Renewal theory is the branch of
probability theorythat 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.
A renewal process is a generalisation of the
Poisson process. In essence, the Poisson process is a continuous-time Markov processon the positive integers (usually starting at zero) which has independent identically distributed"holding times" at each integer (exponentially distributed) before advancing (with probability 1) to the next integer:. 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 IIDproperty of the holding times is retained).
Let be a sequence of
independent identically distributed random variables such that
We refer to the random variable as the "th" "holding time".
Define for each "n" > 0 :
each referred to as the "th" "jump time" and the intervals
being called "renewal intervals".
Then the random variable given by
is called a renewal process.
One may choose to think of the "holding times" as the time elapsed before a machine breaks for the "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" record the successive times at which the machine breaks and the "renewal process" records the number of times the machine has so far had to be repaired at any given time .
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 be a sequence of IID random variables ("rewards") satisfying
Then the random variable
is called a renewal-reward process. Note that unlike the , each 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" (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 . 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" are the successive (random) financial losses/gains resulting from successive eggs ("i" = 1,2,3,...) and records the total financial "reward" at time "t".
Properties of renewal processes and renewal-reward processes
We define the renewal function:
The elementary renewal theorem
The renewal function satisfies
Below, you find that the strong law of large numbers for renewal processes tell us that
To prove the elementary renewal theorem, it is sufficient to show that is uniformly integrable.
To do this, consider some truncated renewal process where the holding times are defined by where is a point such that which exists for all non-deterministic renewal processes. This new renewal process is an upper bound on and its renewals can only occur on the lattice . Furthermore, the number of renewals at each time is geometric with parameter . So we have
The elementary renewal theorem for reward renewal processes
We define the reward function:
The renewal function satisfies
The renewal equation
The renewal function satisfies
where is the cumulative distribution function of and is the corresponding probability density function.
Proof of the renewal equation
:We may iterate the expectation about the first holding time:
:But by the
law of large numbersfor renewal processes)
: (strong law of large numbers for renewal-reward processes)
:First consider . By definition we have:
:for all and so
:for all "t" ≥ 0.
:Now since we have:
almost surely(with probability 1). Hence:
:almost surely (using the strong law of large numbers); similarly:
:Thus (since is sandwiched between the two terms)
:Next consider . We have
:almost surely (using the first result and using the law of large numbers on ).
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.
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:
where "F""S" is the cumulative distribution function of the IID holding times "Si".
Proof of the inspection paradox
Observe that the last jump-time before "t" is ; and that the renewal interval containing "t" is . Then
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?
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 . 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 0.5"t". So the overall expected lifetime of the machine is:
and the expected cost "W" per machine is:
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:
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.
Continuous-time Markov process
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