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Suppose I have a room with two light bulbs. The average life of a bulb is $1 $ year, and it is a Poisson process (negative exponential). When a bulb burns out, it is replaced after exactly $24$ hours. If both bulbs burn out within the same $24$ hour period, the room will be dark.

My question is, what is the mean time until darkness?

I know that the difference between two similar independent exponential variables is a double-exponential distribution. And if you take the absolute value of the difference, then you just get another exponential distribution. But I get confused after that.

What if the room has three bulbs?

No, this is not homework. I am trying to approximate the MTBF of an EMC Isilon cluster with $(n+2)$ scale-out.

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  • $\begingroup$ I think it is related to this post math.stackexchange.com/questions/181968/… $\endgroup$
    – John Henckel
    Commented Feb 19, 2014 at 22:28
  • $\begingroup$ I think this boils down to calculating the expectation of the minimum of a collection of independent Poisson processes. I'll just do the case with two light bulbs. By the memoryless property, conditioned on the event that one of the light bulb has blown out, the probability that the other light bulb blows out within one day is still Poission. But the important part is that it is independent. So each time one of your light bulbs blow out, you get to flip an independent coin to see if you stop. If you don't stop then you just restart your process. $\endgroup$
    – Ben
    Commented Feb 19, 2014 at 22:45
  • $\begingroup$ So if $N$ is the minimum of your two Poisson processes and $K$ is the number of times you have to flip this coin to see if your process keeps going, then the ammount of time you have to wait is $$ \sum_{i=1}^K (N_i +1) $$ where $N_i$ are independent copies of $N$. The expected value of this is just $$ \mathbb{E} K \mathbb{E} (N+1). $$ $K$ is geometric so its expectation is easy but I'm not sure about $N$. $\endgroup$
    – Ben
    Commented Feb 19, 2014 at 22:49

1 Answer 1

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How accurate do you want it ? Look at the dark periods of bulb 1. By independence, the number of events of bulb 2 that fall in the first dark period is poisson(1/365). In terms of dark periods, you are waiting for the first non-zero event. This gives you expected time to $366*(1/(1-e^{-1/365})$ until bulb 1 ends the double darkness. It is a slight overstimate of what you wanted. If you want to do better, I think because of the exponential distribution the distribution of time within the the 24 hr dark when bulb 2 burns out is truncated exponential, as it must be like excess over the boundary, so you can correct the last period using that..

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  • $\begingroup$ Thanks for the help! You think the expected time will be approx 133773 years? That seems too large to me. You got me thinking, though. The "half-life" of the bulb is 365 days. That means that on average each bulb will be off for 12 hours per year. If we look at the p.d.f of bulb 2 at only the 12h per year, it is the same as stretching the time axis by a factor of 730. So I think the expected time for the second bulb to fail while the first is off will be 730 times longer. Thus expected time to darkness is 730 years. With three bulbs it is 730 * 730 = 532900 years. Does that seem right? $\endgroup$
    – John Henckel
    Commented Feb 19, 2014 at 20:43
  • $\begingroup$ days${}{}{}{}{}$ $\endgroup$
    – mike
    Commented Feb 20, 2014 at 12:40

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