1 Continuous-Time Chains pp-1. 2 Markovian Streams pp-1. 3 Regular Streams pp-1. 4 Poisson Streams pp-2. 5 Poisson Processes pp-5

Transcription

1 Stat667 Random Processes Poisson Processes This is to supplement the textbook but not to replace it. Contents 1 Continuous-Time Chains pp-1 2 Markovian Streams pp-1 3 Regular Streams pp-1 4 Poisson Streams pp-2 5 Poisson Processes pp-5 6 Random Streams pp-6 7 Supperposition of Poisson Streams pp-1 8 Decomposition of a Stream pp-11 9 Compound Poisson Processes pp-14 1 Spatial Poisson Processes pp Examples pp-18

2 1 Continuous-Time Chains Continuous-time chains can be studied partially by looking into the embedded discrete-time chains. This method, while greatly simplies the study and provides certain insight to the understanding of the chain, suffers the lost of information of the behavior of the chain in the time domain. In order to fully understand the chain, the time that transitions take place must be considered. 2 Markovian Streams DEFINITION 2.1 (STREAM (OF EVENTS)) A sequence of similar events occur in time is called a stream (of events). E E 1 E 2 E 3 T = T 1 T 2 T 3 t DEFINITION 2.2 (MARKOVIAN STREAM) A stream of events in which its development to the future is independent of its past is called a Markovian stream. That is, if a stream is memoryless, it is Markovian. REMARK 2.3 (NONOVERLAPPING INTERVALS) The developments within any two nonoverlapping time intervals of a Markovian stream are independent. 3 Regular Streams DEFINITION 3.1 (EXPLOSION) A stream is said to have an explosion at a particular time T < if, approaching T, events occur more and more intense and by time T, an infinite number of occurrences have already taken place. EXAMPLE 3.2 (A BOUNCING BALL) A ball bouncing on the floor, let the state of the system be the number of bounces it has made. A physically reasonable assumption can be made that the time (in seconds) between the nth bounce and the (n + 1)th bounce is 2 n. Then x n = n and the time the nth bounce taking place is T n = n 1 = n 1. Clearly T n < 2 and T n 2 = T as n. The process hence explodes at time T = 2. DEFINITION 3.3 (REGULAR STREAM) A stream having no explosion is called a regular stream. pp-1

3 THEOREM 3.4 (NECESSARY AND SUFFICIENT CONDITION OF REGULAR STREAM) A stream is regular if and only if, within an infinitesimal interval (t, t+h], there can be at most one occurrence of the event. That is, a stream is regular if and only if P (two or more occurrences within (t, t + h]) = o(h). Hence, denoting we have m i (t, h) = P (exactly i occurrences within (t, t + h]), m (t, h) = 1 λ(t)h + o(h) m 1 (t, h) = λ(t)h + o(h) m i (t, h) = o(h), i > 1. 4 Poisson Streams DEFINITION 4.1 (POISSON STREAM) A stream that is Markovian and regular is called a Poisson stream. The expected number of occurrences within (t, t + h] is Hence, [1 λ(t)h + o(h)] + 1 [λ(t)h + o(h)] + io(h) = λ(t)h + o(h). E{number of occurrences within (t, t + h]} λ(t) = lim. h h i=2 DEFINITION 4.2 (OCCURRENCE RATE) For a Poisson stream, λ(t), the expected number of occurrences per unit time at time t is called the occurrence rate (or instantaneous rate (of occurrence) or intensity) of the chain. DEFINITION 4.3 (AVERAGE OCCURRENCE RATE) For a Poisson stream, the average occurrence rate during (t, t + τ] is λ(t, τ) = E{number of occurrences within (t, t + τ]} τ DEFINITION 4.4 (TIME-HOMOGENEOUS POISSON STREAMS) A Poisson stream is called timehomogeneous if λ(t) = λ t. Consequently, λ = λ(t) = λ(t, τ), t > and τ >. Unless otherwise mentioned, assume hereafter time-homogeneity. DEFINITION 4.5 (REMAINING LIVES) The remaining life (or residual life) of a (time-homogeneous) Poisson stream, denoted r(t) is the time (from t) to the next occurrence. pp-2

