Difference between Poisson processes and Poisson distribution

Question

We suppose that a factory has on average 3 call per minutes. What is the probability to have 3 call in 4 minutes? I'm always confuse. Should I use a Poisson random variable or a stochastic process? i.e. if $X$ is the number of call in $4$ minutes, does $X\sim Poiss(4\lambda )$ or do I have to use a Poisson process $(X_t)$ s.t. $X_t\sim Poiss(\lambda t)$? Well, at the end, will have that if $Y\sim Poiss(4\lambda )$, then $$\mathbb P(X_4=3)=\mathbb P(Y=3),$$

but can someone explain me when I have to use a Poisson process or just a Poisson distribution?

Answer 1

It's of course not a Poisson process. Here it's a Poisson distribution $\mathcal P(3)$. To be more precise, if $X_i\sim \mathcal P(3)$ are independents where $X_i$ denote the number of call during $1$ minute, then your r.v. should be $X_1+...+X_4\sim \mathcal P(12).$

• In general, a Poisson distribution allow you to answer to the question : What is the probability to receive $k$ calls per unit of time ?

• A Poisson process can do much better since indeed it answer to the question : What is the probability to receive $k$ call per unit of time ? But also to the question : what is the probability to receive a call between $t$ and $t+h$ ? A simple Poisson random variable only see the number of call globally (i.e. on a fix interval of time) but have no clue on what happen locally. I.e. if $X$ denote the number of call during $1$ minutes, it see what happen on intervals of the form $[a,a+1]$, but for example, it has no clue of what happen in $[a+\frac{1}{2}, a+\frac{1}{2}+\frac{1}{100}]$, whereas a Poisson process does.

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But notice that a Poisson process of rate $\lambda =3$ will also model the situation if you are interested on question such that : In average, how long time is there between two calls ? Or : After how long will I receive $10$ calls ? But in your actual problem, such a question doesn't arise, so using a poisson process is really not necessary. But indeed, you can answer solve your problem using Poisson processes.

Answer 2

You can think both in terms of Poisson process or Poisson distribution.

You are given that the rate of the process $\lambda$ is $3$ per minute, and asked about the probability of having 3 calls (arrivals) in 4 minutes; you can use directly the formula for the probability of having $k$ arrivals in time $\tau$ for Poisson process with rate $\lambda$, $$P(k,\tau)=\frac{(\lambda\tau)^ke^{-\lambda\tau}}{k!}$$

Alternatively, you can think that the number of calls (arrivals) in 4 minutes is Poisson random variable with parameter $\lambda = 3\cdot 4$ (average number of arrivals in 4 minutes), and use the formula for Poisson distribution, $$P(k)=\frac{\lambda^ke^{-\lambda}}{k!}$$