Let $\{X_t\}_{t\ge 0}$ be a Poisson Process with parameter $\lambda$. Suppose that each event is type 1 with probability $\alpha$ and type 2 with probability $1-\alpha$. Let $\{X^{(1)}_t\}_{t\ge 0}$ the number of type 1 events up until time $t$ and $\{X^{(2)}_t\}_{t\ge 0}$ the number of type 2 events up until time $t$
Prove that $\{X^{(1)}_t\}_{t\ge 0}$ and $\{X^{(2)}_t\}_{t\ge 0}$ are Poisson Processes with parameter $\lambda \alpha$ and $\lambda(1-\alpha)$ respectively
Furrthermore prove that for each $t\ge 0$ the random variables $\{X^{(1)}_t\}_{t\ge 0}$ and $\{X^{(2)}_t\}_{t\ge 0}$ are independent
My attempt: In order to prove that they are poisson process I will use the next definition:
An stochastic process $\{Y_t\}_{t\ge 0}$ is a poisson process iff:
a) $Y_0=0$
b) It has independent increments
c) $Y_{t+s}-Y_{s}$~$Poisson(\lambda t)$ for any values $s\ge 0$ and $t>0$
a) For any $t\ge 0$ we have: $X_t=X^{(1)}_t+X^{(2)}_t$; we know that $\{X_t\}_{t\ge 0}$ is a poisson process hence $X_0=0$ $\Rightarrow X^{(1)}_0+X^{(2)}_0=0 \Rightarrow X^{(1)}_0=0$ and $X^{(2)}_0=0$
b)Let $n\in \mathbb N$ In this part I need to prove that for any $n$ arbitrary times $0<t_1\le t_2\le…\le t_n$ and states $x_1,…,x_n$
$$P[X^{(1)}_{t_1}=x_1,X^{(1)}_{t_2}-X^{(1)}_{t_1}=x_2,…,X^{(1)}_{t_n}-X^{(1)}_{t_{n-1}}=x_n]=P[X^{(1)}_{t_1}=x_1]P[X^{(1)}_{t_2}-X^{(1)}_{t_1}=x_2]…P[X^{(1)}_{t_n}-X^{(1)}_{t_{n-1}}=x_n]$$
I don´t know how to Formally prove this part, and I don´t think this is trivial. Any help would be highly appreciated
c) $$P[X^{(1)}_t=k]=\sum_{i=k}^\infty P[X^{(1)}_t=k|X_t=i]P[X_t=i]=\sum_{i=k}^\infty \binom{i}{k}\alpha^i(1-\alpha)^{i-k}{e^{-\lambda t}(\lambda t)^i \over i!}={e^{-\lambda \alpha t}(\lambda \alpha t)^k\over k!}$$
Know I need to compute $P[X^{(1)}_{t+s}-X^{(1)}_t=n]=\sum_{j=0}^\infty P[X^{(1)}_{t+s}-X^{(1)}_t=n|X^{(1)}_s=j]P[X^{(1)}_s=j]$
This part is also giving me trouble because I dont´know what to do from here
I would really apreciate if you can help me with this problem. Also I hope that this won´t be marked as a duplicate because I haven´t seen a formal proof about the splitting poisson process.
Best Answer
I will be using $p$ instead of $\alpha$.
We have the poisson process
$$X(t) = X_1(t) + X_2(t) $$
First we calculate the joint probability
$$P[X_1(t) = k, X_2(t) = m] = \sum_{n=0}^{\infty} P[X_1(t) = k, X_2(t) = m \mid X(t) = n]P[X(t) = n]$$
Note that
$$P[X_1(t) = k, X_2(t) = m \mid X(t) = n] = 0 \:\:\: \text{when}\:\: n \neq k+m$$
Now
$$P[X_1(t) = k, X_2(t) = m] = P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m]P[X(t) = k+m]$$
$$= P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m]e^{-\lambda t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$
Now, given that $k+m$ events occurred, since each event has probability $p$ of being a type $1$ event and probability $1-p$ of being a type $2$ event, it follows that
$$P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m] = \binom{k+m}{k} p^k(1-p)^m$$
Thus,
$$P[X_1(t) = k, X_2(t) = m] = \binom{k+m}{k} p^k(1-p)^m e^{-\lambda t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$
$$= \frac{(k+m)!}{k!m!} p^k(1-p)^m e^{-\lambda t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$
$$= e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t} \frac{[\lambda (1 - p)t]^m}{m!} \:\:\:\:\:\:\:\:\:(1)$$
Then
$$P(X_1 = k) = \sum_{m=1}^{\infty} P[X_1(t) = k, X_2(t) = m]$$
$$=e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t} \sum_{m=1}^{\infty} \frac{[\lambda (1 - p)t]^m}{m!}$$
$$= e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t} e^{\lambda(1-p) t}$$
$$=e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} \:\:\:\:\:\: (2)$$
which indicates that $X_1(t)$ is a poisson process with rate $\lambda p$.
Similarly, we can obtain
$$P(X_2(t) = m) = \sum_{k=1}^{\infty} P[X_1(t) = k, X_2(t) = m]$$
$$= e^{-\lambda(1-p) t} \frac{[\lambda (1 - p)t]^m}{m!} \:\:\:\:\:\: (3)$$
and so $X_2(t)$ is a poisson process with rate $\lambda (1-p)$.
Finally, from equations $(1)$, $(2)$ and $(3)$
$$P[X_1(t) = k, X_2(t) = m] = P[X_1(t) = k]P[X_2(t) = m]$$
Hence, $X_1(t)$ and $X_2(t)$ are independent.