4 THEOREM 4.6 (REMAINING LIFE IS EXPONENTIALLY DISTRIBUTED) The remaining life r(t) is distributed as E(λ). The above theorem holds because P (r(t) > s) = P (zero occurrence within (t, t + s]) = lim P ( k k j=1(zero occurrence within (t + (j 1)s k k = lim P (zero occurrence within (t + (j 1)s, t + js ]) k k k j=1 = lim (1 λs k k + o( 1 k ))k = e λs., t + js k ])) DEFINITION 4.7 (LIVES OF A STREAM) The duration l i between two consecutive occurrences E i 1 and E i in a stream is called the ith life. That is l i = T i T i 1 = ith life = interoccurrence time. A graphical view of lives of a Poisson stream is given below: E l 1 E 1 l 2 E 2 l 3 E 3 l 4 E 4 t T = T 1 T 2 T 3 T 4 THEOREM 4.8 (LIVES AND REMAINING LIVES OF POISSON STREAMS) The following are equivalent: a. a stream is a (time-homogeneous) Poisson stream; b. the remaining life is exponentially distributed; c. the lives are exponentially distributed. Note that the equivalence of (a) and (b) is already established above. Also (b) implies (c) by observing l n = r(t n 1 ). It remains to show that (c) implies (b). Observing the follorwing figure l n θ r(t) T n 1 = t θ t T n t+s T n+1 time we have then P (r(t) < s l n > θ) = P (r(t) < s and l n > θ) P (l n > θ) = 1 e λs. pp-3 = P (θ < l n < θ + s) P (l n > θ) = e λθ e λ(θ+s) e λθ

5 DEFINITION 4.9 (SHORTER REMAINING LIFE) Consider two completely independent Poisson streams: S X having rate λ X and remaining life r X (t), and S Y having rate λ Y and remaining life r Y (t). Define R 2 (t) = min(r X (t), r Y (t)) as the shorter remaining life of the two streams. If the event (R 2 (t) = r X (t)) (r X (t) < r Y (t)) occurs then stream S X is said to have a shorter remaining life than stream S Y does. THEOREM 4.1 (DISTRIBUTION OF SHORTER REMAINING LIFE) The distribution of R 2 (t) follows E(λ X + λ Y ) and λ X P (r X (t) < r Y (t)) =. λ X + λ Y To verify the above theorem first note that P (R 2 (t) > s) = P (r X (t) > s and r Y (t) > s) = P (r X (t) > s)p (r Y (t) > s) = e λ Xs e λ Y s = e (λ X+λ Y )s. So R 2 (t) is exponentially distributed with rate λ X + λ Y. Now, consider an infinitesimal duration (t, t + h]. Denote O ij = the event that exactly i coourrences in stream S X and exactly j occurrences in stream S Y. Employing a continuous version of the first-step analysis by conditioning on the occurrences of the two streams within (t, t + h]: P (r X (t) < r Y (t)) = i= P (r X (t) < r Y (t) O ij )P (O ij ) j= = 1 (λ X h(1 λ Y h) + o(h)) + ((1 λ X h)λ Y h + o(h)) +P (r X (t) < r Y (t)) ((1 λ X h) (1 λ Y h) + o(h)) + P (r X (t) < r Y (t) O ij )P (O ij ) o(h) i>1 or j>1 = λ X h + (1 λ X h λ Y h)p (r X (t) < r Y (t)) + o(h). Hence, P (r X (t) < r Y (t)) = λ X λ X + λ Y + o(h) (λ X + λ Y )h h λ X λ X + λ Y. REMARK 4.11 (SHORTEST REMAINING LIFE) Consider k independent Poisson streams S i with rate λ i and remaining life r i (t), i = 1,..., k. Let R k (t) be the shortest remaining life of these streams from time t. That is, R k (t) = min 1 i k r i(t). pp-4

6 Then R k (t) follows E(λ) where In addition, λ = λ λ k. P (r j (t) = R k (t)) = λ j λ. 5 Poisson Processes DEFINITION 5.1 (COUNTING PROCESSES) A counting process {m(t) : t } is a continuoustime chain which counts the number of occurrences of the event in a stream within (, t]. Define m(). The following is observed: REMARK 5.2 (PROPERTIES OF COUNTING PROCESSES) For a counting process the state space is {, 1, 2, }; it is nondecreasing, i.e., m(t) m(s) if t < s; it increases by one whenever an event occurs. DEFINITION 5.3 (POISSON PROCESSES) A counting process is called a (time-homogeneous) Poisson process if m(t) is Poisson distributed with parameter λt t >. The parameter λ is called the rate of the Poisson process. THEOREM 5.4 (PROPERTIES OF POISSON PROCESSES) A Poisson process has INDEPENDENT INCREMENTS: counting process increases independently in nonoverlapping intervals (Markovian); STATIONARY INCREMENTS: time-homogeneous; POISSON SEGMENTS: m(t + τ) m(t) P(λτ), τ > and t > ; i.e., the number of occurrences of events within any time segment is Poisson distributed with mean occurrence rate λ (length of the segment). REMARK 5.5 (POISSON PROCESS STREAM IS REGULAR) The stream associated with a Poisson process is regular and has exponentially distributed life since the tail probability of the remaining life is P (r(t) > τ) = P (no occurrence during (t, t + τ]) = e λτ. pp-5

