Dead-time counters and replacement models

A Type 1 counter records the arrivals of radioactive particles. Suppose that the arrival process is Poisson with intensity $\lambda$, and that the counter is locked for a dead period of fixed length $T$ after each detected arrival. Show that the detection process $\widetilde{N}$ is a renewal process with interarrival time distribution $\widetilde{F}(x) = 1-e^{-\lambda (x-T)}$ if $x \geq T$. Find an expression for $\mathbb {P}\left(\widetilde{N}(t) \geq k\right)$.

Particles arrive at a Type 1 counter in the manner of a renewal process $N$; each detected arrival locks the counter for a dead period of random positive length. Show that

where $F_{L}$ is the distribution function of a typical dead period.

Particles are detected by a Type 2 counter of the following sort. The incoming particles constitute a Poisson process with intensity $\lambda$. The $j$ th particle locks the counter for a length $Y_{j}$ of time, and annuls any after-effect of its predecessors. Suppose that $Y_{1}, Y_{2}, \ldots$ are independent of each other and of the Poisson process, each having distribution function $G$. The counter is unlocked at time 0.

Let $L$ be the (maximal) length of the first interval of time during which the counter is locked. Show that $H(t) = \mathbb {P}\left(L > t\right)$ satisfies

Solve for $H$ in terms of $G$, and evaluate the ensuing expression in the case $G(x) = 1-e^{-\mu x}$ where $\mu > 0$.

A specific component in a cryptometer has an $\mathrm{Exp}(\mu )$-distributed lifetime, $\mu > 0$. If replacement is made as soon as a component fails, and if $X(t) = \#$ failures during $(0,t] = \#$ replacements during $(0,t]$, then $\left\{ X(t), t \geq 0\right\}$ is, of course, a Poisson process. Let $\left\{ V_n, n \geq 1\right\}$ be these usual interreplacement times, and suppose, instead, that the $n$th component is replaced: