Renewal theory

The renewal theory (Engl. Renewal theory ) is a specialized field of probability theory and deals with processes that after each reaching the initial state again behave like the start of the experiment .

motivation

A motivating example is the regular renewal of the incandescent lamp used to operate a luminaire , which must be replaced after each failure. The service life of an incandescent lamp is described by a random variable , the distribution of which is the same for all incandescent lamps in question, i.e. represents a known, characteristic property. It is also assumed that these lifetimes are independent of one another. It is now of interest how often the light bulb has to be changed on average; that is, one asks how many renewals are to be made before a given operating time.

Very similar tasks are obtained for more general maintenance work or for customer service times that appear at a dispatch point after a predetermined distribution and form a queue there. Here, the renewal theory provides information for optimal maintenance intervals or optimal staff availability at service points. If one understands the occurrence of a maintenance case as a case of damage, it becomes immediately understandable that the renewal theory is also important in actuarial mathematics .

definition

At each renewal time jumps by at least one unit. ${\displaystyle S_{n}}$ ${\displaystyle N}$

A renewal process is given by a sequence of independent, identically distributed non-negative random variables , with the probability being based on the underlying probability space . ${\displaystyle (T_{n})_{n\in \mathbb {N} }}$ ${\displaystyle P(T_{n}=0)<1}$ ${\displaystyle P}$

${\displaystyle S_{n}:=T_{1}+\ldots +T_{n}}$

is called the -th renewal time , where the constant function is also 0. The sequence is called the renewal sequence , an interval is called the renewal cycle and by definition has the length , which is therefore also called the cycle time. After all, you bet for ${\displaystyle n}$ ${\displaystyle S_{0}}$ ${\displaystyle (S_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (S_{n-1},S_{n}]}$ ${\displaystyle T_{n}}$ ${\displaystyle t\in [0,\infty )}$

${\displaystyle N(t):=|\{n\in \mathbb {N} |\,S_{n}\leq t\}|}$,

the number of all for whom the -th renewal time has not yet exceeded the point in time . The stochastic process so defined is called the renewal process . ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle t}$ ${\displaystyle (N(t))_{t\in [0,\infty )}}$

Remarks

Interpretation of the definition

These definitions are immediately understandable with the light bulb example above. models the operating time of the -th incandescent lamp, is the total lighting time produced by incandescent lamps one after the other, and finally the number of incandescent lamp changes required up to that point in time . The condition ensures that a newly installed incandescent lamp will certainly not fail again immediately; only then does it make sense to consider regular renewals over time. Similar interpretations for maintenance work, services or damage cases are obvious. ${\displaystyle T_{n}}$ ${\displaystyle n}$ ${\displaystyle S_{n}}$ ${\displaystyle n}$ ${\displaystyle N(t)}$ ${\displaystyle t}$ ${\displaystyle P(T_{n}=0)<1}$

Delayed renewal process

A variant that is frequently used is the so-called delayed renewal process, in which the distribution may deviate from the joint distribution of the others . This is necessary if you do not know the initial situation and therefore have to make a different assumption, or if, for example in the case of maintenance work, the original components are different from the spare parts that are regularly replaced. The actual renewal process only begins after what the term explains as a delayed renewal process. ${\displaystyle T_{1}}$ ${\displaystyle T_{n},n\geq 2}$ ${\displaystyle T_{1}}$ ${\displaystyle T_{1}}$

Withdrawal process

As a rule, with the occurrence of each renewal period, payments are associated, which can also be negative in the event of costs. Therefore, in addition to the data given in the above definition, a sequence of independent and identically distributed random variables is also considered, which represent the payouts at the nth renewal time . The total payout by the time is then ${\displaystyle (R_{n})_{n\in \mathbb {N} }}$ ${\displaystyle S_{n}}$ ${\displaystyle t}$

${\displaystyle R(t):=\sum _{n=1}^{N(t)}R_{n}}$.

The stochastic process is called the payout process associated with the renewal process . ${\displaystyle R(t)_{t\in [0,\infty )}}$

Many applications are about optimizing the data associated with this payout process. In a variant not considered here, it can be replaced by a function that develops during the -th renewal cycle , so that the above is the payout accumulated during the cycle. This avoids the jumps occurring during the renewal times. ${\displaystyle R_{n}}$ ${\displaystyle n}$ ${\displaystyle R_{n}}$

Basics

There is a renewal process as described above, the distribution function of the cycle times . The average cycle time is positive, otherwise it would almost certainly be 0, which would contradict the requirement . The following applies to the expected value of the renewal process ${\displaystyle F}$ ${\displaystyle T_{n}}$ ${\displaystyle \mu =E(T_{n})}$ ${\displaystyle T_{n}}$ ${\displaystyle P(T_{n}=0)<1}$

