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Renewal theory is the branch of probability theory that generalizes Poisson processes for arbitrary holding times. Applications include calculating the expected time for a monkey who is randomly tapping at a keyboard to type the word Macbeth and comparing the long-term benefits of different insurance policies.
## Renewal processes

### Introduction

A renewal process is a generalisation of the 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 the same informal spirit, we may define a renewal process to be the same thing, except that the holding times take on a more general distribution. (Note however that the IID property of the holding times is retained).
### Formal definition

### Interpretation

## Renewal-reward processes

### Interpretation

## Properties of renewal processes and renewal-reward processes

We define the renewal function:### The elementary renewal theorem

The renewal function satisfies#### Proof

### The elementary renewal theorem for reward renewal processes

We define the reward function:### The renewal equation

The renewal function satisfies#### Proof of the renewal equation

### Asymptotic properties

#### Proof

### The inspection paradox

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.#### Proof of the inspection paradox

## 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.#### Solution

We may model the lifetime of the n machines 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.## See also

Let $S\_1\; ,\; S\_2\; ,\; S\_3\; ,\; S\_4\; ,\; S\_5,\; ldots$ be a sequence of independent identically distributed random variables such that

- $0\; <\; mathbb\{E\}[S\_i]\; <\; infty.$

We refer to the random variable $S\_i$ as the "$i$th" holding time.

Define for each n > 0 :

- $J\_n\; =\; sum\_\{i=1\}^n\; S\_i,$

each $J\_n$ referred to as the "$n$th" jump time and the intervals

- $[J\_n,J\_\{n+1\}]$

being called renewal intervals.

Then the random variable $(X\_t)\_\{tgeq0\}$ given by

- $X\_t\; =\; sum^\{infty\}\_\{n=1\}\; mathbb\{I\}\_\{\{J\_n\; leq\; t\}\}=sup\; left\{,\; n:\; J\_n\; leq\; t,\; right\}$

is called a renewal process.

One may choose to think of the holding times $\{\; S\_i\; :\; i\; geq\; 1\; \}$ as the time elapsed before a machine breaks for the "$i$th" time since the last time it broke. (Note this assumes that the machine is immediately fixed and we restart the clock immediately.) Under this interpretation, the jump times $\{\; J\_n\; :\; n\; geq\; 1\; \}$ record the successive times at which the machine breaks and the renewal process $X\_t$ records the number of times the machine has so far had to be repaired at any given time $t$.

However it is more helpful to understand the renewal process in its abstract form, since it may be used to model a great number of practical situations of interest which do not relate very closely to the operation of machines.

Let $W\_1,\; W\_2,\; ldots$ be a sequence of IID random variables (rewards) satisfying

- $mathbb\{E\}|W\_i|\; <\; infty.,$

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.

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.

- $m(t)\; =\; mathbb\{E\}[X\_t].,$

- $lim\_\{t\; to\; infty\}\; frac\{1\}\{t\}m(t)\; =\; 1/mathbb\{E\}[S\_1].$

Below, you find that the strong law of large numbers for renewal processes tell us that

- $lim\_\{t\; to\; infty\}\; frac\; \{X\_t\}\{t\}\; =\; frac\{1\}\{mathbb\{E\}[S\_1]\}.$

To prove the elementary renewal theorem, it is sufficient to show that $left\{frac\{X\_t\}\{t\};\; t\; geq\; 0right\}$ is uniformly integrable.

To do this, consider some truncated renewal process where the holding times are defined by $overline\{S\_n\}\; =\; a\; mathbb\{I\}\{S\_n\; >\; a\}$ where $a$ is a point such that $0\; <\; F(a)\; =\; p\; <\; 1$ 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

- $$

- $g(t)\; =\; mathbb\{E\}[Y\_t].,$

The renewal function satisfies

- $lim\_\{t\; to\; infty\}\; frac\{1\}\{t\}g(t)\; =\; frac\{mathbb\{E\}[W\_1]\}\{mathbb\{E\}[S\_1]\}.$

- $m(t)\; =\; F\_S(t)\; +\; int\_0^t\; m(t-s)\; f\_S(s),\; ds$

where $F\_S$ is the cumulative distribution function of $S\_1$ and $f\_S$ is the corresponding probability density function.

- We may iterate the expectation about the first holding time:

- $m(t)\; =\; mathbb\{E\}[X\_t]\; =\; mathbb\{E\}[mathbb\{E\}(X\_t\; mid\; S\_1)].$

- But by the Markov property

- $mathbb\{E\}(X\_t\; mid\; S\_1=s)\; =\; mathbb\{I\}\_\{\{t\; geq\; s\}\}\; left(1\; +\; mathbb\{E\}[X\_\{t-s\}]\; right).$

- So

- $$

- as required.

