Let $X_n,Y_n$ be independent and $X_n\sim X>0,\ Y_n\sim Y>0$ s.t $$\mathbb{E}[X]=a>0\ , \mathbb{E}[Y]=b>0$$ Lets think at of light bult are lit till they are burned out. At first, the first light bulb is lit for $X_1$ time and then burned, then we wait till the light bulb is changed $Y_1$ time. then we have light for $X_2$ time and so for.
Denote $R_t$ to be the time there is light at $[0,t]$ time. prove $$\frac{R_t}{t} = \frac{a}{a+b} \ \ \ a.s$$
I tried solving this using a similiar approach as polya urn problem using red and green balls in a bag but got stuck since this time I dont know the probability there is a light at a giving moment.
I tried to show $\frac{R_t}{t}$ is a martingale and if I did I finished.
Can somebody give me a hint?
Thank you,
Here is the intuition; martingales are not needed. I guess $X,Y\geq 0$ a.s. We have that $\frac{1}{n}\sum_{k\leq n}X_k(\omega)\to a,\,\frac{1}{n}\sum_{k\leq n}Y_k(\omega)\to b$ for all $\omega \in \Omega^*$ where $P(\Omega^*)=1$ (that is, we have almost sure convergence) by law of large numbers. Let $\omega \in \Omega^*$. Consider the sequence $$Z_n(\omega)=\frac{\frac{1}{n}\sum_{k\leq n}X_k(\omega)}{\frac{1}{n}\sum_{k\leq n}X_k(\omega)+\frac{1}{n}\sum_{k\leq n}Y_k(\omega)}=\frac{\sum_{k\leq n}X_k(\omega)}{t_{n}(\omega)}$$ We have that $t_n(\omega)=\sum_{k\leq n}X_k(\omega)+\sum_{k\leq n}Y_k(\omega)$ is the total time until the end of the $n$-th bulb change waiting time, while $\sum_{k\leq n}X_k(\omega)$ is the total time of light for the first $n$ light bulbs used. We can see that $Z_n(\omega)\to \frac{a}{a+b}$ by continuous mapping. You can also see that $R_{t_n(\omega)}=\sum_{k\leq n}X_k(\omega)$. So we have $$\frac{R_{t_n(\omega)}}{t_n(\omega)}\to \frac{a}{a+b}$$ for all $\omega \in \Omega^*$.