7 REMARK 5.6 (VERIFICATION OF POISSON STREAMS) First pick a convenient time interval, then test the (null) hypothesis that the number of occurrences in this interval is Poisson. Caution should be exercised to interpret the outcome of the hypothesis test: it only shows that the data do not violate the Poisson stream assumption if the null hypothesis is not rejected. DEFINITION 5.7 (ERLANG DISTRIBUTIONS) From the definition of lives of a Poisson stream, T k = time to kth occurrence = l 1 + l l k has an Erlang distribution with parameters k and λ. REMARK 5.8 (ERLANG DISTRIBUTION IS A GAMMA DISTRIBUTION) Erlang distribution is a special case of the more general family of distribution, namely, the gamma distribution with parameters α = k and λ. Hence the p.d.f., the mean and the variance of an Erlang distribution with parameters k and λ are f(t) = λk t k 1 e λt Γ(k) = λk t k 1 e λt, t > (k 1)! E(T k ) = k λ Var(T k ) = k λ 2. 6 Random Streams Poisson streams are also referred to as random streams. THEOREM 6.1 (CONDITIONAL UNIFORM OCCURRENCES) For a Poisson process m(t), given that m(t) = n (i.e., given that there are n occurrences within (, t]), the n (ordered) occurrence times T 1, T 2,..., T n have the same joint distribution as the order statistics corresponding to n independent uniform (U(, t)) random variables over the interval (, t). That is, f T1,...,T n m(t)(t 1,, t n n) = n! t n, < t 1 < t 2 < < t n < t. ( ] T 1 T 2 T 3 T n t time REMARK 6.2 (UNORDERED UNIFORM) Intutively, we usually say that under the condition that n occurrences of events have taken place within (, t], the times at which events occur, considered as unordered random variables, are taken from a sample of size n from U(, t]. pp-6

8 REMARK 6.3 (TESTING POISSON) The above theorem may also be used to test the hypothesis that a given counting process is a Poisson process. This may be done by observing the process for a fixed time t. If in this time period we observed n occurrences and if the process is Poisson, then the unordered occurrence times would be independently and uniformly distributed on (, t]. Hence, we may test if the process is Poisson by testing the hypothesis that the n occurrence times come from a uniform (, t] population. This may be done by standard statistical procedures such as the Kolmogorov-Smirov test. EXAMPLE 6.4 (AN INFINITE SERVER POISSON QUEUE) Suppose that customers arrive at a service station according to a Poisson process with rate λ. Assume, upon arrival, the customer is immediately served by one of an infinite number of servers. Assume also that the service times are independent having common distribution G. Let X(t) denote the number of customers in the system at time t. X(t) = # of time t T n T 3 T 1 T 2 To determine the distribution of X(t), condition on m(t), the total number of customers who have arrived by t. By the Law of Total Probability we have P (X(t) = j) = λt (λt)n P (X(t) = j m(t) = n)e. n! n= Now, the probability that a customer who arrives at time x (the time x, given that m(t) = n, is selected uniformly from (, t] according to the above theorem) will still be present at time t is 1 G(t x) = P (service time exceeds t x). Hence, given that m(t) = n, the probability that an arbitrary one of these customers still being present at time t is given by p = t [1 G(t x)] 1 t dx = t 1 G(x) dx, t independently of others. Consequently, { ( n ) P (X(t) = j m(t) = n) = j p j (1 p) n j, j =, 1,, n,, for j > n, t time and thus P (X(t) = j) = n=j ( n )p j (1 p) n j λt (λt)n e j n! pp-7 λpt (λpt)j = e. j!

