# Independence of Thinning of a Poisson process

Consider a process $X_t=\sum_{k=1}^{\infty} \mathbf{1}_{\{T_k\le t\}}$, and a new process $\hat{X_t}=\sum_{k=1}^{\infty} z_k \mathbf{1}_{\{T_k\le t\}}$, where $P\{z_k=0\}=p, P\{z_k=1\}=1-p$, and $\{z_k\}$ are iid , and $T_k=\sum_{i}^{k}\tau_j$ ,and $P(\tau_j\le t)=1-e^{-\lambda t}$. How can I show that $X_t$ and $\hat{X_t}$ are independent.

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 They are not.  – Did Dec 12 '11 at 17:26 do you mind explaining why? I am not too sure, thanks – John K Dec 12 '11 at 17:29 Why do you think they are? What did you try to show they are? What would you try to show they are not? – Did Dec 12 '11 at 17:43 Is it possible you wanted $\sum_{k=1}^\infty (1-z_k)\mathbf{1}_{\{T_k\le t\}}$ to be your definition of $X_t$? – Michael Hardy Dec 12 '11 at 20:30 ....if so, I think then you'd have independence. – Michael Hardy Dec 12 '11 at 20:31

## 1 Answer

$X_t=0 \implies \hat{X_t}=0$

That is because you are using the same sequence of random times.

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Perhaps you would consider expanding upon your answer? – mixedmath Dec 13 '11 at 4:33
What for?   – Did Jan 12 '12 at 10:06