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Say I have a really weird die, with numbers 1 to 200. The die is not biased. The random variable X is defined as the number of times in which the number 200 is rolled. Say the number of trials is 500. Therefore, X is binomially distributed with parameters (500, 0.005). How can one change this problem such that we would actually need a Poisson distribution to model the situation? With that being said, let me point out that I don't know what the Poisson distribution distributes. What is the random variable whose distribution I'm looking at? How will the Poisson distribution also be looking at the probability of success? I simply don't know how to create an analogy between it and the binomial distribution.

[Note: I am aware that the Poisson distribution can be used to approximate the binomial distribution in scenarios such as the one pointed out above, since the number of trials and the probability of success is quite small.]

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Let's start with the Binomial distribution, $$ P(k; N,q) = {N \choose k} q^k (1-q)^{N-k}. $$

Let $\lambda = N q.$ Then we can write this as, $$ \begin{split} P(k;N,q) &= \frac{N!}{k! (N-k)!} q^k (1-q)^{N-k} \\ &= \frac{q^k}{k!} (1-q)^{N-k} \prod_{j=N-k+1}^N j \\ &= \frac{\lambda^k}{k!} (1 - \frac{\lambda}{N})^{N-k} \frac{1}{N^k} \prod_{j=N-k+1}^N j, \end{split} $$ where in the third line I substituted $q = \lambda/N.$

Now suppose we take a limit, $N \rightarrow \infty$ and $q \rightarrow 0$ such that $N q = \lambda$ remains constant. In this limit, for finite $k,$ the above expression becomes

$$ \begin{split} P(k; \lambda) &= \lim_{N \rightarrow \infty}\frac{\lambda^k}{k!} (1 - \frac{\lambda}{N})^{N-k} \frac{1}{N^k} \prod_{j=N-k+1}^N j \\ &= \lim_{N \rightarrow \infty}\frac{\lambda^k}{k!} (1 - \frac{\lambda}{N})^N \frac{N^k}{N^k} \\ &= \frac{\lambda^k}{k!} \lim_{N \rightarrow \infty} (1 - \frac{\lambda}{N})^N \\ &= \frac{\lambda^k}{k!} e^{- \lambda}, \end{split} $$ where in the second line, I used the fact that $k$ is finite to set $N-k \rightarrow N$ in the limit (which sets every factor in the product to $N$), and in the fourth line I made use of the famous definition of the exponential function.

So what does this mean? My original definition for $\lambda$ was $\lambda = N q$ which is the expected value of the outcome of the binomial distribution, i.e. $\mathbb{E}_k [P(k; N,q)] = N q.$ This is intuitively obvious, as the expected number of times we get a "success" is simply the probability of a success on one trial, multiplied by the number of trials. Because the Poisson distribution, $P(k; \lambda),$ is simply a limit of the Binomial distribution, as I have just demonstrated, $\lambda$ takes the same interpretation, i.e. it is the expected value of the outcome $k.$ Because the limit is where $N \rightarrow \infty$ and $q \rightarrow 0,$ it seems appropriate that the limit is for large $N$ and small $q$, while taking the value of $\lambda = N q.$ (In your example, $\lambda = 2.5.$)

As with the Binomial distribution, the Poisson distribution is a probability distribution over positive integers and $0.$ The reason it has one less parameter (only $\lambda$ as opposed to both $N$ and $q$) is because the Binomial distribution is bounded above by $k=N$, whereas the Poisson is unbounded above. This is reflective of the fact that the Poisson distribution is a limit of the Binomial distribution in which the number of trials is essentially infinite, so the value of $k$ is unbounded. This is a good approximation (even if the amount of trials is finite) if, in the Poisson distribution, the probability of getting an outcome equal to or greater than the true finite number of trials is negligible. In your example, 99.9% of the cumulative Poisson distribution lies in the first nine (0 to 8) observations of $k.$ So even though there are "possible" values of $k$ in the Poisson distribution that are nonsensical (where $k > N$), in the scenarios where the Poisson distribution is a good approximation, the probability of getting those nonsensical results is negligible.

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