Random Stat667 Processes
This is to supplement the textbook but not to replace it.
Poisson Processes
Contents
1 2 3 4 5 6 7 8 9 Continuous-Time Chains Markovian Streams Regular Streams Poisson Streams Poisson Processes Random Streams Supperposition of Poisson Streams Decomposition of a Stream Compound Poisson Processes pp-1 pp-1 pp-1 pp-2 pp-5 pp-6 pp-10 pp-11 pp-14 pp-15 pp-18
10 Spatial Poisson Processes 11 Examples
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
D EFINITION 2.1 (S TREAM ( OF EVENTS )) A sequence of similar events occur in time is called a stream (of events).
E0 T0 = 0
q
E1 T1
q
E2 T2
q
E3 T3
q
t
D EFINITION 2.2 (M ARKOVIAN 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. R EMARK 2.3 ( NONOVERLAPPING INTERVALS ) The developments within any two nonoverlapping time intervals of a Markovian stream are independent.
3
Regular Streams
D EFINITION 3.1 (E XPLOSION ) 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 innite number of occurrences have already taken place. E XAMPLE 3.2 ( A BOUNCING BALL ) A ball bouncing on the oor, 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 2n . Then xn = n and the time the nth bounce taking place is Tn = 1 + 1 1 1 + + n1 = 2 n1 . 2 2 2
Clearly Tn < 2 and Tn 2 = T as n . The process hence explodes at time T = 2. D EFINITION 3.3 (R EGULAR S TREAM ) A stream having no explosion is called a regular stream.
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T HEOREM 3.4 ( NECESSARY AND SUFFICIENT CONDITION OF REGULAR STREAM ) A stream is regular if and only if, within an innitesimal 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 mi (t, h) = P (exactly i occurrences within (t, t + h]), we have m0 (t, h) = 1 (t)h + o(h) m1 (t, h) = (t)h + o(h) mi (t, h) = o(h), i > 1.
4
Poisson Streams
D EFINITION 4.1 (P OISSON S TREAM ) A stream that is Markovian and regular is called a Poisson stream. The expected number of occurrences within (t, t + h] is
0 [1 (t)h + o(h)] + 1 [(t)h + o(h)] +
i=2
io(h) = (t)h + o(h).
Hence, (t) = lim E{number of occurrences within (t, t + h]} . h0 h
D EFINITION 4.2 (O CCURRENCE R ATE ) For a Poisson stream, (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. D EFINITION 4.3 (AVERAGE O CCURRENCE R ATE ) For a Poisson stream, the average occurrence rate during (t, t + ] is (t, ) = E{number of occurrences within (t, t + ]}
D EFINITION 4.4 (T IME -H OMOGENEOUS P OISSON S TREAMS ) A Poisson stream is called timehomogeneous if (t) = t. Consequently, = (t) = (t, ), t > 0 and > 0. Unless otherwise mentioned, assume hereafter time-homogeneity. D EFINITION 4.5 (R EMAINING L IVES ) 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
T HEOREM 4.6 (R EMAINING L IFE I S E XPONENTIALLY D ISTRIBUTED ) The remaining life r(t) is distributed as E(). The above theorem holds because P (r(t) > s) = P (zero occurrence within (t, t + s]) = = =
k
lim P (k (zero occurrence within (t + j=1
k
(j1)s ,t k
+
js ])) k
k
lim
P (zero occurrence within (t +
j=1
(j1)s ,t k
+
js ]) k
= e
k s
lim (1 .
s 1 + o( ))k k k
D EFINITION 4.7 (L IVES OF A S TREAM ) The duration Ei1 and Ei in a stream is called the ith life. That is
i
i
between two consecutive occurrences
= Ti Ti1 = ith life = interoccurrence time.
