Poisson paradigm: Why is $\lambda$, the rate of occurrence of events, equal to the sum of the probabilities of all the events that occur?

Question

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.

Answer 1

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 $p_i$.

Answer 2

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 $\lambda$.

Answer 3

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.

Answer 4

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 $\log(e^{-\lambda}) = -\lambda$.