Lecture 08: Key Renewal Theorem and Applications
1 Key Renewal Theorem
For eachδ >0 andn∈N, we define intervalsI(n,δ) = [(n−1)δ,nδ)that partition the positive axis R+= [0,∞). Leth:R+7→Rbe a function bounded over finite intervals, denoting
m(h,n,δ) =inf{h(u):u∈I(n,δ)} m(h,n,δ) =sup{h(u):u∈I(n,δ)}.
A functionh:R+7→Risdirectly Riemann integrableand denoted byh∈Dif the partial sums obtained by summing the infimum and supremum ofh, taken over intervals obtained by partitioning the positive axis, are finite and both converge to the same limit, for all finite positive interval lengths. That is,
σδ ,lim
δ→0δ
∑
n∈N
m(h,n,δ) =lim
δ→0δ
∑
n∈N
m(h,n,δ),σδ. If both limits exist and are equal, then the integral value is equal to the limit.
We compare the definitions of directly Riemann integrable and Riemann integrable functions.
For a finite positiveM, a functiong:[0,M]→Ris Riemann integrable if lim
δ→0δ
∑
n≤M/δ
m(g,n,δ) =lim
δ→0δ
∑
n≤M/δ
m(g,n,δ).
In this case, the limit is the value of the integral. Forhdefined onR+, Z
u∈R+
h(u)du= lim
M→∞
Z M 0
h(u)du, if the limit exists. For many functions, this limit may not exist.
A directly Riemann integrable function overR+ is also Riemann integrable, but the converse need not be true. For instance, consider the following Riemann integrable function
h(t) =
∑
n∈N
1
t∈
n− 1
(2n2),n+ 1 (2n2)
is Riemann integrable, butδ∑n∈Nm(h,n,δ)is always infinite for everyδ >0.
Proposition 1.1. Following are sufficient conditions for a function h to be directly Riemann integrable.
(a) If h is non-negative, continuous, and has finite support.
(b) If h is non-negative, continuous, bounded, andσδ is bounded for someδ. (c) If h is non-negative, monotone non-increasing, and Riemann integrable.
(d) If h is non-negative and bounded above by a directly Riemann integrable function.
Proposition 1.2 (Tail Property). If h is non-negative, directly Riemann integrable, and has bounded integral value, then
t→∞limh(t) =0.
Theorem 1.3 (Key Renewal Theorem). Consider a renewal process with renewal function m(t), and the mean and the distribution of inter-renewal times being denoted byµand F respectively. If F is non-lattice and F(∞) =1, then for any h∈D, we have
t→∞lim Z ∞
0
h(t−x)dm(x) = 1 µ
Z ∞
0
h(t)dt. (1)
If F is lattice with period d and∑k∈N0h(t+kd)converges, then
n→∞lim Z ∞
0
h(t+nd−x)dm(x) = d µ
∑
n∈N0
h(t+kd).
Proposition 1.4 (Equivalence). Blackwell’s theorem and key renewal theorem are equivalent.
Proof. Let’s assume key renewal theorem is true. We selecthas a simple function with value unity on interval[0,a]and zero elsewhere. That is,
h(x) =1{x∈[0,a]}. It is easy to see that this function is directly Riemann integrable.
To see how we can prove the key renewal theorem from Blackwell’s theorem, observe from Black- well’s theorem that,
t→∞lim dm(t)
dt
(a)=lim
a→0lim
t→∞
m(t+a)−m(t)
a = 1
µ .
where in(a)we can exchange the order of limits under certain regularity conditions. We defer the formal proof for a later stage.
Key renewal theorem is very useful in computing the limiting value of some functiong(t), probability or expectation of an event at an arbitrary timet, for a renewal process. This value is computed by conditioning on the time of last renewal prior to timet.
2 Alternating renewal processes
Alternating renewal processes form an important class of renewal processes, and model many interesting applications. We find one natural application of key renewal theorem in this section.
Let{(Zn,Yn), n∈N}be aniidrandom process, whereYnandZnare not necessarily independent. A renewal process where each inter-renewal timeTnconsist ofontimeZnfollowed byofftimeYn, is called analternating renewal process. We denote the distributions for on, off, and renewal periods byH,G, andF, respectively. Let
P(t) =Pr{on at timet}.
To see that the alternating renewal process is indeed a renewal process, it needs to be established that {Tn:n∈N}is aniidsequence. However, this follows trivially from the fact that{f(Yn,Zn):n∈N}is an
iidsequence whenever{(Zn,Yn),n∈N}is aniidsequence. Let f(a,b) =a+bto see that{Tn=Yn+Zn: n∈N}is aniidsequence.
