Poisson Process and Related Distributions Prepared by: SUHAD NATOOR Supervision of ASSIST. PROF.DR. SAHAND DANESHVAR

Outlines Interarrival time Theorem 4.2, 4.3, 4.4 Examples Purely (completely) Poisson process Theorem 4.5 Further interesting properties of Poisson process Random modification of X (residual time of X) Poisson process and geometric distribution (poisson count process) example

Interarrival time With a Poisson process, {N(t), t≥0}, where N(T) denotes the number of occurrences of an event E by epoch t, let consider an associated random variable X: the interval between two successive occurrences of E. We proceed to show that X has a negative exponential distribution. Theorem: the interval between two successive occurrences of a Poisson process {N(t), t ≥ 0} having parameter λ has a negative exponential distribution with mean 1/λ. Proof:

Theorem 4.3. The intervals between successive occurrences (interarrival times) of a Poisson process (with mean λt ) are identically and independently distributed random variables (i.i.d) which follow the negative exponential law with mean 1/λ.

Theorem 4.4. If the intervals between successive occurrences of an event E are independently distributed with a common exponential distribution with mean 1/λ, then the events E from a Poisson process with mean λt. We can see theorem 4.3 and 4.4 are conversed and they give a characterization of Poisson Process Proof :

Proof

Notes Poisson process has independent distributed interrival times and Gamma distributed waiting times

Example 1 Suppose that the customers arrive at a counter in accordance with a Poisson process with mean rate of 2 per minute(λ =2/minute). Then the interval between any two successive arrivals follows exponential distribution with mean 1/λ =1/2 minute. The probability that the interval between two successive arrivals is:

Example 2 Suppose that customers arrive at a counter independently from two different sources. Arrivals occur in accordance with a Poisson process with mean rate of λ per hour from the first source and µ per hour from the second source. since arrivals at the counter (form either source) constitute a Poisson process with mean (λ+µ) per hour, the interval between any tow successive arrivals has a negative exponential distribution with mean 1/(λ+µ ) per hours.

Example 3 For example, if taxis arrive at a spot from the north at the rate of 1 per minute and from the south at the rate of 2 per minute in accordance with two independent Poisson processes. The interval between arrival of two taxis has a negative exponential distribution with mean 1/3 minute, the probability that a lone person will have to wait more than a given time t can be found.

Results Poisson type of occurrences are also called purely random events. The Poisson process is called a purely (or completely) random process. The reason is: The occurrence is equally likely to happen anywhere in (0,T) given that only one occurrence has taken place in that interval. We state this by the following theorem:

Theorem 4.5 Given that only one occurrence of a Poisson process N(T) has occurred by epoch T, then the distribution of the time interval γ in (0,T) in which it occurred is uniform in (0,T), i.e.

Result It may be said that a Poisson process distributes points at random over the infinite interval (0, ∞ ) in the same way as the uniform distribution distributes points at random over a finite interval (a,b).

Further Interesting Properties Of Poisson Process We have shown that the interval Xi(=ti+1-ti) between two successive occurrences Ei, Ei+1 (i≥1) of a Poisson process with parameter λ has an exponential distribution with mean 1/λ , further, the following result holds. For a Poisson process with parameter λ , the interval of time X up to the first occurrence also follows an exponential distribution with mean 1/ λ. For,

In other words, the relation (2. 1) does not depend on i nor ti In other words, the relation (2.1) does not depend on i nor ti. As in the following slide:

Another Important Property Suppose that the interval X is measured from an arbitrary instant of time ti+ γ (γ arbitrary) in the interval (ti, ti+1) and not just the instant ti of the occurrence of Ei, and Y is the interval up to the occurrence of Ei+1 measured from ti+γ, ,i.e Y= ti+1 – (ti+ γ), Y is called random modification of X or residual time of X, it follows that:

Another Important Property if X is exponential then its random modification Y has also exponential distribution with the same mean. For a Poisson process with parameter λ, the interval upto the occurrence of the next event measured from any start of time (not necessarily from the instant of the previous occurrence) is independent of the elapsed time (since the previous instant of occurrence) and is a random variable having exponential distribution with mean 1/ λ.

Example Suppose that the random variable N(t) denotes the number of fish caught by an angler in (0,t). Under certain ideal conditions such as: The number of fish available is very large The angler stands in no better chance of catching fish than others The number of fish likely to nibble at one particular instant in the same as at another instant , The process {N(t), t≥0} may be considered as a Poisson process.

Example The interval upto the first catch, as also the interval between two successive catches has the same exponential distribution. So olso is the time interval upto the next catch (from as arbitrary instant γ) which is independent of the elapsed time since the last catch to that instant γ. The long time spent since the last catch gives “no premium for waiting "so far as the nest catch is concerned.

Example Suppose that E and F occur independently and in accordance with Poisson process with parameters a and b respectively.

Questions?