EDRP lecture 7. Poisson process. Pawe J. Szab owski

Transcription

1 EDRP lecture 7. Poisson process. Pawe J. Szab owski 2007

2 Counting process Random process fn t ; t 0g is called a counting process, if N t is equal total number of events that have happened up to moment t. Examples of counting processes are: 1. number of people, that entered some magazine before or at the moment t. N t denotes here this number of people, 2. number of people that were born before or at the moment t, N t denotes here this number of people. 3. number of goals that was collected by a football player up to the moment t..

3 Counting process Let us notice that every counting process has to satisfy the following conditions (i) N t 0, (ii) values of N t are integers, (iii) for s < t, N t N s is equal to the number of events that happened in time interval (s, t >. I Counting processes are sometimes called streams (of calls)

4 Counting process De nition Counting process is called with independent increments, if in disjoint time intervals increments of the process are independent. Fact In example no 1 probably assumption of independence of increments is satis ed. On the other hand this assumption is not satis ed in example no. 2.

5 Counting process De nition Counting process has stationary increments if number of events, that have happened in the time interval (s, t > depends only on the length of the interval i.e. on the quantity t s De nition Counting process fn t ; t 0g is called single if a) 8t > s 0 : P (N t N s = 1) = λ (t s)+o (jt sj), b) 8t > s 0 : P (N t N s 2) = o (jt sj).

6 Poisson process De nition Counting process fn t ; t 0g is called, Poisson process with intensity λ, if:and i) N 0 = 0, ii) has independent and stationary increments iii) number of events in the interval of the length t has Poisson distribution with parameterλt, i.e. 8t, s 0 : P (N t+s N s = k) = e λt (λt) k. k!

7 Poisson process Theorem The only counting process, single with stationary, independent increments is Poisson process. Proof. Let us denote g(t) = E exp ( vn t ). We have g(t + h) = E exp ( vn t+h ) = E exp ( vn t ) E exp( v (N t+t N t )) = g (t) E exp ( vn h ). We have mad use of independence of increments and their stationarity.

8 Poisson process Proof. Further we have: E exp ( vn h ) = E (exp ( vn h ) jn h = 0) P (N h = 0) +E (exp ( vn h ) jn h = 1) P (N h = 1) +E (exp ( vn h ) jn h 2) P (N h 2) = 1 λh + o 1 (h) + e v (λh + o 2 (h)) + o 3 (h) = 1 λh + λhe v + o (h).

9 Poisson process Proof. Hence g (t + h) h g (t) Let h! 0. We ll get then: = g (t) λ e v 1 + o (h) h. g 0 (t) = g (t) λ e v 1. and consequently: g (t) = exp λt e v 1. This is a Laplace transform of Poisson distribution with parameterλt.

10 Time intervals between calls Let be given Poisson process fn t ; t 0g and let T 1 a moment of the occurrence of the rst event. Further let for n > 0, T n denotes time interval between n 1 th and n -th event. Sequence ft i g i1 is called a sequence of interarrival times. Let us notice that ft 1 > tg = fn t = 0g P (T 1 > t) = P (N t = 0) = exp ( λt). Similarly we have: P (T 2 > t) = E (P (T 2 > tjt 1 )). But P (T 2 > tjt 1 = s) = P (lack of events in time interval (s, s + t > jt 1 = s) = P (lack of events in time interval (s, s + t >) = exp( λt). In the last equalities we used independence and satisfactorily of increments. Thus we have:

11 Fact Sequence ft n g n1 of interarrival times consists of i.i.d. random variables having exponential distributions with parameter λ. Further let us denote by S n moment of arrival of the n th call or otherwise a waiting time for the n th call. It is easy to deduce that n S n = T i, i=1 hence S n has gamma distribution with parameters n and λ. Other words, that distribution of S n has density equal to: f Sn (t) = λ exp( λt) (λt)n 1 (n 1)! ; t 0.

