Lifetime evidence, scatter and extrapolation

Lifetimes form a distribution

Nominally similar specimens can fail at different times because of defects, processing and environmental variation. Report failed specimens and runouts, sample origin and test conditions. A mean lifetime alone does not describe the lower tail relevant to reliability.

Consider the two invented complete records below. Every failure time is known, and each group contains five specimens. Both sums are 500 h, so both arithmetic means are 100 h. Yet group B contains failures at 10 and 20 h, while the earliest failure in A is at 90 h. The lower tail means the short-lifetime end of a distribution; an average does not locate it.

At 50 h, all five A specimens remain unfailed and only three of five B specimens do. Those observed fractions are 1.00 and 0.60. They describe these chosen records; five observations do not establish a population minimum or a reliable rare-failure percentile. A specimen population means the units whose behaviour the inference is intended to describe, rather than merely the list already tested.

Scatter is not necessarily detector error. Variation in a crack-initiating defect, a local surface condition or processing state can produce physically different paths to the same endpoint. Improving measurement precision does not remove that material variation. Conversely, mixed test conditions can create apparent scatter that should not be attributed to one homogeneous population.

Define what the lifetime measures

A lifetime is a time or cycle count from a defined start to a defined event. First detectable cracking, a specified crack extension, loss of a functional property and complete separation are different endpoints. So are hours under sustained load and cycles under repeated load. Changing the event, the clock or the loading schedule changes the quantity, even if the worksheet still calls it life.

Let T be a specimen’s time to the stated endpoint and S(t) = P(T > t) its population survival function: the probability of remaining short of that endpoint beyond time t. For a complete sample, the empirical estimate at t is the number with T > t divided by the number tested. The count is an observation; S(t) is an unknown population quantity being estimated. Survival here refers to the chosen test event, not necessarily safe or useful component operation.

Condition the comparison on the material and history it describes: source lot or sampling process, geometry, surface, processing, stress/strain control, temperature, environment and endpoint. A random sample from one controlled process may support inference about that process. It does not automatically represent a different production route, a larger component, an aged surface or a changing service load. Several readings from one specimen are not several independent specimens.

Survival after an age is also conditional. For t₂ > t₁ and S(t₁) > 0, P(T > t₂ given T > t₁) = S(t₂)/S(t₁), because surviving beyond t₂ already includes surviving beyond t₁. This asks about units that have reached t₁ without the endpoint, under the stated history. It does not reset their age or prove that the remaining life distribution is identical to that of a new unit.

A runout records a limit, not a failure time

Now use a different original teaching cohort of eight specimens, all entering observation at t = 0 under the same sustained-load conditions. The endpoint is complete separation, continuously detected; times are treated as exact for this illustration. Observation of specimen C ended at 150 h without separation under a monitoring-allocation plan fixed in advance; it remained under the specified loading, but no later outcome was recorded. Four others remained unfailed when the scheduled test ended at 400 h.

The record for C says T > 150 h; it does not say T = 150 h or establish what happened by 400 h. This is right censoring: observation ends before the event is seen. An unfailed specimen stopped at a planned test limit is often called a runout. Its exposure is evidence, even though its eventual failure time is unknown. Keep both the recorded time and whether it is an event or a censoring time.

Only the three failures have complete lifetimes. Their mean, 200 h, is the mean among observed failures, not the mean lifetime of all eight or of the population. Replacing the five censored times with failures would give an average recorded time of 293.75 h, but that is not the average time to separation either. A statistical method must use what was known without converting the limits into invented events.

If a specimen is intact at one inspection and separated at the next, its event time is instead known only within that interval. Do not insert the later inspection time as an exact failure without acknowledging the change. The calculation below assumes exact observed event times and right censoring; interval-censored data need an appropriate extension.

Use the specimens still being observed

At each event time, count the risk set: specimens still under observation and not yet failed just before that time. If n_j are at risk and d_j fail at that time, the observed conditional survival fraction is 1 − d_j/n_j. Multiply these fractions over event times up to t to form the Kaplan–Meier, or product-limit, estimate Ŝ(t). The hat means estimate, not a known population probability. Before the first event the estimate is 1.

The multiplication follows the conditional-probability chain. To remain unfailed through successive event times, a unit must first survive the earlier interval and then survive the next given that earlier survival. The product is not an assumption that those survival events are independent. Censoring does not make the curve drop: it removes a unit from later risk sets, after contributing its known earlier observation.

