7 Poisson random measures

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1 Advanced Probability M03) 48 7 Poisson random measures 71 Construction and basic properties For λ 0, ) we say that a random variable X in Z + is Poisson of parameter λ and write X Poiλ) if PX n) e λ λ n /n! We also write X Poi0) to mean X 0 and write X Poi ) to mean X Proposition 711 Addition property) Let N k, k N, be independent random variables, with N k Poiλ k ) for all k Then ) N k Poi λ k k Proposition 712 Splitting property) Let N, Y n, n N, be independent random variables, with N Poiλ), λ < and PY n j) p j for all j 1,, k and all n Set N N j 1I {Ynj} n1 Then N 1,, N k are independent random variables with N j Poiλp j ) for all j Let,, µ) be a σ-finite measure space A Poisson random measure with intensity µ is a map M : Ω Z + satisfying, for all sequences A k : k N ) of disjoint sets in, i) M k A k ) k MA k), ii) MA k ), k N, are independent random variables, iii) MA k ) Poi µa k ) ) for all k by Denote by the set of integer-valued measures on and define Set σ X A : A ) X : Z +, X A : Z +, A Xm, A) X A m) ma) Theorem 713 There exists a unique probability measure µ on, ) such that X is a Poisson random measure with intensity µ k

2 Advanced Probability M03) 49 Proof Uniqueness) For disjoint sets A 1,, A k and n 1,, n k Z +, set A { m } : ma 1 ) n 1,, ma k ) n k Then, for any measure µ making X a Poisson random measure with intensity µ, µ A ) k e µaj) µa j ) nj /n j! j1 Since the set of such sets A is a π-system generating, this implies that µ is uniquely determined on xistence) Consider first the case where λ µ) < There exists a probability space Ω, F, P ) on which are defined independent random variables N and Y n, n N, with N Poiλ) and Y n µ/λ for all n Set MA) def N 1I {Yn A}, A 71) n1 It is easy to check, by the Poisson splitting property, that M is a Poisson random measure with intensity µ More generally, if,, µ) is σ-finite, then there exist disjoint sets k, k N, such that k k and µ k ) < for all k We can construct, on some probability space, independent Poisson random measures M k, k N, with M k having intensity µ k Set MA) def k N M k A k ), A It is easy to check, by the Poisson addition property, that M is a Poisson random measure with intensity µ The law µ on is then a measure with the required properties 72 Integrals with respect to a Poisson random measure Theorem 724 Let M be a Poisson random measure on with intensity µ and let g be a measurable function on If µ) is finite or g is integrable, then X gy) Mdy) is a well-defined random variable with e iux) { exp e iugy) 1 ) µdy)} Moreover, if g is integrable, then so is X and X) gy) µdy), Var X) gy) 2 µdy)

3 Advanced Probability M03) 50 Proof Assume for now that λ µ) < Then M) is finite as so X is well defined If g 1I A for some A, then X MA), so X is a random variable This extends by linearity and by taking limits to all measurable functions g Since the value of e iux ) depends only on the law µ of M on, we can assume that M is given as in 71) Then so e iux ) e iux N n ) e iugy1)) n e e iux N n ) PN n) n0 n0 e iugy) µdy) λ ) ne λ λ n /n! exp{ iugy) µdy) ) n λ e iugy) 1 ) µdy)} If g is integrable, then formulae for X) and Var X) may be obtained by a similar argument It remains to deal with the case where g is integrable and µ) Assume for now that g 0, then X is obviously well defined We can find 0 g n g with µ g n > 0) < for all n The conclusions of the theorem are then valid for the corresponding integrals X n Note that X n X and X n ) µg) < for all n It follows that X is a random variable and, by dominated convergence, X n X in L 1 P) Further, using the estimate e iux 1 ux, we can obtain the desired formulae for X by passing to the limit Finally, for a general integrable function g, we have gy) Mdy) gy) µdy) so X is well defined Also X X + X, where X ± gy) Mdy) {±g>0} and X + and X are independent Hence the formulae for X follow from those for X ± We now fix a σ-finite measure space,, K) and denote by µ the product measure on 0, ) determined by µ 0, t] A ) tka), t 0, A Let M be a Poisson random measure with intensity µ and set M M µ Then M is a compensated Poisson measure with intensity µ Proposition 725 Let g be an integrable function on Set def X t gy) Mds, dy) 0,t]

4 Advanced Probability M03) 51 Then X t is a cadlag martingale with stationary independent increments Moreover, { e iuxt ) exp t e iugy) 1 iugy) ) Kdy)}, Xt 2 ) t gy) 2 Kdy) Theorem 726 Let g L 2 K) and let g n : n N ) be a sequence of integrable functions such that g n g in L 2 K) Set def g n y) Mds, dy) X n t 0,t] Then there exists a cadlag martingale X t such that X n s X s 2) 0 for all t 0 Moreover, X t has stationary independent increments and e iuxt) { exp t e iugy) 1 iugy) ) Kdy)} The notation 0,t] gy) Mds, dy) is used for X t even when g is not integrable with respect to K Of course X t does not depend on the choice of approximating sequence g n ) This is a simple example of a stochastic integral Proof Fix t > 0 By Doob s L 2 -inequality and Proposition 725, Xs n Xs m 2) 4 Xt n Xt m ) 2) 4t g n g m ) 2 Kdy) 0 as n, m Hence Xs n converges in L 2 for all s t For some subsequence we have X n k s Xs nj 0 as as j, k The uniform limit of cadlag functions is cadlag, so there is a cadlag process X s ) such that X n k s X s 0 as Since Xs n converges in L 2 for all s t, X s ) is a martingale and so by Doob s L 2 -inequality Xs n X s 2) 4 Xt n X t ) 2) 0

5 Advanced Probability M03) 52 Note that e iug 1 iug u 2 g 2 /2 Hence, for s < t we have ) ) e iuxt Xs) Fs M lim e iuxn t Xn s ) Fs M n { lim exp t s) e iug ny) 1 iug n y) ) Kdy)} n { exp t s) e iugy) 1 iugy) ) Kdy)} which shows that X t has stationary independent increments with the claimed characteristic function