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Poisson Process

Review Session 2EE384X

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Point Processes

Supermarket model : customers arrive (randomly), get served, leave the store

Need to model the arrival and departure processes

Server Queue

Arrival Process Departure Process

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What does Poisson Process model?

Start time of Phone calls in Palo Alto Session initiation times (ftp/web

servers) Number of radioactive emissions (or

photons) Fusing of light bulbs, number of

accidents Historically, used to model packets

(massages) arriving at a network switch (In Kleinrock’s PhD thesis, MIT 1964)

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Properties

Say there has been 100 calls in an hour in Palo Alto

We expect that : The start time of each call is independent of the others The start time of each call is uniformly distributed over

the one hour The probability of getting two calls at exactly the same

time is zero

Poisson Process has the above properties

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Notation

0

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Notation

A(t) : Number of points in (0,t] A(s,t) : Number of points in (s,t]

Arrival points : Inter-arrival times:

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A(0)=0 and each jump is of unit magnitude Number of arrivals in disjoint intervals are

independent For any

the random variables are independent.

Number of arrivals in any interval of

length is distributed Poisson()

Poisson Process- Definition

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Basic Properties

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Stationary Increments

The number of arrivals in (t,t+] does not depend on t

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Orderliness

The probability of two or more arrivals in an interval of length gets small as

Arrivals occur “one at a time”

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Poisson Rate

Probability of one arrival in a short interval is (approx) proportional to interval length

Poisson process is like a continuous version of Bernoulli IID

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Additional Properties

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Inter-arrival times

Inter-arrival times are Exponential() and independent of each other

0

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Points to the left and right is a fixed point closest point to the right (left) of

Apparent Paradox: Inter-arrival = sum two exp (why?)

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Merging two Poisson processes

Merging two independent Poisson processes with rates 1 and 2 creates a Poisson process with rate 1+2

A(0)=A1(0)+A2(0)=0 Number of arrivals in disjoint intervals are independent Sum of two independent Poisson rv is Poisson

merge

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Sum of two Poisson rv

Characteristic function:

So

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Splitting a Poisson process

For each point toss a coin (with bias p): With probability p the point goes to A1(t)

With probability 1-p the point goes to A2(t)

A1(t) and A2(t) are two independent Poisson processes with rates

Split

:Poisson process with rate

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proof

Define A1(t) and A2(t) such that: A1(0)=0 A2(0)=0

Number of points in disjoint intervals are independent for A1(t) and A2(t) They depend on number of points in disjoint intervals of A(t)

Need to show that number of points of A1 and A2 in an interval of size are independent Poisson(1) and Poisson(2)

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Dividing a Poisson rv

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Dividing a Poisson rv (cont)

So

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Uniformity of arrival times

Given that there are n points in [0,t], the unordered arrival times are uniformly distributed and independent of each other.

0

Ordered variables

Unordered variables

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Single arrival case

0

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General case

It is the n order statistics of uniform distribution.