The table below gives Ŝ = 7/8 after 100 h, then (7/8)(5/6) after 200 h and (7/8)(5/6)(4/5) after 300 h. At 400 h four observation lines end without another failure. The last estimate remains 7/12 ≈ 0.5833; the lack of later observations does not turn it into zero. Nor does its final horizontal segment establish survival beyond 400 h.

To interpret this estimate for the target population, the remaining observed units must represent the relevant failure risk. Here censoring is stipulated to be unrelated to impending separation within the matched group. Removing the most damaged specimens early would violate that useful assumption: the remaining risk sets could appear too durable. Calling a removal censored does not by itself make it uninformative. The small cohort also leaves sampling uncertainty; the curve is an estimate without a confidence band in this illustration.

No observed failures does not establish perfect survival

For a separate simple experiment, inspect n independently sampled specimens at one fixed time τ, with no losses before τ and one common survival probability p = S(τ). Choose n and τ before seeing outcomes. Each unit supplies one pass/fail outcome. If all n pass, the empirical fraction is 1. The probability of that all-pass record under a proposed p is p^n, because the specimens’ outcomes are assumed independent.

Invert that probability for a one-sided 95% exact-binomial lower confidence bound after all pass. Let α = 0.05 and solve p_L^n = α, giving p_L = α^(1/n). Values below p_L make this all-pass record even less likely than 5% under the assumed sampling model. The lower bound is about the population fraction passing this specified time and test, not a bound on an individual eventual failure time.

For ten out of ten passing 200 h, p_L = 0.05^(1/10) ≈ 0.741. For thirty out of thirty under the same assumptions it is about 0.905. Both observed fractions are 1, but the inference has different precision. This is a particular one-sided binomial construction, not a universal testing requirement or a substitute for defining the population and endpoint.

A 95% confidence procedure is designed so that its bounds cover the fixed population parameter in at least 95% of repeated samples under its assumptions. It is not a 95% posterior probability assigned to p, a statement that every part survives, or evidence about times beyond τ. Common-batch dependence, selective sampling, changed conditions or an unrecorded early withdrawal can invalidate the simple calculation. The all-pass formula is not the general formula when some units fail.

Agreement at a short-time anchor leaves different futures

A confidence bound addresses uncertainty within an inference model. Extrapolation also asks whether that model has the right form outside the observed range. To isolate that issue, pretend the population value S(100 h) = 0.90 were known exactly, which is stronger information than an empirical nine out of ten. Even that value does not specify the rest of the survival curve.

Let z = t/(100 h), a dimensionless time. Consider two illustrative assumed curves: S_A(t) = 0.90^z and S_B(t) = 0.90^(z²). Both start at 1, decrease toward zero, and equal 0.90 at 100 h. They are mathematical alternatives, not fits to the preceding eight-specimen record or measured laws for a named material. Their shapes encode different changes of failure risk with age.

At 200 h they predict 0.81 and 0.6561; at 1000 h they predict about 0.3487 and 0.0000266. No choice between them follows from their shared anchor. Actual event times at additional ages, larger representative samples and mechanism evidence can constrain the shape. Using a more complete short record may already reject a candidate; matching one summary is weaker than fitting and checking that full record.

Selecting the curve with the more attractive long-time answer does not add evidence. Nor does a narrow coefficient confidence interval quantify every alternative shape or an unobserved mechanism change. Report which part of a graph is observed, which is estimated within its range and which is an extension. The distinction remains necessary when a short fit uses many closely spaced readings from only a few specimens.

Acceleration changes more than the clock

Higher stress or temperature can shorten tests, but may change the dominant mechanism. Fit models over a justified range and examine independent evidence for the proposed extrapolation. Separate measured survival from predicted lifetime and include the inference uncertainty.

A simple physical-acceleration model assumes that the same mechanism and statistical response can be represented by a constant change of time scale. A lifetime quantile is an age by which a stated fraction has reached the endpoint; the median is the 50% quantile. Define AF = t_use/t_test for corresponding lifetime quantiles under two stated conditions. If that assumption holds, S_use(t) = S_test(t/AF). Thus AF = 5 maps a 40 h test quantile to a 200 h use-condition quantile. It does not turn one observed failure into the guaranteed service life of every specimen.

The assumption has testable consequences. More than one quantile or stress level can examine whether the same time scaling preserves the distribution’s shape. Fracture paths, deformation history and chemical or structural observations can examine whether the mechanism is comparable. If heating introduces a different cracking route, one factor may mix different processes rather than speed the relevant one. Different failure mechanisms need not share an acceleration factor.

