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5. Stochastic Processes (1)

5. Stochastic Processes (1)

5. processes (1)

Contents

• Basic concepts • Poisson process

5. Stochastic processes (1)

lect05.ppt S-38.1145 - Introduction to Teletraffic Theory – Spring 2006 1 2

5. Stochastic processes (1) 5. Stochastic processes (1)

Stochastic processes (1) Stochastic processes (2)

• Consider some quantity in a teletraffic (or any) system • Definition: A (real-valued) X = (Xt | t ∈ I) is a • It typically evolves in time randomly collection of random variables Xt

– Example 1: the number of occupied channels in a telephone link – taking values in some (real-valued) set S, Xt(ω) ∈ S, and at time t or at the arrival time of the nth customer – indexed by a real-valued (time) parameter t ∈ I. – Example 2: the number of packets in the buffer of a statistical multiplexer • Stochastic processes are also called random processes at time t or at the arrival time of the nth customer (or just processes) • This kind of evolution is described by a stochastic process • The index set I ⊂ℜis called the parameter of the process – At any individual time t (or n) the system can be described by a • The value set S ⊂ℜis called the of the process – Thus, the stochastic process is a collection of random variables • Note: Sometimes notation Xt is used to refer to the whole stochastic process (instead of a single random variable related to the time t)

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Stochastic processes (3) Summary

• Each (individual) random variable Xt is a mapping from the sample • Given the sample point ω ∈ Ω Ω ℜ space into the real values : – X(ω) = (Xt(ω) | t ∈ I) is a real-valued (of t ∈ I) t ∈ I Xt : Ω → ℜ, ω a Xt (ω) • Given the time index , – Xt = (Xt(ω) | ω ∈ Ω) is a random variable (as ω ∈ Ω) X • Thus, a stochastic process can be seen as a mapping from the • Given the sample point ω ∈ Ω and the time index t ∈ I, I Ω into the set of real-valued functions ℜ (with t ∈ I as – Xt(ω) is a real value an argument): X : Ω → ℜI , ω a X (ω)

• Each sample point ω ∈ Ω is associated with a real-valued function X(ω). Function X(ω) is called a realization (or a path or a trajectory) of the process.

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Example Traffic process

• Consider traffic process X = (Xt | t ∈ [0,T]) in a link between two channel-by-channel call holding telephone exchanges during some time [0,T] occupation time 6 X t 5 – t denotes the number of occupied channels at time 4 3 • Sample point ω ∈ Ω tells us 2 channels 1 – what is the number X0 of occupied channels at time 0, – what are the remaining holding times of the calls going on at time 0, time call arrival times – at what times new calls arrive, and blocked call – what are the holding times of these new calls. nr of channels occupied • From this , it is possible to construct the realization X(ω) of 6 5 the traffic process X 4 3 – Note that all the in the process is included in the sample point ω 2 1

– Given the sample point, the realization of the process is just a (deterministic) channels of nr 0 function of time time

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Categories of stochastic processes Examples

• Reminder: • Discrete-time, discrete-state processes – Parameter space: set I of indices t ∈ I – Example 1: the number of occupied channels in a telephone link th – State space: set S of values Xt(ω) ∈ S at the arrival time of the n customer, n = 1,2,... • Categories: – Example 2: the number of packets in the buffer of a router output link th – Based on the parameter space: at the arrival time of the n customer, n = 1,2,... • Discrete-time processes: parameter space discrete • Continuous-time, discrete-state processes Continuous-time processes • : parameter space continuous – Example 3: the number of occupied channels in a telephone link – Based on the state space: at time t > 0 Discrete-state processes • : state space discrete – Example 4: the number of packets in the buffer of router output link Continuous-state processes • : state space continuous at time t > 0 • In this course we will concentrate on the discrete-state processes (with either a discrete or a continuous parameter space (time)) – Typical processes describe the number of customers in a queueing system (the state space being thus S = {0,1,2,...})

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Notation Distribution

• For a discrete-time process, • The stochastic characterization of a stochastic process X is made – the parameter space is typically the set of positive , I = {1,2,…} by giving all possible finite-dimensional distributions t n X , X (ω) – Index is then (often) replaced by : n n P{X ≤ x ,K, X ≤ x } t1 1 tn n • For a continuous-time process, where t1,…, tn ∈ I, x1,…, xn ∈ S and n = 1,2,... I = [0, T] – the parameter space is typically either a finite interval, , or all non- • In general, this is not an easy task because of dependencies between negative real values, I = [0, ∞) the random variables Xt (with different values of time t) – In this case, index t is (often) written not as a subscript but in parentheses: X(t), X (t;ω) • For discrete-state processes it is sufficient to consider of the form P{X = x ,K, X = x } t1 1 tn n – cf. discrete distributions