9 Or, in other words, X(t) is Poisson distributed with mean occurrence rate λ t [1 G(x)]dx. Or, {X(t), t } is a time-inhomogeneous Poisson process with instantaneous rate λ t [1 G(x)]dx. If the service distribution G follows an exponential distribution with mean service rate µ then the instantaneous rate becomes λ (1 µ e µt ) and the whole system is said to be an M/M/ queueing system which stands for Memoryless arrival time/memoryless service time/infinite number of servers which is so by recalling that the lives and the remaining lives of Poisson streams are exponentially distributed (memoryless). The figure below shows a sample path of {X(t)} from an M/M/ with arrival rate λ = 3 and service rate µ = 2: X(t) time, t arrivals departures If, instead, G follows general distribution (non-exponential) then the system is called an M/G/ queueing system. EXAMPLE 6.5 (AN ELECTRONIC COUNTER) Electrical pulses having random amplitudes arrived at a counter in accordance with a Poisson process with rate λ. With initial amplitude A, the amplitude decays in time, independent of its initial amplitude, and is modeled as Ae αt a time t later. Suppose the initial amplitudes of the pulses are independent having a common distribution function F. Denote the initial amplitudes of the pulses A 1, A 2,.... The total amplitude at time t is then m(t) A(t) = A n e α(t Tn). n=1 Note that A(t) is a continuous-time continuous state space random process. A sample path of A(t) is shown below: pp-8

10 A(t) t time We shall determine the distribution of A(t) by calculating its moment generating function M A(t) (u) = E m(t) M A(t) m(t) (u) = E[e ua(t) λt (λt)n m(t) = n]e. n! Conditioning on m(t) = n, the unordered arrival times are i.i.d. U(, t] distributed. Hence, given that m(t) = n, A(t) has the same distribution as It follows then Now, n= n A j e α(t Yj), Y j s are i.i.d. U(, t]. j=1 E[e ua(t) m(t) = n] = E and hence ( t E[e ua(t) m(t) = n] = [ { exp u n j=1 }] A j e α(t Y j) = ( E[exp{uA 1 e α(t Y1) }] ) n. E [ exp { ua 1 e α(t Y 1) } Y 1 = y ] = M A1 ( ue α(t y) ) ( M ) ) n ( A1 ue α(t y) 1 1 t dy = t Therefore, ( 1 t ( M A(t) (u) = E m(t) M ) m(t) ( A1 ue αy 1 dy) = Φ m(t) t t { [ ( 1 t ( = exp λ 1 M ) ) ]} A1 ue αy dy t { = exp λ t [ ( 1 M )] } A1 ue αy dy. t pp-9 t M A1 ( ue αy ) dy) n. t ( M ) ) A1 ue αy dy

11 The moments of A(t) may then be calculated by differentiation. For example, the mean function of the process {A(t), t } µ A (t) = E[A(t)] = M A(t)() = λ t = λe(a 1) (1 e αt ). αt t M A 1 ()e αy dy 7 Supperposition of Poisson Streams Consider k independent Poisson streams S i having respective rates λ i and remaining lives r i (t) at time t, for 1 i k. DEFINITION 7.1 (SUPERPOSITION OF POISSON STREAMS) A stream S X is called the superposition of S i s if an event in S X occurs whenever an event occurs in at least one of S i s. Denote r X (t) the remaining life of stream S X at time t. THEOREM 7.2 (SUPPERPOSED STREAM) The remaining life of stream S X is the shortest remaining life of S i s, i.e., r X (t) = min 1 i k r i(t). Moreover, stream S X is also a Poisson stream with rate λ X = k i=1 λ i. An example of the superposition of two Poisson streams is illustrated in Figure in the textbook. REMARK 7.3 (NEXT OCCURRENCE) As remarked in 4 (Remark 4.11), the probability that next occurrence of an event in stream S X comes from stream S i is λ i /λ X. In addition, the time until the next occurrence of an event in stream S X is independent of whether it comes from any particular stream S j. That is, see Exercise 8.21 in the textbook. P (r X (t) < τ and r j (t) = r X (t)) = λ j λ X [ 1 e λ X τ ] t, τ >, Now, consider the Poisson processes associated with the streams. Let {m i (t)} be the Poisson processes defined on stream S i, 1 i k, and {m X (t)} the Poisson process defined on stream S X. It follows then the theorem below can be established. THEOREM 7.4 (SUM OF POISSON PROCESSES) The sum of independent Poisson processes is also a Poisson process with rate equals to the sum of rates of the independent Poisson processes. The following figure shows sample paths of two independent Poisson processes with rates 1 and 2, respectively, and the sum process: pp-1

