# Renewal theory

Renewal theory is the branch of probability theory that generalizes compound Poisson process for arbitrary holding times. Applications include calculating the best strategy for replacing worn-out machinery in a factory (example below) and comparing the long-term benefits of different insurance policies.

## Renewal processes

### Introduction

The renewal process is a generalization of the compound Poisson process. In essence, the Poisson process is a continuous-time Markov process on the positive integers (usually starting at zero) which has independent identically distributed holding times at each integer $i$ (exponentially distributed) before advancing (with probability 1) to the next integer: $i+1$ . In a compound Poisson process, the jump size need not be from i to i + 1, but is a random variable, and those random variables are independent and identically distributed. The exponential distribution of the holding times must be a memoryless exponential distribution if the number of jumps in each time interval is to have a Poisson distribution with expected value proportional to the length of the interval.

In a renewal process, the holding times need not have a memoryless distribution; rather, the process loses its memory only when one holding period ends and the next begins. That means the conditional probability distribution of the future of the process, given the past, is the same every time such a "renewal" occurs, and thus does not depend on the past. However that the independence and identical distribution (IID) property of the holding times is retained.

### Formal definition

Let $S_{1},S_{2},S_{3},S_{4},S_{5},\ldots$ be a sequence of positive independent identically distributed random variables such that

$0<\operatorname {E} [S_{i}]<\infty .$ We refer to the random variable $S_{i}$ as the "$i$ -th" holding time.

$\operatorname {E} [S_{i}]$ is the expectation of $S_{i}$ .

Define for each n > 0 :

$J_{n}=\sum _{i=1}^{n}S_{i},$ each $J_{n}$ is referred to as the "$n$ -th" jump time and the intervals

$[J_{n},J_{n+1}]$ being called renewal intervals.

Then $(X_{t})_{t\geq 0}$ is given by random variable

$X_{t}=\sum _{n=1}^{\infty }\operatorname {\mathbb {I} } _{\{J_{n}\leq t\}}=\sup \left\{\,n:J_{n}\leq t\,\right\}$ where $\operatorname {\mathbb {I} } _{\{J_{n}\leq t\}}$ is the indicator function

$\operatorname {\mathbb {I} } _{\{J_{n}\leq t\}}={\begin{cases}1,&{\text{if }}J_{n}\leq t\\0,&{\text{otherwise}}\end{cases}}$ $(X_{t})_{t\geq 0}$ represents the number of jumps that have occurred by time t, and is called a renewal process.

### Interpretation

If one considers events occurring at random times, one may choose to think of the holding times $\{S_{i}:i\geq 1\}$ as the random time elapsed between two consecutive events. For example, if the renewal process is modelling the numbers of breakdown of different machines, then the holding time represents the time between one machine breaking down before another one does.

## Renewal-reward processes

Let $W_{1},W_{2},\ldots$ be a sequence of IID random variables (rewards) satisfying

$\operatorname {E} |W_{i}|<\infty .\,$ Then the random variable

$Y_{t}=\sum _{i=1}^{X_{t}}W_{i}$ is called a renewal-reward process. Note that unlike the $S_{i}$ , each $W_{i}$ may take negative values as well as positive values.

The random variable $Y_{t}$ depends on two sequences: the holding times $S_{1},S_{2},\ldots$ and the rewards $W_{1},W_{2},\ldots$ These two sequences need not be independent. In particular, $W_{i}$ may be a function of $S_{i}$ .

### Interpretation

In the context of the above interpretation of the holding times as the time between successive malfunctions of a machine, the "rewards" $W_{1},W_{2},\ldots$ (which in this case happen to be negative) may be viewed as the successive repair costs incurred as a result of the successive malfunctions.

An alternative analogy is that we have a magic goose which lays eggs at intervals (holding times) distributed as $S_{i}$ . Sometimes it lays golden eggs of random weight, and sometimes it lays toxic eggs (also of random weight) which require responsible (and costly) disposal. The "rewards" $W_{i}$ are the successive (random) financial losses/gains resulting from successive eggs (i = 1,2,3,...) and $Y_{t}$ records the total financial "reward" at time t.

