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A light bulb burns for an amount of time having distribution F with mean μF then burns out. A janitor comes at times of a rate Poisson process to check the bulb and will replace the bulb if it is burnt out.

Question: What is the limiting fraction of visits on which the bulb is working?

Attempt:

We seek to determine the fraction of visits on which the bulb is working is the number of visits a janitor makes in time t with the bulb working to the total number of visits the janor makes.

I am unable to proceed further in reducing this problem. The author suggested the use of the strong laws of large numbers for which I have not the remote clue why.

The strong laws of large numbers says this:

Let $x_{1},\cdot \cdot \cdot$ be I.I.D with $E\left ( x_{i} \right )=\mu$ and let $S_{n}=x_{1}+\cdot \cdot \cdot x_{n}$. Then, with probability one: $\frac{S_{n}}{n}\rightarrow \mu$ as $n\rightarrow \infty$

Any help is appreciated.

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Let $L_1$ be the first bulb's lifetime. Due to the memoryless property of the exponential distribution, the time until the next inspection has $\mathsf{Exp}(\lambda)$ distribution. The bulb is then replaced, completing the first cycle of an alternating renewal process. Let $R(t)$ be the number of bulbs replaced by time $t$, then the key renewal theorem yields $$ \lim_{t\to\infty} \frac{R(t)}t = \frac1{\mu_F+1/\lambda}. $$ By the strong law of large numbers, we have $$ \lim_{t\to\infty} \frac{N(t)}t = \lambda. $$ Then the limiting fraction of visits on which the bulb is replaced is given by $$ \lim_{R(t)}{N(t)} = \frac{1/\lambda}{\mu_F+1/\lambda}, $$ and so the limiting fraction of visits on which the bulb is working is $$ \frac{\mu_F}{\mu_F+1/\lambda}. $$

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