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Exercise

In a factory the number of accidents follows a Poisson process, at the rate of two accidents per week. We ask:

a) Probability of occurrence: some accident, in a week.

b) Probability of occurrence: four accidents, in the course of two weeks.

c) Probability of occurrence: two accidents in one week, and two more in the following week.

Solution

$X$~$P(2) \Longrightarrow f_X(x)=\dfrac{e^{-2}\cdot2^x}{x!}$

  • a) $P(X≥1)=1-f_X(0)=0.864664716...$
  • b) $\lambda = 4 \Longrightarrow f_X(4) = 0.195366814...$
  • c) What does this mean? What should I really calculate?
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    Well, first of all, I don't think it's clear whether, in each part, they mean "exactly" or "at least". For part a you appear to assume that they mean "at least" but for part b you appear to assume they mean "exactly". Once you decide what is meant, the answer for part c is just the square of the result for one week only (as the number of occurrences for one week is independent of the number the next). – 2017-02-08

2 Answers 2

1

Let $X_{(t;t+\Delta t]}$ be the count of incidents in an interval starting at time $t$ and duration $\Delta t$, where the incidents belong to a Poisson Process with rate of occurrence $2$ per week.   Measure time units in weeks.

$\mathsf P(X_{(t;t+\Delta t]}= k) = \dfrac{(2\Delta t)^k e^{-2\Delta t}}{k!}$

a) Probability of occurrence: some accident, in a week.

$\mathsf P(X_{(t;t+1]} \geq 1)=1-\dfrac{2^0 e^{-2}}{0!}$ as you had (assuming calculations okay)

b) Probability of occurrence: four accidents, in the course of two weeks.

$\mathsf P(X_{(t;t+2]}=4) = \dfrac{4^4e^{-4}}{4!}$ as you had (assuming calculations okay)

c) Probability of occurrence: two accidents in one week, and two more in the following week.

(I) The counts of incidents in disjoint intervals of a Poisson Process, are independently distributed.

(II) The distribution for a count of incidents in an interval is not determined by the start of the interval, only its length.

$$\begin{align}\mathsf P(X_{(t;t+1]}=2\cap X_{(t+1;t+2]}=2)~&=~\mathsf P(X_{(t;t+1]}=2)\cdot\mathsf P(X_{(t+1;t+2]}=2) \tag I \\[1ex] &=~\mathsf P(X_{(0;1]}=2)^2 \tag{II} \\[1ex] &=~{\left(\dfrac{2^2e^{-2}}{2!}\right)}^2\end{align}$$

1

They are two separate Poisson random variables, I think. Week 1 is $X_1\sim P(2)$, week 2 is $X_2\sim P(2)$. You want $P(X_1=2 \& X_2=2)$.

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    By definition, the occurrences are independent. If this does not imply that they are also disjoint, then what you say would be calculated as the product of both probabilities. Or am I wrong? – 2017-02-08
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    Why would they be disjoint? "If this week there had been 2 accidents, then for sure there will not be 2 accidents next week"? Then they wouldn't be independent! – 2017-02-08
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    Therefore, since they are not disjoint, and are independent: Is the probability of occurring $ X_1 $ and $ X_2 $ the product of their probabilities? – 2017-02-08
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    Yes, it is. I hope it helps! And, as the other responder said, be careful if you want to do $X\geq x$ or $X=x$ in each part. – 2017-02-08
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    But if it says "4 accidents happen in the course of 2 weeks". Then it is not "at least 4" or "up to 4". Or am I wrong? – 2017-02-08
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    I am not saying it is wrong! I mean, like @lulu said, in part a) you were considering "at least". Now I see you have edited it. – 2017-02-08
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    That's because I trust the translator. I hate you Google haha. – 2017-02-08