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Suppose I have a light-bulb whose risk of fizzling out and dying is constant over a time scale $[0,1]$. That is, suppose the time of light bulb death $\tau$ satisfies $$Pr(\tau\in[t,t+dt]\ \Big| \ \tau>t) = a\cdot dt,$$ where we have a conditional probability expressing the fact that there is zero probability that the light-bulb can die twice.

What is the expected time that the light-bulb will go out?

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    You should probably add "lightbulb" in the title. Its was scary when i read it on unanswered page. *O* – 2017-01-29
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    It's ok now, friend. – 2017-01-29
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    New friend, \*o*/ – 2017-01-29

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The condition $$Pr(\tau\in[t,t+dt]\ \Big| \ \tau>t) = a\cdot dt$$ means that $\tau$ has exponential distribution with parameter $a$.

The expected value for a an exponential distribution with parameter $\lambda$ is $\frac{1}{\lambda}$, so the answer is $\frac{1}{a}$.

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    Thanks. Do you know where I mind find a derivation of this fact? It's not intuitive to me. – 2017-01-29
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    Nice and smooth. It would have been better to add [some references](https://en.wikipedia.org/wiki/Exponential_distribution) – 2017-01-29
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    If $F_x(t) = Pr(x > t)$, try to show that $F' = a F$. @Lepidopterist – 2017-01-29