Renewal Theory Definitions, Limit Theorems, Renewal Reward Processes, Alternating Renewal Processes, Age and Excess Life Distributions, Inspection Paradox Chapter 7

Continuous Time Markov Chain: Exponential times between transitions Relax counting process Poisson Process: Counting process iid exponential times between arrivals Relax exponential interarrival times Renewal Process: Counting process iid times between arrivals Chapter 7

Counting Process A stochastic process {N(t), t  0} is a counting process if N(t) represents the total number of events that have occurred in [0, t] Then {N(t), t  0} must satisfy: N(t)  0 N(t) is an integer for all t If s < t, then N(s)  N(t) For s < t, N(t) - N(s) is the number of events that occur in the interval (s, t]. Chapter 7

Renewal Process A counting process {N(t), t  0} is a renewal process if for each n, Xn is the time between the (n-1)st and nth arrivals and {Xn, n  1} are independent with the same distribution F. The time of the nth arrival is with S0 = 0. Can write and if m = E[Xn], n  1, then the strong law of large numbers says that Note: m is now a time interval, not a rate; 1/ m will be called the rate of the r. p. Chapter 7

Fundamental Relationship It follows that where Fn(t) is the n-fold convolution of F with itself. The mean value of N(t) is Condition on the time of the first renewal to get the renewal equation: Chapter 7

Exercise 1 Is it true that: Chapter 7

Exercise 3 If the mean-value function of the renewal process {N(t), t  0} is given by Then what is P{N(5) = 0} ? Chapter 7

Exercise 6 Consider a renewal process {N(t), t  0} having a gamma (r,l) interarrival distribution with density Show that Hint: use the relationship between the gamma (r,l) distribution and the sum of r independent exponentials with rate l to define N(t) in terms of a Poisson process with rate l. Chapter 7

Limit Theorems With probability 1, Elementary renewal theorem: Central limit theorem: For large t, N(t) is approximately normally distributed with mean t/m and variance where s2 is the variance of the time between arrivals; in particular, Chapter 7

Exercise 8 A machine in use is replaced by a new machine either when it fails or when it reaches the age of T years. If the lifetimes of successive machines are independent with a common distribution F with density f, show that the long-run rate at which machines are replaced is the long-run rate at which machines in use fail equals Hint: condition on the lifetime of the first machine Chapter 7

Renewal Reward Processes Suppose that each time a renewal occurs we receive a reward. Assume Rn is the reward earned at the nth renewal and {Rn, n  1} are independent and identically distributed (Rn may depend on Xn). The total reward up to time t is If then and Chapter 7

Age & Excess Life of a Renewal Process The age at time t is A(t) = the amount of time elapsed since the last renewal. The excess life Y(t) is the time until the next renewal: SN(t) t A(t) Y(t) What is the average value of the age Chapter 7

Average Age of a Renewal Process Imagine we receive payment at a rate equal to the current age of the renewal process. Our total reward up to time s is and the average reward up to time s is If X is the length of a renewal cycle, then the total reward during the cycle is So, the average age is Chapter 7

Average Excess or Residual Now imagine we receive payment at a rate equal to the current excess of the renewal process. Our total reward up to time s is and the average reward up to time s is If X is the length of a renewal cycle, then the total reward during the cycle is So, the average excess is (also) Chapter 7

Inspection Paradox Suppose that the distribution of the time between renewals, F, is unknown. One way to estimate it is to choose some sampling times t1, t2, etc., and for each ti, record the total amount of time between the renewals just before and just after ti. This scheme will overestimate the inter-renewal times – Why? For each sampling time, t, we will record Find its distribution by conditioning on the time of the last renewal prior to time t Chapter 7

Inspection Paradox (cont.) SN(t) SN(t)+1 t-s t Chapter 7

Inspection Paradox (cont.) SN(t) SN(t)+1 t-s t For any s, so where X is an ordinary inter-renewal time. Intuitively, by choosing “random” times, it is more likely we will choose a time that falls in a long time interval. Chapter 7