1 Stationary point processes

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1 Copyright c 22 by Karl Sigman Stationary point processes We present here a brief introduction to stationay point processes on the real line.. Basic notation for point processes We consider simple point processes, sequences of points on the real line, ψ = {t n : n }; t < t < t 2 <, with n t n =. {N(t) : t } is the counting process; N(t) = the number of points that lie in the interval (, t], and N() def =. Interarrival times are defined by T n = t n+ t n, n, and we can alternatively describe ψ by its interarrival time representation {t, {T n : n }}; it completely determines ψ and visa versa. Now suppose that t =. If t n t < t n+, then N(t) = n, and so in general we can write t N(t) t < t N(t)+ ; t N(t) is the last point before or at time t, and t N(t)+ is the first point strictly after time t. We thus define A(t) def = t N(t)+ t, the forward recurrence time; it is the time until the next point strictly after time t and is also called the excess. Similarly, we define B(t) = t t N(t), the backwards recurrence time; it is the time since the last point before or at time t and is also called the age (If t >, then B(t) def = for t < t ). Finally, S(t) def = B(t) + A(t) = t N(t)+ t N(t) = T N(t) is called the spread, it is the length of the interarrival time covering t. Note that graphing A(t) yields a sequence of right triangles, where at each point t n it jumps with height T n, and linearly decreases at rate reaching at time t n+. B(t) is the mirror image; at time t n it starts at and linearly increases at rate until time t n+ at which time it is of height T n. Finally, graphing S(t) yields squares of height and length T n during t [t n, t n+ ). For any s, ψ s = {t n (s) : n } is the point process shifted into the future to make time s the origin. t (s) = A(s), the excess or forward recurrence time, and t n (s) = t N(s)+n s, n. T n (s) = t n+ (s) t n (s). We can also shift to a particular point, t j, yielding ψ (j) = {t n (t j ) : n } which has its intial point t (t j ) = and t n (t j ) = t j+n t j, n. T n (t j ) = T j+n, n. Thus ψ (j) is the point process with a point at the origin and interarrival times {T j+n : n }..2 Random point processes When the t n are rvs, then the point process is called a random point process; but we will just say it is a point process (pp) for short. We say that two point processes ψ = {t () n } and ψ 2 = {t (2) n } have the same distribution (e.g., are identically distributed) if all the finite joint distributions of {t () n } are the same as for {t (2) n }. For example (t (), t () 5, t () 9 ) must have the same joint distribution as (t (2), t (2) 5, t (2) 9 ); P (t () x, t () 5 x 5, t () 9 x 9 ) = P (t (2) x, t (2) 5 x 5, t (2) 9 x 9 ), for all x, x 5, x 9. Equivalently, all finite joint

2 distributions of the interarrival time representations are the same, {t (), {T n () : n }} and {t (2), {T n (2) : n }}. Definition. ψ is called time stationary if ψ s has the same distribution for all s. ψ is called point stationary if ψ (j) has the same distribution for all j. Thus, under time stationarity, shifting by any s does not change the distribution; it has the same distribution for all s, namely the distribution of ψ itself: ψ d s ψ. Similarly, under point stationarity, shifting to any point, t j, does not change the distribution: ψ d (j) ψ. The following is immediate from the definitions of stationarity: Proposition. A point process ψ = {t n : n } is point stationary iff P (t = ) = and the interarrival time sequence {T n : n } is stationary, whereas it is time stationary iff P (t = ) = and the counting process {N(t) : t } has stationary increments. Note that the Poisson process is point stationary if we place a point t at the origin, whereas it is time stationary if we remove the point from the origin. A renewal process (with iid interarrival times distributed as F ) will be point stationary if the initial delay t =. A renewal process will be time stationary if the delay t is distributed as F e, the equilibrium distribution of F. (This fact will be proved later.).3 Functionals of point processes Let f(ψ) be any non-negative function of ψ. Examples. f(ψ) = t, in which case f(ψ s ) = t (s) = A(s). (Note that f(ψ (j) ) = because ψ (j) has its initial point at the origin by definition). 2. f(ψ) = I{t > x}, in which case f(ψ s ) = I{A(s) > x} 3. f(ψ) = I{t x, t y,..., t k y k }in which case f(ψ s ) = I{A(s) x, t (s) y,..., t k (s) y k }. 4. f(ψ) = T in which case f(ψ (j) ) = T j. 5. f(ψ) = I{T x} in which case f(ψ (j) ) = I{T j x} The indicator examples above all are special cases of f(ψ) = I{ψ E} where E denotes a special set of point processes. For example E = {all point processes : t > x}, 2

