If I have two Poisson processes, $X$ and $Y$, each with rate $\lambda$, then what is the rate of $Z$ where $Z=X-Y$.
Is it $2 \lambda$? and would this differ if $X$ and $Y$ had different rates?
Thank you.
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$Z$ is not a Poisson process. It does not have support over $\mathbb{N}$. – Learner Apr 11 '13 at 14:40
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@Learner Thank you for the clarification. – Kabamaro Apr 11 '13 at 14:58
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$Z$ would no longer be a Poisson process, because a Poisson process only jumps up. $Z$ can now be a negative.
The jump rate of $Z$ will be $2\lambda$ with the $Q$ matrix with each row being
$[ ....0, \lambda, -2\lambda, \lambda, 0, ...]$
If the rates are different, then it is just
$[ ....0, \lambda, -(\lambda+\gamma), \gamma, 0, ...]$
Lost1
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Thank you very much for the clarification. If I had $Z=XY$, then what would be the jump rate in that case? I am trying to understand how the jump rate changes based on the components of Z. Thanks again. – Kabamaro Apr 11 '13 at 14:57
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@Kabamaro you mean Z=X+Y? it would be $2\lambda$. $Z=XY$ is not a Poisson process. It doen't event jump even in step size of 1 – Lost1 Apr 11 '13 at 14:58
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I assume in this case Z would be Poisson with rate $2\lambda$ and jump rate $2\lambda$. Is that correct? – Kabamaro Apr 11 '13 at 14:59
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"I assume in this case Z would be Poisson with rate 2λ and jump rate 2λ"
this doesn't make much sense Poisson with rate 2λ means its jump rate is 2$\lambda$
– Lost1 Apr 11 '13 at 14:59 -
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