Statistical Prediction Based on Censored Life Data. Luis A. Escobar Department of Experimental Statistics Louisiana State University.

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Statistical Prediction Based on Censored Life Data Overview Luis A. Escobar Department of Experimental Statistics Louisiana State University and William Q. Meeker Department of Statistics Iowa State University October 19, 22 Problem background and motivation, general prediction problem. Probability prediction, naive statistical prediction, and coverage probability. Calibrating statistical prediction intervals and pivotal methods. Predicting the number of future field failures Single cohort Multiple cohorts Concluding remarks. 1 2 Introduction Motivation: Prediction problems are of interest to consumers, managers, engineers, and scientists. A consumer would like to bound the failure time of a product to be purchased. Managers want to predict future warranty costs. Engineers want to predict the number of failures in a future life test. Engineers want to predict the number of failures during the following time period (week, month, etc.) of an ongoing life test experiment. Related Literature Surveys and methods: Hahn and Nelson (1973), Patel (1989), Hahn and Meeker (1991). Analytical frequentist theory: (1984). Cox (1975), Atwood Simulation/bootstrap frequentist theory: Beran (199), Bai, Bickel, and Olshen (199), Efron and Tibshirani (1993). Log-location-scale distributions with failure (Type II) censored data frequentist approach: Faulkenberry (1973), Lawless (1973), Nelson and Schmee (1979), Engelhardt and Bain (1979), Mee and ushary (1994). Likelihood theory: albfleisch (1971). Bayesian theory: Geisser (1993). 3 4 New-Sample Prediction Based on previous (possibly censored) life test data, one could be interested in: Time to failure of a new item. Within-Sample Prediction Predict future events in a process based on early data from the process. Followed n units until t c and observed r failures. Data are first r of n failure times: t (1) <...<t (r). Want to predict: r Failures n Units at Start Number of additional failures in interval [t c,t w). t c Future Unit(s)? n Units at Start r Failures tc? tw Time until k failures in a future sample of m units. Number of failures by time t w in a future sample of m units. Time of next failure. Time until k additional failures. 5 6

Needed for Prediction Probability Prediction Interval (θ nown) In general to predict one needs: A statistical model to describe the population or process of interest. This model usually depends on a set of parameters θ. Information on the values of the parameters θ. This information could come from laboratory test. field data. An exact 1(1 α)% probability prediction interval is (ignoring any data) PI(α) =[T, T ] = [t α/2, t 1 α/2 ] where t p = t p(θ) isthepth quantile of T. Probability of coverage: Pr[T PI(α)] = T Pr(T T ) = Pr(t α/2 T t 1 α/2 ) = 1 α. α/2 1 α α/2 Nonparametric new-sample prediction also possible (e.g., Chapter 5 of Hahn and Meeker 1991). t α/2 t 1 α/2 7 8 Example 1: Probability Prediction for Failure Time of a Single Future Unit Based on nown Parameters Statistical Prediction Interval (θ Unknown) Assume cycles to failure follows a lognormal distribution with known parameters µ =4.16,σ =.5451 A 9% probability prediction interval is PI(α) =, [T T ]=[t α/2, t 1 α/2 ] = [exp(4.16 1.645.5451), exp(4.16 + 1.645.5451)] = [26.1, 157.1]. Then Pr(T T T )=Pr(26.1 T 157.1) =.9. With misspecified parameters, coverage probability may not be.9. Objective: Want to predict the random quantity T based on alearningsample information (DATA). The random DATA leads parameter estimate θ and prediction interval PI(α), =[T T ]. Thus, [T T ]andt have a joint distribution that depends on a parameter θ. Probability of coverage: PI(α) is an exact 1(1 α)% prediction interval procedure if Pr[T PI(α)] = Pr(T T T )=1 α. First we consider evaluation, then specification of PI(α). 9 1 Coverage Probabilities Concepts One-Sided and Two-Sided Prediction Intervals Conditional coverage probability for the interval: For fixed DATA (and thus fixed θ and [T, T ]): CP[PI(α) θ; θ] = Pr(T T T θ; θ) = F ( T ; θ) F (T ; θ) Unknown given [T, T ] because F (t; θ) depends on θ. Combining lower and upper 1(1 α/2)% prediction bounds gives an equal-probability two-sided 1(1 α)% prediction interval. If Pr(T T<) = 1 α/2 and Random because [T, T ] depends on θ. Pr( <T T ) = 1 α/2, Unconditional coverage probability for the procedure: then CP[PI(α); θ] = Pr(T T T ; θ) { = CP[PI(α) E θ θ; θ] }. In general CP[PI(α); θ] 1 α. When CP[PI(α); θ] does not depend on θ, PI(α) is an exact procedure. Pr(T T T )=1 α. α/2 1 α α/2 T ~ T ~ 11 12

