Stochastic Proceses Poisson Process

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1 Stochastic Proceses Poisson Process Giovanni Pistone Revised June 9, / 31

2 Plan 1. Poisson Process (Formal construction) 2. Wiener Process (Formal construction) 3. Infinitely divisible distributions (Lévy-Khinchin formula) 4. Lévy processes (Generalities) 5. Stochastic analysis of Lévy processes (Generalities)

3 References Billingsley Brémaud DD1 DDG Patrick Billingsley, Probability and measure, second ed., Wiley Series in Probability and Mathematical Statistics: Probability and Mathematical Statistics, John Wiley & Sons Inc., New York, 1986 Pierre Brémaud, An introduction to probabilistic modeling, Undergraduate texts in mathematics, Springer, New York, 1997 Didier Dacunha-Castelle and Marie Duflo, Probabilités et statistiques. tome 1: Problèmes à temps fixe, Collection Mathématiques Appliqées pour la Matrise, Masson, Paris, 1982 Didier Dacunha-Castelle, Marie Duflo, and Valentine Genon-Catalot, Execices de probabilités et statistiques. tome 2: Problèmes à temps mobile, Collection Mathématiques Appliqées pour la Matrise, Masson, 1984 Pintacuda Nicolò Pintacuda, Probabilità, Decibel-Zanichelli, Padova, 1994 R Ross Sage Williams R Core Team, R: A language and environment for statistical computing, R Foundation for Statistical Computing, Vienna, Austria, 2014 Sheldon M. Ross, Introduction to Probability Models, 10th ed., Academic Press, San Diego, CA, 2010 W. A. Stein et al., Sage Mathematics Software (Version 5.9), The Sage Development Team, 2013, David Williams, Probability with martingales, Cambridge Mathematical Textbooks, Cambridge University Press, Cambridge, 1991.

4 Random variables Random variable Let (S, S) and (R, B) be measurable spaces, and X : S R a function. The function X is a (R, B)-random variable if X 1 (B) S. If C { B is a generating set } of B, and X 1 (C) S, then X is a random variable. In fact, let E = B B X 1 (B) S. Then E is a sub-σ-algebra of B, and C E, then E = B, see [Williams 3.2.b]. The σ-algebra generated by X is X 1 (B). From C { B, we have X 1 (C) X 1 (B), } hence σ(x 1 (C)) X 1 (B) = X 1 (σ(c)). Let E = B B X 1 (B) σ(x 1 (C)). As E is a σ-algebra containing C and contained in in B, then E = B and X 1 (B) = σ(x 1 (C)). Checking Measurability Let (S, S) be a measurable space and X : S R. If C is a family of subsets of R and X 1 (C) S, then 1. X is a random variable in (S, σ(c)); 2. σ(x ) = X 1 (σ(c)) = σ(x 1 (C)). In particular, if S R and X : S x x R is the immersion, then X 1 (C) = C S = {C S C C}. 1 / 31

5 2. Stochastic Process 1 Trajectory: definition Let I be a real interval, the set of times. Let D be a set of real functions u : I R, the set of trajectories. Given t I, the mapping Π t : D R defined by Π t (u) = u(t) is the evaluation of the trajectory at t. Let D = σ(π t : t I) be the σ-algebra generated by the evaluations. The measurable space (D, D) is a space of trajectories. 2 Stochastic process: definition Given a probability space (Ω, F, P) and a space of trajectories (D, D), a stochastic process on (Ω, F, P) with trajectories in D is a random variable X : Ω D, X 1 : D F. The distribution of X is P X 1. 3 Stochastic process: distribution 1. If t 1,..., t n I, then (X t1,..., X tn ) is random variable in R n whose distribution is called finite-dimensional distribution at t 1,..., t n I. 2. Finite dimensional distributions characterize the distribution of the stochastic process. 2 / 31

