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1. Poisson processes Stochastic Process By TMJA Cooray

2. Introduction • The probability models we have discussed so far involves random variables occurring in discrete sequence. • That is both the random variable and the time parameter have been discrete. • Here we shall consider random variables in continuous time. MA (4030) Level 4-Poisson Processes

3. We introduce the basic stochastic model for events that occur at random moments in continuous time. • This model is called the Poisson Process • Examples: telephone calls arriving, emission of radio active atoms, occurrence of accidents, cosmic rays arriving etc. MA (4030) Level 4-Poisson Processes

4. Let the discrete random variable X(t): represents the size of the population being considered at any instant of time t. • Denote the probability that the population size at time t is n, as: MA (4030) Level 4-Poisson Processes

5. Memory less random variables • There’s a bin full of integrated circuit chips of a particular type. • Consider the random variable :The total operating time X of a single chip. • Assume that all the chips were manufactured under identical conditions ,so that the distribution of X is the same for all the chips. there are no faulty chips in the bin • X is a continuous r.v. and P[X>0]=1. MA (4030) Level 4-Poisson Processes

6. Consider the following scenario: • Select a chip at random from the bin install it and turn it on at time t=o. the probability that it is still operating at time t is P[X>t]. • Unplug the switch while its still working , after t hours of use. • Take the same chip and install it. • What is the pr. that chip will still be working after s hours of use in the second installation? • The conditional pr. P[X>s+t|X>t]=? MA (4030) Level 4-Poisson Processes

7. For many types of electronic equipments • P[X> s+t |X>t]= P[X>s] • A r.v. X≥0 satisfying the above equation for all s≥0 and t≥0 is said to have no memory. • Theorem :If X≥0 is a r.v. satisfying the above equation then X is an exponential r.v. with parameter . Conversely, every exponential r.v. satisfies the above equation . MA (4030) Level 4-Poisson Processes

8. Consider a very small time interval Δt, Probability of no failure during time t+Δt, when it has worked for time t • Probability of one failure during time t+Δt, when it has worked for time t Probability of more than one failure no failure during time t+Δt, when it has worked for time t • P[X>t+ Δt |X>t]= P[X> Δt] =e - Δt, • P[X<t+ Δt |X>t]= P[X< Δt] =1-e - Δt, MA (4030) Level 4-Poisson Processes

9. Poisson Process • Let X(t) be the total number of events (telephone calls) recorded up to time t. • This total number of calls is the population size in question. • It is usual to assume that the events involved occur at“random” • That is the chance of a new call arriving in any short interval is independent, not only of the previous state of the system ,but also of present state of the system.. MA (4030) Level 4-Poisson Processes

10. We can therefore assume that • The chance of a new addition to the total count ( a new call) during a very short interval of time Δt can be written as Δt +o(Δt ), where  is average number of calls per unit time (which is some suitable constant characterizing the intensity of calls). • The chance of two or more calls during the very short interval of time Δt can be written as o(Δt ), • The chance of no call during a very short interval of time Δt can be written as 1- Δt +o(Δt ), MA (4030) Level 4-Poisson Processes

11. Then the probability of getting n calls up to time t +Δt are :, • Ignoring the small terms ( higher powers of small Δt ). • If n>0 ,this can arise in 2 ways. : n calls up to time (0,t) and no calls in time (t,t+Δt) : n-1 calls up to time (0,t) and one call in time (t,t+Δt) MA (4030) Level 4-Poisson Processes

12. Now it follows from 2, For n=0, n=0 at time t+Δt , only if n=0 at time t and no new calls are coming in Δt .In this case we can write: MA (4030) Level 4-Poisson Processes

13. If we start at time 0 with a new telephone or with a new record, • The set of differential- difference equations (3) and (4), together with the boundary condition (5) determines the probability distribution Pn(t). • Solving (4) log(p0(t))=- t+c using (5) c=0 giving , p0(t)= e -t ---------(6) Substituting this in (3),with n,=1 , gives p1(t)=  t e -t MA (4030) Level 4-Poisson Processes

14. Repeating this procedure the following general formula can be obtained. This is simply a Poisson distribution with parameter t MA (4030) Level 4-Poisson Processes

15. This set of differential-difference equations given in (3) can be solvedusing generating functions also. • Recall the probability generating functions p(x,t) for the probability distribution pn(t) • where p(x,t) is a function of t as well as x. MA (4030) Level 4-Poisson Processes

16. MA (4030) Level 4-Poisson Processes

17. (9) involves differentiation in t only and,can be integrated directly in conjunction with (10), to obtain • this is immediately identifiable as The probability generating function of a Poisson distribution with parameter t, the individual probabilities given by (7) MA (4030) Level 4-Poisson Processes

18. Partial differential-difference equation (10) can be written for the moment generating function M(,t),merely by putting x=e, giving • In different situations one generating function may be easily soluble than the other. MA (4030) Level 4-Poisson Processes

19. Solution of linear Partial differential equations. • For many processes ,,these generating functions are linear functions of the differential operators • in order to solve them ,the main results that we shall require are given as follows. MA (4030) Level 4-Poisson Processes

20. Consider the linear partial differential equation: • subject to some appropriate boundary conditions ,where P,Q and R may be functions of x,y and z.first step is to form the subsidiary equations given by : MA (4030) Level 4-Poisson Processes

21. We can find two independent integrals of these subsidiary equations ,writing them in the form , u(x,y,z)= constant, v(x,y,z)=constant ...(16) • The most general solution of (13) is now given by (u,v)=0 • u=(v) ........ (17) • where  and  are arbitrary functions.The precise form of these functions can be determined by using boundary conditions. MA (4030) Level 4-Poisson Processes

22. Note:relationship between Poisson and (negative) exponential • If all the assumptions hold as for a Poisson process : • probability of no occurrence during a time interval (0,t) = p0(t)= e -t • That is the time interval between two consecutive occurrences ( a r.v.T) is >t. • P[T>t]= e -t ,Then P[T t]=1- e -t MA (4030) Level 4-Poisson Processes

23. which is the cumulative distribution F(t) of the r.v. T [the time interval between two consecutive occurrences ( or the inter arrival time)]. • Then P[T t] =F(t)=1- e -t • to find the corresponding density function differentiating F(t), we get F’(t)=f(t)= e -t , 0<t • which is the negative exponential distribution. MA (4030) Level 4-Poisson Processes