A graphical view of lives of a Poisson stream is given below:
E0 T0 = 0
q
l1
E1 l2 E2 T1
q
l3
E3 l4 E4 T3
q
T2
q
T4
q
t
T HEOREM 4.8 (L IVES AND R EMAINING L IVES OF P OISSON S TREAMS ) 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 n = r(Tn1 ). It remains to show that (c) implies (b). Observing the follorwing gure
t ln r(t) Tn
q
T n1 = t
q
t+s
T n+1
q
time
we have then P (r(t) < s|
n
> ) =
P (r(t) < s and n > ) P ( < = P ( n > ) P( s = 1e . pp-3
n n
< + s) e e(+s) = > ) e
D EFINITION 4.9 (S HORTER R EMAINING L IFE ) Consider two completely independent Poisson streams: SX having rate X and remaining life rX (t), and SY having rate Y and remaining life rY (t). Dene R2 (t) = min(rX (t), rY (t)) as the shorter remaining life of the two streams. If the event (R2 (t) = rX (t)) (rX (t) < rY (t)) occurs then stream SX is said to have a shorter remaining life than stream SY does. T HEOREM 4.10 (D ISTRIBUTION lows E(X + Y ) and
OF
S HORTER R EMAINING L IFE ) The distribution of R2 (t) folX . X + Y
P (rX (t) < rY (t)) = To verify the above theorem rst note that
P (R2 (t) > s) = P (rX (t) > s and rY (t) > s) = P (rX (t) > s)P (rY (t) > s) = eX s eY s = e(X +Y )s . So R2 (t) is exponentially distributed with rate X + Y . Now, consider an innitesimal duration (t, t + h]. Denote Oij = the event that exactly i coourrences in stream SX and exactly j occurrences in stream SY . Employing a continuous version of the rst-step analysis by conditioning on the occurrences of the two streams within (t, t + h]:
P (rX (t) < rY (t)) =
i=0 j=0
P (rX (t) < rY (t)|Oij )P (Oij )
= 1 (X h(1 Y h) + o(h)) + 0 ((1 X h)Y h + o(h)) +P (rX (t) < rY (t)) ((1 X h) (1 Y h) + o(h)) +
i>1 or j>1
P (rX (t) < rY (t)|Oij )P (Oij ) o(h)
= X h + (1 X h Y h)P (rX (t) < rY (t)) + o(h). Hence, P (rX (t) < rY (t)) = X o(h) X + . X + Y (X + Y )h h0 X + Y
R EMARK 4.11 (S HORTEST R EMAINING L IFE ) Consider k independent Poisson streams Si with rate i and remaining life ri (t), i = 1, . . . , k. Let Rk (t) be the shortest remaining life of these streams from time t. That is, Rk (t) = min ri (t).
1ik
pp-4
Then Rk (t) follows E() where = 1 + + k . In addition, P (rj (t) = Rk (t)) = j .
5
Poisson Processes
D EFINITION 5.1 (C OUNTING P ROCESSES ) A counting process {m(t) : t 0} is a continuoustime chain which counts the number of occurrences of the event in a stream within (0, t]. Dene m(0) 0. The following is observed: R EMARK 5.2 ( PROPERTIES OF COUNTING PROCESSES ) For a counting process the state space is {0, 1, 2, }; it is nondecreasing, i.e., m(t) m(s) if t < s; it increases by one whenever an event occurs. D EFINITION 5.3 (P OISSON P ROCESSES ) A counting process is called a (time-homogeneous) Poisson process if m(t) is Poisson distributed with parameter t t > 0. The parameter is called the rate of the Poisson process. T HEOREM 5.4 ( PROPERTIES OF P OISSON A Poisson process has
INDEPENDENT INCREMENTS : PROCESSES )
counting process increases independently in nonoverlapping inter-
vals (Markovian);
STATIONARY INCREMENTS:
time-homogeneous;
P OISSON SEGMENTS: m(t + ) m(t) P( ), > 0 and t > 0; i.e., the number of occurrences of events within any time segment is Poisson distributed with mean occurrence rate (length of the segment). R EMARK 5.5 (P OISSON P ROCESS S TREAM IS R EGULAR ) 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) > ) = P (no occurrence during (t, t + ]) = e .