6 E1 Check Use [Williams.3]: Def 3.1 applies to general range, not just to real valued functions. { The family of events of the form ω S X t1 B 1,..., X tn B n }, for B 1,..., B n B, is a π-system that generates X 1 (D). Note that it is enough to take the B s is a π-system for R. E2 Consider the polynomials l 0 (x) = 1, l 1 (x) = 3(2x 1), l 2 (x) = 5(6x 2 6x + 1); check the orthonormality on [0, 1]. Define L i (t) = t 0 l i(x)dx. Given Z 0, Z 1, Z 2 iid N(0, 1), define ( the stochastic ) process W (2) = L 0 Z 0 + L 1 Z 1 + L 2 Z 2. Compute Cov W s (2), W (2) t and the marginal distribution of (W (2) s, W (2) t ) 3 / 31

7 Note ( that W (2) is a stochastic process whose trajectories are polynomials of degree up to 3, and E W (2) ) t = E (L 0 (t)z 0 + L 1 (t)z 1 + L 2 (t)z 2 ) = l 0 (x)l 0 (x) dx = 1 2 dx = l 1 (x)l 1 (x) dx = ( 1 3(2x 1)) 2 dx = 3 (4x 2 4x + 1) dx = l 0 (x)l 1 (x) dx = 1( 3(2x 1)) dx = l 2 (x)l 2 (x) dx = ( 1 5(6x 2 6x + 1)) 2 dx = 5 (36x 4 72x x 2 12x + 1) dx = l 0 (x)l 2 (x) dx = 1( 5(6x 2 6x + 1)) dx = l 1 (x)l 2 (x) dx = (( 3(2x 1)))(( 5(6x 2 6x + 1))) dt = ( Cov W (2) s, W (2) ) t = E ((L 0 (s)z 0 + L 1 (s)z 1 + L 2 (s)z 2 )(L 0 (t)z 0 + L 1 (t)z 1 + L 2 (t)z 2 )) = (W (2) s, W (2) t ( ) N 0, To be continued in the chapter on Wiener process. n L i (s)l j (t) E ( ) n Z i Z j = L i (s)l i (t) i,j=0 [ ni=0 L 2 i (s) ni=0 ]) L i (s)l i (t) ni=0 L i (s)l i (t) ni=0 L 2 i (t) i=0 4 / 31

8 Exponential distribution The exponential distribution Exp(λ) has support [0, + [ and it is a model for random times. The parameter λ > 0 is the intensity. The positive random variable X has exponential distribution with intensity λ (λ > 0) if its density is f X (x) = λ exp( λx)(x > 0). The survival function is 1 if x < 0 and for x 0 its value is R X (x) = exp( λx). The distribution function is 0 if x < 0 and for x 0 its value is F X (x) = 1 R X (x) = 1 exp( λx). The quantile function, {x F X (x) u} = {x Q X (u) x}, 0 < u < 1, is Q X (u) = 1 λ ln(1 u), ]0, 1[. The expected value is E (X ) = + R 0 X (x)dx = + exp( λx)dx = 1 0 λ. The moment generating function is M X (t) = E (exp(tx )) = if t < λ. The λ λ t variance is Var (X ) = 1 λ. If X 2 1,..., X n are iid Exp(λ), then T = X X n is Γ(n, λ), with density f T (t) = λn t n 1 (n 1)! e λt (t > 0). [ If p is a proposition, the (p) is the truth value, i.e. the indicator function.] 5 / 31

9 The exponential distribution is memory-less Easy version Let X Exp(λ). For each a, t 0, P (X > t + a X > a) = P (X > t). More sophisticated version Let X Exp(λ) and let G be a σ-algebra independent of X. If A and B are positive G-measurable random variables, then for all t > 0, E (B(X > t + A)) = P (X > t) E (B(X > A)). Proofs: Independence implies (see [Williams 8.3-4]) ( + ) E (B(X > t + A)) = E B(x > t + A) λe λx ( dx = E Be λ(t+a)) = e λt E (Be λa) 0 The easy version is A = a, B = 1, G trivial. 6 / 31

