Poisson processes and their properties

Poisson processes and their properties Poisson processes. collection {N(t) : t [, )} of rando variable indexed by tie t is called a continuous-tie stochastic process, Furtherore, we call N(t) a Poisson process if (a) Starting with N(), the process N(t) takes a non-negative integer, 1, 2,... for all t > ; (b) the increent N(t + s) N(t) is surely nonnegative for any s > ; (c) the increents N(t 1 ), N(t 2 ) N(t 1 ),, N(t n ) N(t n 1 ) are independent for any < t 1 < t 2 < < t n 1 < t n ; (d) the increent N(t + s) N(t) has the distribution which is dependent on the value s > but independent of t >. Counting processes. stochastic process satisfying (a) and (b) is called a counting process in which N(t) represents the total nuber of events counted up to tie t. Properties (c) and (d) are respectively called independent and stationary increents. Particularly by applying (a) and (d) together, we obtain for all k, 1, 2,.... P (N(t + s) N(t) k) P (N(s) k) (2.1) rrival tie and sojourn tie. Events counted by a Poisson process N(t) are called Poisson events. Let T n denote the tie when the n-th Poisson event occurs. Here we call T n a arrival tie (also referred as occurrence tie ), and define the sojourn tie W n (or interarrival tie ) by W n : T n T n 1 for n 1, 2,..., where T for convenience. Properties of sojourn tie. We can observe that the event {W n > s} for a sojourn tie is equivalently expressed by the event {N(T n 1 + s) N(T n 1 ) } in ters of counting process, and that it has the probability that N(s). This will justify the following properties of sojourn tie: (a) The sojourn ties W 1, W 2,... are independent; (b) the sojourn tie W n has the distribution which is independent of n. Survival function. Consider the probability K(s) : P (W 1 > s). The function K(s) is known as a survival function. Then we obtain K(t + s) P (W 1 > t + s) P (N(t), N(t + s) N(t) ) P (N(t) )P (N(t + s) N(t) ) P (N(t) )P (N(s) ) P (W 1 > t)p (W 1 > s) K(t)K(s) Page 1 Special lecture/june 216

Then the sojourn tie W 1 ust have an exponential distribution with paraeter λ; see Proble 1 below. Proble 1. Let X be a non-negative rando variable satisfying for s, t, P (X > t + s X > s) P (X > t). (2.2) The above property is generally referred as eoryless property. By copleting the following questions, we will show that X ust be an exponential rando variable. (a) Let K(t) P (X > t) be the survival function of X. Show that the eoryless property iplies that K(s + t) K(s)K(t) for s, t. (b) Let κ K(1). rgue that κ >. (c) Show that K ( ) 1 1 n κ n and K ( n ) κ n. Therefore, we can obtain K(t) κ t for any t. By letting λ ln κ, we can write K(t) e λt, and therefore, we can find the pdf f(t) λe λt, t, for X. Proble 2. Custoers arrive at a service facility according to a Poisson process of rate λ. Let N(t) be the nuber of custoers that have arrived up to tie t, and let T 1, T 2,... be the successive arrival ties of the custoers. (a) Find E(T 4 t N(t) 3). (b) Find E(T 4 N(t) 3). (c) Find E(T 5 N(t) 3). Poisson distribution. Since W 1, W 2,... are independent and exponentially distributed with the coon paraeter λ, the arrival tie T n n k1 W k has the gaa distribution with paraeter (n, λ). s we have already seen in the previous lecture note, the discrete rando variable N(t) with a fixed tie t has the Poisson distribution with paraeter λt, λt (λt)k P (N(t) k) e for k, 1, 2,.... Proble 3. Let N(t) be a Poisson process with rate λ, and let < s < t. (a) Find P (N(t) N(s) 1). (b) Find P (N(s), N(t) 1). (c) Find P (N(s) 1 N(t) 1). Page 2 Special lecture/june 216

