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My notes say the following about the Poisson paradigm:

Let $A_1, A_2, \dots, A_n$ be events with $p_j = P(A_j)$, where $n$ is large, the $p_j$ are small, and the $A_j$ are independent or weakly dependent. Let

$$X = \sum_{j = 1}^n I(A_j)$$

count how many of the $A_j$ occur. Then $X$ is approximately $Pois(\lambda)$ with $\lambda = \sum_{j = 1}^n p_j$.

$\lambda = \sum_{j = 1}^n p_j$ is the sum of all of the probabilities of the events that occur. And $\lambda$ is the rate of occurrence of events. But I'm wondering why $\lambda$, the rate of occurrence of events, would be equal to the sum of the probabilities of all the events that occur? I'm not seeing how this makes sense.

I would greatly appreciate it if people could please take the time to clarify this.

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The Pointer
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4 Answers4

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The $i$th event has probability $p_i$ of occurring. Let $X_i$ be $1$ if the event occurs and $0$ otherwise. Then $E(X_i)=p_i$.

$\lambda$ is the expected number of events that will occur. This is the sum of all the $E(X_i)$ and therefore the sum of the $p_i$.

  • Thanks for the answer. "$\lambda$ is the expected number of events that will occur"; how does this translate into also being the "rate of occurrence of events"? – The Pointer Sep 01 '19 at 17:42
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    My understanding of this topic (I'm on holiday at the moment and cannot easily access an authoritative source) is that the parameter $\lambda$ is called the event rate but is defined to be the expected number of events in the designated interval. I think this peculiar use of words is why your question is very worth asking. –  Sep 01 '19 at 17:52
  • Ahh, that clears up a lot of confusion that I had. Your answer is simple, yet clearly and directly answers the question. – The Pointer Sep 01 '19 at 18:49
  • You're very welcome. –  Sep 01 '19 at 18:58
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    Yes, I am sure it's a reasonable name but there is arguably a linguistic issue. If I am discussing a Poisson distribution with a student and we have, say, 4 incidences per day then I want us both to be clear that $\lambda$ is 4. I don't want the student to say $\lambda$ is 1 every 6 hours which would be a natural way of stating a rate. However, let's not argue about it - I think we all understand the maths! –  Sep 01 '19 at 19:56
  • P.S. Just taken a (not very random) sample. Asked about the rate of taking medication a majority do seem to say 'x a day' rather than 'every y hours'. –  Sep 01 '19 at 20:09
  • I'm a little bit perplexed why the second paragraph of my answer did not resonate with the OP in the same way as this answer did, seeing as we're basically saying the same thing (select $\lambda$ such that the expected value of the Poisson approximant agrees with the expected value of the random variable being approximated). Still, I'm glad they got it cleared up, even though I really don't think this view fully explains why any Poisson approximant is reasonable. It seems to only explain why Poisson($\lambda$) is "the best Poisson approximant". – Ian Sep 16 '19 at 01:59
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If $A_i$ occurs with probability $p_i$, then the expected number of $A_i$ occurring is also $p_i$. Since expected values are additive under all circumstances, $E[X]=\sum p_i$. So if this is a Poisson distribution, then necessarily one with the correct expected value, which is $\lambda$.

Hagen von Eitzen
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    Thanks for the answer. "If $A_i$ occurs with probability $p_i$, then the expected number of $A_i$ occurring is also $p_i$"; I don't see how that makes sense? If $A_j$ occurs with probability $p_j = 0.5$, then it seems to me that that doesn't necessarily mean that the expected number of $A_j$ occurring is $0.5$? – The Pointer Aug 27 '19 at 19:56
  • @ThePointer perhaps you see it indirectly: If I pay you 1 dollar if $A_i$ occurs (and nothing otherwise), then your expected win is ... ? – Hagen von Eitzen Aug 28 '19 at 18:27
  • The expected win is 1 dollar. But how does this answer my question? – The Pointer Aug 28 '19 at 19:01
  • @ThePointer If you get $1 when $A_i$ occurs and otherwise nothing then your expected earnings are $P(A_i)$. Mathematically, $E[I(A)]=P(A)$. This is basically what I was getting at in the second part of my answer. – Ian Aug 28 '19 at 22:01
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In the independent case, it is basically like the Poisson approximation to the binomial, which you can see by a direct calculation of a limit. In the weakly dependent case, you're basically approximating by pretending the dependence isn't there and then applying the previous form.

