Renewal Theory

In this section we will study a generalization of the Poisson process. There we had a counting porcess with interarrival times that were iid exponential. Now we will allow other distributions for the interarrival times.

Let {X_n;n≥1} be a sequence of nonnegative iid rv's with common cdf F such that F(0)=P(X_n=0)<1. We will interpret X_n as the time between the (n-1)^{st and the n^th} event. The condition F(0)<1 means that there is some chance that the process stays in the same state for some time, it does not (necessarily) immediately jump away. Let m=E[X_n] denote the mean time until the next event, and note that because of the condition 0<m≤∞. Set S₀=0 and S_n=∑_i=1ⁿX_i, so S_n is the time of the n^th event. As the number of events by time t will equal the largest value of n for which the n^th event has occured before or at time t, we have

Definition

{N(t);t≥0}=sup{n: S_n≤t} is called a renewal process

Here we will use the terms event and renewal interchangeably.
Note that by the law of large numbers S_n/n m in probability. Since m>0 this means that S_n must be going to infinity as n goes to infinity. Thus S_n can be less than or equal to any finite t for at most a finite number of n. Hence N(t) must be finite and we can write
{N(t);t≥0}=max{n: S_n≤t}

For the Poisson process we found that {N(t)≥=n}={S_n≤t}. The same equivalence holds here and can in principle be used to find the distribution of N(t):

Now since the {X_i;i≥1} are iid F, it follows that S_n=∑_i=1ⁿX_i has cdf F_n, the n-fold convolution of F with itself. Therefore we find
P(N(t)=n) = F_n(t)-F_n+1(t).
Let m(t)=E[N(t)]. m(t) is called the renewal function, and one of the main issues in renewal theory is to determine its properties. We find

Proposition

m(t)= ∑_i=1ⁿ F_n(t)

Proof Let I_n be the indicator function that the n^th renewal occured in [0,t], then
N(t) = ∑_i=1^∞ I_n
and so

Proposition

m(t)<∞ for all 0≤t<∞

Proof Since P(X_n=0)<1 it follows by the continuity property of probabilities that there exists an a>0 such that P(X_n≥a)>0. Now define a related renewal process {Y_n;n≥1} by

Notice that Y_n/a has a Bernoulli distribution with success parameter p=P(Y_n/a=1) = P(X_n≥a)
Let N'(t)=sup{n:Y₁+..+Y_n≤t}.
Now Y_n takes only values 0 or a, so Y₁+..+Y_n takes only values 0, a, 2a, .., na. So
N'(t) = max{n: na≤t} = n*a for some n*{0,1,2,..}
Also P(N'(ma)=n) = P( Therefore for this new process renewals only take place at times t=na, n=0,2,.., and also the number of renewals at each of these times are independent geometric rv. with mean 1/P(X_n≥a). Thus