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A text I'm reading says the following,

Consider the occurrence of any uncertain event over time or space in such a way that the average occurrence of the event over unit time or space is m. We may take the number of accidents occurring over a time period with m denoting the average number of accidents per month; or we may be interested in the number of defects occurring in a strip if cloth manufactured by a mill, with m denoting the average number of defects per metre. For each of such situations, we see the possibility of dividing the time or space interval into n very small segments such that within a small segment the conditions of the Bernoulli process hold. Thus, one month can be divided into (say) 30 x 24 x 60 intervals of one minute each, so that the probability of occurrence of an accident in any minute = m/(30X 24 X60), and reduces to a very small quantity, so that there is almost no chance of having two accidents occurring in one minute, The independence property of the Bernoulli trial also holds true here, as a one minute interval basically corresponds to a trial. Similar possibilities also exist in the cloth example.

Question: If m is the average occurrence of accidents in a month, then how can average occurrence of accidents in a minute i.e. m/(30 X 24 X 60) be equivalent to probability occurrence of accident in a minute?

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2 Answers 2

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Strictly speaking, it is only approximate. That is, if a probability of an event is very small then the average number of such events over multiple intervals is approximately equal to the probability of one event.

Each event occurs in some small interval. It is possible to have more than one event in a single interval. The average number over many intervals should be close to the the probability of one event in one interval. This average should also be a good estimate of the probability of one event in an interval.

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  • $\begingroup$ Can you please explain once more? What do you mean by ‘average number of such events over multiple intervals’? $\endgroup$
    – Quorthon
    Dec 28, 2023 at 21:33
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    $\begingroup$ Suppose there are 1000 intervals and 7 events occured. The average is the number of events, 7, divided by the number of intervals, 1000. $\endgroup$ Dec 28, 2023 at 21:37
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This text describes the link between the binomial and Poisson distribution, where the latter is a limiting case of the former as $n\rightarrow \infty$. Recall that the probability of seeing $x$ successes out of $n$ Bernoulli trials each with success probability $p$ - exactly what the binomial distribution describes - is:

$$ P(x)=\binom n xp^x(1-p)^{n-x} $$ The expected outcome of this process is $np$ which we can denote by $\lambda$, or $p=\lambda/n$. Rewriting the above in these terms:

$$ P(x)=\binom n x \bigg(\frac{\lambda}{n}\bigg)^x\bigg(1-\frac{\lambda}{n}\bigg)^{n-x} $$

With some expanding of the binomial coefficient and a bit of calculus you can then show (proofs here):

$$ \lim_{n \to \infty}P(x)=\frac{e^{-\lambda}\lambda^x}{x!} $$

This last right-hand side is exactly the Poisson point mass function. The approximation works both ways: a Poisson process can be thought of as a near-infinite number of Bernoulli experiments (which would make $p$ arbitrarily small for any finite $\lambda$ since $n\rightarrow \infty$), and you can approximate the expected value of a binomial distribution with very large $n$ through a Poisson distribution.

Showing an example calculation in R that this holds, let's say you have 1 accident per month:

## Probability of accident in a given minute
p <- 1/(30*24*60)

## Probability of 2 accidents in 1 minute - essentially zero
p**2
> 5.36e-10

## Probability of 1 accident in a given month, binomial
dbinom(1, 30*24*60, p)
> 0.3678837

## Probability of 1 accident in a given month, Poisson
dpois(1, 1)
> 0.3678794

## Can make interval arbitrarily small, let's use seconds instead
## Much smaller chance still of 2 accidents in one interval
(p/60)**2
> 1.49e-13

## Binomial even closer to Poisson now
dbinom(1, 30*24*60*60, p/60)
> 0.3678795
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  • $\begingroup$ Thanks but question remains that how intuitively, if no. of accidents in a month are 1, then probability of no. of accidents in a minute is 1/(30*24*60). How does minute average equate to the minute probability ? $\endgroup$
    – Quorthon
    Dec 28, 2023 at 21:32
  • $\begingroup$ This is $\lambda=np$ in my answer, approximation gets better the higher $n$, and thereby the smaller the interval and $p$. It's in essence arbitrary, you can always choose a smaller interval that makes the approximation better, that's what I showed with the last lines of code. The same holds for cloth & defects per meter: you can go per milli-, micro-, nanometer, each of those will make the binomial-Poisson approximation better ($n$ larger). Note that it's still the overall average (per month, not per minute) that's approximated: choosing $n$ fixes $p$ and vice versa for given $\lambda$. $\endgroup$
    – PBulls
    Dec 28, 2023 at 21:47

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