3 or or E = {all point processes : t x, t y,..., t k y k }, E = {all point processes : T x}. Taking the expected value of such an indicator yields P (ψ E) and the entire distribution of ψ is determined by considering all such E. We denote the distribution of ψ by P (ψ ) just as, in the case of real-valued random variables X with cdf F (x) = P (X x) we sometimes write F = F ( ). This is done for simplicity and elegance of notation..4 Point and time-stationary versions Given a point process ψ, assuming the its exist, define the distributions of two (new) point processes ψ and ψ as P (ψ ) P (ψ ) def = n n def = t t n P (ψ (j) ), () t P (ψ s )ds. (2) P (ψ ) represents the point-stationary distribution of ψ and P (ψ ) represents the time-stationary distribution of ψ. ψ is called a point-stationary version of ψ and ψ is called a time-stationary version of ψ. We view the distribution P (ψ ) as that obtained by randomly observing the point process way out at a point. By random we mean that we observe at point t J, and consider the shifted pp ψ (J) where J is chosen randomly according to the discrete uniform distribution over the integers {, 2,..., n} and then we let n. J has probability mass function P (J = j) = /n, j {,..., n}, and thus the distribution of ψ (J) is exactly the right-hand-side of () before we take the it. We view the distribution P (ψ ) as that obtained by randomly observing the point process ψ way out in time. By random we mean that we observe at time S, and consider the shifted pp ψ S where S is uniformly distributed over the interval (, t), and then let t. S has density function f(s) = /t, < s < t, and thus the distribution of ψ S is exactly the right-hand-side of (2) before we take the it. Proposition.2 The distribution of ψ is time stationary and the distribution of ψ is point stationary. Proof : We prove the time stationary case, the point stationary case being similar. We will show that ψ u has the same distribution as ψ for any u. Note that P (ψu t ) = P (ψ s+u )ds. t t 3

4 So, for any fixed u, the distribution of ψ u is identical to the distribution obtained by randomly observing ψ u way out in time. But this is the same as observing ψ way out in time as is seen by using the equalities below and letting t : t t t P (ψ s+u )ds = u+t P (ψ s )ds t t u = t u+t P (ψ s )ds t u P (ψ s )ds, as t, since u is fixed, while u+t P (ψ s )ds = ( ) u+t u+t t u+t P (ψ s )ds P (ψ )..5 Examples u P (ψ s )ds.. Renewal Process. Let Ψ = {t n } be a renewal process with interarrival time distribution F (x) def = P (T n x), F () =, and delay t G (independent of the other T n ). Let < λ = {E(T )} <. Clearly, by the i.i.d. structure, Ψ () is a non-delayed version of Ψ, as is Ψ (j) for any j. Thus Ψ becomes point stationary after only shifting to the initial point t (e.g., removing the delay); so construction of a point-stationary version is immediate. We now proceed to the time stationary version. Consider t P (Ψ s E)ds. t t For each s, the distribution of Ψ s is completely determined by the distribution of the forward recurrence time A(s), because by the i.i.d. structure, A(s) is independent of the future interarrival times, {T n (s) : n }, which are i.i.d. F. Letting {X n : n } be i.i.d. F, and independent of Ψ, it follows that for each s, (t (s), T(s)) d (A(s), X, X,..., ), that is, Ψ s is a delayed renewal process with delay t = A(s). As s varies, only the distribution of A(s) is changing; thus, we conclude that the time stationary version is in fact a delayed renewal process with delay t F e, where F e is the distribution obtained by randomly observing A(s) way out in time: t F e (x) = P (A(s) x)ds. t t This distribution is referrred to as the stationary forward recurrence time distribution for the renewal process. By the Renewal Reward Theorem, it can be computed as x F e (x) = λ F (y)dy, where F (x) = F (x) = P (T > x) is the tail of F. 4

