, g =, and since g 2, h f 2, h this gives f, g = f, gh =, hence (f g) 1 1 h, (f g) 1 1 h = 1 2 f 2, h g 2, h + (f g) 1 h, (f g) 1 h = 1 2 f 2, h g 2, h ( f, f g2, h 2 + g, g f 2, h 2 ) 1 2 f 2, h 2 1/2 g 2, h 2 1/ ( g2, h 2 + f 2, h 2 ) (3.7) = 1 4 ( f 2, h 2 1/2 g 2, h 2 1/2 ) 2, which shows that (f g) 1 1 h, (f g) 1 1 h = implies f 2, h 2 = g 2, h 2, and, by the equality (3.7), for some λ IR such that fh = λf, and gh = λg, λ = f 2, h 2 1/2 = g 2, h 2 1/2. Hence we have and h = λ1 {f } λ1 {g }, (3.8) λ 2 fg = h 2 fg, which imply that fg = a.e. on {h 2 = } = {f = h = } {fg }, hence fg =. Finally we note that λ by (3.6) and (3.8). 12

13 The next lemma has been used in the proof of Lemma 3.5. Lemma 3.7 For any f, g, h L 2 (IR + ), the condition (f g) 1 h = is equivalent to fh = gh =. Proof. Assuming again that f, f = g, g = 1, we note that since (f g) 1 h = 1 (f (gh) + g (fh)), (3.9) 2 we have (f g) 1 h, (f g) 1 h = 1 4 ( g2, h 2 + f 2, h f, g fg, h 2 ) 1 4 ( g2, h 2 + f 2, h 2 2 fg, h 2 ) (3.1) 1 4 ( g2, h 2 + f 2, h 2 2 g 2, h 2 1/2 f 2, h 2 1/2 ) = 1 4 ( f 2, h 2 1/2 g 2, h 2 1/2 ) 2. Assuming that (f g) 1 h =, if fg, h 2 then the equality (3.1) implies f = g and fh = gh = by (3.9). On the other hand, if fg, h 2 = we have = (f g) 1 h, (f g) 1 h = 1 4 ( g2, h 2 + f 2, h 2 ), hence fh = gh =. This shows that fh = gh = in all cases, and (f g) 1 h = is equivalent to fh = gh =. 4 Conditional expectations In this section we present some complements on conditional expectations. Next is an adaptation of Proposition 3 in [3] to the Poisson case. Proposition 4.1 Let f n L 2 (IR + ), n 1, and let A be a Borel set of IR + with finite measure. We have E[I n (f n ) I n (1 n A )] = E[I n(f n ) I 1 (1 A )] = f n, 1 n A 1 A, 1 A I n(1 n n A ), which is a polynomial in I 1 (1 A ). 13

14 Proof. For any k 1, by (2.2) we have I 1 (1 A ) k = k l= α l I l (1 l A ), (4.1) where α,..., α k (, ), hence which shows that By (2.2) we also have E[I n (f n )(I 1 (1 A )) k ] = α k 1 {k n} f n, 1 n A = f n, 1 n A 1 A, 1 A n E[(I 1(1 A )) k I n (1 n A )], E[I n (f n ) I 1 (1 A )] = f n, 1 n A 1 A, 1 A n I n(1 n A ). I n (1 n A ) = n l= β l I l (1 l A ), where β,..., β n IR, which implies σ(i n (1 n A )) σ(i 1(1 A )), and E[I n (f n ) I n (1 n A )] = E[E[I n(f n ) I 1 (1 A )] I n (1 n A )] = f n, 1 n A 1 A, 1 A n E[E[I n(1 n A ) I 1(1 A )] I n (1 n A )] = f n, 1 n A 1 A, 1 A n I n(1 n A ). When n = 1 the above result says that E[I 1 (f 1 ) I 1 (1 A )] = f 1, 1 A 1 A, 1 A I 1(1 A ), (4.2) which follows by a direct orthogonal projection argument. In particular when X P(s) and Y P(t s) are independent centered Poisson random variables with means s and t s, X I 1 (1 [,s] ) and Y I 1 (1 [s,t] ) taking f 1 = 1 [,s] and A = [, t], Relation (4.2) follows from the fact that X + s has a binomial distribution with parameter s/t given X + Y + t. 14

15 Finally we note that, contrary to the Wiener case and Proposition 2 of [3], the conditional expectation of an odd order integral given an even order integral is not zero in general, indeed, E[I 3 (f 3 )(I 2 (g 2 )) 2 ] does not vanish in general, from the multiplication formula (2.2). References [1] S. Chivoret and A. Amirdjanova. Product formula and independence criterion for multiple Huang-Cambanis integrals. Statist. Probab. Lett., 76(18): , 26. [2] O. Kallenberg. On an independence criterion for multiple Wiener integrals. Ann. Probab., 19(2): , [3] D. Nualart, A.S. Üstünel, and M. Zakai. On the moments of a multiple Wiener-Itô integral and the space induced by the polynomials of the integral. Stochastics, 25(4):233 24, [4] N. Privault. On the independence of multiple stochastic integrals with respect to a class of martingales. C. R. Acad. Sci. Paris Sér. I Math., 323:515 52, [5] N. Privault. Independence of a class of multiple stochastic integrals. In R. Dalang, M. Dozzi, and F. Russo, editors, Seminar on Stochastic Analysis, Random Fields and Applications (Ascona, 1996), volume 45 of Progress in Probability, pages Birkhäuser, Basel, [6] N. Privault. Stochastic Analysis in Discrete and Continuous Settings, volume 1982 of Lecture Notes in Mathematics. Springer-Verlag, Berlin, 29. [7] J. Rosiński and G. Samorodnitsky. Product formula, tails and independence of multiple stable integrals. In Advances in stochastic inequalities (Atlanta, GA, 1997), volume 234 of Contemp. Math., pages Amer. Math. Soc., Providence, RI, [8] C. Tudor. A product formula for multiple Poisson-Itô integrals. Rev. Roumaine Math. Pures Appl., 42(3-4): , [9] A.S. Üstünel and M. Zakai. On independence and conditioning on Wiener space. Ann. Probab., 17(4): , [1] A.S. Üstünel and M. Zakai. On the structure on independence on Wiener space. J. Funct. Anal., 9(1): ,