# Doubts about Proof of Durrett Theorem 3.7.4. Thinning of Poisson Process

poisson distributionpoisson processprobability theory

I am having trouble understanding Durrett's logic in his proof of the thinning of the Poisson process.

Here is the statement of the Theorem: $$N_j(t)$$ are independent rate $$\lambda P(Y_i = j)$$ Poisson processes. $$N(t)$$ is assumed to be a Poisson process with rate $$\lambda$$ (e.g., the number of cars arrive at a store at time t), and $$N_i(t)$$ is defined as the number of $$i \leq N(t)$$ with $$Y_i = j$$, where $$Y_i$$ is an additionally defined property associated with the arrivals (e.g., the number of passengers in the arrived cars).

In Durrett's proof, he first proved a simple case where this additional property $$Y_i$$ is binary. He defined $$P(Y_i = 1) = p$$ and $$P(Y_i = 2) = 1 – p$$, which is fine.

What I am having problems with are:

(1) He asserted that $$N_1(t)$$ and $$N_2(t)$$ are Poisson processes without proving it.

This is what I do not understand – isn't it part of the statement to be proved that $$N_1(t)$$ and $$N_2(t)$$ are Poisson processes? Why is this true?

(2) He proved that if $$X_i = N_i(t + s) – N_i(s)$$, $$X_1 = j$$ and $$X_2 = k$$, there must be $$j + k$$ arrivals between $$s$$ and $$s + t$$, so $$P(X_1 = j, X_2 = k) = e^{\lambda t} \frac{(\lambda t)^{j+k}}{(j+k)!}\frac{(j+k)!}{j!k!}p^j (1-p)^k = e^{\lambda pt} \frac{(\lambda p t)^j}{j!}e^{\lambda(1-p)}t \frac{(\lambda (1-p) t)^j}{j!}$$. Then he asserted that $$X_1 = Poisson(\lambda pt)$$ and $$X_2 = Poisson(\lambda (1-p)t)$$.

My question is, why can you assert that because $$P(X_1 = j, X_2 = k)$$ factors into two Poisson distributions, so $$X_1$$ and $$X_1$$ are independent, $$X_1$$ must be $$Poisson(\lambda pt)$$ and $$X_2$$ must be $$Poisson(\lambda (1-p)t)$$?

Thank you very much.

Here is the statement and the full proof of the theorem: (the questions are highlighted in blue)

(1) You are right- it is premature to state that $$N_1(t)$$ and $$N_2(t)$$ are Poisson processes before proving it. But he does prove it later.

(2) If $$X,Y$$ are random variables taking values in $$0,1,2,3, \ldots$$, and $$P(X=j, Y=k)=p(j)q(k)$$ for all $$k,j$$ where $$p(\cdot)$$ and $$q(\cdot)$$ are probability distributions, then $$P(X=j)=\sum_k p(j)q(k)=p(j)\sum_k q(k) =p(j)$$ and similarly $$P(Y=k)=q(k)\,.$$