12 The above mentioned density could have been obtained noticing that n th call happens before t if and only if, when the number of events before the moment t was at least n. i.e. hence N t n () S n t, F Sn (t + ) = P (S n t) = P (N t n) = e λt (λt) j. j=n j!

13 After di erentiating with respect to t, we get: f Sn (t) = λ j=n e λt (λt) j j! = λe λt (λt) n 1 (n 1)!. + j=n λe λt (λt) j 1 (j 1)!

14 Summing of Poisson processes Let be given Poisson process fn t ; t 0g with intensity λ. Let us suppose, that every event (call) is classi ed as of I or II type. Further let us assume, that call is classi ed as of type I or II with probabilities respectively p and 1 p independently of other events (calls). (For example let us assume, that clients arrive to the shop and every one of them turns out to be a man with probability 1.2 or a women with probability 1/2. Let N (1) t and N (2) t denote respectively numbers of events of type I or II during time interval [0, t]. Let us notice that N t = N (1) t + N (2) t. We have the following proposition: Lemma Both N (1) t and N (2) t are Poisson processes with intensities respectively λp and λ (1 p). Moreover these processes are independent.

15 Proof. Let us calculate joint probability: P N (1) t = n, N (2) t = P k=0 = m N (1) t = n, N (2) t = mjn t = k P (N t = k). Let us notice that rstly in order to have n events of type I and m events of type II we should have jointly n + m events, hence P N (1) t = n, N (2) t = m = P N (1) t = n, N (2) t = mjn t = n + m e λt (λt) n+m (n + m)!.

16 Proof. Now let us notice, that if there were n + m events then number of events of type I has binomial distribution (number of successes among n + m Bernoulli experiments). Thus so P P N (1) t = n, N (2) t = mjn t = n + m = N (1) t = n, N (2) t = m n + m n p n (1 p) m, n + m = p n (1 p) m e λt (λt) n+m n (n + m)! = e λpt (λpt) n e λ(1 p)t (λ (1 p) t) m. n! m!

17 Proof. Moreover we have = P N (1) t P m=0 = n N (1) t = n, N (2) t = m = e λpt (λpt) n n! e λ(1 p)t (λ (1 p) t) m m=0 m! = e λpt (λpt) n. n! Hence it can be seen, that N (1) t and N (2) t processes. are independent Poisson

18 Fact The fact that processes N 1 and N 2 are Poissonian is not very surprising. One could have expect it. Unexpected seems to be the fact that these processes are independent.

19 Conditional distribution of arrival times Let us start from the moment of arrival of the rst call under condition, that on the interval [0, t] there was one call. Another words we will nd P (T 1 < sjn t = 1). We get then: P (T 1 < sjn t = 1) = P (T 1 < s, N t = 1) P (N t = 1) = P (1 call on (0, s) and 0 calls on [s, t]) P (N t = 1) = P (1 call on (0, s)) P ( 0 calls on [s, t]) P (N t = 1) λ(t s) = λse λs e λte λt = s t. Hence this distribution turns out to be uniform on the interval < 0, t >. This result can be generalized. However in order to do this one needs to introduce a notion of order statistics.

20 Digression on order statistics De nition Let fx 1,..., X n g be a simple random sample. Random vector fy 1,..., Y n g satisfying conditions: Y 1 Y 2,..., Y n with probability 1 and de ned in the following way : Y i = and -th (with respect to the magnitude) value of fx 1,..., X n g will be called vector of order statistics of the vector fx 1,..., X n g Fact Coordinates of the vector fy 1,..., Y n g will be denoted traditionally in the following way fx 1:n,..., X n:n g. We need also the following two simple lemmas:

21 Digression on order statistics Lemma If the random sample fx 1,..., X n gis simple (i.e. random variables fx i g n i=1 are independent with identical distributions) and the density of X 1 is equal f (x), then the joint density of order statistics fx 1:n,..., X n:n g is equal: for y 1 y 2... y n. g (y 1,..., y n ) = n! n i=1 f (y i ), Proof. Let us notice, that in order to observe Xo 1:n = y 1,..., X n:n = y n any of n! permutations nx (1),..., X (n) of variables fx 1,..., X n g has to assume values fy 1,..., y n g.

