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Example 3.4.2. (timeaverage Age) Let Z (t) be the age of a renewal process at time
t where age is deﬁned as the interval from the most recent arrival before (or at) t until t, 108 CHAPTER 3. RENEWAL PROCESSES
1/ε ❅
❅
❅
❅
❅Y (t)
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅
❅✲
1/ε
❅❅❅❅❅❅❅❅❅✛
❅❅❅❅❅❅ Figure 3.7: Average Residual life is dominated by large interarrival intervals. Note the
large intervals of time (within the large triangles) over which Y (t) is large, and small
aggregate intervals over which it is small. i.e., Z (t) = t − SN (t) (see Figure 3.8). We notice that the age process, for a given sample
function of the renewal process, is almost the same as the residual life process—the isosceles
right triangles are simply turned around. Thus the same analysis as before can be used to
show that the time average of Z (t) is the same as the timeaverage of the residual life,
£§
Rt
Z (τ ) dτ
E X2
lim τ =0
=
W.P.1.
(3.25)
t→1
t
2E [X ]
°
° Z (t)° °
°
°
°
°
°
°
°
°
°°
°
°
°
t
°°
°°
° S1 S2 S3 S4 °
°
° S5 S6 Figure 3.8: Age at time t: For any given sample function of the renewal process, the
sample function of age increases linearly with a slope of 1 from the beginning to the
end of each interrenewal interval. Example 3.4.3. (timeaverage Duration) Let X (t) be the duration of the interrenewal
interval containing time t, i.e., X (t) = XN (t)+1 = SN (t)+1 − SN (t) (see Figure 3.9). It is
clear that X (t) = Z (t) + Y (t), and thus the timeaverage of the duration is given by
£§
Rt
E X2
τ =0 X (τ ) dτ
lim
=
W.P.1.
(3.26)
t→1
t
E [X ]
Again, long intervals are heavily weighted in this average, so that the timeaverage duration
is at least as large as the mean interrenewal interval and often much larger. 3.4.1 General renewalreward processes In each of these examples, and in many other situations, we have a random function of
time (i.e., Y (t), Z (t), or X (t)) whose value at time t depends only on where t is in the 3.4. RENEWALREWARD PROCESSES; TIMEAVERAGES 109 X (t) ✛ X5 ✲ t
S1 S2 S3 S4 S5 S6 Figure 3.9: Duration X (t) = XN (t) of the interrenewal interval containing t.
current interrenewal interval (i.e., on the age Z (t) and on the duration X (t) of the current
interrenewal interval). We now investigate the general class of reward functions for which
the reward function at time t depends only on the age and the duration at t, i.e., the reward
R(t) at time t is given explicitly as a function R(Z (t), X (t)) of the age and duration at t.
For the three examples above, the function R is trivial. That is, the residual life, Y (t), is
given by X (t) − Z (t) and the age and duration are given directly.
Rt
We now ﬁnd the timeaverage value of R(t), namely, limt→1 1 0 R(τ ) dτ . As in examples
t
3.4.1 to 3.4.3 above, we ﬁrst want to look at the accumulated reward over each interrenewal
period separately. Deﬁne Rn as the accumulated reward in the nth renewal interval,
Z Sn
Z Sn
Rn =
R(τ ) d(τ ) =
R[Z (τ ), X (τ )] dτ .
(3.27)
Sn−1 Sn−1 For residual life (see Example 3.4.1), Rn is the area of the nth isosceles right triangle in
Figure 3.6. In general, since Z (τ ) = τ − Sn−1 ,
Z Sn
Z Xn
Rn =
R(τ −Sn−1 , Xn ) dτ =
R(z , Xn ) dz .
(3.28)
Sn−1 z =0 Note that Rn depends only on the value of Xn and the form of the function R(Z, X ). From
this, it is clear that {Rn ; n ≥ 1} is essentially7 a set of IID random variables. For residual
2
life, R(z , Xn ) = Xn − z , so the integral in (3.28) is Xn /2, as calculated by inspection before.
In general, from (3.28), the expected value of Rn is given by
Z1 Zx
E [Rn ] =
R(z , x) dz dFX (x).
(3.29)
x=0 z =0 Rt Breaking 0 R(τ ) dτ into the reward over the successive renewal periods, we get
Zt
Z S1
Z S2
Z SN (t)
Zt
R(τ ) dτ =
R(τ ) dτ +
R(τ ) dτ + · · · +
R(τ ) dτ +
R(τ )dτ
0 0 S1 N (t) = X n=1
7 Rn + Z SN (t)−1 SN (t) t R(τ ) dτ . (3.30) SN (t) One can certainly deﬁne functions R(Z, X ) for which the integral in (3.28) is inﬁnite or undeﬁned for
some values of Xn , and thus Rn becomes a defective rv. It seems better to handle this type of situation
when it arises rather than handling it in general. 110 CHAPTER 3. RENEWAL PROCESSES The following theorem now generalizes the results of Examples 3.4.1, 3.4.2, and 3.4.3 to
general renewalreward functions.
Theorem 3.6. Let {R(t); t > 0} ≥ 0 be a nonnegative renewalreward function for a
renewal process with expected interrenewal time E [X ] = X . If X < 1 or E [Rn ] < 1, then
with probability 1
Z
1t
E [Rn ]
lim
R(τ ) dτ =
.
(3.31)
t→1 t τ =0
X
Proof: Using (3.30), the accumulated reward up to time t can be bounded between the
accumulated reward up to the renewal before t and that to the next renewal after t,
Rt
PN (t)
PN (t)+1
Rn
n=1 Rn
τ =0 R(τ ) dτ
≤
≤ n=1
.
(3.32)
t
t
t
The left hand side of (3.32) can now be broken into
PN (t)
n=1 Rn t = PN (t) Rn N (t)
.
N (t)
t n=1 (3.3...
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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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