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Unformatted text preview: in t, and thus has either a ﬁnite or inﬁnite limit as t → 1.
For each n, Pr {N (t) ≥ n} ≥ 1/2 for large enough t, and therefore E [N (t)] ≥ n/2 for all
such t. Thus E [N (t)] can have no ﬁnite limit, and limt→1 E [N (t)] = 1.
For any given t > 0, the random variable N (t) is the number of renewal epochs in the
interval (0, t]. The random variable SN (t) is then the epoch at which renewal N (t) occurs,
i.e., the latest renewal epoch before or equal to time t. Similarly SN (t)+1 is the ﬁrst arrival
epoch after time t (see Figure 3.2). Thus we have the inequalities
SN (t)
SN (t)+1
t
≤
<
.
N (t)
N (t)
N (t) (3.1) From lemma 3.1, limt→1 N (t) = 1 with probability 1. Assuming that X < 1, the strong
law of large numbers (Theorem 1.5) asserts that limn→1 Sn /n = X with probability 1. For
any sample function (i.e., sample point ω ), SN (t) (ω )/N (t, ω ) runs through the same sequence
of values with increasing t as Sn (ω )/n runs through with increasing n. Thus letting ≠0 be
the set of sample points of ω for which both limn→1 Sn /n = X and limt→1 N (t) = 1, we 3.2. STRONG LAW OF LARGE NUMBERS FOR RENEWAL PROCESSES 95 have limt→1 SN (t) /N (t) = X for all sample points in ≠0 . In the same way
lim t→1 SN (t)+1
SN (t)+1 N (t) + 1
= lim
=X
t→1 N (t) + 1
N (t)
N (t)
Slope =
Slope = N (t)
t for all ω ∈ ≠0 . (3.2) N (t)
SN (t) ✘✘✘
❅
❄ ✘✘✘
SN (t)+1
❅ ✘✘✘
❘
❅
❆
✘✘✘
❆ ✘✘✘ ❆
✘✘
t
✘✘✘ N (t) 0 SN (t)+1 Slope = N (t) S1 SN (t) Figure 3.2: Comparison of N (t)/t with N (t)
SN (t) and N (t)
SN (t)+1 . Since t/N (t) in (3.1) lies between two random variables both converging to X for all sample
points in ≠0 , we see that limt→1 t/N (t) = X for all sample functions in ≠0 , i.e., with
probability 1. Since X must be greater than 0, it follows that limt→1 N (t)/t = 1/X for all
sample points in ≠0 . This proves the following strong law for renewal processes.
Theorem 3.1 (Strong Law for Renewal Processes). For a renewal process with mean
interrenewal interval X , limt→1 N (t)/t = 1/X with probability 1.
This theorem is also true if the mean interrenewal interval is inﬁnite; this can be seen by a
truncation argument (see Exercise 3.3). We could also prove a™weak law for N (t) (i.e., we
©
could show that for any ≤ > 0, limt→1 Pr N (t)/t − 1/X  ≥ ≤ = 0). This could be done
by using the weak law of large numbers for Sn (Theorem 1.3) and the fact that the event
Sn ≤ t is the same as N (t) ≥ n. Such a derivation is tedious, however, and illustrates that
the strong law of large numbers is often much easier to work with than the weak law. We
shall not derive the weak law here, since the strong law for renewal processes implies the
weak law and it is the strong law that is most often useful.
Figure 3.3 helps give some appreciation of what the strong law for N (t) says and doesn’t say.
The strong law deals with timeaverages, limt→1 N (t, ω )/t, for individual sample points ω ;
these are indicated in the ﬁgure as horizontal averages, one for each ω . It is also of interest
to look at time and ensembleaverages, E [N (t)/t], shown in the ﬁgure as vertical averages.
N (t, ω )/t is the timeaverage number of renewals from 0 to t, and E [N (t)/t] averages also
over the ensemble. Finally, to focus on arrivals in the vicinity of a particular time t, it is of
interest to look at the ensembleaverage E [N (t + δ ) − N (t)] /δ .
Given the strong law for N (t), one would hypothesize that E [N (t)/t] approaches 1/X as
t → 1. One might also hypothesize that limt→1 E [N (t + δ ) − N (t)] /δ = 1/X , sub ject
to some minor restrictions on δ . These hypotheses are correct and are discussed in detail
in what follows. This equality of timeaverages and limiting ensembleaverages for renewal
processes carries over to a large number of stochastic processes, and forms the basis of
ergodic theory. These results are important for both theoretical and practical purposes.
It is sometimes easy to ﬁnd time averages (just like it was easy to ﬁnd the timeaverage 96 CHAPTER 3. RENEWAL PROCESSES N (t, ω )/t from the strong law of large numbers), and it is sometimes easy to ﬁnd limiting
ensembleaverages. Being able to equate the two then allows us to alternate at will between
time and ensembleaverages.
Ensemble Average at t Time and ensemble
Average over (0, τ ) (1/δ )E [N (t + δ ) − N (t)] N (t,ω3 )
t ✲ Time Ave.
at ω1 0
N (t,ω2 )
t ✲ Time Ave.
at ω2 0
N (t,ω1 )
t 0 ✲ Time Ave.
at ω3 τ t Figure 3.3: Time average at a sample point, time and ensemble average from 0 to a
given τ , and the ensembleaverage in an interval (t, t + δ ]. Note that in order to equate timeaverages and limiting ensembleaverages, quite a few
conditions are required. First, the time average must exist in the limit t → 1 with probability one and have a ﬁxed value with probability one; second, the ensembleaverage...
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This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.
 Spring '09
 R.Srikant

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