pp-5
R EMARK 5.6 (V ERIFICATION OF P OISSON S TREAMS ) 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. D EFINITION 5.7 (E RLANG D ISTRIBUTIONS ) From the denition of lives of a Poisson stream, Tk = time to kth occurrence = has an Erlang distribution with parameters k and . R EMARK 5.8 (E RLANG D ISTRIBUTION IS A G AMMA D ISTRIBUTION ) Erlang distribution is a special case of the more general family of distribution, namely, the gamma distribution with parameters = k and . Hence the p.d.f., the mean and the variance of an Erlang distribution with parameters k and are k tk1 et k tk1 et = , t>0 (k) (k 1)! k E(Tk ) = k Var(Tk ) = 2 . f (t) =
1
+
2
+ +
k
6
Random Streams
Poisson streams are also referred to as random streams. T HEOREM 6.1 (C ONDITIONAL U NIFORM O CCURRENCES ) For a Poisson process m(t), given that m(t) = n (i.e., given that there are n occurrences within (0, t]), the n (ordered) occurrence times T1 , T2 , . . . , Tn have the same joint distribution as the order statistics corresponding to n independent uniform (U (0, t)) random variables over the interval (0, t). That is, fT1 ,...,Tn |m(t) (t1 , , tn |n) = n! , 0 < t1 < t2 < < tn < t. tn
q q q q
0
(
T1
T2
T3
Tn
]
t
time
R EMARK 6.2 ( UNORDERED UNIFORM ) Intutively, we usually say that under the condition that n occurrences of events have taken place within (0, t], the times at which events occur, considered as unordered random variables, are taken from a sample of size n from U (0, t].
pp-6
R EMARK 6.3 ( TESTING P OISSON ) 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 xed 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 (0, t]. Hence, we may test if the process is Poisson by testing the hypothesis that the n occurrence times come from a uniform (0, t] population. This may be done by standard statistical procedures such as the Kolmogorov-Smirov test. E XAMPLE 6.4 (A N I NFINITE S ERVER P OISSON Q UEUE ) Suppose that customers arrive at a service station according to a Poisson process with rate . Assume, upon arrival, the customer is immediately served by one of an innite 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 customers @ time t Tn
q
T1 0
T2
T3
q q q
q
time t
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) =
n=0
P (X(t) = j|m(t) = n)et
(t)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 (0, 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
t
p=
0
1 [1 G(t x)] dx = t
t 0
1 G(x) dx, t
independently of others. Consequently, P (X(t) = j|m(t) = n) = and thus P (X(t) = j) =
n=j
pj (1 p)nj , j = 0, 1, , n, 0, for j > n,
n j
n j n j nj t (t) pt (pt) p (1 p) e =e . j n! j!
pp-7
Or, in other words, X(t) is Poisson distributed with mean occurrence rate 0 [1 G(x)]dx. Or, t {X(t), t 0} is a time-inhomogeneous Poisson process with instantaneous rate 0 [1 G(x)]dx. If the service distribution G follows an exponential distribution with mean service rate then the instantaneous rate becomes (1 et ) and the whole system is said to be an M/M/ queueing system which stands for M emoryless arrival time/M emoryless service time/innite number of servers which is so by recalling that the lives and the remaining lives of Poisson streams are exponentially distributed (memoryless). The gure below shows a sample path of {X(t)} from an M/M/ with arrival rate = 3 and service rate = 2:
q
t
4
q
q
q
X(t)
3
q
q
2
q
0
1
q
arrivals departures
0 time, t
If, instead, G follows general distribution (non-exponential) then the system is called an M/G/ queueing system. E XAMPLE 6.5 (A N E LECTRONIC C OUNTER ) Electrical pulses having random amplitudes arrived at a counter in accordance with a Poisson process with rate . With initial amplitude A, the amplitude decays in time, independent of its initial amplitude, and is modeled as Aet 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 A1 , A2 , . . .. The total amplitude at time t is then
m(t)
A(t) =
n=1
An e(tTn ) .
Note that A(t) is a continuous-time continuous state space random process. A sample path of A(t) is shown below:
pp-8
0 0 t i me
A (t ) 5 10
15
t
We shall determine the distribution of A(t) by calculating its moment generating function
MA(t) (u) = Em(t) MA(t)|m(t) (u) =
n=0
E[euA(t) |m(t) = n]et
(t)n . n!
