The Coefficients and the Poisson Distribution

Now let’s consider an approximation for the binomial distribution, where we consider a very large number of trials n and all with a very small individual probability p. When we do this, we move from the typical Binomial process to a Poisson Process. To visualize this, let’s imagine a sequence of the form \{Bi(n;k;p_n)\}_n, where n\to\infty and p_n satisfies the relationship np_n=\lambda \gt 0. From this, we will see that

\displaystyle\lim_{n\to\infty}P\left(Bi(n;k;P_n) \right) = \frac{\lambda^k}{k!}e^{-\lambda}

This is actually not difficult to prove, if we take the probability of a Bernoulli event Bi(n;k;p_n) and multiply and divide by n^k, we get the following reasoning:

So if we calculate the limit when n\to\infty, we will have:

\begin{array} \displaystyle \lim_{n\to\infty} {{n}\choose{k}}p_n^k(1-p_n)^{n-k} &= \lim_{n\to\infty} \underbrace{\frac{n(n-1)\cdots[n-(k-1)]}{n^k}}_{\to 1} \cdot \frac{\overbrace{(np_n)^k}^{\to\lambda^k}}{k!} \overbrace{(1-p_n)^{-k}}^{\to 1} {(1-p_n)^n} \\ \\ &\displaystyle = \frac{\lambda^k}{k!} \lim_{n\to\infty}\left(1 - \frac{\lambda}{n} \right)^n \\ \\ & \displaystyle = \frac{\lambda^k}{k!}e^{-\lambda} \end{array}

From this, the Poisson coefficients, Po(k;\lambda), are defined as follows:

\displaystyle Po(k;\lambda) := \lim_{n\to\infty} {{n}\choose{k}}p^k(1-p_n)^{n-k} = \frac{\lambda^k}{k!}e^{-\lambda}

And a random variable X is said to have a Poisson distribution, X\sim Po(k,\lambda), if it holds that:

P(X=k) = Po(k;\lambda)

Approximate Poisson Processes

Spatial Poisson Process

Suppose we have a container of volume V with a liquid containing n tiny particles uniformly mixed. Here we assume the liquid is well stirred and the particles do not interact with each other, neither attracting nor repelling. These are assumptions that can be formalized through the following statements:

• Hypothesis of Spatial Homogeneity: The probability of finding a particle in a region D of the liquid depends solely on the volume of that region.
• No-Interaction: The events “the j-th particle is in the region D,” with j=1,2,…,n are all n-independent.
• No-Overlap: Two particles cannot occupy the same place in space.

If we are given a region D with a volume v, the probability of the event “there are k particles in D” depends solely on v; let’s call g_k(v) such an event. Let h(v) be the probability that the particle is inside a region of volume v. If D_1 and D_2 are two disjoint regions of volume v_1 and v_2 respectively, then if D=D_1\cup D_2, has a volume v, then v=v_1+v_2. And since D_1 and D_2 are disjoint (D_1\cap D_2 = \emptyset ), it follows that

h(v) = h(v_1) + h(v_2)

If V is the volume of the entire liquid, then

h(V) = 1

And consequently:

h(v) =\displaystyle \frac{v}{V}

From here we have that the event g_k(v) is actually a Bernoulli-type event with p=v/V and is given by:

g_k(v) =B(n;k;p=v/V)

However, most practical situations of this kind involve a large number of particles n and the regions considered tend to be small relative to the size of the system, so that the conditions for applying the Poisson approximation are met and it holds that:

\displaystyle P(g_k(v)) = \lim_{\begin{matrix}n\to\infty\\ v/V=c \end{matrix}}P(B(n;k;p=v/V)) =\displaystyle \frac{(cv)^k}{k!}e^{-cv}

Temporal Poisson Process

Suppose we are recording the number of particles emitted by a radioactive substance starting at time t=0, and from that, we will calculate the probability that exactly k particles are emitted in the interval [0,t[ under the following assumptions:

• Invariance: The conditions of the experiment do not change over time.
• No-Memory: What happened in [0,t[ does not affect what happens in [t,t'[.
• Isolated Events: Particles are emitted one at a time.

If we compare the assumptions of the temporal process with those of the spatial process, we will see that they are formally equivalent. Just as the probability of finding a particle in a region does not depend on the location of the region but only on its size, the probability of observing the emission of a particle does not depend on the moment chosen to measure, but only on the observation interval. The no-memory assumption is analogous to the no-interaction assumption of spatial processes: what happened at one time does not affect what happens at other times. And finally, isolated events imply that at any given moment, only one particle can be emitted, just as a place in space can only be occupied by one body at a time.

Thus, if we define the event “k particles are emitted in a time interval t,” its probability of occurring will be an event of the form g_k(t), that is:

P(g_k(t)) =\displaystyle \frac{(ct)^k}{k!} e^{-ct}

Temporal and Spatial

Both processes, spatial and temporal, are formally equivalent. They only vary in how they are interpreted for practical purposes. A quick way to make this distinction clearer is by observing the role played by the constant “c” that appears in both cases. For the exponential function to be well-defined, its argument must be dimensionless; however, it contains units of time or space depending on whether we are dealing with temporal or spatial processes. This problem is fixed precisely by the constant c. We have:

Po(k;\lambda)=\displaystyle \frac{\lambda^k}{k!}e^{-\lambda}=\left\{\begin{matrix} {Taking\,\lambda = \rho v } & \longmapsto &\displaystyle \frac{(\rho v)^k}{k!}e^{-\rho v} & {Spatial\,Process} \\ {Taking\,\lambda = \nu t } & \longmapsto &\displaystyle \frac{(\nu t)^k}{k!}e^{-\nu t} & {Temporal\,Process} \end{matrix} \right.

• If c=\rho, it is a spatial density (number of things per unit space), therefore it defines a spatial Poisson process.
• If c=\nu, it is a temporal density (or frequency, number of occurrences per unit time), therefore it defines a temporal Poisson process.

Practical examples where the Poisson distribution is used

1. The checkout at a supermarket serves an average of 2 customers every 9 minutes. Create a table showing the probabilities of serving between 1, 2, 3, and so on, up to 5 people in a 5-minute time frame.
2. A veterinary clinic has the capacity to serve a maximum of 12 clients per day. If they receive an average of 9 clients each day, what is the probability that on any given day, the clinic’s capacity will be exceeded?
3. A certain locality has a population density of 10 people per 1000 square meters. What is the probability that in an area of 60 square meters, we find less than 15 people?
4. A chicken wants to cross the road. Walking in a straight line, it takes 58 seconds. If the road has a traffic flow of 3 vehicles per minute, and if a vehicle passes while the chicken is attempting to cross, it will certainly be run over with fatal results. What is the probability that the chicken will make it to the other side alive?