Distribution of arrival ties. Consider the sall probability that a Poisson event occurs in the tie interval [t, t + dt). By noticing that e λdt 1, we can find such probability as P (N(dt) 1) λ dt and call λ an arrival rate. Let f 1 (s) be the conditional density of the arrival tie T 1 given N(t) 1. Then we can copute the infinitesial probability P (s < T 1 s + ds N(t) 1) f 1 (s) ds by P (s < T 1 s + ds N(t) 1) P (N(s), N(s + ds) N(s) 1, N(t) N(s + ds) N(t) 1) P (N(s) )P (N(ds) 1)P (N(t s ds) ) P (N(t) 1) e λt λds e λt λt 1 t ds (2.3) Thus, f 1 (s) is the unifor pdf on (, t). General cases. Suppose that T 1, T 2,..., T n are the arrival ties up to the n-th occurrence in a Poisson process. Then we can generalize the notion of infinitesial probability to a joint distribution of the arrival ties T 1, T 2,..., T n, and express an infinitesial probability by P (s 1 < T 1 s 1 + ds 1,..., s n < T n s n + ds n N(t) n) f(s 1, s 2,..., s n ) ds 1 ds 2 ds n, (2.4) where f(s 1, s 2,..., s n ) is the joint density function. Proble 4. Show that the joint density function of (2.4) is given by f(s 1, s 2,..., s n ) n! t n for < s 1 < s 2 < < s n < t. Joint distribution of arrival ties. Let U 1, U 2,..., U n be independent and identically distributed (iid) unifor rando variable on (, t), and let U (1) < U (2) < < U (n) be the order statistics of U i s. Then the joint infinitesial probability is given by P (s 1 < U (1) s 1 + ds 1, s 2 < U (2) s 2 + ds 2,..., s n < U (n) s n + ds n ) n! t n ds 1 ds 2 ds n. Thus, conditioned upon N(t) n, the joint distribution of the arrival tie T 1, T 2,..., T n are identical to that of the order statistics U (1), U (2),..., U (n) of iid unifor rando variables on (, t). Proble 5. Let T 1, T 2,..., T n be the arrival ties in a Poisson process N(t) as in Proble 4. (a) Find E(T i N(1) n) for i 1,..., n. (b) Find E(T 1 + T 2 + + T n N(1) n). Page 3 Special lecture/june 216

Proble 6. Let N(t) be a Poisson process with rate λ, representing the nuber of custoers entering a store. Each custoer spend a duration in the store randoly. Then let X(t) denote the nuber of custoers reaining in the store at tie t. ssuing that the durations of custoer are independent rando variables and identically distributed as the pdf g(v), deterine the distribution for X(t). Proble 7. lpha particles are eitted according to a Poisson process with rate λ. Each particle exists for a rando duration and is annihilated. Suppose that the lifeties of particle are independent rando variables, and are identically and exponentially distributed with paraeter β. Find the expected nuber of particles existing at tie t. Superposition of Poisson processes. The oent generating function (gf) M X (t) for a Poisson rando variable X with paraeter λ can be calculated as M X (t) e λ(et 1). Suppose that X and Y are independent Poisson rando variables with respective paraeter λ and µ. Then the gf for X + Y is given by M X+Y (t) M X (t) M Y (t) e (λ+µ)(et 1), which iplies that X + Y is also a Poisson rando variable with paraeter λ + µ. Now consider independent two Poisson processes N(t) and M(t) having the respective arrival rate λ and µ. Then the superposition L(t) N(t) + M(t) becoes a Poisson process with rate λ + µ. Rando sus. Let X 1, X 2,... be iid rando variables with ean µ E[X i ] and variance σ 2 Var(X i ) for all i. Suppose that a discrete rando variable N takes a nonnegative integer value, 1, 2,..., and is independent of the rando variables X 1, X 2,.... Then we define a rando su by N Z X i. i1 For the rando variable Z observe that E(Z N) µn and Var(Z N) E((Z µn) 2 N) σ 2 N. We can copute E[Z] E[E(Z N)] µe[n], Var(Z) Var(E(Z N)) + E[Var(Z N)] µ 2 Var(N) + σ 2 E[N] Copound Poisson processes. Now suppose that N(t) is a Poisson process with arrival rate λ, independent of X 1, X 2,.... Then we can introduce a copound Poisson process by N(t) Z(t) X i. The copound Poisson process is essentially the rando su. Since E[N(t)] Var[N(t)] λt, we obtain E[Z(t)] µλt and Var(Z(t)) (µ 2 + σ 2 )λt. Non-hoogeneous Poisson processes. For an interval on [, ) we can introduce the total nuber N() of Poisson events occurring on the tie interval, and calculate its expected Page 4 Special lecture/june 216 i1

nuber by E[N()] λ dt. Suppose that the stationary increent of property (d) does not hold [but (a) (c) does], and that its expectation is rather deterined by E[N()] () : τ(t) dt with nonnegative intensity function τ(t). Then the rando variable N() has the Poisson distribution with paraeter (), and the process N(t) is called a non-hoogeneous Poisson process. Spatial Poisson processes. Consider a ulti-diensional space S. We can introduce a Poisson point process on S by extending a concept of non-hoogeneous Poisson process. For a subset of S let N() denote the total nuber of Poisson events each of which occurs at a point on the subset. If (i) the rando variables N( 1 ), N( 2 ),..., N( n ) are independent for every sequence 1, 2,..., n of utually disjoint subsets and (ii) the expected nuber E[N()] is given by E[N()] () : τ(x) dx with nonnegative intensity function τ(x) on S. Then the rando variable N() has the Poisson distribution with paraeter (). In particular when τ(x) λ, we call the process hoogeneous. Proble 8. The nuber of bacteria distributed throughout a volue of liquid can be considered as a Poisson spatial process. Suppose that the intensity function τ(x).6 (organiss per 3 ) Find the probability that ore than two bacteria are detected in a 1 3 volue. Marked point process. Consider a Poisson process N(t) with arrival rate λ, and iid rando variables X 1, X 2,... having a coon pdf g(x) on a space S. Let T 1, T 2,... denote arrival tie of the Poisson process. Then we can define a arked Poisson point process by introducing Poisson events at points (T 1, X 1 ), (T 2, X 2 ),... on the space [, ) S. For any subset of [, ) S, the nuber N() of Poisson events on becoes a non-hoogeneous Poisson point process with E[N()] () : λ g(x) dx dt Proble 9. Custoers arrive according to a Poisson process with rate λ 8. Suppose that each custoer is independently classified as high priority with probability.2, or low priority with probability.8. Then find the probability that three high priority and five low priority custoers arrive by t 1. Page 5 Special lecture/june 216