Conditioning on m(t) = n, the unordered arrival times are i.i.d. U (0, t] distributed. Hence, given that m(t) = n, A(t) has the same distribution as
n
Aj e(tYj ) , Yj s are i.i.d. U (0, t].
j=1
It follows then
n
E[euA(t) |m(t) = n] = E exp u
j=1
Aj e(tYj )
= E[exp{uA1 e(tY1 ) }] .
n
Now, E exp uA1 e(tY1 ) |Y1 = y = MA1 ue(ty) and hence
t
E[e Therefore,
uA(t)
|m(t) = n] =
0
MA1 ue
(ty)
1 dy t
n
=
1 t
t
n
MA1 ue
0
y
dy
.
MA(t) (u) = Em(t)
1 t
t
m(t)
MA1 uey dy
0
= m(t)
1 t
t
MA1 uey dy
0
= exp 1 = exp t
t
1 t
t
MA1 uey dy
0
1 MA1 uey dy .
0
pp-9
The moments of A(t) may then be calculated by differentiation. For example, the mean function of the process {A(t), t 0} A (t) = E[A(t)] = MA(t) (0) = E(A1 ) = (1 et ). t t
t
MA1 (0)ey dy
0
7
Supperposition of Poisson Streams
Consider k independent Poisson streams Si having respective rates i and remaining lives ri (t) at time t, for 1 i k. D EFINITION 7.1 (S UPERPOSITION OF P OISSON S TREAMS ) A stream SX is called the superposition of Si s if an event in SX occurs whenever an event occurs in at least one of Si s. Denote rX (t) the remaining life of stream SX at time t. T HEOREM 7.2 (S UPPERPOSED S TREAM ) The remaining life of stream SX is the shortest remaining life of Si s, i.e., rX (t) = min ri (t).
1ik
Moreover, stream SX is also a Poisson stream with rate X = k i . An example of the superi=1 position of two Poisson streams is illustrated in Figure 8.14.1 in the textbook. R EMARK 7.3 ( NEXT OCCURRENCE ) As remarked in 4 (Remark 4.11), the probability that next occurrence of an event in stream SX comes from stream Si is i /X . In addition, the time until the next occurrence of an event in stream SX is independent of whether it comes from any particular stream Sj . That is, P (rX (t) < and rj (t) = rX (t)) = see Exercise 8.21 in the textbook. Now, consider the Poisson processes associated with the streams. Let {mi (t)} be the Poisson processes dened on stream Si , 1 i k, and {mX (t)} the Poisson process dened on stream SX . It follows then the theorem below can be established. T HEOREM 7.4 (S UM OF P OISSON P ROCESSES ) 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 gure shows sample paths of two independent Poisson processes with rates 1 and 2, respectively, and the sum process: j 1 eX t, > 0, X
pp-10
m 1(t ) m 2(t ) m X (t )
0 0 time
1
# occurrences 2 3 4 5
6
7
t
8
Decomposition of a Stream
Consider a Poisson stream S having rate . Upon an occurrence in S, a Bernoulli trial independent of the stream is performed. An occurrence in stream SX is triggered if the outcome is a success (with probability p), otherwise, an occurrence in stream SY is triggered. D EFINITION 8.1 (D ECOMPOSITION compose the stream S. Now, in any duration , P (m occurrences in SX and k occurrences in SY ) = 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) ( )m+k m+k m k = e p q (m + k)! m (p )m p (q )k q = e e m! k! = P (m occurrences in SX )P (k occurrences in SY ). Hence the following theorem holds. pp-11
OF A
S TREAM ) The two streams SX and SY are said to de-
T HEOREM 8.2 (D ECOMPOSED S TREAMS ) The two streams SX and SY are Poisson streams with rates p and q, respectively. D EFINITION 8.3 (M ARKED P OISSON P ROCESSES ) Suppose T1 , T2 , are the ordered of times occurrences of a Poisson stream S, and Y1 , Y2 , are independent random variables, independent of the stream, having common distribution function G. The sequence of pairs (T1 , Y1 ), (T2 , Y2 ), is called a marked Poisson process. It follows then from the discussion above: T HEOREM 8.4 (D ECOMPOSED P OISSON P ROCESSES ) Let {m(t), t 0} 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 SX and SY are
m(t) m(t)
m1 (t) =
k=1
Yk and m0 (t) = m(t) m1 (t) =
k=1
(1 Yk ), respectively.