12 # occurrences m 1 (t) m 2 (t) m X (t) t time 8 Decomposition of a Stream Consider a Poisson stream S having rate λ. Upon an occurrence in S, a Bernoulli trial independent of the stream is performed. An occurrence in stream S X is triggered if the outcome is a success (with probability p), otherwise, an occurrence in stream S Y is triggered. DEFINITION 8.1 (DECOMPOSITION OF A STREAM) The two streams S X and S Y are said to decompose the stream S. Now, in any duration τ, P (m occurrences in S X and k occurrences in S Y ) = P (m + k occurrences in S; and m successes in m + k trials) = P (m + k occurrences in S)P (m successes in m + k trials) ( ) = (λτ)m+k m + k (m + k)! e λτ p m q k m ( ) ( ) (λpτ) m (λqτ) = e λpτ k e λqτ m! k! = P (m occurrences in S X )P (k occurrences in S Y ). Hence the following theorem holds. pp-11

13 THEOREM 8.2 (DECOMPOSED STREAMS) The two streams S X and S Y are Poisson streams with rates λp and λq, respectively. DEFINITION 8.3 (MARKED POISSON PROCESSES) Suppose T 1, T 2, are the ordered times of occurrences of a Poisson stream S, and Y 1, Y 2, are independent random variables, independent of the stream, having common distribution function G. The sequence of pairs (T 1, Y 1 ), (T 2, Y 2 ), is called a marked Poisson process. It follows then from the discussion above: THEOREM 8.4 (DECOMPOSED POISSON PROCESSES) Let {m(t), t } be the Poisson process corresponding to the Poisson stream S, and upon occurrences the Bernoulli trials are incurred. Then the Poisson processes corresponding to the two streams S X and S Y are m(t) m(t) m 1 (t) = Y k and m (t) = m(t) m 1 (t) = (1 Y k ), respectively. k=1 Moreover, {m 1 (t)} is a Poisson process with rate λp and {m (t)} is a Poisson process with rate λq. In addition, the two decomposed processes are independent. REMARK 8.5 (GENERAL DECOMPOSED POISSON PROCESSES) Consider G as from a discrete distribution having possible values, 1, 2, with probability function P (Y n = k) = a k > for k =, 1, with k= a k = 1. Define k=1 m(t) m k (t) = I (Yn=k), for k =, 1, 2,. n=1 Then {m (t)}, {m 1 (t)}, are independent Poisson processes with rates λa, λa 1,, respectively. The following is am example for which a i >, i =, 1, 2, 3 and a + a 1 + a 2 + a 3 = Y 3 = 3 Y 9 = 3 m 3 (t) Y 5 = 2 Y 8 = 2 Y 1 = 1 Y 4 = 1 Y 6 = 1 Y 1 = 1 m 2(t) m 1 (t) Y 2 = Y 7 = m (t) time T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 T 9 T 1 pp-12

14 EXAMPLE 8.6 (CUSTOMERS PURCHASES) Customers enter a store according to a Poisson process of rate λ = 8 per hour. According to the past experience, each customer, independent of others, buys something with probability p =.4 and leaves without making a purchase with probability q = 1 p =.6. What is the probability that during the first hour 9 people enter the store with only 4 making purchases and 5 do not? To see this, first note that N 1 = m 1 (1), the number of customers who make purchases during the first hour, and N = m (1), the number of customers who do not make a purchase are independently Poisson distributed with respective rates λp = 3.2 and λq = 4.8. Now, P (4 make purchases and 5 do not) = P (N 1 = 4, N = 5) = P (N 1 = 4)P (N = 5) = 3.24 e 3.2 = (.1781) (.1747) =.311. EXERCISE 8.7 (FOR MARKED POISSON PROCESSES) 4! 4.85 e 4.8 5! a. Customers demanding service at a central processing station arrive according to a Poisson process of rate λ = 1. Each demand, independent of others, is classified as urgent with probability p u =.1, high priority with probability p h =.2, and low priority with probability p l =.7. What is the probability that 3 urgent, 6 high priority, and 12 low priority demands arise in the first two units of time? b. Shocks occur to a system according to a Poisson process of rate λ. Each shock causes some damage Y to the system, and these damages accumulate. Denote m(t) the number of shocks up to time t. Assume the damages Y i s are independently and identically distributed. The total damage up to time t is m(t) D(t) = Y i. (a) Suppose P (Y > y) = e θy for y >. Determine the mean and the variance of D(t). (b) Suppose now P (Y > n) = q n for n =, 1, where < q < 1. The system continues to function as long as the total damage is strictly less than some critical value a, and fails in the contrary circumstance. Determine the mean time to system failure. c. Alpha particles are emitted from a fixed mass of material in accordance with a Poisson process of rate λ. Each particle exists for a random duration and is then annihilated. Assume the lifetimes Y i s of these particles are i.i.d. having common distribution function G. Determine the distribution of N(t), the number of particles existing at time t. d. A tour boat that makes trips through the Houston ship channel leaves every T minutes from Pier 1. Tourists arrive at the pier according to a Poisson process with rate λ. The time that a tourist runs out of patience waiting for the tour boat follows a exponential distribution with rate µ. Running out of patience for waiting, he/she simply leaves without taking the tour boat. Determine the expected number of tourists present at each boat departure epoch. pp-13 i=1