5 In general, constructing point and time stationary versions is not as easy as for renewal processes. But a lot can be learned by studying the simple example when the interarrival time sequence is alternating {T n } = {, 2,, 2,, 2,...}. For then it is immediate that Ψ is given by t = and {T n} = {, 2,, 2,..., w.p..5; 2,, 2,,..., w.p..5. Randomly choosing a point t j, we would be equally likely to be at the beginning of a length or a length 2 interarrival time; this yields the.5 mixture for Ψ. For ψ, let U be uniformly distributed on (, ). Then {t, {T n : n }} = { U, 2,, 2..., w.p. /3; 2U,, 2,..., w.p. 2/3. Note how t has a continuous distribution given as a /3, 2/3 mixture of U and 2U. Also observe (verify) that this mixture is in fact the equilibrium distribution, F e, of F where F is distributed as T n: F (x) = P (T n x), P (T n = ) = P (T n = 2) =.5.: If you randomly select a time s way out in the infinite future then you will land within some interarrival time; with probability /3 it will be of length and have a remaining length distributed as U (the equilibrium distribution of ), with probability 2/3 it will be of length 2 and have a remaining length distributed as 2U Unif(, 2) (the equilibrium distribution of 2). 2 From ψ to ψ Our objective here is to derive the inversion formula for expressing the distribution of time stationary (and ergodic) ψ in terms of point stationary (and ergodic) ψ : P (ψ ) = E{ T I{ψ s }ds } E(T ) Analogous to Renewal Reward Theory, this formula says that. (3) A time average can be expressed as an expected value over a cycle divided by an expected cycle length, where now a cycle length is an interarrival time. Cycles from ψ are stationary and thus identically distributed, so we pick the first one (starting from t = ) for simplicity and denote it by T = T in the formula. (Note that T = t t = t.) The inversion formula can be re-written as P (ψ ) = λe { T I{ψs }ds }. 5

6 Recall that by definition, to obtain ψ from ψ, we take ψ and randomly observe it way out in time : P (ψ t ) = P (ψ t s )ds, t so we need to show that the it above yields the right-hand-side of (3) To do so we use functional notation and proceed with a sample-path analysis as follows: t f(ψ t s)ds N(t) t j f(ψ t s)ds, (4) where the actual difference between them is asymptotically negligeable (as t ) if for example f is a bounded function. Defining Y j = t j t j t j f(ψ s)ds, {Y j : j } forms a stationary (and ergodic) sequence because ψ is assumed point stationary (and ergodic). Thus with probability, t f(ψ t t s)ds = ( N (t))( ) N (t) t t N (t) t j t j f(ψ s)ds (5) = λe(y ) (6) = λe { T f(ψs)ds } (7) = E{ T f(ψs)ds } (8) E(T ) by the SLLN for stationary ergodic sequences. If f is a bounded function (such as an indicator which is always bounded by ), we can use the dominated convergence theorem which allows us to take expected values inside (of the left-hand-side of (5)) : t t t Using f(ψ) = I{ψ } then yields (3). 3 From ψ to ψ E{f(ψ s)}ds = E{ T f(ψ s)ds } E(T ) Our objective here is to derive the inversion formula for expressing the distribution of point stationary (and ergodic) ψ in terms of time stationary (and ergodic) ψ :. 6

