Suppose we have a stochastic process $X_t$ of a light using a single light bulb. When the light bulb burns out it is immedieatly replaced with a new one. Suppose that the time between failures is given by the distribution: $$f(t) = p_1 \frac{t}{a^2}e^{-t/a}+p_2 \frac{t}{b^2}e^{-t/b}$$ where $a,b,p_1,p_2 > 0 $ and $p_1+p_2=1$.
Assuming a light bulb just burned out, the long term of burn outs per days(out unit of time) is just $1/E(X)$.
Now assume we are in a random point in time. I want to find the probability that the first repair occurs within $n$ days and find the expected number of burn outs in $m$ days. Furthermore, I would like to calculate the covariance between the last repair and the time to the next repair.
The first question can be set up as $$P(N(\tau+n)-N(\tau)\geq 1) = 1-P(N(\tau+n)-N(\tau)= 0) $$ where $\tau$ is a random point in time and $N(t)$ is the underlying counting process. Now since the underlying process is not poisson I'm having a hard time evaluating the distribution of $N(\tau+n)-N(\tau)$.
The second is similarly just $$E(N(\tau+m)-N(\tau)) $$ where again knowing the distribution of $N(\tau+m)-N(\tau)$ or the distribution of each individually has to be known.
Finally, the sojourn times can be used for the final question but I'm not sure how to set it up.
To present the idea, I will simplify the question to avoid some technical calculations. Let's instead assume the time between failures is a mixture of exponentially distributed random variables:
$ f(t) = p_1 \lambda_1e^{-\lambda_1 t}+ p_2 \lambda_2 e^{-\lambda_2 t}$
The idea is that, you have two types of light bulbs, and at time of repair, you randomly (with $(p_1, 1-p_1=p_2)$) pick one type of light bulb. Let's use $Y$ to represent the type of light bulbs selected. So marginally, $P(Y=1)=p_1$, $P(Y=2) = p_2$.
Now, at an arbitrary (but deterministic) time $\tau$, let $N(\tau)$ be the number of failures before $\tau$, and let $S_n$ be the time when the $n-th$ failure occurs. Now, the first repair time after $\tau$, given all the information up to time $\tau$ can be considered as finding:
$P(S_{N(\tau)+1}-\tau\le t|\mathcal F_{\tau})$
But the only relevant information we need here is the (probability of) type of light bulb that is used in the last repair, and this piece of information can be derived using the time between last repair and $\tau$, i.e. $\tau - S_{N(\tau)}$.
Let $Y$ be the type of light bulb used in the last repair, and $\gamma$ be the time of failure since the last repair. We know that,
$P( \gamma>\tau-S_{N(\tau)} | Y=1) = e^{-\lambda_1 (\tau-S_{N(\tau)})} $
$P( \gamma>\tau-S_{N(\tau)} | Y=2) = e^{-\lambda_2 (\tau-S_{N(\tau)})} $
By Bayes formula
$P(Y=1| \gamma>\tau-S_{N(\tau)}) = \frac{p_1 e^{-\lambda_1(\tau-S_{N(\tau)})}}{p_1 e^{-\lambda_1(\tau-S_{N(\tau)})}+p_2 e^{-\lambda_2(\tau-S_{N(\tau)})}}\triangleq q(\tau-S_{N(\tau)})$
So, time to next repair (starting from $\tau$) follow the distribution:
$g(t) = q(\tau-S_{N(\tau)}) \lambda_1e^{-\lambda_1 t}+ (1-q(\tau-S_{N(\tau)}) )\lambda_2 e^{-\lambda_2 t}$
To answer the original question, the same idea applies, but additional calculations are needed to derive the corresponding $q$ using Bayes formula，and also the conditional distribution of time to fail after surviving $\tau- S_{N(\tau)}$ for each type of light bulb (It is no longer "memoryless"). But all these should be straightforward.