# Strong Markov property explained.

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

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 $N_t$.
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