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Dependence Stationarity

• The most simple (but not so interesting) example of a stochastic • Definition: Stochastic process X is stationary if all finite-dimensional process is such that all the random variables Xt are independent of distributions are invariant to time shifts, that is: each other. In this case P{X ≤ x ,K, X ≤ x } = P{X ≤ x ,K, X ≤ x } P{X ≤ x ,..., X ≤ x } = P{X ≤ x }LP{X ≤ x } t1+∆ 1 tn +∆ n t1 1 tn n t1 1 tn n t1 1 tn n for all ∆, n, t ,…, t and x ,…, x • The most simple non-trivial example is a discrete state Markov 1 n 1 n process. In this case • It follows (by choosing n = 1) that all (individual) random variables Xt of t ∈ I P{X = x ,..., X = x } = a are identically distributed, i.e. for all t1 1 tn n P{X ≤ x} = F(x) P{X = x }⋅ P{X = x | X = x }LP{X = x | X = x } t t1 1 t2 2 t1 1 tn n tn−1 n−1 • This is related to the so called : This is called the stationary distribution of the process. – Given the current state (of the process), the (of the process) does not depend on the past (of the process), i.e. how the process has arrived to the current state 13 14

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Stochastic processes in teletraffic theory Arrival process

• In this course (and, more generally, in teletraffic theory) various • An arrival process can be described as th stochastic processes are needed to describe –a (τn | n = 1,2,...) where τn tells the arrival time of the n – the arrivals of customers to the system (arrival process) customer (discrete-time, continuous-state) state process – the state of the system ( ) • non-decreasing: τn+1 ≥τn kaikilla n (thus non-stationary!)

• Note that the latter is also often called as traffic process • typically it is assumed that the interarrival times τn −τn-1 are independent and identically distributed (IID) ⇒ renewal process • then it is sufficient to specify the interarrival time distribution • exponential IID interarrival times ⇒ Poisson process –a counter process (A(t) | t ≥ 0) where A(t) tells the number of arrivals up to time t (continuous-time, discrete-state) • non-decreasing: A(t+∆) ≥ A(t) for all t,∆≥0 (thus non-stationary!) • independent and identically distributed (IID) increments A(t+∆) − A(t) with ⇒ Poisson process

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State process Contents

• In simple cases • Basic concepts – the state of the system is described just by an • Poisson process • e.g. the number X(t) of calls or packets at time t – This yields a state process that is continuous-time and discrete-state • In more complicated cases, – the state process is e.g. a vector of integers (cf. loss and queueing network models) • Typically we are interested in – whether the state process has a stationary distribution – if so, what it is? • Although the state of the system did not follow the stationary distribution at time 0, in many cases state distribution approaches the stationary distribution as t tends to ∞

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Bernoulli process Definition of a Poisson process

• Definition: with success p is an infinite • Poisson process is the continuous-time counterpart of a Bernoulli process series (Xn | n = 1,2,...) of independent and identical random (τ | n = 1,2,...) τ experiments of Bernoulli type with joint success probability p – It is a point process n where n tells tells the occurrence time of the nth event, (e.g. arrival of a client) • Bernoulli process is clearly discrete-time and discrete-state – “failure” in Bernoulli process is now an arrival of a client – Parameter space: I = {1,2,…} • Definition 1: A point process (τn | n = 1,2,...) is a Poisson process with – State space: S = {0,1} intensity λ if the probability that there is an event during a short time (t, t+h] λh + o(h) • Finite dimensional distributions (note: Xn’s are IID): interval is independently of the other time intervals – o(h) o(h)/h → 0 h → 0 P{X = x ,..., X = x } = P{X = x }LP{X = x } refers to any function such that as 1 1 n n 1 1 n n – new events happen with a constant intensity λ: (λh + o(h))/h →λ n – probability that there are no arrivals in (t, t+h] is 1 −λh + o(h) xi 1−xi ∑i xi n−∑i xi = ∏ p (1− p) = p (1− p) • Defined as a point process, Poisson process is discrete-time and i=1 continuous-state – Parameter space: I = {1,2,…} • Bernoulli process is stationary (with Bernoulli(p) as the stationary – State space: S = (0, ∞) distribution) 19 20 5. Stochastic processes (1) 5. Stochastic processes (1)

Poisson process, another definition Poisson process, yet another definition (1)

• Consider the interarrival time τn −τn-1 between two events (τ0 = 0) • Consider finally the number of events A(t) during time interval [0,t] – Since the intensity that something happens remains constant λ, the ending – In a Bernoulli process, the number of successes in a fixed interval would of the interarrival time within a short period of time (t, t+h], after it has follow a . As the “time slice” tends to 0, this approaches lasted already the time t, does not depend on t (or on other previous a Poisson distribution. arrivals) – Note that A(0)=0 – Thus, the interarrival times are independent and, additionally, they have the • Definition 3: A counter process (A(t)| t ≥ 0) is a Poisson process with memoryless property. This property can be only the one of exponential λ distribution (of continuous-time distributions) intensity if its increments in disjoint intervals are independent and follow a Poisson distribution as follows: • Definition 2: A point process (τ | n = 1,2,...) is a Poisson process n A(t + ∆) − A(t) ∼ Poisson(λ∆) with intensity λ if the interarrival times τn −τn−1 are independent and identically distributed (IID) with joint distribution Exp(λ) • Defined as a counter process, Poisson process is continuous-time and discrete-state – Parameter space: I = [0, ∞) – State space: S = {0,1,2,…} 21 22

