Lecture-08: Key Renewal Theorem and Applications
1 Key Renewal Theorem
Theorem 1.1 (Key Renewal Theorem). Consider a an aperiodic and recurrent renewal process with renewal function m(t), and the mean and the distribution of inter-renewal times being denoted byµand F respectively.
For any directly Riemann integrable function 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). (2)
Example 1.2 (Renewal Equation).
Proposition 1.3 (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]}. (3)
It is easy to see that this function is directly Riemann integrable. Conversely, it follows from Blackwell theorem that
t→∞lim dm(t)
dt
(a)=lim
a→0lim
t→∞
m(t+a)−m(t)
a = 1
µ. (4)
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 function g(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 applica- tions. 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. 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 aniidsequence 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 withnth inter-renewal timeTnfor eachn∈N, thenth renewal instant is denoted by Sn=∑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}. (5)
1
It is easy to see thatW is a regenerative process with regenerative sequenceS. We denote the probability of the process being on at timetby
P(t) =Pr{W(t) =1}. (6)
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). (7)
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).¯ (8) Hence, we can write the renewal equation forP(t)as
P(t) =H(t) + (F¯ ∗P)(t). (9)
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]. (10) Proof. SinceHis the distribution function of the non-negative random variableZn, it follows that
t→∞lim
H(t) =¯ 0, and Z ∞
0
H(t)dt¯ =EZn. (11)
Applying key renewal theorem to Theorem??, we get the result.
A Directly Riemann Integrable
For eachδ>0 andn∈N, we define intervalsI(n,δ) = [(n−1)δ,nδ)that partition the positive axisR+= [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)}.
The infinite sums are denoted byσδ ,δ∑n∈Nm(h,n,δ)andσδ ,δ∑n∈Nm(h,n,δ). A functionh:R+7→Ris directly 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,
lim
δ→0(σδ−σδ) =0 (12)
If both limits exist and are equal, then the integral value is equal to the limit.
Remark1. 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). (13)
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, (14)
if the limit exists. For many functions, this limit may not exist.
Remark2. 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)
(15) is Riemann integrable, butδ∑n∈Nm(h,n,δ)is always infinite for everyδ >0.
2
Proposition A.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 A.2 (Tail Property). If h is non-negative, directly Riemann integrable, and has bounded integral value, then
t→∞limh(t) =0. (16)
B Chebyshev’s sum inequality
Lemma B.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)]. (17) Proof. LetY be a random variable independent ofX and with the same distribution. Then,
(f(X)−f(Y))(g(X)−g(Y))≥0. (18)
Taking expectation on both sides the result follows.
3
