There are many bulbs and each bulb has an expectation of lifespan of $1$ hour, and we could assume that the lifespan conforms to exponential distribution. Once a bulb runs out of its life, we change it with a new one. And we observe the bulb at the time point of exactly $10000$th hour, what's the expectation of lifespan of this observed bulb?
The answer says that the lifespan is $2$ hours, which seems extremely confusing to me. I can imagine that the expectation should increase, but I have no idea where this "$2$ times" comes from.
Can anyone explain? I have basic college Calculus and Probability knowledge so formulas are OK.
What I've tried
I think we should first calculate this probability:
$$ P(\text{i+1th bulb's life is x | the first i bulbs' life doesn't exceed 10000 and the first i+1 bulbs' life exceed 10000}) $$
It's difficult to directly solve this, so I used Bayes formula transforming it to:
$$ \frac{P(\text{the first i bulbs' life doesn't exceed 10000 and the first i+1 bulbs' life exceed 10000 | i+1th bulb's life is x})*P(\text{i+1th bulb's life is x})}{P(\text{the first i bulbs' life doesn't exceed 10000 and the first i+1 bulbs' life exceed 10000})} $$
Using this I derived a formula but failed to calculate it:
$$ \sum_{i=0}^{\inf} \int_0^{10000}\frac{(F_{\Gamma(i,1)}(10000) - F_{\Gamma(i,1)}(10000-x))e^{-x}}{\int_0^{10000}(F_{\Gamma(i,1)}(10000)-F_{\Gamma(i,1)}(10000-t))e^{-t}dt}xdx $$
Wolframalpha can't calculate this complicated formula so I guess there must be some special properties underlying this formula to simplify it.
By the memoryless property of the exponential distribution, if we observe the bulb at some time $t>0$, the expected remaining lifetime is equal to the original expected lifetime of the bulb. The expected total lifetime of the bulb, when observed at time $t$, is equal to the expected age plus the expected remaining lifetime. Informally, as $t\to\infty$, the expected age converges to the expected remaining lifetime, and hence the expected total lifetime to twice that of the original expected lifetime.
A slightly stronger result may be derived from renewal theory; namely that the distribution of the age at time $t$ is a truncated exponential distribution, and hence as $t\to\infty$ the age distribution converges to the same as the remaining lifetime distribution, so that the limiting total lifetime distribution is exponential with mean twice that that of the original lifetime distribution. The math involved here is quite a bit beyond "basic college and probability knowledge" though, so I will spare the details.