10 E3 Same assumptions: E (B(X > t + A) G ) = Be λ(t+a). Use [Williams 9.10]. E4 Compute E ((X > t) σ({x > a} : a s)), s t. Let (S, S, P) the probability space where X is defined and let X t = (X t) be the stochastic processes that has a unit jump at the random time X. For each time t, the σ algebra σ(x t ) is {, S, {X t}, {X > t}}, with generator {X > t}. The σ-algebra F s = σ(x a : a s) has generators C = {{X > a} a s}, which is a π-system as {X > a 1 } {X > a 2 } = {X > max(a 1, a 2 )}, see [Williams 1.6]. A bounded random variable B is F s -measurable if, and only if, B = g(x )(X s) + c(x > s), with g Borel bounded and c constant. In fact the class H = {g(x )(X s) + c(x > s) g B b, c R} is a monotone class containing the indicators of C because (X > a) = (X > a)(x s) + (X > a), see [Williams 3.14]. Let us compute g and c such that E ((X > t) F s ) = g(x )(X s) + c(x > s), see [Williams 9.2]. For all s a, E ((X > a)(x > t)) = E ((X > a)(g(x )(X s) + c(x > s))), a s, hence exp ( λt) = E ((a < X s)g(x )) + c exp ( λs), a s. If a = s, exp ( λt) = c exp ( λs), then c = exp ( λ(t s)). It follows s E ((a < X s)g(x )) = g(x) λe λx dx = 0, s s a so that g = 0 and E ((X > t) F s ) = exp ( λ(t s)) (X > s). 7 / 31

11 E5 Compute E (X t X s F s ), s t. From the previous computation, E (X t X s F s ) = 1 E ((X > t) F s ) X s = 1 exp ( λ(t s)) (X > s) X s = 1 exp ( λ(t s)) (1 X s ) X s = (exp ( λ(t s)) 1)(X s 1) 0, then (X t, F t ) t 0 is a sub-martingale, see [Williams 10.3]. E6 Show that A = λ 0 (X u 1) du is an absolutely continuous compensator of X, that is it is a process with absolutely continuous trajectories and such that M = X A is a martingale [CD1]. Having checked that it is correct to exchange integration with conditional expectation, E (A t A s F s ) = ( t ) t t E λ (X u 1) du F s = λ E ((X > u) F s ) du = λ e λ(u s) (X > s) du = s s s ( t ) λ e λ(u s) du (X > s) = e λ(u s) t (X > s) = E (Xt Xs Fs ), s s so that E (X t A t F s ) = X s A s. 8 / 31

12 E7 Compute E (φ(x t ) F s ) to show that the process X is a Markov process. Compute the transitions. As φ(x t ) = φ(0)(x > t) + φ(1)(x t) = φ(1) + (φ(0) φ(1)(x > t), the conditional expectation is E (φ(x t ) F s ) = φ(1)+(φ(0) φ(1)) E ((X > t) F s ) = φ(1)+(φ(0) φ(1))e λ(t s) (X > s) = P s,t φ(x s ), with P s,t φ(x) = { ( φ(0)e λ(t s) + φ(1) 1 e λ(t s)) if x = 0 φ(1) if x = 1. The transitions are computed as P (Xs = x, Xt = y) E ((Xs = x)(xt = y)) E ((Xs = x) E ((Xt = y) Fs )) P (X t = y X s = x) = = = P (X s = x) P (X s = x) P (X s = x) and the previous formula for φ = 1 y. E8 If X 1, X 2,... is an infinite sequence of independent random variables with common distribution Exp(λ) and T 0 = 0, T 1 = X 1, T 2 = X 1 + X 2, T 3 = X 1 + X 2 + X 3... i.e., T 0 = 0 and T n = T n 1 + X n, n 1, then T n + for n a.s. 9 / 31

13 For all a > 0 a (λt) n 1 P (T n a) = λ 0 (n 1)! e λt dt 0, n, ( ) as (λt) n 1 /(n 1)! is decreasing if n λa. Cf. [Williams 12.5]. n 10 / 31