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Poisson process, yet another definition (2) Three ways to characterize the Poisson process

• One dimensional distribution: A(t) ∼ Poisson(λt) • It is possible to show that all three definitions for a Poisson process are, – E[A(t)] =λt, D2[A(t)] =λt indeed, equivalent • Finite dimensional distributions (due to independence of disjoint intervals): A(t) P{A(t1) = x1,..., A(tn ) = xn} = P{A(t1) = x1}P{A(t2) − A(t1) = x2 − x1}L

P{A(tn ) − A(tn−1) = xn − xn−1} τ4−τ3

• Poisson process, defined as a counter process is not stationary, but it τ τ τ τ has stationary increments 1 2 3 4 – thus, it doesn’t have a stationary distribution, but independent and identically distributed increments no event with prob. 1−λh+o(h) event with prob. λh+o(h)

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Properties (1) Properties (2)

• Property 1 (Sum): Let A1(t) and A2(t) be two independent Poisson • Property 2 (Random sampling): Let τn be a Poisson process with processes with intensities λ1 and λ2. Then the sum (superposition) intensity λ. Denote by σn the point process resulting from a random process A1(t) + A2(t) is a Poisson process with intensity λ1 +λ2. and independent sampling (with probability p) of the points of τn. Then • Proof: Consider a short time interval (t, t+h] σn is a Poisson process with intensity pλ. – Probability that there are no events in the superposition is • Proof: Consider a short time interval (t, t+h] (1− λ1h + o(h))(1− λ2h + o(h)) =1− (λ1 + λ2)h + o(h) – Probability that there are no events after the random sampling is (1− λh + o(h)) + (1− p)(λh + o(h)) =1− pλh + o(h) – On the other hand, the probability that there is exactly one event is – On the other hand, the probability that there is exactly one event is (λ1h + o(h))(1− λ2h + o(h)) + (1− λ1h + o(h))(λ2h + o(h)) p(λh + o(h)) = pλh + o(h) = (λ1 + λ2)h + o(h) λ λ1 pλ λ2 λ +λ 1 2 25 26

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Properties (3) Properties (4)

• Property 3 (Random sorting): Let τn be a Poisson process with • Property 4 (PASTA): Consider any simple (and stable) teletraffic (1) X(t) intensity λ. Denote by σn the point process resulting from a random model with Poisson arrivals. Let denote the state of system at time t Y and independent sampling (with probability p) of the points of τn. (continuous-time process) and n denote the state of the system seen (2) n Denote by σn the point process resulting from the remaining points. by the th arriving customer (discrete-time process). Then the (1) (2) X(t) Then σn and σn are independent Poisson processes with stationary distribution of is the same as the stationary distribution intensities λp and λ(1 − p). of Yn. • Proof: Due to property 2, it is enough to prove that the resulting two • Thus, we can say that processes are independent. Proof will be ignored on this course. – arriving customers see the system in the stationary state – PASTA= “Poisson Arrivals See Time Avarages” λ • PASTA property is only valid for Poisson arrivals λp – and it is not valid for other arrival processes λ(1-p) – consider e.g. your own PC. Whenever you start a new session, the system is idle. In the continuous time, however, the system is not only idle but also busy (when you use it). 27 28 5. Stochastic processes (1) 5. Stochastic processes (1)

Example (1) Example (2)

• Connection requests arrive at a server according to a Poisson process • Consider the system described on the previous slide. λ=5 with intensity requests in a minute. • A new connection request has just arrived at the server. What is the • What is the probability that exactly 2 new requests arrive during the probability that it takes more than 30 seconds before the next request next 30 seconds? arrives? – The number of new arrivals during this time interval follows Poisson λ∆ = (5/60)⋅30 = 2.5 – Consider the process as a point process. The interarrival time follows distribution with the parameter , i.e. exponential distribution with parameter . A(t + 30) − A(t) ∼ Poisson(2.5) P{τi+1 −τi ≥ 30} =1− P{τi+1 −τi ≤ 30} – Thus −5/ 60⋅30 −2.5 2 = e = e = 0.082 P{A(t + 30) − A(t) = 2} = 2.5 e−2.5 = 0.257 2! – Consider the process as a counter process, cf. slide 29. Now we can restate the question above as ”What is the probability that there are no arrivals during 30 seconds?” 0 P{A(t + 30) − A(t) = 0} = 2.5 e−2.5 = e−2.5 = 0.082 0! 29 30

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