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We have a system in which events happen after each other. Events are i.i.d. An event, shown by random variable $X$, follows exponential distribution with $E(X)=\frac{1}{\lambda}$. We suppose the system runs as long as no interval-time between two events is larger than $\tau$.

So, we can show the time-arrival between events as follows:

$t_1,t_2,t_3,\dots,t_{n-1},t_n$

in which $t_i < \tau\ 1\le i \le (n-1)$ and $t_n > \tau$

Now, the question is how we can find the distribution density function for the time-interval between the beginning and end of the system over $t$ as $t=\sum^{n-1}_{i=1}t_i + t_n$

I know that such as distribution density functions is like: $f(t) = \sum^{\infty}_{n=1}f(t|n)P(t<\tau)^{(n-1)}(1-p(t<\tau))$

in which $p(t<\tau) = 1 - e^{-\lambda\tau}$

In this regard, what $f(t|n)$ would be?

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    $\begingroup$ The most convenient approach to compute the PDF of the end time $T$ might be to note that, conditioning on the time $X$ of the first event, $$T=X\mathbf 1_{X>t}+(X+T')\mathbf 1_{X<t}=X+T'\mathbf 1_{X<t},$$ where $T'$ is distributed like $T$, and independent of $X$. Turning to Laplace transforms, this shows that $u(s)=E(e^{-sT})$ solves $$u(s)=E(e^{-sX}(1+e^{-sT'}\mathbf 1_{X<t}))=E(e^{-sX}\mathbf 1_{X>t})+u(s)E(e^{-sX}\mathbf 1_{X<t}),$$ hence... $\endgroup$ – Did Jun 3 '16 at 7:56
  • $\begingroup$ thanks for your reply. what is $1$ in your equations. $\endgroup$ – Alireza Montazeri Gh Jun 3 '16 at 8:05
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    $\begingroup$ Indicator function: $\mathbf 1_A(\omega)=1$ if $\omega\in A$ and $\mathbf 1_A(\omega)=0$ otherwise. $\endgroup$ – Did Jun 3 '16 at 9:48
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Let $$\sigma = \inf\{n>0:T_n>\tau \}, $$ then $$\mathbb P(\sigma = 1) = \mathbb P(T_1>\tau)=e^{-\lambda\tau}$$ and $$\mathbb P(\sigma = n) = \mathbb P(T_n>\tau,T_{n-1}\leqslant\tau,\ldots,T_1\leqslant\tau)=\left(1-e^{-\lambda\tau}\right)^{n-1}e^{-\lambda\tau},\ n\geqslant 1. $$ Let $S_n=\sum_{k=1}^n T_k$, then the lifetime of the system is given by $S_\sigma$ - a geometric sum of exponential random variables. By conditioning on $\sigma$ we may show that $S_\sigma\sim\mathsf{Exp}(\lambda e^{-\lambda\tau})$, and so $S_\sigma$ has density $$ f_{S_\sigma}(t) = \lambda e^{-\lambda\tau}e^{-\lambda e^{-\lambda\tau}t}\ \mathsf1_{(0,\infty)}(t). $$

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