14 11. Poisson process: construction Definition Let X 1, X 2,... be iid Exp(λ) on the probability space (S, S, P) and T 0 = 0 and T n = T n 1 + X n, n 1. Let (D, D) be the space of trajectories on the set of times [0, + [ which are 0 at t = 0 and take the value n = 0, 1, 2... on the interval [t n, t n+1 [ where t 0 < t 1 < t 2... and t n as n. The Poisson process with jump times (T n ) n is defined at each ω {lim n T n = + } as the trajectory in (D, D) with jumps T 1 (ω), T 2 (ω),..., that is N t (ω) = n n (T n (ω) n < T n+1 (ω)) = # {T k (ω) T k t} = (T n t). n=0 11 / 31

15 12. Poisson process: properties from Def (0) 1. Each N t is Poisson distributed with mean λt: P (N t = n) = (λt)n n! e λt. 2. Each increment N t N s, s < t, is Poisson distributed with mean λ(t s): P (N t N s = n) = (λ(t s))n n! e λ(t s). 3. Disjoint increments are independent: for all natural n and reals t 1 < T 2 < < t n, the random variables N tj N tj 1, j = 1,..., n and independent, hence 4. the sequence N t1, N t2,..., N tn has the Markov property E ( φ(n tn ) Ntn 1, N tn 2,... ) = φ(n + N tn 1 ) (λ(t n t n 1 ) n e λ(tn tn 1) n! n=0 12 / 31

16 1. P (N t = n) = E ((T n t)(t T n < X n+1 )) = E ((T n t) exp ( λ(t T n))) = t0 e λ(t x) λn x n 1 e λx λt (λt)n dx = e. (n 1)! n! 2.a P (N s = n, N t = n) = E ((T n s)(t T n < X n+1 )) = E ((T n s) exp ( λ(t T n))) = s0 e λ(t x) λn x n 1 e λx λt (λs)n dx == e = Poisson(n, λ). (n 1)! n! P (N t N s = 0) = n=0 P (N s = n, N t = n) = λt (λs)n n=0 e = e λ(t s) = Poisson(0; λ(t s)). n! 2.b P (N s = n 1, N t = n) = E ( (T n 1 s < T n t)(t T n < X n+1 ) ) = E ( (T n 1 s < T n t) exp ( λ(t T ) n)) = e λt E ((T n 1 s)e λt ) n 1 (s T n 1 X n < t T n 1 )e λxn = ( e λt E (T n 1 s)e λt n 1 t T ) ( n 1 e λx λe λx dx = λ(t s)e λt E (T s T n 1 s)e λt ) n 1 = n 1 λ(t s)e λt s 0 e λx λn 1 x n 2 (n 2)! e λx dx = λ(t s) (λs)n 1 e λt. (n 1)! P (N t N s = 1) = n=1 P (N s = n 1, N t = n) = n=1 λ(t s) (λs)n 1 e λt = λ(t s)e λ(t s) = (n 1)! Poisson(1; λ(t s)). 2.c P (N s = n k, N t = n) = E ( (T n k s < T n k+1 )(T n t)(t T n < X n+1 ) ) = E ( (T n k s < T n k+1 )(T n t) exp ( λ(t T ) n)) = ) e λt E ((T n k s < T n k+1 )(T n t)e λtn = ( e λt E (T n k s < T n k+1 )e λt n k+1 (X n k X n 1 t T n k+1 )e λ(x n k+2 + +X n 1 )) = ( e λt E (T n k s)(t < T n k+1 )e λt n k+1 t T n k+1 0 e λy λk 1 y k 2 ) e λy dy = (k 2)! ( λt e λk 1 (k 1)! E (T n k s)(t < T n k+1 )e λt n k+1 (t T n k+1 ) k 1) 13 / 31

17 14. Poisson process: filtration Let N be a Poisson process with jump times T 1, T 2,.... We say that N is the counting process of the sequence T 1, T 2,... We have {N t n} = {T n t}. σ-algebras generated by a Poisson process 1. The σ-algebra σ(n t ) has the generating π-system {{T n t} n = 1, 2,...}. 2. The σ-algebra F t = σ(n s : s t) has generating π-system C t = {{T n1 s 1,..., T sm s m } m N, n 1 n m, s 1 < < s m t} 3. A random variable Y is F t measurable if, and only if, Y = G k (N t = k) k=0 where G k is a σ(t 1,..., T k )-measurable random variable, k Z / 31

