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
    $\begingroup$ (+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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.