How long do light bulbs last when some never fail?
Tonight's idea
A lab switches on 8 bulbs and watches for 6 months. Four burn out. One is unplugged at month 3 because its socket is needed. Three are still glowing at the end. For those four you know only that the lifetime is longer than some number. Such a story is censored: cut short before its ending.
Averaging only the failures keeps just the short lives. To use everything, follow the survival curve \(S(t)\), the estimated fraction of bulbs lasting beyond time \(t\). Start at \(S = 1\). Each time bulbs fail, look only at the bulbs at risk (still on test just before that moment) and multiply:
\(S_{\text{new}} = S_{\text{old}} \times \dfrac{\text{at risk} - \text{failures}}{\text{at risk}}\)
A censored bulb causes no step down. It quietly leaves the at-risk group, so it counts in every fraction up to its removal and in none after. This is the Kaplan-Meier estimate (1958), a standard tool in medical trials. It assumes that being removed has nothing to do with how close a bulb was to failing.
Watch one: the 8 bulbs
| failure at month | 1 | 2 | 4 | 5 |
|---|---|---|---|---|
| at risk just before | 8 | 7 | 5 | 4 |
| factor | \(7/8\) | \(6/7\) | \(4/5\) | \(3/4\) |
| \(S\) after | \(0.875\) | \(0.875 \times 6/7 = 0.75\) | \(0.75 \times 4/5 = 0.6\) | \(0.6 \times 3/4 = 0.45\) |
Month 3 has no column: nothing failed. But at month 4 only 5 are at risk, not 6, because C has gone. So an estimated 45% last beyond 5 months, a rough figure from 8 bulbs.
Your turn
- Draw the survival curve for the 8 bulbs, with \(t\) from 0 to 6 across and \(S\) from 0 to 1 up. It is a staircase: flat between failures, dropping straight down at each one. Mark the removal at month 3 with a small circle on the curve. Read off \(S(3)\) and \(S(4.5)\).
- Three shortcuts, each wrong. (a) Average the lifetimes of the four bulbs that failed. (b) Count bulb C as a failure at month 3, then give the fraction of the 8 lasting beyond month 5. (c) Count bulb C as still working at month 6, and do the same. Say what is wrong with each, and compare (b) and (c) with 0.45.
- The estimated median lifetime is the first time the curve is at or below 0.5. (a) Find it for the 8 bulbs. (b) By when has an estimated quarter of the bulbs failed? (c) Why does this test give no time by which three quarters have failed?
- More made-up data: 10 bulbs, tested for 8 months. Failures at months 2, 2, 5, 7 and 7. Two bulbs are removed at month 3. Three still work at month 8. (a) Find \(S\) after each failure time, handling failures at the same time together. (b) Find the median lifetime. (c) Show that taking the two month 7 failures one after the other gives the same factor.
Stretch: why there is no average
For lifetimes with nothing censored, the mean lifetime equals the area under the survival staircase (check it for three bulbs failing at months 1, 2 and 6). Find the area under the 8-bulb staircase from month 0 to month 6. Then explain why this is not an estimate of the mean lifetime, and what extra information would be needed to get one.
Answers
- The curve is level at 1 until month 1, then 0.875 until month 2, 0.75 until month 4, 0.6 until month 5, and 0.45 from month 5 to the end at month 6. \(S(3) = 0.75\): the removal causes no drop. \(S(4.5) = 0.6\).
- (a) \((1 + 2 + 4 + 5)/4 = 3\) months. It uses only the short lives, and three bulbs have already lasted twice that long. (b) \(3/8 = 0.375\): too gloomy, since C was working when it left. (c) \(4/8 = 0.5\): too rosy, since nobody saw C reach month 6. Kaplan-Meier lands between, at 0.45. Of the 5 bulbs on test after month 3, 3 lasted beyond month 5, so it in effect counts C as \(3/5\) of a survivor: \((3 + 0.6)/8 = 0.45\).
- (a) Month 5, where the curve drops from 0.6 to 0.45. (b) Month 2, the first time \(S\) is at or below 0.75. (c) The curve stops at 0.45 when the test ends. It never gets down to 0.25, so the data cannot say when it would.
- (a) Month 2: 10 at risk, 2 fail, \(S = 8/10 = 0.8\). The removals leave 6 at risk. Month 5: 1 fails, \(S = 0.8 \times 5/6 = 2/3 \approx 0.67\). Month 7: 5 at risk, 2 fail, \(S = 2/3 \times 3/5 = 0.4\). (b) Month 7, the first time \(S\) is at or below 0.5. (c) One at a time: \(4/5 \times 3/4 = 3/5\), the same as \((5 - 2)/5\). The 4s cancel, and that always happens with ties.
Stretch. Three bulbs failing at 1, 2 and 6 have mean 3, and the staircase area is \(1 + \tfrac23 + 4 \times \tfrac13 = 3\). For the 8 bulbs the area to month 6 is \(1 + 0.875 + 2 \times 0.75 + 0.6 + 0.45 = 4.425\) months. But the curve is still at 0.45 there, and the rest of the area lies beyond month 6, where nothing was seen. If F, G and H had all failed at month 7 the area would be \(4.425 + 0.45 = 4.875\); if they had lasted until month 30 it would be \(4.425 + 24 \times 0.45 = 15.225\). Both fit the test equally well. You would need to keep watching until the curve reaches 0, or else assume a model for how bulbs fail. Here the median needs neither, because the curve did get below 0.5. That is why reports of this kind often quote a median.