15 9 Compound Poisson Processes In a Poisson stream having rate λ, whenever an event occurs there is a reward Y (or cost). These rewards are independent and identically distributed random variables having common distribution G, mean µ, and variance σ 2. Let {m(t)} be the Poisson process defined on the stream. Consider the process {n(t), t } which is defined by m(t) n(t) = Y i. i= DEFINITION 9.1 (COMPOUND POISSON PROCESSES) The process {n(t)} as described above is called a compound Poisson process. Note that m 1 (t) described in the previous section when each Y is a Bernoulli trial is a compound Poisson process. THEOREM 9.2 (MEAN AND VARIANCE OF A COMPOUND POISSON PROCESS) E[n(t)] = µλt and Var[n(t)] = E(Y 2 )λt. EXAMPLE 9.3 (OF COMPOUND POISSON PROCESSES) a. (RISK THEORY) Suppose claims arrive at an insurance company according to a Poisson process with rate λ. Denote Y k the magnitude of claim k. Then A(t) = m(t) k=1 Y k represents the cumulative amount claimed up to time t. b. (STOCK PRICES) Suppose that transactions in a certain stock take place according to a Poisson process with rate λ. Denote Y k the market price change of the stock between the (k 1)th and the kth transactions. Let X(t) = m(t) k=1 Y k represent the total price change up to time t. The convolution notation below can be used to denote the distribution of the sum of independent random variables: { 1 for y, G () (y) = for y < n G (n) (y) = P ( Y k y) = G (n 1) (y z)dg(z). We have then P (X(t) x) = k=1 = n= P (m(t) Y k x m(t) = n ) (λt) n e λt n! k=1 (λt) n e λt G (n) (x). n! n= pp-14

16 1 Spatial Poisson Processes DEFINITION 1.1 (POINT PROCESSES) Let T R n and the parameter space A = {A : A T }. A point process in T is a random process {m(a)} indexed by the sets A in A taking nonnegative integers as possible values where m(a) = # of points in A with certain characteristics. Also for A B =, A B A we have m(a B) = m(a) + m(b). EXAMPLE 1.2 (OF POINT PROCESSES) a. spatial distribution of stars or galaxies; b. spatial distribution of plants or animals; c. spatial distribution of bacteria on a slide; d. spatial distribution of defects on a surface or in a volume; e. spatial distribution of oil in a region. DEFINITION 1.3 ((HOMOGENEOUS) POISSON POINT PROCESSES) of intensity λ > if a. m(a) takes only nonnegative integer values, and < P (m(a) = ) < 1 if < A < ; b. the distribution of m(a) depends on A only through its size A (independent of shape and location) and P (m(a) = 1) = λ A + o( A ), P (m(a) 2) = o( A ) as A ; c. if A 1,..., A n are mutually exclusive regions then m(a 1 ),..., m(a n ) are independent and n m( A i ) = i=1 n m(a i ). i=1 THEOREM 1.4 (PROPERTIES OF POISSON POINT PROCESS) a. For A A, m(a) P(λ A ). b. For finite mutually exclusive sets A 1,..., A n in A, we have that m(a 1 ),..., m(a n ) are independent. That is, the outcome in one region A does not influence or be influenced by the outcomes in other regions nonoverlapping the region A. pp-15

