Section 10.5

Renewal–reward processes

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Problem 10.5.1

If X(t)X(t) is an irreducible positive recurrent Markov chain, and u(⋅)u(\cdot ) is a bounded function on the integers, show that

1t∫0tu(X(s))ds→ a.s. ∑i∈Sπiu(i) \frac{1}{t} \int _{0}^{t} u(X(s)) d s \xrightarrow {\text{ a.s. }} \sum _{i \in S} \pi _{i} u(i)

where π\mathbf{\pi } is the stationary distribution of X(t)X(t).

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Problem 10.5.2

Let M(t)M(t) be an alternating renewal process, with interarrival pairs {Xr,Yr:r≥1}\left\{ X_{r}, Y_{r}: r \geq 1\right\}. Show that

1t∫0tI{M(s) is even }ds→ a.s. E[X1]E[X1]+E[Y1] as t→∞ \frac{1}{t} \int _{0}^{t} I_{\left\{ M(s) \text{ is even }\right\} } d s \xrightarrow {\text{ a.s. }} \frac{\mathbb {E}\left[X_{1}\right]}{\mathbb {E}\left[X_{1}\right]+\mathbb {E}\left[Y_{1}\right]} \quad \text{ as } t \rightarrow \infty

Is this limit valid for an arbitrary joint distribution of independent pairs (Xi,Yi)\left(X_{i}, Y_{i}\right)?

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Problem 10.5.3

Let C(s)C(s) be the current lifetime (or age) of a renewal process N(t)N(t) with a typical interarrival time XX. Show that

1t∫0tC(s)ds→ a.s. E[X2]2E[X] as t→∞ \frac{1}{t} \int _{0}^{t} C(s) d s \xrightarrow {\text{ a.s. }} \frac{\mathbb {E}\left[X^{2}\right]}{2 \mathbb {E}\left[X\right]} \quad \text{ as } t \rightarrow \infty

Find the corresponding limit for the excess lifetime.

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Problem 10.5.4

Let jj and kk be distinct states of an irreducible discrete-time Markov chain XX with stationary distribution π\pi. Show that

P(Tj<Tk∣X0=k)=1/πkE[Tj∣X0=k]+E[Tk∣X0=j] \mathbb {P}\left(T_{j} < T_{k} \mid X_{0} = k\right) = \frac{1 / \pi _{k}}{\mathbb {E}\left[T_{j} \mid X_{0} = k\right]+\mathbb {E}\left[T_{k} \mid X_{0} = j\right]}

where Ti=min⁡{n≥1:Xn=i}T_{i} = \min \left\{ n \geq 1: X_{n} = i\right\} is the first passage time to the state ii. [Hint: Consider the times of return to jj having made an intermediate visit to kk.]

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Problem 10.5.5

Use the result of Exercise (10.5.2) to show that the limiting distribution function as t→∞t \rightarrow \infty of the total life D(t)D(t) of a renewal process is

FD(y)=∫0y1μxf(x)dx F_{D}(y) = \int _{0}^{y} \frac{1}{\mu } x f(x) d x

where ff is the density function of a typical interarrival time XX, and μ=E[X]\mu = \mathbb {E}\left[X\right]. The integrand is called the size-biased (or length-biased) density for XX.

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Problem 10.5.6
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(a)

Let XX be exponentially distributed with parameter λ\lambda. Show that

E[min⁡{X,d}]=1λ(1−e−λd),d≥0 \mathbb {E}\left[\min \left\{ X, d\right\} \right] = \frac{1}{\lambda }\left(1-e^{-\lambda d}\right), \quad d \geq 0
(b)

John's garage offers him the choice of two tyre replacement plans.

  1. The garage undertakes to replace all the tyres of his car at the normal price whenever one of the tyres requires replacing.

  2. The garage undertakes to replace all the tyres of his car at 5%5 \% of the normal price two years after they have last been replaced. However, if any tyre needs replacing earlier, the garage will then replace all the tyres at a price that is 5%5 \% higher than the normal price. Assuming that a new tyre has an exponentially distributed lifetime with mean 8 years, determine the long run average cost per year under the two options. Which option should John choose if his car is new?

Problem 10.5.7

A machine MM is repaired at time t=0t = 0, and its uptime after any repair is exponentially distributed with parameter λ\lambda, at which point it breaks down (assume the usual independence). Following any repair at time TT, say, it is inspected at times T,T+m,T+2m,…T, T+m, T+2 m, \ldots, and instantly repaired if found to be broken (the inspection schedule is then restarted). Show that the long run proportion of time that MM is working (the 'uptime ratio') is m−1∫0me−λxdxm^{-1} \int_{0}^{m} e^{-\lambda x} d x.

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