Lecture 7. Poisson and lifetime processes in risk analysis

Transcription

1 Lecture 7. Poisson and lifetime processes in risk analysis Jesper Rydén Department of Mathematics, Uppsala University Statistical Risk Analysis Spring 2014

2 Example: Life times of ball bearings In an experiment, the life times of ball bearings were recorded (million revolutions). The 22 observations, sorted in order: 17.88, 28.92, 33.00, 41.52, 42.12, 45.60, 48.48, 51.84, 51.96, 54.12, 55.56, 67.80, 68.64, 68.88, 84.12, 93.12, 98.64, , , , ,

3 Example: Life times of ball bearings Millions of cycles to fatigue Empirical distribution function, 22 observations of lifetimes for ball bearings. Parametric model: Weibull distribution?

4 Failure-intensity function Definition. For a r.v. T 0 there is a function Λ(t), called the cumulative failure-intensity function, such that R(t) := P(T > t) = e Λ(t), t 0. If T has a density, then R(t) = P(T > t) = 1 F (t) = exp ( t λ(s) ds ), 0 where the function λ(s), called the failure-intensity function, satisfies Λ(t) = t 0 λ(s) ds.

5 Failure-intensity function If the distribution of T is continuous, with pdf f (t), the failure intensity can be calculated as It can also be shown that λ(s) = f (s) 1 F (s). P(T s + t T > s) λ(s) = lim, t 0 t that is, for small values of t, λ(s)t is approximately the probability that an item of age s will break within the time period t. Classification of failure-intensity functions: IFR (Increasing Failure Rate), DFR (Decreasing Failure Rate), both mechanisms: Bathtube curve.

6 Examples, failure-intensity functions

7 Example 7.4: Life insurance Let T be a lifetime for a human being. In practice, R(t) is specified by standard tables, based on observed lifetimes of a huge number of people. One example is the Norwegian N-1963 standard. In life-insurance mathematics, a popular choice of λ(s) is the Gompertz-Makeham distribution with the failure-rate function λ(s) = α + βc s, where s is measured in years. For example, for N-1963, the estimates are α = , β = , ĉ = We want to solve the following problems: (i) Calculate the probability that a person will reach the age of at least seventy. (ii) A person is alive on the day he is thirty. Calculate the conditional probability that he will live to be seventy.

8 Estimation procedures The Nelson Aalen estimator estimates the cumulative failure-intensity function Λ(t) = t 0 λ(s) ds. Introduce the following notation: t i : time points for failures, d i : number of failures at time t i, n i : number of items at risk at time t i, i.e. number of items not yet failed prior to failure time t i. The estimator is given by and thus Λ (t) = t i t d i n i, R (t) = exp( Λ (t)). Note that R (t) 1 F n (t). If censoring is present, the values of n i are affected.

9 Log-rank test The log-rank test compares intensities, λ 1 (t) and λ 2 (t), in two groups (1 and 2). The aim is to test the hypothesis H 0 : λ 1 (t) = λ 2 (t). Consider the time points for events t 1, t 2,..., t D, both groups considered. Introduce the following notation: d i1 : number of events in group 1 at times t i, d i2 : number of events in group 2 at times t i, d i : d i = d i1 + d i2, n i1 : number of items in group 1 at risk at time t i, i.e. number of items not yet failed prior to failure time t i, n i2 : number of items in group 2 at risk at time t i, i.e. number of items not yet failed prior to failure time t i, n i : n i = n i1 + n i2.

10 Log-rank test, ctd. Test quantity: where ( Q = 1 D s 2 d i1 i=1 s 2 = D i=1 D i=1 d i n i ni d i n i d i n i1 n i ) 2, ni1n i2 n i 1. The test is similar to the χ 2 test and is as follows: If Q χ 2 α(1), reject H 0. (Note that since for X N(0, 1), X 2 χ 2 (1); hence, χ 2 α(1) = λ 2 α/2.)

11 Poisson Point Process (PPP) Definition. If the time intervals, T 1, T 2,..., between occurences of an event are independent, exponentially distributed variables with common failure intensity λ, then the times 0 < S 1 < S 2 <... when the event occurs form a Poisson point-process with intensity λ. S 1 S 2 S 3 S 4 T 1 T 2 T 3 T 4 t

12 Properties of a PPP Let λ be the intensity of a PPP. Then The time to the first event, T, is exponentially distributed: P(T > t) = e λt. The times between events, T i, are independent and exponentially distributed: P(T i > t) = e λt. The number of events, N(s, t), is Poisson distributed: P(N(s, t) = n) = (λt)n e λt. n! The number of events in disjoint time intervals are independent and (obviously) Poisson distributed.

13 A general Poisson process Let N(B) denote the number of events (or accidents) occuring in a region B. Consider the following list of assumptions: (A) More than one event can not happen simultaneously. (B) N(B 1 ) is independent of N(B 2 ) if B 1 and B 2 are disjoint. (C) Events happen in a stationary (in time) and homogeneous (in space) way, more precisely, the distribution of N(B) depends only on the size B of the region; for example N(B) Po(λ B ). The process for which we can motivate that (A B) are true is called a Poisson process. It is a stationary process with constant intensity λ if (A C) holds.

14 Poisson process in the plane B 2 B 1 B N(B) = 11, N(B 1 ) = 2, N(B 2 ) = 3.

15 Example 7.20: Pines Consider Japanese black pines in a square sampling region in a natural region. Area: m 2. In this area, 65 pines were found. Estimation of intensity? Poisson distribution valid?

16 Superposition (decomposition) of Poisson processes + Superpos. Decompos.

17 Superposition theorem Assume that we have two independent Poisson point-processes, Si I and Si II, with intensities λ I and λ II, respectively. Consider a point process S i which is a union of the point processes Si I and Si II. The point process of S i is a superposition of the two processes and is a PPP itself, with intensity λ = λ I + λ II.

18 Example: Transports (Problem 7.13) Lorries travel over a bridge. Assume that the times S i of arrivals form a Poisson process with intensity per day. Consider a scenario B = A lorry transports hazardous material. Let s assume that the scenario is independent of the stream of lorries. (This would not be the case if a chemical company is usually sending a convoy of lorries with hazardous material to the same destination). Statistics: with probability p = 0.08, a lorry transports hazardous material; with probability q = 0.92, other material is transported. 1. Find the probability that during a week (Monday Friday), more than lorries will travel over the bridge. 2. Find the probability that during a week (Monday Friday), there are more than 820 transports of hazardous materials.