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$N(t),t>0\quad$ is a poisson process with rate $\lambda$=2.5 and we need to solve $Cov[N(3),N(6)-N(1)]$

There is a worked example for this but it seems to be skipping some steps which is not obvious to me (mainly the first two steps, the variance part I follow), if somebody can help out it would be greatly appreciated:

$=Cov[N(3)-N(1),N(6)-N(1)]\\=Cov[N(3)-N(1),N(3)-N(1)]\\=Var[N(3)-N(1)]\\=Var[N(2)]\\=5$

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  • $\begingroup$ What´s the meaning of $N(3), N(1),...$ ? $\endgroup$ – callculus May 15 '17 at 16:43
  • $\begingroup$ Welcome to math.SE. Please see math.meta.stackexchange.com/questions/588/… for information on how to attract quality answers. $\endgroup$ – mlc May 15 '17 at 16:44
  • $\begingroup$ @callculus, I have edited the post, hopefully this makes it clearer $\endgroup$ – Dashawn May 15 '17 at 16:49
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$$C:=\mathrm{Cov}[N(3),N(6)-N(1)]=\mathrm{Cov}[N(3)-N(1)+N(1), N(6)-N(1)]$$ The linearity of covariance leads to $$C=\mathrm{Cov}[N(3)-N(1), N(6)-N(1)]+\underbrace{\mathrm{Cov}[N(1), N(6)-N(1)]}_{0}$$ The second summand equals to $0$ since $N(1)$ and $N(6)-N(1)$ are independent.

Do the same at the second step: $N(6)-N(1)=(N(6)-N(3))+(N(3)-N(1))$ and use independence.

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  • $\begingroup$ You are welcome! If you wish, you can also consider the possibility to "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – NCh May 15 '17 at 17:02

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