A more abstract way of viewing it is that the expected number of occurrences is definitely $\sum_{i=1}^n p_i$ regardless of any assumptions about the dependence, so whatever approximation that you make should at least have the same expected number of occurrences as the actual expected number. Since the Poisson only has one parameter anyway, this criterion locks down which Poisson you can use.

Ian
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  • Thanks for the answer, but it is not clear to me that this answers my question. – The Pointer Aug 27 '19 at 20:14
  • @ThePointer If the $A_j$ are independent events, then the indicators are independent Bernoulli random variables, and a sum of independent Bernoulli random variables is "basically" a binomial. Technically it's not, but you can think about how much error is really committed if you shuffle probabilities of the $A_j$ around in such a way as to preserve the sum while making all the probabilities the same. – Ian Aug 27 '19 at 20:38
  • @ThePointer The second perspective should make perfect sense, the question is about how much error is really committed in doing that (which can be seen to sometimes be a lot, for example think about the case where the events are mutually exclusive, so that $\sum_j I(A_j)$ is Bernoulli. This is obviously not weak dependence, but it shows that arbitrary dependence can't be permitted.) – Ian Aug 27 '19 at 20:39
  • But I'm not asking about the independence/dependence aspect, I'm asking why $\lambda$, the rate of occurrence of events, would be equal to the sum of the probabilities of all the events that occur. It still isn't clear to me how your answer addresses this question; it sounds like it's addressing something else. – The Pointer Aug 27 '19 at 20:45
  • @ThePointer It's not; look into the Poisson approximation to the binomial to see a more concrete example where you can see it more easily. (The result in the end is that a Binomial($n,p$) with $n$ large and $p$ small is well-approximated by a Poisson($np$).) – Ian Aug 27 '19 at 20:45
  • I did, but it is not clear what the Poisson approximation to the binomial has to do with my question. Again, I don't see how anything in your answer or comments directly addresses my question. – The Pointer Aug 28 '19 at 19:04
  • @ThePointer Do you see how you get a binomial when the $A_j$ are independent and the $p_j$ are the same? The general case is essentially analogous to this one. – Ian Aug 28 '19 at 19:06
  • I do see how we get a binomial when the $A_j$ are independent and the $p_j$ are the same. But I do not see the words binomial or independent in my question. Are you sure you're understanding my question? It seems like you're commenting on something the author has said, rather than my question. – The Pointer Aug 28 '19 at 19:30
  • @ThePointer "Independent" is indeed in your question, as it must be. What isn't there is the assumption that the $p_j$ are the same. But if you can at least understand how this works in the binomial case then we can talk in more detail about why it makes sense to approximate your distribution by a binomial with success probability equal to the average of the $p_j$. – Ian Aug 28 '19 at 21:08
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In a time interval of length $\frac{1}{\sum_i p_i}$, you expect one of the $A_i$'s to occur. Therefore, in a time interval of length $1$, you expect $\sum_i p_i$ of the $A_i$'s to occur. We thus say the event rate is $\sum_i p_i$. When discussing $\lambda$, the rate of occurrence of events, you look at a time interval of length $1$, but, as I did in my first sentence, you could ask "what time interval yields an expected number of $1$ $A_i$ occurring?", which is of course (and intuitively) equivalent (think of the analogy of frequency and period).

Another heuristic derivation of $\lambda$ being $\sum_i p_i$ is as follows. The probability no $A_i$ occurs is, on the one hand, $e^{-\lambda}$, and on the other is $\prod_i (1-p_i)$ (assuming independence). And just note that $\log\left(\prod_i (1-p_i)\right) = \sum_i \log(1-p_i) \approx \sum_i -p_i$ and $\log(e^{-\lambda}) = -\lambda$.

mathworker21
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