# Difference between Poisson processes and Poisson distribution

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?

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.

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.

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, $$P(k)=\frac{\lambda^ke^{-\lambda}}{k!}$$