18 15. Proofs for slide 14 items 1,2 1. The sets [a, + [, a R form a π-system for (R, B), hence {{N t a} a R} is a π-system of σ(n t ). As N t actually takes values in Z +, the system contains set of the form {N t n} = {T n t} for n = 0, 1, The π-system for F t contains sets which are finite intersections of elements of the form {T n s} for all n and s t, that is, for example, {T n1 t 1, T n2 t 2,..., T nm t m } for all m, n 1,..., n m N and t 1,..., t m R +. If we order the t i s in increasing order, consider the π-system property for each time, and the increasing order of the T n s we obtain the stated subclass. 15 / 31

19 16. Proof for slide 14 item 3 We use the monotone class theorem [Williams 3.14]. Consider the set of random variables H = G k (N t = k) k=0 G k L (G k ), k Z +, with G k = σ(t 1,..., T k ). It is a vector space, it contains the constants, and it is closed for increasing limits. In fact, on each element of the partition {N t = k}, k Z + is a generic class of bounded random variables. Moreover, it is an ring, as G k (N t = k) G k (Nt = k) = G k G k (Nt = k). k=0 k=0 k=0 Each indicator function of the form (T n s), n Z +, s t, belongs to H. In fact, (T n s)(n t = k) = (T n s, T k t, T k+1 > t) = { (T n s)(n t = k) if n k, 0 if n > k, hence (T n s) = (T n s)(n t = k) = k n k G k (N t = k), with G k = 0 for k < n and G k = (T n s) for k n. As H is closed for product, each element of the π-system C t is included, and the result follows for bounded random variables. The general case is obtained by point-wise limits of bounded random variables. 16 / 31

20 17. Poisson process: conditioning Conditioning to F t Let N be a Poisson process on (S, S) with jumps T 1, T 2,..., and let a time s 0 be given. 1. For each integrable random variable Y there exists a sequence Y k = g k (T 1,..., T k ), k = 0, 1,..., such that E (Y F s ) = Y k (N s = k). k=0 2. Each Y n = g n (T 1,..., T n ), n = 0, 1,... is characterized by E (Y (N s = n)g) = e λs E ( Y n (T n s)e λtn G ), G L (T 1,..., T n ). 17 / 31

21 18. Exercise E9 Prove the previous theorem See [Williams 9.2]. The conditional expectation E (Y F s ) is a random variable measurable for F s, then E (Y F s ) = k=0 Y k (N s = k). Note that G n(n s = n), G n = g ( Y 1,..., Y n), G n indicator, n = 0, 1,..., is a π-system for F s. Then we want E (YG n(n s = n)) = E (E (Y F s ) G n(n s = n)) = E Y k (N s = k) G n(n s = n) = E (Y ng n(n s = n)). k=0 As (N s = n) = (T n s < T n+1 ) = (T n s < T n + X n+1 ) = (T n s)(s T n < X n+1 ), from the independence of (T 1,..., T n) and X n+1, E (Y ng n(n s = n)) = E (Y ng n(t n s)(s T n < X n+1 )) = ( E Y ng n(t n s)e λ(s Tn)) = e λs ( E Y ng n(t n s)e λtn ). 18 / 31

22 19. Markov property, independent increments Key result 1. The Poisson process is a Markov process i.e., for each bounded φ: R R and times s < t we have with transition ˆφ(n) = k=n E (φ(n t ) F s ) = ˆφ(N s ), (λ(t s))k n φ(k) (k n)! = (λ(t s))m φ(n + m). m! m=0 2. It follows that the Poisson process has independent homogeneous increments i.e., for each bounded φ: R R and times s < t we have E (φ(n t N s ) F s ) = E (φ(n t s )). 19 / 31