## Properties of renewal processes and renewal-reward processes

We define the renewal function as the expected value of the number of jumps observed up to some time $t$ :

$m(t)=\operatorname {E} [X_{t}].\,$ ### The elementary renewal theorem

The renewal function satisfies

$\lim _{t\to \infty }{\frac {1}{t}}m(t)={\frac {1}{\operatorname {E} [S_{1}]}}.$ #### Proof

$\lim _{t\to \infty }{\frac {X_{t}}{t}}={\frac {1}{\operatorname {E} [S_{1}]}}.$ To prove the elementary renewal theorem, it is sufficient to show that $\left\{{\frac {X_{t}}{t}};t\geq 0\right\}$ is uniformly integrable.

To do this, consider some truncated renewal process where the holding times are defined by ${\overline {S_{n}}}=a\operatorname {\mathbb {I} } \{S_{n}>a\}$ where $a$ is a point such that $0 which exists for all non-deterministic renewal processes. This new renewal process ${\overline {X}}_{t}$ is an upper bound on $X_{t}$ and its renewals can only occur on the lattice $\{na;n\in \mathbb {N} \}$ . Furthermore, the number of renewals at each time is geometric with parameter $p$ . So we have

{\begin{aligned}{\overline {X_{t}}}&\leq \sum _{i=1}^{[at]}\operatorname {Geometric} (p)\\\operatorname {E} \left[\,{\overline {X_{t}}}^{2}\,\right]&\leq C_{1}t+C_{2}t^{2}\\P\left({\frac {X_{t}}{t}}>x\right)&\leq {\frac {\operatorname {E} \left[X_{t}^{2}\right]}{t^{2}x^{2}}}\leq {\frac {\operatorname {E} \left[{\overline {X_{t}}}^{2}\right]}{t^{2}x^{2}}}\leq {\frac {C}{x^{2}}}.\end{aligned}} ### The elementary renewal theorem for renewal reward processes

We define the reward function:

$g(t)=\operatorname {E} [Y_{t}].\,$ The reward function satisfies

$\lim _{t\to \infty }{\frac {1}{t}}g(t)={\frac {\operatorname {E} [W_{1}]}{\operatorname {E} [S_{1}]}}.$ ### The renewal equation

The renewal function satisfies

$m(t)=F_{S}(t)+\int _{0}^{t}m(t-s)f_{S}(s)\,ds$ where $F_{S}$ is the cumulative distribution function of $S_{1}$ and $f_{S}$ is the corresponding probability density function.

#### Proof of the renewal equation

We may iterate the expectation about the first holding time:
$m(t)=\operatorname {E} [X_{t}]=\operatorname {E} [\operatorname {E} (X_{t}\mid S_{1})].\,$ But by the Markov property
$\operatorname {E} (X_{t}\mid S_{1}=s)=\operatorname {\mathbb {I} } _{\{t\geq s\}}\left(1+\operatorname {E} [X_{t-s}]\right).\,$ So
{\begin{aligned}m(t)&=\operatorname {E} [X_{t}]\\[12pt]&=\operatorname {E} [\operatorname {E} (X_{t}\mid S_{1})]\\[12pt]&=\int _{0}^{\infty }\operatorname {E} (X_{t}\mid S_{1}=s)f_{S}(s)\,ds\\[12pt]&=\int _{0}^{\infty }\operatorname {\mathbb {I} } _{\{t\geq s\}}\left(1+\operatorname {E} [X_{t-s}]\right)f_{S}(s)\,ds\\[12pt]&=\int _{0}^{t}\left(1+m(t-s)\right)f_{S}(s)\,ds\\[12pt]&=F_{S}(t)+\int _{0}^{t}m(t-s)f_{S}(s)\,ds,\end{aligned}} as required.

### Asymptotic properties

$(X_{t})_{t\geq 0}$ and $(Y_{t})_{t\geq 0}$ satisfy

$\lim _{t\to \infty }{\frac {1}{t}}X_{t}={\frac {1}{\operatorname {E} [S_{1}]}}$ (strong law of large numbers for renewal processes)
$\lim _{t\to \infty }{\frac {1}{t}}Y_{t}={\frac {1}{\operatorname {E} [S_{1}]}}\operatorname {E} [W_{1}]$ (strong law of large numbers for renewal-reward processes)

almost surely.

