Poisson process: compute E[N(3) |N(2),N(1)]

[kingwinner]

note: N(t) is the number of points in [0,t] and N(t1,t2] is the number of points in (t1,t2].

Let {N(t): t≥0} be a Poisson process of rate 1.
Evaluate E[N(3) |N(2),N(1)].

If the question were E[N(3) |N(2)], then I have some idea...
E[N(3) |N(2)]
=E[N(2)+N(2,3] |N(2)]
=E[N(2)|N(2)] + E{N(2,3] |N(2)}
=N(2)+ E{N(2,3]} (independent increments)
=N(2) + 1
since N(2,3] ~ Poisson(1(3-2)) =Poisson(1)

But for E[N(3) |N(2),N(1)], how can I deal with the extra N(1)?

Thanks for any help! :)

 

[bpet]

But for E[N(3) |N(2),N(1)], how can I deal with the extra N(1)?

Apply the Markov property

 

[kingwinner]

Apply the Markov property

hmm...what is Markov property? How do we use it here?
Does it mean that E[N(3) |N(2),N(1)] = E[N(3)]?

Also, is Poisson process considered as a Markov process? Why or why not?

Can someone please explain more?

Any help is greatly appreciated!

 

[bpet]

Can someone please explain more?

The wikipedia article is a good place to start.

 

[kingwinner]

http://en.wikipedia.org/wiki/Continuous-time_Markov_process
"Markov property states that at any times s > t > 0, the conditional probability distribution of the process at time s given the whole history of the process up to and including time t, depends only on the state of the process at time t."

So according to this, it does not depend on ANY of its past history, and therefore E[N(3) |N(2),N(1)] = E[N(3)]?