Unformatted text preview: ) − N (t)]
1
=
.
δ
X2 (3.74) If {Xi ; i ≥ 2} are arithmetic with span d and X1 is arithmetic with span md for some
positive integer m, then
lim Pr {renewal at t = id} = i→1 3.7.1 d
.
X2 (3.75) Delayed renewalreward processes We have seen that the distribution of the ﬁrst renewal epoch has no eﬀect on the time
or ensembleaverage behavior of a renewal process (other than the ensemble dependence
on time for an arithmetic process). This carries over to reward functions with almost no
change. In particular, the extended version of Theorem 3.6 is as follows: 3.7. DELAYED RENEWAL PROCESSES 125 Theorem 3.12. Let {N (t); t≥0} be a delayed renewal counting process, let Z (t) = t−SN (t) ,
let X (t) = SN (t)+1 − SN (t) , and let R(t) = R(Z (t), X (t)) be a reward function. Assume that
Z1 Zx
E [Rn ] =
R(z , x) dz dFXn (x) < 1 for al l n.
x=0 z =0 Then, with probability one,
1
lim
t→1 t Z t R(τ )dτ = τ =0 E [Rn ]
for n ≥ 2.
X2 (3.76) We omit the proof of this since it is a minor variation of that of theorem 3.6. Finally, since
Blackwell’s theorem holds for delayed renewal processes, Eq. (3.54), giving the ensembleaverage reward for nonarithmetic processes, follows as before, yielding
lim E [R(t)] = t→1 3.7.2 E [Rn ]
.
X2 (3.77) Transient behavior of delayed renewal processes Let m(t) = E [N (t)] for a delayed renewal process. As in (3.5), we have
m(t) = 1
X n=1 Pr {N (t) ≥ n} = 1
X n=1 Pr {Sn ≤ t} . (3.78) For n ≥ 2, Sn = Sn−1 + Xn where Xn and Sn−1 are independent. From the convolution
equation (1.12),
Zt
Pr {Sn ≤ t} =
Pr {Sn−1 ≤ t − x} dF (x) for n ≥ 2.
(3.79)
x=0 For n = 1, Pr {Sn ≤ t} = G(t). Substituting this in (3.78) and interchanging the order of
integration and summation,
m(t) = G(t) +
= G(t) +
= G(t) + Z t 1
X x=0 n=2
Zt X
1
x=0 n=1
Zt
x=0 Pr {Sn−1 ≤ t − x} dF (x)
Pr {Sn ≤ t − x} dF (x) m(t − x)dF (x) ; t ≥ 0. (3.80) This is called the renewal equation and is a generalization of (3.6). [9], Section 11.1, Theorem
3.1, shows that it has a unique solution.
There is another useful integral equation very similar to (3.80) that arises from breaking up
b
b
Sn as the sum of X1 and Sn−1 where Sn−1 = X2 + · · · + Xn . Letting m(t) be the expected
b
number of renewals in time t for an ordinary renewal process with interarrival distribution F , 126 CHAPTER 3. RENEWAL PROCESSES a similar argument to that above, starting with Pr {Sn ≤ t} =
yields
m(t) = G(t) + Z n
o
b
Pr Sn−1 ≤ t − x dG(x)
0 Rt t x=0 m(t − x)dG(x).
b (3.81) This equation brings out the eﬀect of the initial renewal interval clearly, and is useful in
computation if one already knows m(t).
b Frequently, the most convenient way of dealing with m(t) is through transforms. Following
the same argument as that in (3.7), we get Lm (r) = (1/r)LG (r) + Lm (r)LF (r). Solving, we
get
Lm (r) = LG (r)
.
r[1 − LF (r)] (3.82) We can ﬁnd m(t) from (3.82) by ﬁnding the inverse Laplace transform, using the same
procedure as in Example 3.3.1. There is a second order pole at r = 0 again, and, evaluating
the residue, it is 1/L0 (0) = 1/X 2 , which is not surprising in terms of Blackwell’s theorem.
F
We can also expand numerator and denominator of (3.82) in a power series, as in (3.8).
The inverse transform, corresponding to (3.9), is
£ 2§
E X2
t
X1
m(t) =
+
−
+ ε(t) for t → 0.
X2
2X 2
X2 (3.83) where limt→1 ε(t) = 0. 3.7.3 The equilibrium process Consider an ordinary nonarithmetic renewal process with an interrenewal interval X of
distribution F (x). We have seen that Rthe distribution of the interval from t to the next
y
renewal approaches FY (y ) = (1/E [X ]) 0 [1 − F (x)]dx as t → 1. This suggests that if we
look at this renewal process starting at some very large t, we should see a delayed renewal
process for which the distribution G(x) of the ﬁrst renewal is equal to the residual life
distribution FY (x) above and subsequent interrenewal intervals should have the original
distribution F (x) above. Thus it appears that such a delayed renewal process is the same as
the original ordinary renewal process, except that it starts in “steadystate.” To verify this,
we show that m(t) = t/X 2 is a solution to (3.80) if G(t) = FY (t). Substituting (t − x)/X 2
for m(t − x), the right hand side of (3.80) is
Rt
0 [1 − F (x)]dx
X2 + Rt 0 (t − x)dF (x)
X2 = Rt
0 [1 − F (x)]dx
X2 + Rt
0 F (x)dx
X2 = t
.
X2 where we have used integration by parts for the ﬁrst equality. This particular delayed
renewal process is called the equilibrium process, since it starts oﬀ in steady state, and thus
one need not worry about transients. 3.8. SUMMARY 3.8 127 Summary Sections 3.1 to 3.3 give the central results about renewal processes that form the basis
for many of the subsequent chapters. The chapter starts with the strong law for renewal
processes, showing that the time average rate of renewals, N (t)/t, approaches 1/X with
probability 1 as t → 1. This, combined with the strong law of large numbers in Chapter
1, is the basis for most subs...
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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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