intuition - Relation between Poisson Distribution and Process?

I think I'm confused by the fact that they have similar names and both involve i.i.d. exponential RVs. From what I understand,

Poisson distribution: the probability of a particular number of events in a fixed interval. And the inter-arrival times are i.i.d. exponential RVs?

Poisson process: is a sequence of events such that their arrival times are i.i.d. exponential RVs.

So, what's the difference between these? Is there any relation between the two? If someone could give me a bit of a dumbed down intuitive understanding that would probably go a lot farther for me than a very technical treatment. Any help is appreciated, thanks.

Answer

There most certainly is a relation between the two!

In general, a stochastic process is simply an indexed collection of random variables defined on a probability space $(\Omega, \mathcal F,\mathbb P)$, e.g. $N=\{N(t):t\in I\}$ for some index set $I$. A Poisson process is one of the most well-known examples of an arrival process, which satisfies the following properties (with probability $1$, or "almost surely"):

1. $N(0) = 0$


2. If $s
   
3. The map $t\mapsto N(t)$ is right-continuous.


4. If $\lim_{t\to t_0^-} N(t)
   

The fourth condition describes what are called jumps, that is, the random times at which arrivals occur. If we set $T_0=0$ and

$$T_{n+1} = \inf\{t>T_n : N(t)>N(T_n)\}$$
for $n=0,1,2,\ldots$ then $T_n$ are the arrival times of the process $N(t)$. The times between arrivals $T_{n+1}-T_n$ are called interarrival times. A Poisson process is an arrival process which also satisfies the following conditions:

5. For any $s,t\geqslant 0$ and $0\leqslant u_1<\cdots
   
6. For any $s,t\geqslant 0$, $N(s+t)-N(t)$ is independent of $t$.

These are referred to as independent increments and stationary increments, respectively. From these conditions (and assuming that $N(t)$ is not identically zero) it can be shown that there exists $\lambda>0$ such that $\mathbb P(N(t)=k) = \frac{e^{-\lambda t}(\lambda t)^k}{k!}$ for all $t>0$, $k=0,1,2,\ldots$ (for brevity, I will omit the argument, which can be found in any text on stochastic processes). This $\lambda$ is the rate or intensity of the Poisson process.

From there, we can see that for $s,t> 0$ and $k=0,1,2,\ldots$,

$$\mathbb P(N(s+t)-N(t)=k) = \frac{e^{-\lambda s}(\lambda s)^k}{k!},$$
that is, the distribution of arrivals in a time interval of length $s$ follows a Poisson distribution with rate $\lambda s$. To derive the interarrival distribution, note that for for any $n\geqslant 0$ and $t>0$,
$$\mathbb P(T_{n+1}-T_n>t) = \mathbb P(T_1>t)$$
due to stationary increments, and
$$\mathbb P(T_1>t) = \mathbb P(N(t) = 0) = e^{-\lambda t},$$
hence the time between arrivals is exponentially distributed with rate $\lambda$.