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I want to show that two Poisson processes $X$ and $Y$ are independent. So far, I have been able to prove that $X_{t}$ and $Y_{t}$ are independent for each time $t$. Somehow, one can deduce from this that for $s,t \geq 0$, $X_{s}$ and $Y_{t}$ are independent. Once I have established this, I can then conclude that the $\sigma$-algebras generated by $X$ and $Y$ are independent from which the independence of $X$ and $Y$ follows. Can somebody please explain to me why you can deduce that the processes are independent at different times from the fact that they are independent at the same time? I have the feeling it is somehow related to the independent increments property of Poisson processes but I don't really know how.

Thanks!

Linus
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  • Poisson Processes are memoryless, in the sense that if you condition on $X_s$ then $X_t-X_s$ is independent of $X_s$ for $t>s$. Try looking at $P(X_t,Y_s)$ with $t>s$ and condition on $X_s$. – Alex R. Aug 14 '13 at 01:06
  • Could you state the definition of $X$ and $Y$ (or is it too lengthy)? – saz Aug 14 '13 at 15:08
  • @AlexR., somehow I am doing something wrong. I have tried multiple things but nothing really works. This is what I have at the moment (sorry for the change in notation): \begin{align}\mathbb{P}(X_{s}^{a}=a,X_{t}^{b}=b \mid X_{s}^{b}=c) &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}=b\mid X_{s}^{b}=c)\ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c,X_{s}^{b}=c\mid X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c) \end{align} – Linus Aug 14 '13 at 15:49
  • @saz, they are pretty generic Poisson processes with rates $\lambda p$ and $\lambda (1-p)$ - for $p\in[0,1]$ - respectively. The whole question here is part of the proof that if I thin a Poisson process with rate $\lambda$ into two Poisson processes (as above), then these processes are again Poisson and independent of one another. – Linus Aug 14 '13 at 15:56
  • @Linus Ah, that's quite helpful. – saz Aug 14 '13 at 16:32
  • try multiplying both sides by $P(X_s^b=c) and integrating, to get rid of the conditional probability. – Alex R. Aug 14 '13 at 17:13
  • @AlexR., does this here work? $\begin{align} \mathbb{P}(X_{s}^{a}=a,X_{t}^{b}=b, X_{s}^{b}=c) &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c)\mathbb{P}(X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c,X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}=b,X_{s}^{b}=c)\end{align} Summing over all possible values of $c$ then yields the required result \mathbb{P}(X_{s}^{a}=a,X_{t}^{b}=b) &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}=b). \end{align*}$ – Linus Aug 14 '13 at 17:42
  • \mathbb{P}(X_{s}^{a}=a,X_{t}^{b}=b, X_{s}^{b}=c) &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c)\mathbb{P}(X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c,X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}=b,X_{s}^{b}=c) – Linus Aug 14 '13 at 17:46
  • Sorry guys, I can neither delete nor edit the comments above, somehow my code didn't work. Can an admin please delete the gibberish? Thanks! – Linus Aug 14 '13 at 17:52
  • Here is another try: \begin{align*}\mathbb{P}(X_{s}^{a}=a,X_{t}^{b}=b, X_{s}^{b}=c) &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c)\mathbb{P}(X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c,X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}=b,X_{s}^{b}=c) – Linus Aug 14 '13 at 17:53
  • @AlexR., ok, I multiplied on both sides as you said and got this:\begin{align} \mathbb{P}(X_{s}^{a}=a,X_{t}^{b}=b, X_{s}^{b}=c) &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c)\mathbb{P}(X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}-X_{s}^{b}=b-c,X_{s}^{b}=c) \ &=\mathbb{P}(X_{s}^{a}=a)\mathbb{P}(X_{t}^{b}=b,X_{s}^{b}=c) \end{align} Summing over all the possible values of $c$ completes the result. --- Is that okay? – Linus Aug 14 '13 at 17:53

1 Answers1

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Let $s<t$ and denote by $(W_t)_{t \geq 0}$ the process which you would like to thin. Then, $Y_t-Y_s$ and $X_t-X_s$ are independent from $\mathcal{F}_s := \sigma(W_r; r \leq s)$. Thus, by tower property,

$$\begin{align*} \mathbb{E} \exp(\imath \, \xi \cdot X_s + \imath \, \eta \cdot Y_t) &= \mathbb{E} \bigg[ e^{\imath \, \eta \cdot Y_s+\imath \, \xi \cdot X_s} \cdot \underbrace{\mathbb{E} \left( e^{\imath \, \eta \cdot (Y_t-Y_s)} \mid \mathcal{F}_s \right)}_{\mathbb{E}e^{\imath \, \eta \cdot (Y_t-Y_s)}} \bigg] \\ &= \underbrace{\mathbb{E}e^{\imath \, \eta \cdot (Y_t-Y_s)} \cdot \mathbb{E}e^{\imath \, \eta \cdot Y_s}}_{\mathbb{E}e^{\imath \, \eta \cdot Y_t}} \cdot \mathbb{E}e^{\imath \, \xi \cdot X_s}\end{align*}$$

This proves that $X_s$ and $Y_t$ are independent.

saz
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