Poisson Process Decomposition Problems

Question

Hello I am trying to solve a homework of Poisson Process and I have a few about Poisson decomposition I cant solve.

I have a Poisson process $N(t)$ with $\lambda=100$/hour which decomposes into 2 Poisson processes $N_1(t)$ and $N_2(t)$ with probability $0.7$ and $0.3$ respectively. It basically means that each event on my Poisson process N(t) has two posibilities, being counted with the Poisson process N1(t) or being counted with the poisson process N2(t). A common example is that people arrive to a hospital with a Poisson process N(t) with certain rate. Once there, they can be hospitalized or not with probabilities P and (1-P). So N1(t) counts the hospitalized and N2(t) the ones who were not hospitalized.

I have to solve the following problems.

1) Obtain expression for $E[N(1) | N_2(1)>2]$

2) For $0 < t_1 < t_2 < t$: Solve: $E[N_1(t) | N(t_2) - N(t_1) = n]$

I will be very grateful for any help! Greetings

Answer 1

Notice for $t>0$ we have $N_1(t) \sim \text{Poisson}(70t)$, $N_2(t) \sim \text{Poisson}(30t)$, and $$N(t)=N_1(t)+N_2(t) \sim \text{Poisson}(100t)$$ Moreover, $$E(N(1)|N_2(1)>2)=\sum_{k=3}^{\infty}k \cdot \frac{P(N(1)=k,N_2(1)>2)}{P(N_2(1)>2)}$$ Also note for $k\geq 3$ $$\{N(1)=k\}\cap \{N_2(1)>2\}=\coprod_{j=3}^k\{N_1(1)=k-j\}\cap\{N_2(1)=j\}$$ Since the union on the right hand side is disjoint, $$P(N(1)=k,N_2(1)>2)=\frac{e^{-100}}{k!}\sum_{j=3}^k{k \choose j}30^j70^{k-j}$$ Using the Binomial Theorem, $$P(N(1)=k,N_2(1)>2)=e^{-100}\frac{100^k}{k!}\Bigg(1-(0.7)^k\bigg[1+\frac{3k}{7}+\frac{9k(k-1)}{98}\bigg]\Bigg)$$ Meanwhile, $$P(N_2(1)>2)=1-P(N_2(1)\in \{0,1,2\})=1-481e^{-30}$$ Wolfram Alpha says that $$E(N(1)|N_2(1)>2)=\sum_{k=3}^{\infty}k \cdot \frac{P(N(1)=k,N_2(1)>2)}{P(N_2(1)>2)}=\frac{100(e^{30}-346)}{e^{30}-481}\approx 100$$ For part (b) the expected number of arrivals by $N_1$ on $\big[0,t_1\big]\dot\cup\big[t_2,t\big]$ is $70(t+t_1-t_2)$ and since $$N_1(t_2)-N_1(t_1)|N(t_2) - N(t_1)=n \sim \text{Binomial}\Big(n,\frac{7}{10}\Big)$$ we see that $$E(N_1(t)|N(t_2)-N(t_1)=n)=\frac{7n}{10}+70(t+t_1-t_2)$$