$(X\_t)\_\{tgeq0\}$ and $(Y\_t)\_\{tgeq0\}$ satisfy

- $lim\_\{t\; to\; infty\}\; frac\{1\}\{t\}\; X\_t\; =\; frac\{1\}\{mathbb\{E\}S\_1\}$ (strong law of large numbers for renewal processes)

- $lim\_\{t\; to\; infty\}\; frac\{1\}\{t\}\; Y\_t\; =\; frac\{1\}\{mathbb\{E\}S\_1\}\; mathbb\{E\}W\_1$ (strong law of large numbers for renewal-reward processes)

almost surely.

- First consider $(X\_t)\_\{tgeq0\}$. By definition we have:

- $J\_\{X\_t\}\; leq\; t\; leq\; J\_\{X\_t+1\}$

- for all $t\; geq\; 0$ and so

- $$

- for all t ≥ 0.

- Now since $0<\; mathbb\{E\}S\_i\; <\; infty$ we have:

- $X\_t\; to\; infty$

- as $t\; to\; infty$ almost surely (with probability 1). Hence:

- $frac\{J\_\{X\_t\}\}\{X\_t\}\; =\; frac\{J\_n\}\{n\}\; =\; frac\{1\}\{n\}sum\_\{i=1\}^n\; S\_i\; to\; mathbb\{E\}S\_1$

- almost surely (using the strong law of large numbers); similarly:

- $frac\{J\_\{X\_t+1\}\}\{X\_t\}\; =\; frac\{J\_\{X\_t+1\}\}\{X\_t+1\}frac\{X\_t+1\}\{X\_t\}\; =\; frac\{J\_\{n+1\}\}\{n+1\}frac\{n+1\}\{n\}\; to\; mathbb\{E\}S\_1cdot\; 1$

- almost surely.

- Thus (since $t/X\_t$ is sandwiched between the two terms)

- $$

- almost surely.

- Next consider $(Y\_t)\_\{tgeq0\}$. We have

- $frac\{1\}\{t\}Y\_t\; =\; frac\{X\_t\}\{t\}\; frac\{1\}\{X\_t\}\; Y\_t\; to\; frac\{1\}\{mathbb\{E\}S\_1\}cdotmathbb\{E\}W\_1$

- almost surely (using the first result and using the law of large numbers on $Y\_t$).

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:

- $mathbb\{P\}(S\_\{X\_t+1\}\; >\; x)\; geq\; mathbb\{P\}(S\_1>x)\; =\; 1-F\_S(x)$

where F_{S} is the cumulative distribution function of the IID holding times S_{i}.

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

- $$

as required.

What is his optimal replacement policy?

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:

- $mathbb\{E\}S\; =\; mathbb\{E\}[S\; mid\; mbox\{fails\; before\; \}\; t]\; cdot\; mathbb\{P\}[mbox\{fails\; before\; \}\; t]\; +\; mathbb\{E\}[S\; mid\; mbox\{does\; not\; fail\; before\; \}\; t]\; cdot\; mathbb\{P\}[mbox\{does\; not\; fail\; before\; \}\; t]$

- $=\; frac\{t\}\{2\}left(0.5tright)\; +\; frac\{2-t\}\{2\}left(t\; right)$

and the expected cost W per machine is:

- $mathbb\{E\}W\; =\; mathbb\{E\}(W\; mid\; mbox\{fails\; before\; \}\; t)\; cdot\; mathbb\{P\}(mbox\{fails\; before\; \}\; t)\; +\; mathbb\{E\}[W\; mid\; mbox\{does\; not\; fail\; before\; \}\; t).mathbb\{P\}(mbox\{does\; not\; fail\; before\; \}\; t)$

- $=\; frac\{t\}\{2\}(2600\; )\; +\; frac\{2-t\}\{2\}\; (200\; )\; =\; 1200t\; +\; 200.,$

So by the strong law of large numbers, his longterm average cost per unit time is:

- $$

then differentiating with respect to t:

- $$

this implies that the turning points satisfy:

- $$

= 4800t - 1200t^2 -4800t - 800 + 2400t^2 + 400t

- $$

and thus

- $$

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.

- Poisson process
- Continuous-time Markov process
- Semi-Markov process
- Queueing theory
- Ruin theory
- Campbell's theorem
- Little's lemma

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Last updated on Thursday August 21, 2008 at 01:28:28 PDT (GMT -0700)

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