${\displaystyle m(t):=E(N(t))=\sum _{n=1}^{\infty }P(S_{n}\leq t)=\sum _{n=1}^{\infty }F^{*n}}$,

where the n-fold convolution is with itself. One calls the mean value function of the renewal process in an obvious way. Using the condition one can show that finite and therefore is finite almost everywhere. This also results in the limit value behavior expected for a meaningful renewal process ${\displaystyle F^{*n}}$ ${\displaystyle m}$ ${\displaystyle P(T_{n}=0)<1}$ ${\displaystyle m(t)}$ ${\displaystyle N(t)}$

${\displaystyle S_{n}\to \infty }$ almost certainly for ${\displaystyle n\to \infty }$

${\displaystyle N(t)\to \infty }$almost certainly for . ${\displaystyle t\to \infty }$

A much more precise statement can be made about the growth of : ${\displaystyle N(t)}$

${\displaystyle {\frac {N(t)}{t}}\to {\frac {1}{\mu }}}$almost certainly for . ${\displaystyle t\to \infty }$

This statement also applies below the expected value, that is

${\displaystyle {\frac {m(t)}{t}}\to {\frac {1}{\mu }}}$almost certainly for , ${\displaystyle t\to \infty }$

which is also known as the simple renewal theorem. This confirms the intuition that the number of renewals expected in the long term corresponds to the reciprocal of the expected duration between two renewals. The corresponding results are available for the associated payout process:

${\displaystyle {\frac {R(t)}{t}}\to {\frac {E(R_{1})}{\mu }}}$almost certainly for , ${\displaystyle t\to \infty }$

that is, in the long-term average, the payout per time is equal to the average payout of a renewal cycle divided by the average cycle length.

The Poisson process as a renewal process

The simplest case is when they are exponentially distributed with one parameter . Then a Poisson process to the parameter , i.e. H. is Poisson distributed to the parameter . In this case, and the renewal theorem becomes trivial, because the mean cycle time is because this is the expected value of an exponential distribution. ${\displaystyle T_{n}}$ ${\displaystyle \lambda }$ ${\displaystyle (N(t))_{t\in [0,\infty )}}$ ${\displaystyle \lambda }$ ${\displaystyle N(t)}$ ${\displaystyle \lambda t}$ ${\displaystyle m(t)=E(N(t))=\lambda t}$ ${\displaystyle \textstyle \mu ={\frac {1}{\lambda }}}$

application

To illustrate the terms introduced above, we consider the following strategy for replacing incandescent lamps, whose random lifetimes are given by independent and identically distributed random variables with a distribution function . We change at the latest after a time to be determined , which leads to costs in the amount of , and only earlier if the light bulb actually fails, which in addition to additional costs in the amount of . The renewal cycle therefore has the random length ${\displaystyle X_{n}}$ ${\displaystyle F}$ ${\displaystyle T}$ ${\displaystyle a}$ ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle T_{n}=\min(X_{n},T)}$.

At the time of the renewal one then has costs with an expected value ${\displaystyle n}$ ${\displaystyle R_{n}}$

${\displaystyle E(R_{n})=E(R_{1})=aP(X_{n}>T)+(a+b)P(X_{n}\leq T)=a(1-F(T))+(a+b)F(T)=a+bF(T)}$.

The average cycle time is

${\displaystyle \mu =E(T_{n})=\int _{0}^{\infty }P(T_{n}>t)\,\mathrm {d} t=\int _{0}^{T}P(X_{n}>t)\,\mathrm {d} t=\int _{0}^{T}(1-F(t))\,\mathrm {d} t}$.

In the long term, therefore, there are costs per time of

${\displaystyle \lim _{t\to \infty }{\frac {R(t)}{t}}={\frac {E(R_{1})}{\mu }}={\frac {a+bF(T)}{\int _{0}^{T}(1-F(t))\,\mathrm {d} t}}}$.

To determine the optimal change interval with known costs and known distribution functions , one must determine the minimum place of this expression as a function of . This is particularly easy if has a continuous density, because then and the integral are differentiable as a function of the upper limit , that is, the optimization methods of analysis can be used. ${\displaystyle T}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle F}$ ${\displaystyle T}$ ${\displaystyle F}$ ${\displaystyle F}$

See also

• Blackwell Renewal Kit

literature

• Ming Liao: Applied Stochastic Processes , CRC Press 2013, ISBN 1-4665-8933-7 , Chapter 3: Renewal Processes