23 Ex.10 Proof of 1. We apply the formula for conditioning to the r.v. Y = φ(n t ), that is E (φ(n t ) F s ) = k=0 Y k (N s = k), with ) the characterization E (φ(n t )(N s = n)g n) = e λs E (Y n(t n s)e λtn G n, G n = g n(t 1,..., T n). The RES is E (φ(n t )(N s = n)g n) = φ(k) E ((N t = k)(n s = n)g n) = φ(k) E ((N t = k)(n s = n)g n) k=0 because N s N t. We compute each term of the RES. For k = n we have k=n E ((N t = n)(n s = n)g n) = E ((T n s)(t < T n+1 )G n) = ( E ((T n s)(t T n < X n+1 )G n) = E (T n s)e λ(t Tn)Gn ) = e λt E ((T n s)e λtn ) G n. If k = n + 1, E ((N t = n + 1)(N s = n)g n) = E ((N s = n)(t n+1 t < T n+2 )G n) = E ((N s = n)(t n+1 t < T n+1 + X n+2 )G n) = E ((N s = n)(t n+1 t)(t T n+1 < X n+2 )G n) = e λt E ((N s = n)(t n+1 t)e λt ) n+1 G n = e λt E ((T n s < T n + X n+1 )(T n + X n+1 t)e λ(tn+x n+1 ) ) G n = e λt E ((T n s)e λtn (s T n < X n+1 )(T n t)(x n+1 t T n)e λx ) n+1 G n = e λt E ((T n s)e λtn (s T n < X n+1 t T n)e λx ) n+1 G n = λ(t s)e λt E ((T n s)e λtn ) G n 20 / 31

24 If k = n + h, h > 1, E ((N t = n + h)(n s = n)g n) = E ( (N s = n)(t n+h t < T n+h+1 )G n ) = E ( ) ( ) (N s = n)(t n+h t < T n+h + X n+h+1 )G n = E (Ns = n)(t n+h t)(t T n+h < X n+h+1 )G n = e λt E ((N s = n)(t n+h t)e λt ) n+k G n = e λt E ((N s = n)(t n+1 + (X n X n+h ) t)e λ(t n+1 +(X n+2 + +X n+h )) ) G n = e λt E ((N s = n)e λt n+1 (T n+1 t)(x n X n+h t T n+1 )e λ(x n+2 + +X n+h ) ) G n = ( ( t Tn+1 e λt E (N s = n)e λt n+1 (T n+1 t) e λx λh 1 x h 2 ) ) e λx dx G n = 0 (h 2)! λ h 1 λ h 1 (h 1)! e λt E λ h 1 (h 1)! e λt E (h 1)! e λt E ((N s = n)e λt n+1 (T n+1 t)(t T n+1 ) h 1 G n ) = ((T n s < T n+1 )e λt n+1 (T n+1 t)(t T n+1 ) h 1 G n ) = ((T n s)e λtn (s T n < X n+1 t T n)e λx n+1 (t T n X n+1 ) h 1 G n ) = λ h 1 ( ( ) ) t Tn (h 1)! e λt E (T n s)e λtn e λx (t T n x) h 1 λe λx dx G n = s Tn (λ(t s)) h h! e λt E ((T n s)e λtn ) G n 21 / 31

25 Collecting all cases E (φ(n t )(N s = n)g n) = e λt (λ(t s))k n E φ(k) (T n s)e λtn G n. (k n)! k=n It follows Y n = k=n φ(k) (λ(t s))k n (k n)! and (λ(t s))k n E (φ(n t ) F s ) = (N s = n) φ(k) = ˆφ(n)(N s = n) = ˆφ(N s ) n=0 (k n)! k=n n=0 Ex.11 Proof of / 31

26 E12Telegrapher process. Define Y t = t 0 ( 1)Nu du. Is it independent increments? A sub-martingale? A Markov process?

27 Poisson process as a Lévy process Equivalent definition (1) A counting process N(t), t 0 is a Poisson process with intensity λ > 0 if, and only if, 1. N(0) = 0 2. The process has independent increments. 3. The number of events occurring in the time interval ]s, t], with length t s, has Poisson distribution with mean λ(t s): λ(t s) (λ(t s))n P (N(t) N(s) = n) = e n! n = 0, 1, 2 R In particular, such a process has stationary increments and is continuous in probability. 23 / 31