#### Proof

First consider $(X_{t})_{t\geq 0}$ . By definition we have:
$J_{X_{t}}\leq t\leq J_{X_{t}+1}$ for all $t\geq 0$ and so
${\frac {J_{X_{t}}}{X_{t}}}\leq {\frac {t}{X_{t}}}\leq {\frac {J_{X_{t}+1}}{X_{t}}}$ for all t ≥ 0.
Now since $0<\operatorname {E} [S_{i}]<\infty$ we have:
$X_{t}\to \infty$ as $t\to \infty$ almost surely (with probability 1). Hence:
${\frac {J_{X_{t}}}{X_{t}}}={\frac {J_{n}}{n}}={\frac {1}{n}}\sum _{i=1}^{n}S_{i}\to \operatorname {E} [S_{1}]$ almost surely (using the strong law of large numbers); similarly:
${\frac {J_{X_{t}+1}}{X_{t}}}={\frac {J_{X_{t}+1}}{X_{t}+1}}{\frac {X_{t}+1}{X_{t}}}={\frac {J_{n+1}}{n+1}}{\frac {n+1}{n}}\to \operatorname {E} [S_{1}]\cdot 1$ almost surely.
Thus (since $t/X_{t}$ is sandwiched between the two terms)
${\frac {1}{t}}X_{t}\to {\frac {1}{\operatorname {E} [S_{1}]}}$ almost surely.
Next consider $(Y_{t})_{t\geq 0}$ . We have
${\frac {1}{t}}Y_{t}={\frac {X_{t}}{t}}{\frac {1}{X_{t}}}Y_{t}\to {\frac {1}{\operatorname {E} [S_{1}]}}\cdot \operatorname {E} [W_{1}]$ almost surely (using the first result and using the law of large numbers on $Y_{t}$ ).

A curious feature of renewal processes is that if we wait some predetermined time t and then observe how large the renewal interval containing t is, we should expect it to be typically larger than a renewal interval of average size.

Mathematically the inspection paradox states: for any t > 0 the renewal interval containing t is stochastically larger than the first renewal interval. That is, for all x > 0 and for all t > 0:

$\operatorname {P} (S_{X_{t}+1}>x)\geq \operatorname {P} (S_{1}>x)=1-F_{S}(x)$ where FS is the cumulative distribution function of the IID holding times Si.

#### Proof of the inspection paradox The renewal interval determined by the random point t (shown in red) is stochastically larger than the first renewal interval.

Observe that the last jump-time before t is $J_{X_{t}}$ ; and that the renewal interval containing t is $S_{X_{t}+1}$ . Then

{\begin{aligned}\operatorname {P} (S_{X_{t}+1}>x)&{}=\int _{0}^{\infty }\operatorname {P} (S_{X_{t}+1}>x\mid J_{X_{t}}=s)f_{J_{X_{t}}}(s)\,ds\\[12pt]&{}=\int _{0}^{\infty }\operatorname {P} (S_{X_{t}+1}>x|S_{X_{t}+1}>t-s)f_{J_{X_{t}}}(s)\,ds\\[12pt]&{}=\int _{0}^{\infty }{\frac {\operatorname {P} (S_{X_{t}+1}>x\,,\,S_{X_{t}+1}>t-s)}{\operatorname {P} (S_{X_{t}+1}>t-s)}}f_{J_{X_{t}}}(s)\,ds\\[12pt]&{}=\int _{0}^{\infty }{\frac {1-F(\max\{x,t-s\})}{1-F(t-s)}}f_{J_{X_{t}}}(s)\,ds\\[12pt]&{}=\int _{0}^{\infty }\min \left\{{\frac {1-F(x)}{1-F(t-s)}},{\frac {1-F(t-s)}{1-F(t-s)}}\right\}f_{J_{X_{t}}}(s)\,ds\\[12pt]&{}=\int _{0}^{\infty }\min \left\{{\frac {1-F(x)}{1-F(t-s)}},1\right\}f_{J_{X_{t}}}(s)\,ds\\[12pt]&{}\geq \int _{0}^{\infty }(1-F(x))f_{J_{X_{t}}}(s)\,ds=1-F(x)=\operatorname {P} (S_{1}>x),\\[12pt]\end{aligned}} since both ${\frac {1-F(x)}{1-F(t-s)}}$ and $1$ are greater than or equal to $1-F(x)$ for all values of s.