Measured lifetime scatter, uncertainty in fitted parameters, uncertainty in AF and uncertainty in the model’s form are different contributions. A small residual in a short test does not remove the others. An accelerated programme can be useful when it connects those checks to the target material, history and endpoint; its result remains a conditioned projection until appropriate use-condition evidence tests it.

Worked examples

Try the reasoning

1. In the eight-specimen record, a spreadsheet enters every last-observed time as a failure time and obtains 293.75 h. Another deletes all censored rows and obtains 200 h. Which, if either, is the observed mean lifetime of the full cohort?

   Show the answer and reasoning

   Neither. The first is the average of recorded event-or-observation-limit times; the second is the mean of the three observed failures. Five eventual lifetimes were not observed.

   Runouts and the early withdrawal supply lower limits on T. They must not be silently turned into failures or removed as if they contributed no observation. The whole-cohort mean cannot be computed exactly from this record.

2. Both complete five-specimen groups have mean lifetime 100 h. Calculate the observed fraction with T > 50 h in each. Does the minimum observed lifetime estimate a guaranteed minimum for all future specimens?

   Show the answer and reasoning

   The fractions are 1.00 for A and 0.60 for B. Neither the mean nor the sample minimum guarantees every future specimen’s outcome.

   The same average can hide very different short-lifetime observations. A small complete sample has no observations of the untested population’s rarer tail.

3. A new matched six-specimen cohort starts at t = 0. One fails at 50 h, another is withdrawn unfailed at 75 h for a reason stipulated unrelated to its condition, and one fails at 100 h. Three remain observed and unfailed through 150 h. Calculate the product-limit estimate after 100 h and explain whether it drops at 75 h.

   Show the answer and reasoning

   Ŝ after 100 h is (5/6)(3/4) = 5/8 = 0.625. It does not drop at the 75 h withdrawal.

   Six are at risk before the first failure. After that failure and the withdrawal, four are at risk before the second. The withdrawal changes the later denominator without inventing a failure.

4. An operator withdraws specimens showing the largest early cracks and records them merely as unfailed right-censored cases for a later complete-separation endpoint. A survival curve then looks excellent. What assumption is in doubt, and what information must remain available?

   Show the answer and reasoning

   Removal is related to the developing failure risk, so the remaining risk sets may be selectively durable. Preserve the removal rule, observed crack states, times and endpoint distinctions; do not assume the ordinary uninformative-censoring interpretation.

   They may indeed be unseparated at withdrawal, but the reason for losing observation carries information. A calculation cannot repair that missing selection mechanism by renaming it random censoring.

5. A supplied conditioned population model has S(100 h) = 0.90 and S(300 h) = 0.60. What is the model’s probability of reaching beyond 300 h for a unit known to have reached beyond 100 h under that same history? Does the result establish a fresh 200 h lifetime distribution?

   Show the answer and reasoning

   The conditional probability is 0.60/0.90 = 2/3 ≈ 0.6667. It does not reset the unit to a new age-zero state.

   The later-survival event is a subset of the earlier one. This ratio uses the supplied model; memoryless remaining life would be an additional property that these two values do not establish.

6. Thirty independent specimens all pass the same fixed-time checkpoint, with no early losses. Calculate the one-sided 95% lower survival bound from the all-pass formula. Would thirty repeated readings of one specimen justify that number instead?

   Show the answer and reasoning

   p_L = 0.05^(1/30) ≈ 0.90497. Thirty readings of one specimen are not thirty independent fixed-time specimen outcomes.

   The empirical fraction is still one, but a larger independent representative sample supports a tighter bound. Independence and the target population matter as well as the displayed count.

7. One fatigue dataset ends at first detected cracking and another at complete separation, using otherwise similar nominal cycles. A report pools their cycle counts as if all were the same lifetime. Identify the problem before choosing any distribution fit.

   Show the answer and reasoning

   The datasets use different events. First-crack and separation lives cannot be pooled as one endpoint without a justified relation and the associated observation/measurement information.

   A precise fitted curve would still answer an ill-defined question. Define the endpoint, start, clock, specimen state and loading history before comparing or modelling the counts.

What to carry forward

• Define the endpoint, clock, population and conditions before comparing lifetime evidence.
• An average does not locate the early-failure tail or guarantee every specimen.
• Censored observations retain exposure information without supplying exact failure times.
• Empirical fractions, product-limit estimates, confidence bounds and assumed curves answer different questions.
• A short fit and an acceleration factor need evidence about shape and mechanism before supporting an extension.