17 c. If B A then P (m(b) = 1 m(a) = 1) = B A. d. Given that m(a) = n 1, the n points are uniformly distributed over A. e. Suppose A > and m(a) = n 1. Consider a partition {A 1,..., A k } of A. Then m(a 1 ),..., m(a k ) m(a) = n is That is, multinomially distributed with parameters n and A 1 A,..., A k A. P (m(a 1 ) = n 1,, m(a k ) = n k m(a) = n) { n! A 1 n 1 A k n k = n 1! n k, if n = k! A n i=1 n i, n i s are nonnegative integers,, otherwise. EXAMPLE 1.5 (APPLICATIONS OF POISSON POINT PROCESSES) a. (IN ASTRONOMY) Consider stars distributed in space according to a 3-D Poisson point process of intensity λ. For x, y R 3, let the light intensity exerted at x by a star located at y be α g(x, y, α) = x y = α 2 (x 1 y 1 ) 2 + (x 2 y 2 ) 2 + (x 3 y 3 ), 2 where α is a random parameter depending on the intensity of the star at y. The intensities associated with stars are i.i.d. random variables having common mean µ α and variance σα. 2 Assume the total intensity exerted at the point x due to light created by different stars accumulates additively. Denote W (x, A) the total light intensity at x due to light created from all sources located in region A. Then W (x, A) = m(a) i=1 g(x, y i, α i ) = m(a) i=1 α i x y i 2, where y i is the location of the ith star in A. Since the stars distributed in space follow a Poisson point process, y i is uniformly distributed in A and hence E ( x y i 2) = 1 1 A x y dy. 2 We have then Now, E[g(x, y, α)] = E(α)E ( x y i 2) = µ α 1 A A x y dy. 2 E[W (x, A)] = E[m(A)]E[g(x, y, α)] = λµ α pp-16 A A 1 x y 2 dy.

18 b. Customer arrivals at a service station follow a Poisson process of unknown rate. Suppose it is known that 1 customers have arrived during the first three hours. Denote N i the number of customers arriving during the ith hour, i = 1, 2, 3. Determine the probability that N 1 = 3, N 2 = 4, and N 3 = 3. Answer: 1! 3!4!3! (1 3 ) 3 (1 3 ) 4 (1) 3 = c. Bacteria are distributed throughout a volume of liquid according to a Poisson process of intensity λ =.8 organisms per mm 3. A measuring device counts the number of bacteria in in a 5mm 3 volume of the liquid. Determine the probability that more than three bacteria are in this measured volume. Answer: The number of bacteria in this measured volume follows a Poisson distribution with mean rate of 5λ = 4. Hence the desired probability is 1 e 4 4e 4 8e e 4 = d. For a star in space, denote D the distance to its nearest neighbor. Show that D has the probability density function { } 4λπd f(d) = (4λπd 2 3 ) exp, d > 3 if the stars are distributed in space in accordance with a Poisson point process of intensity λ. e. Of furniture making, defects (such as air bubbles, contaminants, chips) occur over the surface of a varnished tabletop according to a Poisson point process at a mean rate of one defect per top. Two inspectors each check separate halves of a given table, determine the probability that both inspectors find defects. f. A piece of a fibrous composite material is sliced across its circular cross section of radius R, revealing fiber ends distributed across the circular area in accordance with a Poisson point process of rate λ fibers per cross section.denote D the radial distance of a fiber from the center of the circular cross section. Determine the probability density function of D. g. Men and women enter a supermarket according to two independent Poisson processes having respective rates (per minute) of two for men and five for women. (a) Starting at an arbitrary time, calculate the probability that at least two men arrive before the first woman arrives. (b) Determine the probability that at least two men arrive before the third woman arrives. pp-17

19 h. Poisson point processes can be applied to check random variates generated by computer algorithm. Say a sequence U, U 1, U 2,, of unform (,1) random variates is generated and 1, pairs, say, of (U 2n, U 2n+1 ) are plotted on the unit square. If the numbers generated are random, then the points in a fixed number of small disjoint squares within the unit square should behave independent Poisson random variables. 11 Examples EXAMPLE 11.1 (AN OPTIMIZATION PROBLEM) Items arrive at a processing plant in accordance with a Poisson process with rate λ. At a fixed time T, all items are dispatched from the system. To minimize the total waiting time of all items, it is to pick an intermediate time t (, T ), at which all items in the system are also dispatched. Now, if all items are dispatched at time t and T, then the expected total waiting time is which can be reasoned as follows: λt λ(t t)2 2 (1) duration = t expected number of items = λt t duration = T t T time each arrival is U(,t), hence similar reasons expected wait for an item = t 2 therefore, expected total waiting time = λt 2 2 expected total λ(t t)2 waiting time = 2 From Equation 1, the constant dispatch time minimizing the expected total waiting time of all items is T λt 2 with the minimal expected total waiting time of. 2 4 Moreover, not only does T minimize the expected total waiting time, it also maximizes the 2 probability that the wait is less than y for every y. To see this, let t be any intermediate dispatch time. Then the total waiting time is W (T ) = m(t ) n=1 W t(n) and { t T(n), if T W t (n) = (n) t T T (n), if T (n) > t, where T (1), T (2), are the (ordered) arrival times of the items. pp-18

