Poisson process and interrarrival times Certain electrical disturbances occur according to a Poisson process with rate $3$ per hour. These disturbances cause damage to a computer.
Assume that a crash will not happen unless there are two disturbances within $5$ minutes of each other. Calculate the probability that the computer will crash in the incoming 10 minutes.
Let $n$ be the total number of electrical disturbances by time $t$, so this means
$$N(t) = n
$$
Let $X_{i+1}$ be the $i+1$'th inter-arrival time, that is, the time between the $i+1$'th disturbance and the $i$'th disturbance.
Then $X_{i+1} \overset{\text{iid}}{\sim} \mathrm{Exp}(3t)$ for $i\in\{1,2,\ldots,n-1\}$
Then we seek to compute
$$\mathbb{P}(\{\exists i \in\{1,2\ldots,n-1\} : X_{i+1}\leq \frac{1}{12}\}) \\
= 1-\mathbb{P}(\forall i \in\{1,\ldots,n-1\}, X_{i+1} > \frac{1}{12}\})\\
= 1 - \mathbb{P}(X_{i+1}<\frac{1}{12})^{n-1} \\ = 1-e^{-t(n-1)/4}.
$$
where $t = \frac{1}{6}$.
Is this correct? Is there another way of thinking of this?
 A: Your argument seems on the right track, yet might observe further improvement. In fact, should there be more than two disturbances within the first ten minutes, the computer would definitely crash as it is a must that some two disturbances occur within five minutes. Therefore, a correct formula seems to predict
$$
\mathbb{P}\left(\left\{X_{j+1}\le\frac{1}{12}\text{ for some }j\in\left\{1,2,\cdots,n-1\right\}\right\}\right)=1
$$
for all $n>2$.

Here is a possible alternative.
Let $N_t$ be a Poisson process with $N_1\sim\text{Poiss}(\lambda)$. Take "hour" as the unit of time, and $\lambda=3$.
Let $t=0$ be the time at present, with $N_0=0$. Let $t_1=1/12$ be the most inter-arrival time that will trigger a crash (i.e., $5$ minutes). Let $t_2=1/6$ be the duration of the crash-or-not (i.e., $10$ minutes).
Define
$$
\tau=\inf\left\{t>0:N_t>0\right\},
$$
the moment when the first disturbance occurs. Note that for all $t>0$, the first disturbance occurs before the moment $t$ if and only if at least one disturbance has occurred by the moment $t$, for which
$$
\mathbb{P}\left(\tau\le t\right)=\mathbb{P}\left(N_t>0\right).
$$
This implies that
$$
\tau\sim\text{Exp}\left(\lambda\right).
$$
Let $f_{\tau}(t)=\lambda e^{-\lambda t}$ be the probability density of $\tau$.
The target is to figure out $\mathbb{P}\left(\left\{\text{crash before }t_2\right\}\right)$. Note that the computer would crash if and only if either of these mutually exclusive cases takes place.


*

*$\tau\le t_1$ and $N_{\tau+t_1}-N_{\tau}>0$ (the first disturbance occurs within the first five minutes, and upon this occurrence, there is at least one disturbance in the next five minutes).

*$\tau\le t_1$ and $N_{\tau+t_1}-N_{\tau}=0$ and $N_{t_2}-N_{\tau+t_1}>1$ (the first disturbance occurs within the first five minutes, but it does not contribute to the disturbances that cause a crash within the first ten minutes).

*$\tau>t_1$ and $N_{t_2}>1$ (no disturbance occurs in the first five minutes, and more than one disturbance occurs in the last five minutes).