Moreover, {m1 (t)} is a Poisson process with rate p and {m0 (t)} is a Poisson process with rate q. In addition, the two decomposed processes are independent. R EMARK 8.5 (G ENERAL D ECOMPOSED P OISSON P ROCESSES ) Consider G as from a discrete distribution having possible values 0, 1, 2, with probability function P (Yn = k) = ak > 0 for k = 0, 1, with ak = 1. Dene k=0
m(t)
mk (t) =
n=1
I(Yn =k) , for k = 0, 1, 2, .
Then {m0 (t)}, {m1 (t)}, are independent Poisson processes with rates a0 , a1 , , respectively. The following is am example for which ai > 0, i = 0, 1, 2, 3 and a0 + a1 + a2 + a3 = 1.
Y3 = 3 3 2 1 0
q q
Y9 = 3 Y5 = 2 Y4 = 1
q
Y8 = 2
q q
q
m 3(t ) m 2(t ) m 1(t ) m 0(t )
Y1 = 1
q
Y6 = 1 Y7 = 0
q
Y 10 = 1
q
Y2 = 0
q
q
q
q
q
q
q
q
q
q
q
time
T1
T2 T3
T4
T5
T6
T7
T8
T9
T10
pp-12
E XAMPLE 8.6 (C USTOMERS P URCHASES ) Customers enter a store according to a Poisson process of rate = 8 per hour. According to the past experience, each customer, independent of others, buys something with probability p = 0.4 and leaves without making a purchase with probability q = 1 p = 0.6. What is the probability that during the rst hour 9 people enter the store with only 4 making purchases and 5 do not? To see this, rst note that N1 = m1 (1), the number of customers who make purchases during the rst hour, and N0 = m0 (1), the number of customers who do not make a purchase are independently Poisson distributed with respective rates p = 3.2 and q = 4.8. Now, P (4 make purchases and 5 do not) = P (N1 = 4, N0 = 5) = P (N1 = 4)P (N0 = 5) = = (0.1781) (0.1747) = 0.0311. E XERCISE 8.7 ( FOR M ARKED P OISSON P ROCESSES ) a. Customers demanding service at a central processing station arrive according to a Poisson process of rate = 10. Each demand, independent of others, is classied as urgent with probability pu = 0.1, high priority with probability ph = 0.2, and low priority with probability pl = 0.7. What is the probability that 3 urgent, 6 high priority, and 12 low priority demands arise in the rst two units of time? b. Shocks occur to a system according to a Poisson process of rate . 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 Yi s are independently and identically distributed. The total damage up to time t is
m(t)
3.24 e3.2 4.85 e4.8 4! 5!
D(t) =
i=1
Yi .
(a) Suppose P (Y > y) = ey for y > 0. Determine the mean and the variance of D(t). (b) Suppose now P (Y > n) = q n for n = 0, 1, where 0 < 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 xed mass of material in accordance with a Poisson process of rate . Each particle exists for a random duration and is then annihilated. Assume the lifetimes Yi 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 . The time that a tourist runs out of patience waiting for the tour boat follows a exponential distribution with rate . 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
9
Compound Poisson Processes
In a Poisson stream having rate , 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 , and variance 2 . Let {m(t)} be the Poisson process dened on the stream. Consider the process {n(t), t 0} which is dened by
m(t)
n(t) =
i=0
Yi .
D EFINITION 9.1 (C OMPOUND P OISSON P ROCESSES ) The process {n(t)} as described above is called a compound Poisson process. Note that m1 (t) described in the previous section when each Y is a Bernoulli trial is a compound Poisson process. T HEOREM 9.2 ( MEAN AND Var[n(t)] = E(Y 2 )t.
