Splitting Poisson process formal proof

Solution 1:

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.

Solution 2:

Another approach is through the martingale characterization of the Poisson process: A non-decreasing right-continuous process $X(t)$, $t\ge 0$, that increases only by unit jumps, is a rate $\lambda$ Poisson process if and only if $X(t)-\lambda t$ is a martingale. Let the jump times of $X$ be $0<T_1<T_2<\cdots$ and let the "type" random variables be given as an iid sequence $Y_1, Y_2,\ldots$ (independent of $X$) with $\Bbb P[Y_n=1]=\alpha=1-\Bbb P[Y_n=2]$. The process $X^{(1)}$ is then given by $X^{(1)}(t) = \sum_{n=1}^\infty 1_{\{T_n\le t\}}1_{\{Y_n=1\}}$. Enlarge the natural filtration $(\mathcal F^X_t)_{t\ge 0}$ of $X$ to $\mathcal G_t:=\mathcal F^X_t\vee\sigma\{Y_n1_{\{T_n\le t\}}, n=1,2,\ldots\}$. It's not hard to show that $X^{(1)}(t)-\lambda\alpha t$ is a martingale relative to $(\mathcal G_t)_{t\ge 0}$. Thus, $X^{(1)}$ is a Poisson process with rate $\lambda\alpha$. Similarly, $X^{(2)}$ is a Poisson process with rate $\lambda(1-\alpha)$.

As for the independence of $X^{(1)}$ and $X^{(2)}$, fix times $0\le t_1<t_2<\cdots<t_n$ and positive numbers $u_1,\ldots u_n, v_1,\ldots,v_n$, and consider $$ \Bbb E[\prod_{k=1}^n u_k^{X^{(1)}(t_k)}\cdot v_k^{X^{(2)}(t_k)}]. $$ Because $X^{(1)}$ is a Poisson process, it is not hard to write down an explicit expression for the martingale $M_1(t):=\Bbb E[\prod_{k=1}^n u_k^{X^{(1)}(t_k)}\,|\,\mathcal G_t]$, whose initial value is $M_1(0):=\Bbb E[\prod_{k=1}^n u_k^{X^{(1)}(t_k)}]$. In particular, $M_1$ has paths of bounded variation, and $\Delta M_1(t)\not=0$ only if $\Delta X^{(1)}(t)\not=0$. The same is true for the analogously defined $M_2$. Because $M_1$ and $M_2$ have no common jumps (since $X^{(1)}$ and $X^{(2)}$ have no common jumps!), the product $M_1M_2$ is a martingale. Thus $$ \eqalign{ \Bbb E[\prod_{k=1}^n u_k^{X^{(1)}(t_k)}\cdot v_k^{X^{(2)}(t_k)}] &=\Bbb E[M_1(t_n)M_2(t_n)]\cr &=\Bbb E[M_1(0)M_2(0)]=M_1(0)M_2(0)\cr &=\Bbb E[\prod_{k=1}^n u_k^{X^{(1)}(t_k)}]\cdot \Bbb E[\prod_{k=1}^n v_k^{X^{(2)}(t_k)}], } $$ proving the independence of $X^{(1)}$ and $X^{(2)}$.