28 Ex.13 Proof of Poisson (1) is equivalent to Poisson (0). The only if part is already proved. Let us compute the joint finite { distributions of the process } N. For t 1 < < t n the random variable (N t1,..., N tn ) has values in k Z n + k 1 k n and (discrete) density ) ) P (N t1 = k 1,..., N tn = k n = P (N t1 = k 1,..., N tn N tn 1 = k n k n 1 = n j=1 ) P (N tj N tj 1 = k j k j 1 k 0 = 0, t 0 = 0 = n j=1 e λ(t j t j 1 ) (λ(t j t j 1 )) k j k j 1 (k j k j 1 )! n = e λtn (λ(t j t j 1 )) k j k j 1. (k j=1 j k j 1 )! With t j = s s j, k j = h h j, we have s j > 0 and h j 0, for j = 1,..., n, hence ) P (N s1 = h 1,..., N s1 + +sn = h h n = n e λs j (λs j ) h j h j=1 j! n = e λ(s 1 + +sn) (λs j ) h j. h j=1 j! Let us compute the joint distribution of the arrival times. The vector (T 1,..., T n) takes values in {t 0 < t 1 < < t n}, which in turn is the image of R n > under the cumulative sum map 24 / 31

29 sum : x (x 1,..., x x n). As { ]s 1, + [ ]s n, + [ sj > 0 } is a π-system, its cumulative sum image is a π-system {]t 1, + [ ]t n, + [ 0 < t 1 < < t n}. Let us compute the probability of each element ( ) P (T 1 > t 1,..., T n > t n) = P N t1 < 1,..., N tn < n ) = P (N t1 = 0,..., N tn = k n = k 2 kn,k j <j k 2 kn,k j <j e λtn n (λ(t j t j 1 )) k j k j 1 j=2 (k j k j 1 )! This equation uniquely identifies the joint distribution of T 1,..., T n, hence the joint distribution of the inter-times X j = T j T j 1, j = 1,..., n. Assume first n = 1, ( ) P (T 1 > t 1 ) = P N t1 < 0 = e λt 1, P (T 1 > t 1 ) = λe λt 1. t 1 If n = 2, ( ) ( ) ( ) P (T 1 > t 1, T 2 > t 2 ) = P N t1 < 1, N t2 < 2 = P N t2 = 0 + P N t1 = 0, N t2 = 1 = e λt 2 + e λt 1 e λ(t 2 t 1 ) λ(t 2 t 1 ) = e λt 2 (1 + λ(t 2 t 1 )), and P (T 1 > t 1, T 2 > t 2 ) = λe λt 2, ( 1) 2 2 P (T 1 > t 1, T 2 > t 2 ) = λ 2 e λt 2. t 1 t 1 t 2 25 / 31

30 ) If n = 3 we have a summation of P (N t1 = k 1, N t2 = k 2, N t3 = k 3 over the cases k 1 k 2 k 3, k 1 < 1, k 2 < 2, k 3 < 3, that is k k k ) The time t 1 appears in P (N t1 = k 1, N t2 = k 2, N t3 = k 3 only if k 2 = 1, hence P (T 1 > t 1, T 2 > t 2, T 3 > t 3 ) = e λt 3 (λ(t 2 t 1 ) + λ(t 2 t 1 )λ(t 3 t 2 )) = t 1 t 1 e λt 3 (1 + λ(t 3 t 2 )) λ(t 2 t 1 ) = λe λt 3 (1 + λ(t 3 t 2 )) t 1 so that the recursion is apparent and gives ( 1) 3 3 t 1 t 2 t 3 P (T 1 > t 1, T 2 > t 2, T 3 > t 3 ) = λ 3 e λt 3. Variant definition (2) 26 / 31

