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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.

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2 Answers 2

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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.

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    $\begingroup$ Your calculation only shows that the one-dimensional distributions of $X_1$ and $X_2$ are the anticipated Poisson distributions. It does not show, for example, that $X_1$ has independent increments. $\endgroup$ Commented May 9, 2016 at 19:14
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    $\begingroup$ It shows this "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 " Which implies independent increments. $\endgroup$
    – JKnecht
    Commented May 9, 2016 at 19:35
  • $\begingroup$ thanks for your answer @JKnecht! but how can I proof that this implies independent increments? I don´t think that this part is trivial $\endgroup$
    – user128422
    Commented May 10, 2016 at 23:43
  • $\begingroup$ Does this results hold for a general renewal process? see the problem here: math.stackexchange.com/questions/1825150/… $\endgroup$ Commented Jun 13, 2016 at 23:40
  • $\begingroup$ If I am not wrong, the summations should start at 0, because there is no apparent (to me) reason why (for example for the computation just after the word "then") $$X_2 \neq 0 $$ and also because the power series of exp starts at 0. $\endgroup$ Commented May 9, 2020 at 19:44
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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 $X^{(2)}$.

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