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I have got 2 theorems,

Theorem 1 The increment $ (N_{t+u} - N_t)_{u\geq 0} $ of a Poisson process rate $\lambda$ is again a Poisson process rate $\lambda$ and is independent of $(N_s)_{0\leq s \leq t}$

Proof

$ P(N_{t+u} - N_t)= k|N_t=n, T_1=t_1,....T_n=t_n) $

$= P(N_{t+u} - N_t)= k) $

$= p_u (0,k) $

Theorem 2 The strong Markov property

$N = (N_t)_{t\geq0} $is a Poisson process rate $\lambda$

Let T be a stopping time

Define $N^T $ via $N^T_t = N_{T+t} - N_T $

Conditional on T being finite $(N^T_t)_{t\geq 0} $ is a Poisson process rate $\lambda$ which is independent of $ (N_t)_{0\leq t \leq T}$

Can anybody

A) explain the first proof why does it hold?

B) explain exactly how these 2 theorems are different?

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@Shyam I think the problem is that my lecturer has defined a poisson process as a counting process such that $ (N_t)_t\geq 0$ is markov and q(x,x+1) = lambda,q(x,y) = 0 for y not in {x,x+1} (whereas I know most definitions define it as having independent increments etc..) –  Rosie Feb 20 '13 at 11:18

1 Answer 1

The first proof is a direct consequence of the independent increment property of Poisson processes: the number of arrivals in $(t,t+u)$ is independent of $N_t$.

Strong Markov property is a result which states that the Markovian property of Theorem 1 not only holds for deterministic times, but also for some particular random times, which are called stopping times.

Strong Markov is a useful and strong extension of Markovity in Poisson processes, DTMCs and CTMCs.

share|improve this answer
    
Thanks for your reply. But I am still a bit confused. So does the proof prove not that the increments are independent but that they form a poisson process rate lambda? Also if the increments are independent i.e.the number of arrivals in (t,t+u) is independent of $ N_t $ then I can see that $N_{t+u} - N_t$ doesnt depend on $ N_t $ but how do we know that $N_{t+u} - N_t$ doesnt depend on $T_1, ...T_n$? –  Rosie Feb 19 '13 at 12:21
    
@Rosie: are you the same Rosie as the one who asked the question? If so, please visit our account merge page to consolidate your accounts. That way you will be able to comment on answers to your own questions. –  Willie Wong Feb 19 '13 at 13:03

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