22 Proof. And moreover P X (1) 2 (y 1, y 1 + dy 1 ),..., X (n) 2 (y n, y n + dy n ) ' n i=1 f (y i ) dy 1,... dy n.

23 Digression on order statistics Let random variables S 1,..., S n be i.i.d and have uniform distribution U (0, t) on the interval < 0, t >. Then joint distribution of the vector of order statistics (S 1:n,..., S n:n ) has the following density: g (y 1,..., y n ) = n! t n, (1) for y 1 y 2... y n.

24 Theorem Under condition that there were exactly n call on the interval < 0, t > moments of successive calls S 1,..., S n are distributed as the order statistics of n i.i.d. random variables drawn from uniform distribution on< 0, t >, i.e. f S1,...,S n (s 1,..., s n ) = n! t n ; for 0 < s 1 <..., < s n < t. Fact This result can be expressed also in the following way: Given that there were n calls, moments of calls considered as unordered random variables are independent and have the same uniform distributions on the interval < 0, t >.

25 Proof. In order to get joint density of the vector (S 1,..., S n ) under condition N (t) = n let us notice, that for 0 < s 1 <,..., < s n < t an event S 1 = s 1, S 2 = s 2,..., S n = s n, N (t) = n means, that we have for the rst n + 1 interarrival times T 1 = s 1, T 2 = s 2 s 2,..., T n = s n s n 1, T n+1 > t s n. Hence making use of independence of interarrival times we have: f (s 1,..., s n jn) = P (T 1 = s 1,..., T n = s n s n 1, T n > t s n ) P (N (t) = n) = λe λs 1 λe λ(s 2 s 1 )... λe λ(s n s n 1 ) e λ(t s n) e λt (λt) n /n! = n! t n.

26 The above mentioned theorem is used to generalize proposition on summing up Poisson processes. Let us assume now, that upon the arrival every event is classi ed as being of one of k types, and that the probability that an event is classi ed as type i event, i = 1,..., k, depends on the time the event occurs. Speci cally, suppose that if an event occurs at time y then it will be classi ed as type i event, independently of anything that has previously occurred, with probability P i (y), i = 1,..., k where k i=1 P i (y) = 1. We can prove the following useful Theorem:

27 Theorem Let Nt; i t 0, i = 1,..., k denotes number elements of type i that arrived by the time t i.e. that occurred in < 0, t >. Then Nt i are independent Poisson random variables with expectations respectively: Z t ENt i = λ P i (s) ds. 0

28 Proof. Let us compute joint probability P Nt 1 = n 1,..., Nt k = n k. To do so rst note that in order for there to have been n i type i events for i = 1,..., k there must have been a total of k i=1 n i events. Hence, conditioning on N (t) yields P Nt 1 = n 1,..., Nt k = n k = P N 1 t = n 1,..., N k t = n k jn t =! k n i P N t = i=1! k n i. i=1 Now consider an arbitrary call, that occurred in the interval (0, t >. If it had occurred at time s, then the probability that it would be a type i event would be P i (s).

29 Proof. Hence since by appropriate Theorem this event will have occurred at some time uniformly distributed on (0, t >, it follows that the probability that this event will be a type i event is p i = 1 t Z t independently of the other events. 0 P i (s) ds,

30 Proof. Hence P N 1 t = n 1,..., N k t = n k jn t =! k n i i=1 will just equal the multinomial probability of n i type i outcomes for i = 1,..., k when each of k i=1 n i independent trials results in outcome i with probability p 1,..., p k,. That is P N 1 t = n 1,..., N k t = n k jn t =! k n i = i=1 k i=1 n i! k i=1 n i! k i=1 p n i i

31 Proof. Consequently = = P Nt 1 = n 1,..., Nt k = n k k i=1 n i! k k i=1 k i=1 n i! i=1 p n i i e λtp i (λtp i ) n i n i! e λt (λt) k i=1 n i k i=1 n i!.