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Thinning theorem If $N= (N_t)_{t\geq0} $ is a poisson process rate $\lambda$ and it is thinned by removing incidents with probability p independently of each other and the poisson process, then what remains (N~) is a poisson process rate $\lambda (1-p)$ The proof given is as follows

need to prove this by proving N~ is markov and that the Q matrix has $q(n,n+1) = \lambda (1-p) $

Proof of markov is trivial, so consider $q(n,n+1) = lim_{t->0}[p_t(n,n+1)/t]$

$ = lim_{t->0}( P[$no thinning by t$]$P[N_t=1|N_0=0] + $P[N_t>1, $and there's only 1 incident not thinned by time t$|N_0=0])$

= $lim_{t->0} [(1/t)(1-p)\lambda t exp(-\lambda t)] $

As the second limit tends to 0 So $q(n,n+1) = \lambda (1-p) $

The part I don't understand is that 'the 2nd limit is 0' Can anyone help? Thanks

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Got something from the answer below? – Did Mar 10 '13 at 12:37

The probability that $N_t>1$ and there remains exactly $1$ unthinned incident is no more than the probability that $N_t>1$. But $$ {\Bbb P}[N_t>1\mid N_0=0]=1-e^{-\lambda t}(1+\lambda t)=e^{-\lambda t}(\frac{\lambda^2 t^2}{2} + \frac{\lambda^3 t^3}{6} + \cdots)=O(t^2), $$ so $$ \lim_{t\to 0} \frac{{\Bbb P}[N_t>1\mid N_0=0]}{t}=\lim_{t\to 0} O(t)=0. $$ Therefore, the second limit vanishes.

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Does this results hold for a general renewal process? see the problem here:… – Su20200 Jun 13 at 23:40

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