# Understanding how to find the Covariance of this poisson process

$N(t),t>0\quad$ 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:

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

• What´s the meaning of $N(3), N(1),...$ ? – callculus May 15 '17 at 16:43
• Welcome to math.SE. Please see math.meta.stackexchange.com/questions/588/… for information on how to attract quality answers. – mlc May 15 '17 at 16:44
• @callculus, I have edited the post, hopefully this makes it clearer – Dashawn May 15 '17 at 16:49

$$C:=\mathrm{Cov}[N(3),N(6)-N(1)]=\mathrm{Cov}[N(3)-N(1)+N(1), N(6)-N(1)]$$ 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.