Let {X (t) : t ≥ 0 } be what is called a Poisson process defined by X (0) = 0 and P [ X (t) – X (s) = k] = e^-b(t-s) b^k (t-s)^k / k! for k = 0,1 ,..., ∞ and 0 ≤ s ≤ t with [X (t) – X (s)] being independent of [X (s) – X (r)] for 0 ≤ r ≤ s ≤ t. Note that X (t) is always integer valued. This

Question

Let [latex] \{X (t) : t \geq 0 \}[/latex] be what is called a Poisson process defined byq1 X (0) = 0 [/latex] and

[Latex] P[X(t)–X(s)= k] \frac{e^{-b(t-s)}b^{k}(t-s)^{k}}{k!} [/latex] for [latex] k=0,1,…, \infty [/latex] and [latex]0 \leq s \leq t[/latex]

with [latex] [X(t) - X(s)][/latex] being independent of [latex] [X(s)- X(r)][/latex] for [latex] 0 \leq r \leq s \leq t [/latex]. Note that [latex] X(t) [/latex] is always integer valued. This distribution arises in many areas of applied probability, with [latex] \{X(t)\}[/latex] representing the number of occurrences (the count) of some event during the time interval [latex] [0,t][/latex]. It corresponds to having the number of occurrences in two nonoverlapping (i.e., disjoint) time intervals be independent of each other and with their being identically distributed if the time intervals are of the same length. The parameter b represents the mean rate of occurrence. Consider the continuity of [latex] \{ X(t)\}[/latex].

Accepted Answer

Note that any time history of this process is a "stairstep" function. It is zero until the time of first occurrence, stays at unity until the time of second occurrence, and so forth, as shown in the accompanying sketch of the kth sample time history. We can write

[latex] X(t)=\sum\limits_{j=1}^{\infty }{U(t-T_{j})} [/latex]

with the random variable [latex] T_{j}[/latex] denoting the time of [latex] j [/latex]th occurrence. Clearly these time histories are not continuous at the times of occurrence. Furthermore, any time [latex] t [/latex] is a time of occurrence for some of the possible time histories, so there is no time at which we can say that all time histories are continuous.

Let us now find the autocorrelation function for [latex] \{X(t)\} [/latex]. For s ≤ t we can define [latex] \Delta X=X(t)-X(s) [/latex] and write

[latex] \phi _{XX}(t,s)=E([X(s)+\Delta X]X(s))=E(X^{2}(s))+\mu _{\Delta X}\mu _{X}(s)[/latex]

in which the independence of [latex] \Delta X [/latex] and [latex] X(s) [/latex] has been used. To proceed further we need the mean value function of [latex] \{X(t)\}[/latex]:

[latex] \mu _{X(t)}=\sum\limits_{k=0}^{\infty }{k^{2}P[X(t)=k]} =\sum\limits_{k=1}^{\infty }{\frac{e^{-bt}(bt)^{k}}{(k-1)!} }=bt [/latex]

Now we use the fact that the distribution of [latex] \Delta X [/latex] is the same as that of [latex] X(t- s) [/latex] to obtain

[latex] \phi _{XX}(t,s)=\sum\limits_{k=0}^{\infty }{k^{2}P[X(s)=k]+b^{2}(t-s)s} =\sum\limits_{k=0}^{\infty }{k^{2}\frac{e^{-bs}(bs)^{k}}{k!}+b^{2}(ts-s^{2}) } = \sum\limits_{k=2}^{\infty }{k^{2}\frac{e^{-bs}(bs)^{k}}{(k-2)!}+\sum\limits_{k=1}^{\infty }{k^{2}\frac{e^{-bs}(bs)^{k}}{(k-1)!}}+b^{2}(ts-s^{2}) } =bs(bt+1)[/latex]

for [latex] s\leq t [/latex]. The general relationship can be written as [latex] \phi_{XX}(t,s) = b^{2}t s + b \min(t,s)[/latex]. It is easily verified that this function is continuous everywhere on the plane of [latex] (t,s)[/latex] values, so [latex] \{X(t)\} [/latex] is mean-square continuous everywhere. This is despite the fact that there is nowhere at which all its time histories are continuous, nor does it have any continuous time histories, except the trivial one corresponding to no arrivals.

Question 4.13: Let {X (t) : t ≥ 0 } be what is called a Poisson process def...

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