# Poisson Process Thinning Process - Shock Arrival

I don't understand why $$P(N=n | T=t) = P(N_1(T)=n-1| T=t)$$ and it is not taking $$N_2(T)$$ into account.

I think it should be $$P(N=n | T=t) = P(N_1(T)=n-1| T=t) \cdot P(N_2(T)=1|T=t)$$ where $$P(N_2(T)=1|T=t) = \lambda p e^{-\lambda p t}$$.

Where am I getting wrong?

This question has confused me for a while. I finally understood it myself. This is because $$T$$ denote the time of failure, in other words, the condition of $$P(N=n|T=t)$$ already implies $$P(N_2(T)=1|T=t)$$.
• (+1) I was going to comment this but lost Internet connection at the moment—Indeed if $N(t)$ is the number of shocks until failure then $N(t)=n$ implies that by time $t$ we have $n-1$ shocks that did not cause a failure (from $N_1$) and $1$ shock that caused the failure, the most recent shock, from $N_2$. – Nap D. Lover Jul 17 '19 at 0:12