VARIANCE OF A COMPOUND
P OISSON
PROCESS )
E[n(t)] = t and
E XAMPLE 9.3 ( OF COMPOUND P OISSON
PROCESSES )
a. (R ISK T HEORY) Suppose claims arrive at an insurance company according to a Poisson m(t) process with rate . Denote Yk the magnitude of claim k. Then A(t) = k=1 Yk represents the cumulative amount claimed up to time t. b. (S TOCK P RICES) Suppose that transactions in a certain stock take place according to a Poisson process with rate . Denote Yk the market price change of the stock between the (k1)th m(t) and the kth transactions. Let X(t) = k=1 Yk 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: G(0) (y) = G We have then
m(t) (n)
1 for y 0, 0 for y < 0
n
(y) = P (
k=1
Yk y) =
G(n1) (y z)dG(z).
P (X(t) x) =
n=0
P
k=1
Yk x|m(t) = n
(t)n et n!
=
n=0
(t)n et (n) G (x). n! pp-14
10
Spatial Poisson Processes
D EFINITION 10.1 (P OINT P ROCESSES ) Let T Rn 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). E XAMPLE 10.2 ( 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. D EFINITION 10.3 ((H OMOGENEOUS ) P OISSON P OINT P ROCESSES ) of intensity > 0 if a. m(A) takes only nonnegative integer values, and 0 < P (m(A) = 0) < 1 if 0 < |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) = |A| + o(|A|), P (m(A) 2) = o(|A|) as |A| 0; c. if A1 , . . . , An are mutually exclusive regions then m(A1 ), . . . , m(An ) are independent and
n n
m(
i=1
Ai ) =
i=1
m(Ai ).
T HEOREM 10.4 ( PROPERTIES OF P OISSON a. For A A, m(A) P(|A|).
POINT PROCESS )
b. For nite mutually exclusive sets A1 , . . . , An in A, we have that m(A1 ), . . . , m(An ) are independent. That is, the outcome in one region A does not inuence or be inuenced by the outcomes in other regions nonoverlapping the region A. pp-15
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| > 0 and m(A) = n 1. Consider a partition {A1 , . . . , Ak } of A. Then m(A1 ), . . . , m(Ak )|m(A) = n is multinomially distributed with parameters n and That is, P (m(A1 ) = n1 , , m(Ak ) = nk |m(A) = n) =
|A1 |n1 |Ak |nk n! , n1 !nk ! |A|n
|Ak | |A1 | ,..., . |A| |A|
0,
if n = k ni , ni s are nonnegative integers, i=1 otherwise.
E XAMPLE 10.5 (A PPLICATIONS OF P OISSON P OINT P ROCESSES ) a. (IN ASTRONOMY) Consider stars distributed in space according to a 3-D Poisson point process of intensity . For x, y R3 , let the light intensity exerted at x by a star located at y be = , g(x, y, ) = 2 2 + (x y )2 + (x y )2 xy (x1 y1 ) 2 2 3 3 where 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 and variance 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
m(A) m(A)
W (x, A) =
i=1
g(x, yi , i ) =
i=1
i x yi
2
,
where yi is the location of the ith star in A. Since the stars distributed in space follow a Poisson point process, yi is uniformly distributed in A and hence E We have then E[g(x, y, )] = E()E Now, E[W (x, A)] = E[m(A)]E[g(x, y, )] =
A
x yi
2
=
1 |A|
A
1 xy |A|
2
dy.
x yi
2
=
A
1 xy 1 xy
2
dy.
2
dy.
pp-16
b. Customer arrivals at a service station follow a Poisson process of unknown rate. Suppose it is known that 10 customers have arrived during the rst three hours. Denote Ni the number of customers arriving during the ith hour, i = 1, 2, 3. Determine the probability that N1 = 3, N2 = 4, and N3 = 3. Answer: 10! 1 3!4!3! 3
3
1 3
4
1 3
3
= 0.0711.
c. Bacteria are distributed throughout a volume of liquid according to a Poisson process of intensity = 0.8 organisms per mm3 . A measuring device counts the number of bacteria in in a 5mm3 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 5 = 4. Hence the desired probability is 1 e4 4e4 8e4 64 4 e = 0.5665. 6
d. For a star in space, denote D the distance to its nearest neighbor. Show that D has the probability density function f (d) = (4d2 ) exp 4d3 , d>0 3
if the stars are distributed in space in accordance with a Poisson point process of intensity . 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 ...