### Superposition

The superposition of independent renewal processes, or superimposed renewal process, is not generally a renewal process, but it can be described within a larger class of processes called the Markov-renewal processes. However, the cumulative distribution function of the first inter-event time in the superposition process is given by

$R(t)=1-\sum _{k=1}^{K}{\frac {\alpha _{k}}{\sum _{l=1}^{K}\alpha _{l}}}(1-R_{k}(t))\prod _{j=1,j\neq k}^{K}\alpha _{j}\int _{t}^{\infty }(1-R_{j}(u))\,{\text{d}}u$ where Rk(t) and αk > 0 are the CDF of the inter-event times and the arrival rate of process k.

## Example applications

### Example 1: use of the strong law of large numbers

Eric the entrepreneur has n machines, each having an operational lifetime uniformly distributed between zero and two years. Eric may let each machine run until it fails with replacement cost €2600; alternatively he may replace a machine at any time while it is still functional at a cost of €200.

What is his optimal replacement policy?

#### Solution

The lifetime of the n machines can be modeled as n independent concurrent renewal-reward processes, so it is sufficient to consider the case n=1. Denote this process by $(Y_{t})_{t\geq 0}$ . The successive lifetimes S of the replacement machines are independent and identically distributed, so the optimal policy is the same for all replacement machines in the process.

If Eric decides at the start of a machine's life to replace it at time 0 < t < 2 but the machine happens to fail before that time then the lifetime S of the machine is uniformly distributed on [0, t] and thus has expectation 0.5t. So the overall expected lifetime of the machine is:

{\begin{aligned}\operatorname {E} [S]&=\operatorname {E} [S\mid {\text{fails before }}t]\cdot \operatorname {P} [{\text{fails before }}t]+\operatorname {E} [S\mid {\text{does not fail before }}t]\cdot \operatorname {P} [{\text{does not fail before }}t]\\[6pt]&=0.5t({\frac {t}{2}})+t({\frac {2-t}{2}})\end{aligned}} and the expected cost W per machine is:

{\begin{aligned}\operatorname {E} [W]&=\operatorname {E} [W\mid {\text{fails before }}t]\cdot \operatorname {P} ({\text{fails before }}t)+\operatorname {E} [W\mid {\text{does not fail before }}t]\cdot \operatorname {P} ({\text{does not fail before }}t)\\[6pt]&=2600({\frac {t}{2}})+200({\frac {2-t}{2}})=1200t+200.\end{aligned}} So by the strong law of large numbers, his long-term average cost per unit time is:

${\frac {1}{t}}Y_{t}\simeq {\frac {\operatorname {E} [W]}{\operatorname {E} [S]}}={\frac {4(1200t+200)}{t^{2}+4t-2t^{2}}}$ then differentiating with respect to t:

${\frac {\partial }{\partial t}}{\frac {4(1200t+200)}{t^{2}+4t-2t^{2}}}=4{\frac {(4t-t^{2})(1200)-(4-2t)(1200t+200)}{(t^{2}+4t-2t^{2})^{2}}},$ this implies that the turning points satisfy:

{\begin{aligned}0&=(4t-t^{2})(1200)-(4-2t)(1200t+200)=4800t-1200t^{2}-4800t-800+2400t^{2}+400t\\[6pt]&=-800+400t+1200t^{2},\end{aligned}} and thus

$0=3t^{2}+t-2=(3t-2)(t+1).$ We take the only solution t in [0, 2]: t = 2/3. This is indeed a minimum (and not a maximum) since the cost per unit time tends to infinity as t tends to zero, meaning that the cost is decreasing as t increases, until the point 2/3 where it starts to increase.