7 P (ψ ) = E{ N () j= I{ψ(j) }}. (9) E(N ()) Here the interpretation is that A point average can be expressed as the expected value over a unit of time divided by the expected number of points in a unit of time ψ has the same distribution over any one unit of time by time stationarity, so we choose the first unit of time for simplicity in the formula. Recall that by definition, to obtain ψ from ψ, we take ψ and randomly observe it way out at a point : n n n P (ψ(j) ), so we need to show that the it above yields the right-hand-side of (9) We use functional notation and proceed with a sample-path analysis as follows: First observe that by n n n N f(ψ(j)) (t) = t N (t) f(ψ (j)). () Taking the time it along the integer time points n, the it must also be given n N (n) N (n) f(ψ (j)). () But ψ is time stationary; that is, {N (t) : t } has stationary increments. Thus { i : i } forms a stationary (and ergodic) sequence where i = N (i) N (i ), i, the number of points in the unit time interval (i, i], and N () =. So each i has the same distribution as N (), and N (n) = + + n. Let J i = N (i) j=n (i ) f(ψ (j)). Then, again by stationarity of ψ, {J i : i } also form a stationary (and ergodic) sequence with each J i distributed as J (the sum over the first unit of time). Therefore N (n) ni= f(ψ J i N (n) (j)) = ni= (2) i 7 = ni= n J i n ni= i (3)

8 Taking the it as n we use the SLLN for stationary ergodic sequences on both numerator and denominator yielding (with probability one) E(J ) E( ), which is precisely the right-hand-side of (9). Thus we conclude that wp, n n n f(ψ(j)) = E{ N () j= f(ψ(j) )}. E(N ()) If f is bounded, then the dominated convergence theorem yields n n n E{f(ψ(j))} = E{ N () j= f(ψ(j) )}, E(N ()) which when applied to f(ψ) = I{ψ } yields (9). Remark Since E(N ()) = λ the formula can also be expressed as λ E { N () j= 3. Computing expected values f(ψ (j)) }. A more general way of stating the inversion formulas (and a consequence of our method of proof of them) is in terms of functions f = f(ψ): E(f(ψ )) = E{ T f(ψs)ds } (4) E(T ) E(f(ψ )) = E{ N () j= f(ψ(j) )}. (5) E(N ()) For example, when f(ψ) = I{ψ } we get the original formulas. 3.2 Examples Here we explore (4) and see that it is completely analogous to what happens in renewal theory. We will give an application of (5) in Section??. 8

9 . f(ψ) = t, in which case f(ψ s ) = t (s) = A(s) and then (4) yields E(t ) = E{ T A (s)ds } E(T ) = E({T } 2 ) 2E(T ). 2. f(ψ) = I{t > x}, in which case f(ψ s ) = I{A(s) > x} and then (4) yields P (t > x) = E{ T I{A (s) > x}ds } E(T ) = E(T x) + E(T ) = F e (x), the tail of the equilibrium distribution of F (x) = P (T x). Note that the previous example computed the mean of F e. 3. Recall the interevent time representation for ψ, {t, {Tn : n }}. Let us compute the distribution of T = t t the first interarrival time. We use f(ψ) = I{T x} yielding P (T x) = E{ T I{T (s) x}ds }. E(T ) Noting that T (s) = T yields for s [, T ) and hence is constant over the first cycle and finally T I{T (s) x}ds = T I{T x}ds = T I{T x}, P (T x) = λe{t I{T x}}. (6) In general, T and T are dependent rvs so that in general we need to know the joint distribution of them inorder to compute the above expected value. But for a renewal process they are independent in which case E{T I{T x}} = E(T )P (T x) = λ F (x), and so (as we well know) P (T x) = F (x) for a renewal process: For a renewal process all interarrival times are iid distributed as F ; only the delay t is different. Let us apply (6) to the case when {T n : n } = {, 2,, 2,, 2,...}, t =. Since here, the rvs {T n} are discrete, we can consider the probability mass function analog P (T = ) = λe{t I{T = }} 9

10 and P (T = 2) = λe{t I{T = 2}}. Conditioning on the events {T = } and {T = 2} (each with probability.5) yields P (T = ) = 2/3 and P (T = 2) = /3. For example, given T =, T I{T = } =, and given T = 2, T I{T = } = 2. Thus P (T = ) = λ(2)(.5) = λ = 2/3. 4. The above Example can be extended to all finite dimensional distributions. For example we can derive a formula for which the reader can check is given by P (t x, T y,, T k y k ), λe { (T x) + I{T y,..., T k+ y k } }. The beauty of this is that we can completely express all joint distributions of {t, {T n}} in terms of the joint distributions of {T n}.