It suffices to calculate the probability of each of the above three cases.
Note that
\begin{align}
\mathbb{P}\left(\tau\le t_1,N_{\tau+t_1}-N_{\tau}>0\right)&=\mathbb{P}\left(N_{\tau+t_1}-N_{\tau}>0|\tau\le t_1\right)\mathbb{P}\left(\tau\le t_1\right)\\
&=\mathbb{P}\left(N_{\tau+t_1-\tau}>0|\tau\le t_1\right)\mathbb{P}\left(\tau\le t_1\right)\\
&=\mathbb{P}\left(N_{t_1}>0\right)\mathbb{P}\left(\tau\le t_1\right)\\
&=\mathbb{P}\left(N_{t_1}>0\right)\mathbb{P}\left(N_{t_1}>0\right),
\end{align}
and that
\begin{align}
&\mathbb{P}\left(\tau\le t_1,N_{\tau+t_1}-N_{\tau}=0,N_{t_2}-N_{\tau+t_1}>1\right)\\
&=\mathbb{P}\left(N_{\tau+t_1}-N_{\tau}=0,N_{t_2}-N_{\tau+t_1}>1|\tau\le t_1\right)\mathbb{P}\left(\tau\le t_1\right)\\
&=\mathbb{P}\left(N_{\tau+t_1}-N_{\tau}=0|\tau\le t_1\right)\mathbb{P}\left(N_{t_2}-N_{\tau+t_1}>1|\tau\le t_1\right)\mathbb{P}\left(\tau\le t_1\right)\\
&=\mathbb{P}\left(N_{\tau+t_1}-N_{\tau}=0|\tau\le t_1\right)\mathbb{P}\left(N_{t_2}-N_{\tau+t_1}>1,\tau\le t_1\right)\\
&=\mathbb{P}\left(N_{\tau+t_1-\tau}=0|\tau\le t_1\right)\mathbb{P}\left(N_{t_2}-N_{\tau+t_1}>1,\tau\le t_1\right)\\
&=\mathbb{P}\left(N_{t_1}=0\right)\mathbb{P}\left(N_{t_2}-N_{\tau+t_1}>1,\tau\le t_1\right)\\
&=\mathbb{P}\left(N_{t_1}=0\right)\int_0^{t_1}\mathbb{P}\left(N_{t_2}-N_{\tau+t_1}>1|\tau=t\right)f_{\tau}(t){\rm d}t\\
&=\mathbb{P}\left(N_{t_1}=0\right)\int_0^{t_1}\mathbb{P}\left(N_{t_2-\tau-t_1}>1|\tau=t\right)f_{\tau}(t){\rm d}t\\
&=\mathbb{P}\left(N_{t_1}=0\right)\int_0^{t_1}\mathbb{P}\left(N_{t_1-t}>1\right)f_{\tau}(t){\rm d}t,
\end{align}
and that
\begin{align}
\mathbb{P}\left(\tau>t_1,N_{t_2}>1\right)&=\mathbb{P}\left(N_{t_1}=0,N_{t_2}>1\right)\\
&=\mathbb{P}\left(N_{t_1}=0,N_{t_2}-N_{t_1}>1\right)\\
&=\mathbb{P}\left(N_{t_1}=0\right)\mathbb{P}\left(N_{t_2}-N_{t_1}>1\right)\\
&=\mathbb{P}\left(N_{t_1}=0\right)\mathbb{P}\left(N_{t_2-t_1}>1\right)\\
&=\mathbb{P}\left(N_{t_1}=0\right)\mathbb{P}\left(N_{t_1}>1\right).
\end{align}
All these three terms are now able to be figured out by using $N_t\sim\text{Poiss}(\lambda t)$. Thanks to the mutual exclusivity, the sum of them would give the expected probability. The result reads
$$
\mathbb{P}\left(\left\{\text{crash before }t_2\right\}\right)=1-\frac{49}{32\sqrt{e}}\approx 7.125\%.
$$
This result could be verified by numerical simulations. For example, generate a partition for $t\in\left[0,1/6\right]$ with $10^3$ equi-spaced grid points. Generate $10^6$ trajectories for $N_t$ with $t\in\left[0,1/6\right]$ with $N_1\sim\text{Poiss}(3)$. With these settings, a simulation gives that $71359$ trajectories witness at least two disturbances occur within $1/12$ unit of time, meaning that the probability of crash within the first ten minutes would be around $7.1359\%$. This matches the above theoretical result very well.
