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Let $N\left(\cdot\right)$ be a Poisson process with rate $1$ and let $\Lambda\left(t\right)$ be a non-decreasing right-continuous function. Define $N_{\Lambda}\left(t\right)=N\left(\Lambda\left(t\right)\right)$ , I need to show that given a time vector $0=t_{0}<t_{1}<...<t_{n}$ the increments $\left\{ N_{\Lambda}\left(t_{i}\right)-N_{\Lambda}\left(t_{i-1}\right)\,|\,1\leq i\leq n\right\} $ are independent.

One thing that completely baffles me is what happens if there are $i\neq j$ such that $\Lambda\left(t_{i-1}\right)=\Lambda\left(t_{i}\right)=\Lambda\left(t_{j-1}\right)=\Lambda\left(t_{j}\right)$ , in this case how is it possible for the increments to still be independant. For that matter if I take $\Lambda\left(t\right)\equiv1$ which meets the conditions how is it possible that the resulting process has independant increments?

Serpahimz
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  • Interesting question. Is there a reference for this question or did you come up with it? If former, I'd like to know the reference. If not, well, we'll work it out either way...:) – Gautam Shenoy Apr 24 '13 at 13:14
  • It's from an exercise I received but I don't know where the question was originally taken from so I'm afraid I can't provide a better reference. – Serpahimz Apr 24 '13 at 13:17

1 Answers1

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If $\Lambda(s)=\Lambda(t)$, then $N_\Lambda(t)-N_\Lambda(s)=0$ almost surely hence $N_\Lambda(t)-N_\Lambda(s)$ is independent of every random variable.

Did
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