How to understand this Poisson process in terms of the corresponding Binomial distribution in its limit?

Question

I am trying to understand Poisson processes:

Suppose it is observed that on average 180 cars per hour pass a specified Point on a particular road in the morning rush hour. Due to impending road works it is estimated that congestion will occur close to the city center if more than 5 cars pass the point in any one minute. What is the probability of congestion occurring?

and the solution is supposed to go like this:

This is a Poisson process, call it $X$. The average number of cars passing the point in one minute is $\frac{180}{60} = 3$. This implies $\lambda = 3$ (or $\mu = 3$), the Poisson parameter.
Thus, the probability mass function is $$P_X(x) = \frac{e^{-\lambda}\lambda^{x}}{x!} = \frac{e^{-3}\cdot3^{x}}{x!}$$
Hence, the probability of the occurrence of congestion is $$P_X(x>5) = 1 - \sum_{x=0}^{5}{\frac{e^{-3}\cdot3^{x}}{x!}} = 0.0839179$$

Though I somehow came up with the right solution myself without understanding it completely, what I don't understand is how it relates to the corresponding binomial distribution in its limit. What are $p$ and $n$ here? What is the trial and what is considered the success?

Answer 1

The connection between binomial and Poisson distributions is: Taking the limit of a binomial distribution yields a Poisson distribution. Every minute, 3 cars pass by on average. Then on average 1/20 car passes by every second; and there are 60 one-second intervals in 1 minute. So $p=1/20$ and $n=60$. If we choose to sample every 1 ms, then $p=1/20,000$, and $n=60,000$. If we want to consider the extreme case where we divide the interval with infinite numbers, we'll have to take the limit.

Therefore it's a binomial distribution: Bin($n,p$). The average occurrence of an event is $np$, loosely speaking. So we define $\lambda=np$ as the rate of these events. The Probability Mass Function (PMF) of $X\sim$Bin($n,p$) is

\begin{equation} P(X=k)=\left(\begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k} \end{equation} Now we take the limit as $n\rightarrow \infty$

\begin{align} &\lim_{n\rightarrow \infty}P(X=k) \\ =&\lim_{n\rightarrow \infty}\left(\begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k} \\ =&\lim_{n\rightarrow \infty}\frac{n!}{k!(n-k)!} \frac{\lambda^k}{n^k} \left(1-\frac{\lambda}{n}\right)^{n-k} \\ =&\frac{\lambda^k}{k!}\lim_{n\rightarrow \infty}\frac{n!}{(n-k)!n^k} \left(1-\frac{\lambda}{n}\right)^{n-k}, \quad\text{since $\lambda$ and $k$ are constants} \\ =&\frac{\lambda^k}{k!}\lim_{n\rightarrow \infty}\frac{n!}{(n-k)!n^k} \left(1+\frac{-\lambda}{n}\right)^{n}, \quad\text{as $n\rightarrow\infty$, $n-k$ is the same as $n$} \\ =&\frac{\lambda^k}{k!}\lim_{n\rightarrow \infty}\frac{n!}{(n-k)!n^k} \left[\left(1+\frac{1}{-n/\lambda}\right)^{-n/\lambda}\right]^{-\lambda} \\ =&\frac{\lambda^k}{k!}e^{-\lambda}\lim_{n\rightarrow \infty}\frac{n!}{(n-k)!n^k}, \quad\text{by one of $e$'s definition} \\ =&\frac{\lambda^k}{k!}e^{-\lambda}\lim_{n\rightarrow \infty}\frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k} \\ =&\frac{\lambda^k}{k!}e^{-\lambda},\quad\text{as $k$ is constant} \\ \end{align}

This confirms that the limit is a Poisson distribution.