31 A counting process N(t), t 0 is a Poisson process with intensity λ > 0 if 1. N(0) = 0 2. The process has independent and stationary increments. 3. P (N(t) = 1) = λt + o(t) as t 0 4. P (N(t) > 1) = o(t) as t 0 E (N(t)) = λt and for t 0: P (N(t) = 1) = e λt (λt) = λt + o(t) P (N(t) > 1) = 1 e λt e λt (λt) = 1 e λt (1 + (λt)) = o(t) Ex.14 The two definitions are equivalent. 27 / 31

32 ( Let us consider the moment generating function of N(t), i.e. g(t) = E e un(t)). Note g(0) = 1, and let us derive a differential equation for g. g(t + h) = E (e un(t+h)) ( = E e un(t) e u(n(t+h) N(t))) independence ( = E e un(t)) ( E e u(n(t+h) N(t))) stationarity = g(t)g(h) As h 0 g(h) = k e uk P (N(h) = k) = (1 λ + o(h)) + e u (λh + o(h)) + k>1 e uk o(h) = 1 (λh) + e u (λh) + o(h) By letting h 0 in the definition of derivative we obtain the equation g (t) = λ(e u 1)g(t) whose solution is g(t) = exp [ λ ( e u 1 )] This is the moment generating function of the Poisson(λt) distribution. Ex / 31

33 Check the following: prove the last statement; take a subdivision of the interval [0, t] and use the binomial law to count how many contain a jump of N(t); avoid the stationarity condition by introducing a new condition infinitesimal condition. 29 / 31

34 30. Poisson process: simulation The conditional distribution of a random variable X given the event B is the probability measure µ X B of the domain of X such that for all integrable φ it holds E (φ X 1 B ) = P (B) φ(x) µ X B (dx). Note that µ X B is supported by B and the notation E (φ X B ) = φ(x) µ X B (dx). The conditional distribution is identified on a monotone class. Let U 1,..., U n be iid U(0, t). The sorting map τ(t 1,..., t n) = (t (1),..., t (n) ) is almost surely defined and takes values in the open simplex (t) = {s = (s 1,..., s n) 0 < s 1 < < s n < t}. The (t)-valued random variable (U (1),..., U (n) ) = τ(u 1,..., U n) is the order statistics of (U 1,..., U n). The distribution ν of the order statistics is the image under the sorting map of the uniform distribution, (t) φ(s) ν(ds) = (t) φ(s) τ ν(ds) = t n [0,t] n φ τ(t) dt = t n Π P Π 1 ( (t)) φ τ(t) dt = n!t n (t) φ(t) dt, where P is the group of permutation matrices on R n. Past jump times are uniform and independent Let T 1, T 2,... be the arrival times of a Poisson process with intensity λ. The joint distribution of T 1,..., T n, conditional to {N(t) = n} has uniform density f (t 1,..., t n ) = n! t n (0 < t 1 < < t n < t) Such a density is the joint density of the order statistics of n random variables U 1,..., U n IID uniformly distributed on ]0, t[. 30 / 31

35 A simulation lambda <- 1; t <- 10 n <- rpois(1,lambda*t); uniform <- runif(n,0,t) x <- sort(uniform) plot(c(0,x,t),c(0:n,n),type="s",xlab="t",ylab="n") N t 31 / 31

36 Ex.16 Proof. Consider the monotone class {t φ 1 (t 1 ) φ n(t n) φ i bounded}. E (φ 1 (T 1 ) φ n(t n)(n t = n)) = E (φ 1 (T 1 ) φ n(t n)(n t = n)) = E (φ 1 (T 1 ) φ n(t n)(t n t)(t T n < X n+1 )) = e λt E (φ 1 (T 1 ) φ n(t n)(t n t)e λtn ) = e λt E (φ 1 (X 1 ) φ n(x X n)(x X n t)e λ(x 1 + +Xn)) = λ n e λt φ 1 (x 1 ) φ n(x x n)(x x n t) dx 1 dx n = = λ n e λt ]0, [ n (t) = (λt)n e λt n! φ 1 (s 1 ) φ n(s n) dt 1 dt n φ 1 (s 1 ) φ n(s n) n! t n 1 (t) dt 1 dt n. 31 / 31