Proble solutions Proble 1. (a) Observe that P (X > t + s X > s) P (X > t + s) P (X > s) K(t + s) K(s) By applying it to (2.2) we obtain K(s + t) K(s)K(t). (b) The existence of conditional probability in (2.2) requires that P (X > s) > ; thus, κ P (X > 1) >. (c) Since K(1) K ( 1 n ( n) and K ) ( n K 1, ( n) we have K 1 ) 1 n κ n and K ( ) n κ n. Proble 2. Recall that the waiting tie V t T (N(t)+1) t has an exponential distribution with paraeter λ, and that V t is independent of N(t). (a) E(T 4 t N(t) 3) E(T (N(t)+1) t N(t) 3) E(V t N(t) 3) 1 λ (b) E(T 4 N(t) 3) E(T (N(t)+1) N(t) 3) E(t + V t N(t) 3) t + 1 λ (c) E(T 5 N(t) 3) E(T 4 + W 5 N(t) 3) E(T 4 N(t) 3) + E(W 5 N(t) 3) t + 2 λ, where W 5 is the sojourn tie and it is independent of the event that N(t) 3. Proble 3. (a) P (N(t) N(s) 1) P (N(t s) 1) 1 P (N(t s) ) 1 e λ(t s) (b) P (N(s), N(t) 1) P (N(s), N(t) N(s) 1) P (N(s) ) P (N(t) N(s) 1) e λs (1 e λ(t s) ) (c) P (N(s) 1 N(t) 1) e λs (λs) e λ(t s) e λt (λt) s t P (N(s), N(t) N(s) ) P (N(t) 1) Proble 4. We can deonstrate the extension of (2.3) in the case of n 2. P (N(s) ) P (N(t) N(s) ) P (N(t) 1) P (s 1 < T 1 s 1 + ds 1, s 2 < T 2 s 2 + ds 2 N(t) 2) P (N(s 1 ), N(s 1 + ds 1 ) N(s 1 ) 1, N(s 2 ) N(s 1 + ds 1 ), N(s 2 + ds 2 ) N(s 2 ) 1, N(t) N(s 2 + ds 2 ) N(t) 2) P (N(s 1) )P (N(ds 1 ) 1)P (N(s 2 s 1 ds 1 ) )P (N(ds 2 ) 1)P (N(t s 2 ds 2 ) ) P (N(t) 2) e λt (λds 1 )(λds 2 ) e λt (λt) 2 /2! 2! t 2 ds 1 ds 2 Further extension for n 3 is obvious. Proble 5. Let U 1, U 2,..., U n be iid unifor rando variable on (, 1), and let U (1) < U (2) < < U (n) be the order statistics of U i s. (a) E(T i N(1) n) E[U (i) ] i. n+1 Page 6 Special lecture/june 216

(b) E(T 1 + T 2 + + T n N(1) n) n i1 i n+1 n 2 Proble 6. Provided that N(t) n, we call the first n custoers 1, 2,..., n. Let U i be the arrival tie of custoer i, and let V i be the duration of the sae custoer reaining in the store [which is distributed as the pdf g(v)]. The pair (U i, V i ) of rando variables has the joint density function We can calculate p P (U i + V i t) f(u, v) 1 g(v), u t, v t du f(u, v) dv 1 t u t du g(v) dv 1 t u t dz z g(v) dv Then the nuber X(t) of custoers reaining in the store at tie t counts all the events that U i + V i t, i 1,..., n. We obtain ( ) n P (X(t) k N(t) n) p k (1 p) n k k and therefore, P (X(t) k) P (X(t) k N(t) n)p (N(t) n) nk λt (λpt)k e (λ(1 p)t) n k nk (n k)! which iplies that X(t) has a Poission distribution with paraeter λpt λ dz z g(v) dv λpt (λpt)k e Proble 7. This is the iediate application of Proble 6 with g(v) βe βv. Thus, we obtain E[X(t)] λpt λ dz Proble 8. Since () (.6)(1) 6, we obtain P (N() > 2) 1 z 2 k βe βv dv λ β (1 e βt ) () ()k e.938 Proble 9. Consider the arked Poisson point process on [, ) {, 1} with probability ass function g().8 and g(1).2. Let low [, 1] {} and high [, 1] {1}. Then we can calculate the probability as P (N( low ) 5, N( high ) 3) P (N( low ) 5) P (N( high ) 3) 6.4 (6.4)5 1.6 (1.6)3 e e 5! 3!.25 Page 7 Special lecture/june 216