Solution
Bulbs Still Burning at the Deadline
Show the problem again
Six bulbs have independent lifetimes that are exponentially distributed with unknown rate λ (per hour). They are all switched on together and the test is stopped at 12 hours. Three bulbs failed, at 2, 5 and 9 hours. The other three were still working when the test stopped. What is the maximum likelihood estimate of the mean lifetime 1/λ, in hours? Give an exact fraction.
Worked solution
A bulb that fails at time t contributes the density λe^(−λt) to the likelihood. A bulb still working at 12 hours contributes only the survival probability e^(−12λ). The likelihood is therefore λ³ e^(−λT), where T = 2 + 5 + 9 + 3·12 = 52 is the total time on test, censored bulbs included. Maximizing gives λ̂ = (number of failures) / (total time on test) = 3/52, so the estimated mean life is 52/3 hours ≈ 17.3. Throwing away the survivors, or treating 12 hours as their failure time, would both bias the answer; the survivors count for their exposure but not for a failure.
Source: Standard censored-lifetime exercise from reliability statistics. Statement written for AxiomIQ.