For the renewal process with nth inter-renewal time Tn for each n∈N, the nth renewal instant is denoted bySn=∑ni=1Tn. We can define a stochastic process{W(t)∈ {0,1},t>0}that takes values 1 and 0, when the renewal process is in on and off state respectively. In particular, we can write
W(t) =1{A(t)<YN(t)+1}.
It is easy to see thatW is a regenerative process with regenerative sequenceS.
Theorem 2.1 (on probability). Let m be the renewal function associated with the renewal process{Sn: n∈N}with a non-lattice inter-renewal duration distribution F. IfE[Zn+Yn]<∞, then
P(t) =H(t) +¯ Z t
0
H(t¯ −y)dm(y).
Proof. We recall thatP(t) =P{W(t) =1}and compute the kernel
P{W(t) =1,T1>t}=P{Z1>t,T1>t}=P{Z1>t}=H(t).¯ Hence, we can write the renewal equation forP(t)as
P(t) =H(t) + (F¯ ∗P)(t).
Result follows from the solution of renewal equation.
Corollary 2.2 (limiting on probability). IfE[Zn+Yn]<∞and F is non-lattice, then
t→∞limP(t) = E[Zn] E[Yn] +E[Zn].
Proof. SinceHis the distribution function of the non-negative random variableZn, it follows that
t→∞limH(t) =¯ 0, and Z ∞
0
H(t)dt¯ =EZn. Applying key renewal theorem to Theorem 2.1, we get the result.
2.1 Age and excess times
Consider a renewal process with renewal instants{Sn:n∈N}, andiidinter-renewal times{Xn:n∈N} with the common non-lattice distributionF. At timet, the last renewal occurred at timeSN(t), and the next renewal will occur at timeSN(t)+1. Recall that the ageA(t)is the time since the last renewal and the excess timeY(t)is the time till the next renewal. That is,
A(t) =t−SN(t), Y(t) =SN(t)+1−t.
We are interested in finding the limiting distribution of age and excess times. That, is for a fixedx, we wish to compute
t→∞limPr{A(t)≤x}, lim
t→∞Pr{Y(t)≤x}.
Proposition 2.3. Limiting age distribution for a renewal process with non-lattice distribution F is
t→∞limPr{A(t)≤x}= 1 µ
Z x 0
F¯(t)dt.
Proof. We will call this renewal process to be on, when the age is less thanx. That is, we consider an alternative renewal processW such that
W(t) =1{A(t)≤x}.
This is an alternating renewal process with finite probability of off times being zero. Further, we can write thenth on timeZnfor this renewal process as
Zn=min{x,Xn}.
From limiting on probability of alternating renewal process, we get
t→∞limPr{A(t)≤x}=lim
t→∞P{W(t) =1}=Emin{x,X} EX = 1
µ Z x
0
F(t¯ )dt.
Alternative proof. Another way of evaluating limt→∞Pr{A(t)≤x}is to note that{A(t)≤x}={SN(t)≥ t−x}. From the distribution ofSN(t)and the fact that the support of renewal functionm(t)is positive real life,
Pr{A(t)≤x}=Pr{SN(t)≥t−x}= Z ∞
t−x
F¯(t−y)dm(y) = Z x
−∞
F¯(y)dm(t−y) = Z x
0
F(u)dm(t¯ −u).
Applying key renewal theorem, we get the result.
We see that the limiting distribution of age and excess times are identical. This can be observed by noting that if we consider the reversed processes (an identically distributed renewal process),Y(t), the
“excess life time” attis same as the age att,A(t)of the original process.
Proposition 2.4. Limiting excess time distribution for a renewal process with non-lattice distribution F is
t→∞limPr{Y(t)≤x}= 1 µ
Z x 0
F¯(t)dt.
Proof. We can repeat the same proof for limiting age distribution, where we obtain an alternative renewal process by defining on times when the excess time is less thanx. That is, we consider an alternative renewal processW such that
W(t) =1{Y(t)≤x}.
Corollary 2.5. Limiting mean excess time for a renewal process with iid inter-renewal times{Xn:n∈N} having non-lattice distribution F and meanµis
t→∞limEY(t) =E[X2] 2µ .
Proof. One can get the limiting mean from the limiting distribution by integrating its complement. This involves exchanging limit and integration, which can be justified using monotone convergence theorem.
Hence, exchanging integrals using Fubini’s theorem and integrating by parts, we get
t→∞limEY(t) = Z ∞
0
t→∞limP{Y(t)>x}dx= 1 µ
Z ∞
0 Z ∞
x
F(t)dt¯ = 1 2µ
Z ∞
0
F(t)dt¯ 2= 1 2µ
Z ∞
0
t2dF(t).
