0
$\begingroup$

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?
$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – lulu Feb 8 '17 at 1:37
1
$\begingroup$

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.

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

$\endgroup$
1
$\begingroup$

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 $P(X_1=2 \& X_2=2)$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – emi Feb 8 '17 at 1:39
  • $\begingroup$ 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! $\endgroup$ – Anna SdTC Feb 8 '17 at 1:50
  • $\begingroup$ Therefore, since they are not disjoint, and are independent: Is the probability of occurring $ X_1 $ and $ X_2 $ the product of their probabilities? $\endgroup$ – emi Feb 8 '17 at 1:55
  • $\begingroup$ 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. $\endgroup$ – Anna SdTC Feb 8 '17 at 1:56
  • $\begingroup$ 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? $\endgroup$ – emi Feb 8 '17 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.