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Unformatted text preview: ≤ t + b. Thus, applying (3.14) to the truncated
process,
hi
hi
h
i
e
ee
ee
E X (m(t) + 1) ≤ E X (m(t) + 1) = E SN (t)+1 ≤ t + b.
Combining this equation with (3.15), we have upper and lower bounds on m(t),
1
1
m(t)
1
b
1
−≤
≤ h i+ h i− .
E [X ]
t
t
t
e
e
EX
tE X (3.16) hi R
√
b
e
Finally, choose b = t. Then as t → 1, b → 1 also, so that E X = 0 [1−FX (x)] dx→E [X ].
Also, both of the ﬁnal terms on the right of (3.16) approach 0, completing the proof.
Note that this theorem (and its proof ) have not assumed ﬁnite variance. The theorem also
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holds when E [X ] is inﬁnite; E X in (3.16) simply approaches 1 in this case. We have just shown that m(t) has the property that limt→1 [m(t)/t] = 1/E [X ]. Note again
that N [t, ω ]/t is the average number of renewals from 0 to t for a sample function ω , and
m(t)/t is the average of this over ω . Combining with Theorem 3.1, the limiting time and
ensembleaverage equals the timeaverage renewal rate for each sample function except for
a set of probability 0. Another interesting question is to determine the expected renewal
rate in the limit of large t without averaging from 0 to t. That is, are there some values of t
at which renewals are more likely than others for large t? If the interrenewal intervals have
an integer distribution function (i.e., each interrenewal interval must last for an integer
number of time units), then each renewal epoch Sn must also be an integer. This means
that N (t) can increase only at integer times and the expected rate of renewals is zero at all
noninteger times.
An obvious generalization of this behavior for integer valued interrenewal intervals is that
of interrenewals that occur only at integer multiples of some real number d > 0. Such
a distribution is called an arithmetic distribution. The span of an arithmetic distribution
is the largest number d such that this property holds. Thus, for example if X takes on
only the values 0, 2, and 6, its distribution is arithmetic with span 2. Similarly, X takes
on only the values 1/3 and 1/5, then the span is 1/15. The remarkable thing, for our
purposes, is that any interrenewal distribution that is not an arithmetic distribution leads
to a uniform expected rate of renewals in the limit of large t. This result is contained
in Blackwell’s theorem which we state without proof (see Section 11.1, Theorem 1 of [9]).
Recall, however, that for the special case of an interrenewal density that has a rational
Laplace transform, Blackwell’s theorem is a simple consequence of (3.9).
5 This proof can be postponed and read later. 3.4. RENEWALREWARD PROCESSES; TIMEAVERAGES 105 Theorem 3.5 (Blackwell). If a renewal process has an interrenewal distribution that is
nonarithmetic, then for any δ > 0,
lim [m(t + δ ) − m(t)] = t→1 δ
.
E [X ] (3.17) If the interrenewal distribution is arithmetic with span d, then for any integer n ≥ 1
lim [m(t + nd) − m(t)] = t→1 nd
.
E [X ] (3.18) Eq. (3.17) says that for nonarithmetic distributions, the expected number of arrivals in
the interval (t, t + δ ] is equal to δ /E [X ] in the limit t → 1. Since the theorem is true
for arbitrarily small δ , the theorem almost, but not quite, says that dm(t)/dt is equal to
1/E [X ] in the limit t → 1. Unfortunately this latter statement is not true, and one can
see the reason by looking at an example where X can take on only the values 1 and π .
Then no matter how large t is, N (t) can only increase at discrete points of time of the form
k + j π where k and j are nonnegative integers; thus dm(t)/dt is either 0 or 1 for all t.
As t gets larger, however, the jumps in m(t) become both smaller in magnitude and more
closely spaced from one to the next. Thus [m(t + δ ) − m(t)]/δ approaches 1/E [X ] as t → 1
for any ﬁxed δ , no matter how small, but as δ gets smaller, the convergence in t gets slower.
No matter how large one chooses t, [m(t + δ ) − m(t)]/δ does not approach 1/E [X ] as δ → 0.
Since the interrenewal intervals are positive random variables, multiple renewals cannot
occur simultaneously, and thus, as shown in Exercise 3.15, (3.17) implies that the probability
of a renewal in a small interval (t, t + δ ] tends to δ /E [X ] + o(δ ) as t → 1. Thus, for a
nonarithmetic interrenewal distribution, the limiting distribution of renewals in a small
interval (t, t + δ ] satisﬁes
lim Pr {N (t + δ ) − N (t) = 0} = 1 − δ /X + o(δ ) t→1 lim Pr {N (t + δ ) − N (t) = 1} = δ /X + o(δ ) t→1 lim Pr {N (t + δ ) − N (t) ≥ 2} = o(δ ). (3.19) t→1 If we compare this with Eq. (2.17), associating the rate ∏ of a Poisson process with 1/X , we
see that, asymptotically, a renewal process with a nonarithmetic interrenewal distribution
satisﬁes two of the three requirements in deﬁnition 3 of a Poisson process. That is, the
increments are asymptotically stationary and the renewals do not occur simultaneously. If
the renewal process is not Poisson, however, the increments are not independent.
For an arithmetic renewal process with span d, (3.18), with n = 1, states that the p...
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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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