Chapter 3

The Poisson process and its relatives

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Problem 3:1

Let {X(t)}\left\{ X(t)\right\} be a Poisson process with intensity λ>0\lambda > 0. Since the mean of the process is mX(t)=λtm_{X}(t) = \lambda t we can conclude that the Poisson process is not weakly stationary. Is the process Y(t)=X(t)−λtY(t) = X(t) - \lambda t weakly stationary?

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Problem 3:2

In a Poisson process one has observed four events in the interval (0,4](0,4]. Compute the probability that two occurred in (0,1](0,1] and the remaining two in (1,2](1,2].

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Problem 3:3

Fires in passenger trains are assumed to occur according to a Poisson process. During the period January 1, 1986 to December 31, 1994, there were 67 fires in passenger trains in Sweden. The total distance traveled by passenger trains was 544×106544 \times 10^{6} km. Estimate the intensity of passenger train fires (per train km), and construct a two-sided confidence interval with approximate confidence level 95. You may use the normal approximation.

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Problem 3:4

A Geiger counter is used to measure radioactive decay. To estimate the activity in a radioactive sample one can do the following experiment. First, the natural background radiation is measured during a time period TbT_{b} and then, during a time period TsT_{s}, one measures the sample plus background.

Assuming both background and sample radiation are Poisson processes, with intensities λb\lambda_{b} and λs\lambda_{s}, respectively, explain how the sample activity can be estimated. Also, for a fixed total experiment time T=Tb+TsT = T_{b}+T_{s}, find the best choice of TbT_{b} and TsT_{s} that makes the variance of the estimate as small as possible. Finally, construct an approximate confidence interval for λs\lambda_{s}.

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Problem 3:5

Let {X(t)}\left\{ X(t)\right\} be a Poisson process with intensity λ\lambda and denote by T1T_{1} the time of the first event. Then, P(T1≤s)=1−exp⁡(−λs)\mathbb {P}\left(T_{1} \leq s\right) = 1-\exp (-\lambda s). Show that, for 0<s<t0 < s < t,

P(T1≤s∣X(t)=1)=s/t. \mathbb {P}\left(T_{1} \leq s \mid X(t)=1\right) = s/t .
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Problem 3:6

Let {X(t)}\left\{ X(t)\right\} be an inhomogeneous Poisson process with time dependent intensity λ(t)=γt\lambda (t) = \gamma t. Find, with T1T_{1} as in the previous problem, the probabilities P(T1≤s)\mathbb {P}\left(T_{1} \leq s\right) and P(T1≤s∣X(t)=1)\mathbb {P}\left(T_{1} \leq s \mid X(t)=1\right).

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Problem 3:7

The risk for car accidents on a specific road varies with the time of the day; denote by λ(t)\lambda (t) the intensity with which there occurs an accident at a certain road at time tt, 0≤t≤240 \leq t \leq 24.

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(a)

What factors can influence λ(t)\lambda (t), directly or indirectly?

(b)

Assume λ(t)=0.001(2−cos⁡2πt/24)\lambda (t) = 0.001(2-\cos 2\pi t/24), and find the average number of accidents per day, and the probability of no accidents between six in the evening and six in the morning.

Problem 3:8

A muffin with raisins may be viewed as part of a three-dimensional Poisson process with raisins as the points. How many raisins does a muffin need to contain, on the average, in order to make the probability that a muffin contains no raisins at most 0.05? What objection can you find against the use of a Poisson process for raisins muffins?

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Problem 3:9

Germ-grain models are popular as models for spatial structures, for example in ecology and material science. In a simple germ-grain model, points are distributed over a region according to a Poisson process with intensity λ\lambda. With the Poisson points as centers, one places discs (or spheres in R3\mathbb {R}^{3}) with random sizes, independently of each other. One could think of trees of random size located in a forest according to a Poisson process.

A simple question is then: what is the probability that a point chosen at random is covered by at least one disc? Find that probability if the radiuses of discs are independent and uniformly distributed between 0 and a>0a > 0.

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Problem 3:10

Prove that every continuous function m(t)m(t) that satisfies

m(s+t)=m(s)+m(t) m(s+t) = m(s)+m(t)

is of the form m(t)=ktm(t) = kt.

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