Alternatively, one can derive it directly from the regenerative process theory.
Lemma 2.6. Limiting empirical time average of excess time for a renewal process with iid inter-renewal times{Xn:n∈N}having non-lattice distribution F and meanµis almost surely
t→∞lim 1 t
Z t 0
Y(u)du=EX2 2µ .
Proof. Recall that excess times are linearly decreasing in each renewal duration, with valueXnto 0 innth renewal duration of lengthXn. Conditioned onN(t), one can write
Z t 0
Y(u)du=1 2
N(t)
∑
n=1
Xn2+ Z t
SN(t)
(SN(t)+1−u)du.
In particular, one can write the following
∑N(t)n=1Xn2 2N(t)
N(t)
t
≤1 t
Z t 0
Y(u)du≤ ∑N(t)+1n=1 Xn2 2(N(t) +1)
N(t) +1 t
.
Result follows from strong law of large number by taking limits on both sides.
2.2 The Inspection Paradox
DefineXN(t)+1=A(t) +Y(t)as the length of the renewal interval containingt, in other words, the length of current renewal interval.
Theorem 2.7 (inspection paradox). For any x, the length of the current renewal interval to be greater than x is more likely than that for an ordinary renewal interval with distribution function F. That is,
P{XN(t)+1>x} ≥F(x).¯
Proof. Conditioning on the joint distribution of last renewal instant and number of renewals, we can write Pr{XN(t)+1>x}=
Z t 0
Pr{XN(t)+1>x|SN(t)=y,N(t) =n}dF(S
N(t),N(t)).
Now we have,
Pr{XN(t)+1>x|SN(t)=y,N(t) =n}=Pr{Xn+1>x|Xn+1>t−y}=Pr{Xn+1>max(x,t−y)}
Pr{Xn+1>t−y} . Applying Chebyshev’s sum inequality to increasing positive functions f(z) =1{t>x}andg(z) =1{z>
t−y}, we get
Ef(Xn+1)g(Xn+1)/Eg(Xn+1)≥Ef(Xn+1) =F(x).¯ We get the result by integrating over the joint distribution.
We also have a weaker version of inspection paradox involving the limiting distribution ofXN(t)+1. Lemma 2.8. For any x, the limiting probability of length of the current renewal interval being greater than x is larger than the corresponding probability of an ordinary renewal interval with distribution function F. That is,
t→∞limP{XN(t)+1>x} ≥F(x).¯
Proof. Consider an alternating renewal processW for which the on time is the renewal duration if greater thanx, and zero otherwise. That is,
W(t) =1{XN(t)+1>x}.
Hence, each renewal duration consists of either on or off intervals, depending if the renewal duration length is greater thanxor not. We can denotenth on and off times byZnandYnrespectively, where
Zn=Xn1{Xn>x}, Yn=Xn1{Xn≤x}.
From the definition, we have
EW(t) =P{XN(t)+1>x}=P{on at timet}.
From the alternating renewal process theorem, we conclude that
t→∞limPr{XN(t)+1>x}=EX1{X>x}
EX .
The result follows from Chebyshev’s sum inequality applied to positive increasing function f(z) =zand g(z) =1{z>x}.
The inspection paradox states, in essence, that if we pick a pointt, it is more likely that an inter- renewal interval with larger length will containtthan the smaller ones. For instance, ifXiwere equally likely to beε or 1−ε, we see that the mean of any inter arrival length is 1 for any value of ε∈(0,1).
However, for small valuesε, it is more likely that a giventwill be in an interval of length 1−εthan in an interval of lengthε.
Proposition 2.9. If the inter arrival time is non-lattice andE[X2]<∞, we have
t→∞lim
m(t)− t µ
=E[X2] 2µ2 −1.
Proof. From definition of excess timeY(t)and Wald’s lemma for stopping timeN(t) +1 for renewal processes, it follows
ESN(t)+1=E[
N(t)+1 i=1
∑
Xn] =µ(m(t) +1) =t+E[Y(t)].
A Chebyshev’s sum inequality
Lemma A.1. Let f :R→R+and g:R→R+be arbitrary functions with the same monotonicity. For any random variable X , functions f(X)and g(X)are positive and
E[f(X)g(X)]≥E[f(X)]E[g(X)].
Proof. LetY be a random variable independent ofX and with the same distribution. Then, (f(X)−f(Y))(g(X)−g(Y))≥0.
Taking expectation on both sides the result follows.
