# Properties of Poisson Processes Further Investigated

I am reading Introduction to Probability Models by Sheldon M. Ross, and I am having a difficult time comprehending this example. The text explains this section 'Further Properties of Poisson Processes' (skipping the proofs of these propositions and suppositions) majorly with these passages:

"Consider a Poisson process {$$N(t), t \ge 0$$} having rate $$\lambda$$, and suppose that each time an event occurs it is classified as either a type I or type II event. Suppose further that each event is classified as a type I event with probability $$p$$ or a type II event with probability $$1-p$$, independently of all other events.

Let $$N_1(t)$$ and $$N_2(t)$$ denote respectively the number of type I and type II events occuring in $$[0,t]$$. Note that $$N(t) = N_1(t) + N_2(t)$$.

{$$N_1(t), t \ge 0$$} and {$$N_2(t), t \ge 0$$} are both Poisson processes having respective rates $$\lambdap$$ and $$\lambda(1-p)$$. Furthermore, the two processes are independent."

It then gives an example: "If immigrants to area $$A$$ arrive at a Poisson rate of ten per week, and if each immigrant is of English descent with probability $$1/12$$, then what is the probability that no people of English descent will emigrate to area $$A$$ during the month of February?"

The book just states the solution with little to no explanation: "By the previous proposition it follows that the number of Englishmen emigrating to area $$A$$ during the month of February is Poisson distributed with mean $$(4)(10)(1/12) = 10/3$$. Hence the probability desired is $$e^{-10/3}$$."

I, for whatever reason, am having a hard time comprehending this example (I'm sure it's simplistic, which is why I'm upset I'm not mentally grasping it). Could someone explain in detail what was done in this problem. Also, could someone give another example like this one that could maybe help the "light bulb" in my head "light up" and make sense of it. Any help is appreciated!

There's two parts to the solution: since immigrants are $$1/12$$ probability to be English and the rate of immigrants to area A is $$10$$ per week, then the rate of English immigrants to area A is Poisson with rate $$10(1/12)$$ per week by the splitting theorem in the book.

Next, there are 4 weeks in February, you can equivalently consider the rate of English immigrants to area A as Poisson with rate $$4\times10/12$$ per 4 weeks (that is, per February).

Then the probability no English immigrants arrive in February is just $$e^{-4(10)/12}$$.

Firstly they state that the rate of migration is $$10$$ per week. But the question asks about the whole month of February.

So you might be temped to write $$\lambda = 10$$ per week

This is correct, but really we want $$\lambda = 40$$ per month, assuming there are $$4$$ weeks in a month (which is pretty much correct for the month of February).

So the part where it states this: $$(4)(10)(1/12)=10/3$$, the first two numbers on the left hand side refer to $$4$$ weeks per month multiplied by $$10$$ immigrants per week. Which is just $$40$$ immigrants per month.

But what is the $$(1/12)$$? Well for each immigrant there is a $$(1/12)$$ probability of being English. But how does that tie back to the "general theorem"?

We clearly have a Poisson process if we are just looking at the arrival of immigrants at a rate of $$40$$ per month. But we can break this process down and define two types of migration: Type 1, the immigrant is of English descent and Type 2, the immigrant is not of English descent.

Now the total number of migrants (which we are happy to model with a Poisson process) after some amount of time $$t$$ we will denote as $$N(t), t \ge 0$$ with rate $$\lambda = 40$$ per month. However if we also denote $$N_1(t)$$ as the total number of Type 1 (English) immigrants, AND we denote $$N_2(t)$$ as the total number of Type 2 (Non-English) migrants then clearly $$N(t) = N_1(t) + N_2(t)$$. Because the sum of our Non-English immigrants and our English immigrants is just ALL our immigrants.

An interesting property of the Poisson distribution allows this next step (https://en.wikipedia.org/wiki/Raikov%27s_theorem).

We know that $$N$$~$$Pois(40)$$. But $$N = N_1 + N_2$$, which (By Raikov's theorem, assuming that $$N_1$$ and $$N_2$$ are independent of one another) implies that $$N_1$$~$$Pois(\lambda_1)$$ and $$N_2$$~$$Pois(\lambda_2)$$.

But what are the values for $$\lambda_1$$ and $$\lambda_2$$? Well the book is telling you that:

$$\lambda_1 = \lambda p$$

and

$$\lambda_2 = \lambda (1 - p)$$

Where $$p$$ corresponds to the probability of a Type 1 event occurring. Since only a Type 1 event or a Type 2 event can occur, then total probability tells us the probability of a Type 2 event is $$(1 - p)$$. From the above we can see that:

$$\lambda_1 + \lambda_2 = \lambda (p + 1 - p) = \lambda$$

Hopefully the above makes a little more intuitive sense now.

So back to the specific example, we have $$\lambda = 40$$, $$p = \frac{1}{12}$$, therefore $$\lambda_1 = \frac{40}{12}$$.

Thus the number of English migrants, $$N_1$$ ~ $$Pois(\frac{40}{12})$$. We are of course interested in the probability that we get no English migrants:

$$P(N_1 = 0) = e^{-\lambda_1} \frac{\lambda_1^{0}}{0!}$$

$$P(N_1 = 0) = e^{-\frac{40}{12}}$$

Which is the desired result.

• Eloquent and beautifully explained! Really appreciate it! Up-voted. – Trever Dec 3 '18 at 4:21
• You're welcome! – DanielOnMSE Dec 3 '18 at 6:33