20 number of items T (2) W t (2) W t (n) t T (n) T time Now, conditioning on the event that m(t ) = n, the (unordered) arrivals T 1, T 2, are i.i.d. U(, T ). Therefore, W (T ) has the same distribution as m(t ) n=1 Z t(n) where m(t ) P(λT ) and Z t (n) s are i.i.d. and independent of m(t ), having P (Z t (n) < y) = { 1 [min(y, t) + min(y, T t)], T y T 1, y > T. (2) Equation 2 holds for the reasons that follows: To have a waiting time less than y, the item must arrive either between { (t y, t), if t > y (, t), if t < y in which the interval length is min(y, t), or between { (T y, T ), if T y > t (t, T ), if T y < t in which the interval length is min(y, T t). Hence it has to arrive in one of two disjoint intervals having lengths of min(y, t) and of min(y, T t), respectively. The result in Equation 2 then follows since the arrival time of the item is uniformly distributed on (, T ). It follows then from Equation 2 that for fixed y, P (Z t (n) < y) is maximized by t = T. Since 2 the distribution of m(t ) is independent of t and of the Z t (n) s, we have then that P ( m(t ) n=1 Z t(n) < y) = P ( m(t ) n=1 W t(n) < y) is maximized by t = T. Thus, not only does T minimize the expected 2 2 total waiting time, but it also possesses the stronger property of maximizing the probability that the total wait is less than y for every y. EXAMPLE 11.2 (SHOT NOISE) Shot noise process models fluctuations in electrical currents that are due to chance arrivals of electrons to an anode. Assume that electrons arrive at an anode in accordance with a Poisson process with intensity λ. Assume also that an arriving electron produces a current whose intensity x time units after arrival is given by the impulse response function h(x). Examples of impulse response functions include pp-19

21 the power law shot noise: h(x) = x θ, for x > ; the decaying exponential shot noise: triangles, rectangles, etc. h(x) = e θx, for x > ; The intensity of the current at time t is then the shot noise: m(t) I(t) = h(t T k ), k=1 where T k s are the (ordered) arrival times. Note that, conditioning on m(t) = n, I(t) has the same distribution as that of n h(u k ) k=1 where U k s are i.i.d. U(, t]. That is, the (unconditional) distribution of I(t) is the same as that of the random sum m(t) where ε k = h(u k ), k = 1, 2,..., are i.i.d. EXERCISE 11.3 (LAST ONE) k=1 ε k a. Miller Auditorium accepts season ticket subscriptions during an interval (, t] starting time. Each subscription, with probability p i, entails the purchase of i tickets, i = 1, 2, 3, 4, at a cost of c dollars per ticket payable at the time of purchase. A dollar received u time units from now will be discounted with a factor of β (i.e., will have the present value of e βu ). Assume the subscriptions arrive according to a Poisson process with rate λ. Compute the expected present value at time of all revenues received from the ticket sales in (, t]. b. An earthquake caused serious damage in a given locality. The locations of fatality follow a two-dimensional homogeneous Poisson point process with intensity λ =.1 per square mile. What is the probability that there are no fatalities within ten miles radius of the city hall? c. (DATABASE MAINTENANCE) At time, a file in a database is loaded with r records. Additional records will be added in the future according to a Poisson process. For each record in the file, its sueful life follows a distribution G. At the end of an interval of length t, there are two types of records dead and live records. Denote D(t) and L(t), respectively, the pp-2

22 numbers of dead and live records. Then m(t) = D(t) + L(t). (Dead records are logically flagged as deleted with actual pyhsical removal done when the file is compacted at database maintenance time.) The ratio a(t) = E[m(t)] E[L(t)] gives the mean search overhead per live record reference at time t (or the average number of record accesses per reference). Express a(t) for each of cases as follows. Also for each numerical example as given in parentheses, do plots to show the result.: arrival rate useful life λ(t) = λ λ(t) distribution G E(l) case (a) case (b) (r = 5, l = 2 days, λ = 5.58 per day) (r = 5, l = 2 days, ln λ(t) = G case (c) case (d) t sin( πt 15 ).275 cos( πt 15 )) pp-21