We now discuss some variations of the renewal theory that often occur in applications.

# Alternating renewal process

We imagine a machine works, but the breaks down after time $$W_i$$ and then needs to be serviced taking time $$S_i$$. We assume that the sequences $$W=(W_i)_{i \in \mathbb{N}}$$ and $$S = (S_i)_{i \in \mathbb{Z}^{+}}$$ are independent, and that the sequences themselves are i.i.d.ย with common distributions given by $$F_W$$ and $$F_S$$. If we let $$N(t)$$ be the number of repairs completed by time $$t$$, then $$N$$ a renewal process, with inter-arrival times given by $$X_i= W_{i-1} + S_i$$, where with law given by $$F_X = F_W * F_S$$. We are interested in $p(t) = \mathbb{P}(O_t)= \mathbb{P}( \text{the machine is working at time t}),$ and the limiting behaviour of $$p$$.

By considering on the event $$\{W_0 \leq t\}$$, and conditioning on $$X_1$$ we have $\begin{eqnarray*} p(t) &=& \mathbb{P}(O_t, W_0 >t) + \mathbb{P}(O_t, W_0 \leq t)\\ &=& 1- F_W(t) + \mathbb{E}(\mathbb{P}(O_t , W_0 \leq t | X_1)). \end{eqnarray*}$

Let $$\phi(x) = \mathbb{P}(O_t, W_0 \leq t|X_1 =x).$$ Then $$\phi(x) = p(t-x) \mathbf{1}[t \geq x]$$. Thus $\begin{eqnarray*} p(t) &=& 1- F_W(t) + \mathbb{E}\phi(X_1) \\ &=& 1- F_W(t) + \int_0^t p(t-x) dF_X(x). \end{eqnarray*}$ Thus $p = (1- F_W) + p*F_X$

and from the uniqueness of solutions of renewal equations $p = (1- F_W) + (1-F_W)*m,$

where $$m(t) = \mathbb{E}(N(t))$$ is the renewal function.

Thus the key renewal theorem for non-lattice type $$F_X$$, gives that

$p(t) \to \frac{1}{\mathbb{E}(X_1)}\int_0^{\infty} [1-F_w(x)]dx = \frac{\mathbb{E}(W_0)}{\mathbb{E}(W_0) + \mathbb{E}(S_1)},$ as $$t \to \infty$$.

# Renewal-reward process

Given a renewal process $$N$$ with inter-arrival times $$X=(X_i)_{i \in \mathbb{Z} ^{+}}$$, we think of obtaining a random reward (or punishment) $$R_i \in \mathbb{R}$$; we assume that $$R=(R_i)_{i \in \mathbb{Z}^+}$$ is i.i.d., but often $$R$$ will depend on $$X$$. The pair $$(X, R)$$ is sometimes called a renewal reward process. We are interested in the cumulative reward process given by $C(t) = \sum_{i=1} ^{N(t)} R_i.$ The reward function is given by $$c(t) = \mathbb{E} C(t)$$.

Theorem (Law of large numbers for renewal-reward processes): Let $$(X, R)$$ be a renewal process, where the expectations of $$X_1$$ and $$R_1$$ are well-defined and finte. Then as $$t \to \infty$$, we have

• $\frac{C(t)}{t} \to \frac{\mathbb{E}(R_1)}{\mathbb{E}(X_1)} \text{ almost surely,}$ and
• $\frac{c(t)}{t} \to \frac{\mathbb{E}(R_1)}{\mathbb{E}(X_1)}.$

The proof of the first assertion follows from the law of large numbers for renewal processes, and the second assertion requires an argument to allow the interchange of limits. Notice that this is the sort of formula that one might guess is true: the long term average rate of rewards is the expected reward over the expected time if takes for an arrival to occur.

A typical application is as follows. Suppose that a (used) car has a random lifetime with cumulative distribution $$F$$ with density $$f$$ We buy a new one at cost $$C_1$$ if it lasts $$L$$ years, but if it breaks down before $$L$$ years, we have to pay a fee and buy a car costing $$C_2$$; imagine that it breaks down while you are driving, and you have to pay a tow fee to the dump. It is free to get of a car that still runs. What is the long-run average cost of owning a used car.

Since we replace the car at $$L$$ years, regardless of whether it is still running, the inter-arrival times that is not given by $$F$$; if $$X_1$$ is the first replacement, it has cumulative distribution given by $\bar{F}(x) = F(x) \mathbf{1}[x <L] +\mathbf{1}[x \geq L]$ $\mathbb{E}(X_1) = \int_0 ^{\infty} [1-\bar{F}(x)]dx = \int_0 ^{L} [1-F(x)]dx.$ Let $$Y_1$$ have law $$F$$ be the lifetime of the first used car, and let $$R_1$$ be random variable that takes the value $$C_1$$ if $$Y_1 \geq L$$, and the value $$C_2$$ if $$Y_1 < L$$. Notice that $\mathbb{E}{R_1} = C_1(1-F(L)) + C_2F(L).$

The law of large numbers for renewal-reward processes gives that long-run average is $c(t)/t \to \frac{\mathbb{E}(R_1)}{\mathbb{E}(X_1)} = \frac{C_1(1-F(L)) + C_2(F(L))}{ \int_0 ^{L} [1-F(x)]dx}.$

Often one can apply the renewal-reward theorem to prove other theorems by coming up with a suitable reward.

Theorem (Law of large numbers for continuous-time Markov chains): Let $$X$$ be an irreducible continuous-time Markov chain on a finite number of states with stationary distribution $$\pi$$. Let $$s$$ be a state. Then

$\frac{1}{t}\int_0 ^t \mathbf{1}[X(u)=s]du \to \pi(s).$

Proof: Suppose the chain starts at state $$s$$, and consider the renewal process given by inter-arrival times $$Y_i$$ corresponding to how the long chain spends at $$s$$, and the time it takes to return to $$s$$. Let us give ourselves a reward $$R_i$$ when the Markov chain reaches the state $$s$$ for the $$i$$-th-time, which is exactly the amount of time the chain stays at state $$s$$, before leaving. Thus with the renewal-reward process $$(Y, R)$$ the limit

$\lim_{t \to \infty} \frac{C(t)}{t} = \lim_{t \to \infty} \frac{1}{t}\int_0 ^t \mathbf{1}[X(u)=s]du = \frac{\mathbb{E}(R_1)}{\mathbb{E}(Y_1)}$

exists almost surely, since its clear relevant expectations are finite; in particular, we know that $$\mathbb{E}R_1=h_s$$ is the mean holding time at state $$s$$.

Clearly, we would get the same limit if the chain was started at the stationary distribution, and we know that $0 \leq \frac{1}{t}\int_0 ^t \mathbf{1}[X(u)=s]du \leq 1$ and $\mathbb{E}(\mathbf{1}[X(u) =s]) = \mathbb{P}(X(u) = s) = \pi(s).$ Hence by an interchange of limits argument (the bounded convergence theorem) and taking expectations on both sides of the limit, we identify that limit must be $\pi(s) = \frac{h_s}{\mathbb{E} Y_1}.$

Note that we obtained a similar expression for discrete time Markov chains at the end of our first discussion on renewal theory.

# Summary

• We analyzed alternating renewal processes

• We introduced renewal-reward processes and gave a couple of applications