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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?

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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 $P(N_2(T)=1|T=t)$.

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    $\begingroup$ (+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 $N_2$. $\endgroup$ – Nap D. Lover Jul 17 '19 at 0:12

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