Asymptotic Approximation for CP[PI(α); θ] As suggested by Cox (1975) and Atwood (1984): Naive Statistical Prediction Interval When θ is unknown, a naive prediction interval is PI(α) =[T, T ] = [ t α/2, t 1 α/2 ] where t p = t p( θ) is the ML estimate of the p quantile of T. Coverage probability may be far from nominal 1 α, especially with small samples. 13 For the naive lower prediction bound: PI(α) =[T, ] =[ t α, ] =[t α( θ), ], we have CP [ PI(α) θ; θ ] = T<; θ) =g(α, Pr(T θ; θ) CP [PI(α); θ] = [ g(α, E θ θ) ]. Under regularity conditions, using a Taylor expansion of g(α, θ; θ), it follows that CP[PI(α); θ] =α+ 1 k g(α, θ; θ) a i n i=1 θ + 1 k 2 g(α, θ; ( θ) b ij i θ 2n i,j=1 θ i θ +o j θ where a i, b ij are elements of vector a and matrix B defined by (ˆθ θ ) = a(θ)+o(1/n) [ E θ E θ (ˆθ θ)(ˆθ θ) ] = B(θ)+o(1/n). These are, in general, difficult to compute. 14 Prediction interval calibration curve lognormal model CP[PI(alpha_cal);thetahat]..2.4.6.8 1. alpha_upper_cal=.9997 alpha_nom=.5 alpha_lower_cal=.2225 (number of bootstrap samples= 1 )..2.4.6.8 1. alpha_cal 15 Calibrating One-Sided Prediction Bounds To calibrate the naive one-sided prediction bound, find α c, such that CP[PI(α c); θ] ( = Pr T ; T θ ) = Pr [ t αc T ; θ ] =1 α. where = T t αc is the ML estimator of the t αc quantile of T. Can do this analytically or by simulation. When for arbitrary α, CP[PI(α); θ] does not depend on θ, the calibrated PI(α c) procedure is exact. For a two-sided interval, do separately for each tail. 16 Simulation of the Sampling/Prediction Process To evaluate the coverage probability of PI(α )forsome specified <α < 1, do the following: Use the assumed model and ML estimates θ to simulate the sampling and prediction process by computing DATA j and Tj, j =1,...,B for a large number B (e.g., B = 4 or B = 1). For each simulated sample/prediction: Compute ML estimates θ j from simulated DATA j. Use α to compute = T t α from simulated DATA j and j compare with the simulated Tj. The proportion of the B trials having Tj >T gives the Monte Carlo evaluation of j CP [PI(α ); θ] at θ. The Effect of Calibration Result: Beran (199) showed that, under regularity conditions, with PI(α c) being a once-calibrated prediction, CP[PI(α c); θ] =1 α + O ( 1/n 2) and that the order of the approximation can be improved by iterating the calibration procedure. To obtain a PI with a coverage probability of 1(1 α)%, find α c such that CP[PI(α c); θ] =1 α. 17 18

Lognormal probability plot of bearing life test data censored after 8 million cycles with lognormal ML estimates and pointwise 95% confidence intervals Simulation of the bearing life test censored after 8 million cycles (n = 23 and r = 15), lognormal model, histograms of pivotal like Z log(t ) =(log(t ) µ )/ σ and Φ[Z log(t ) ].98.95.9.8 Proportion Failing.7.6.5.4.3.2.1.5.2 1 2 3 4 2 4 6 8 1 1 2 5 1 2 5 Millions of Cycles -1-5 5 1 Z.logT*..2.4.6.8 1. (Z.logT*) 19 2 Prediction interval calibration function for the bearing life test data censored after 8 million cycles, lognormal model Example 2: Prediction Bound for a Single Independent Future T Based on Time-Censored (Type I) Data CP[PI(1-alpha_cal);thetahat].93.95.97.99 number of simulated samples= 1 1-alpha=.95 1-alpha_cal_upper=.967 1-alpha_cal_lower=.964.93.95.97.99 21 Life test run for 8 million cycles; 15 of 23 ball bearings failed. ML estimates of the lognormal parameters are: µ = 4.16, σ =.5451. The naive one-sided lower 95% lognormal prediction bound (assuming no sampling error) is: t.5 = exp[4.16 + ( 1.645)(.5451)] = 26.1. Need to calibrate to account for sampling variability in the parameter estimates. From simulation CP[PI(1.964); θ] =.95 Thus the calibrated lower 95% lognormal prediction bound is = T t.36 = exp[4.16 + ( 1.82)(.5451)] = 24. where z.36 = 1.82. 22 Within-Sample Prediction Predict Number of Failures in Next Time Interval Comparison of Approximate 9% Prediction Intervals for Bearing Life from a Life Test that was Type I Censored at 8 Million Cycles Lognormal Naive [26.1, 157.1] Calibrated [24., 174.4] The sample DATA are singly time-censored (Type I) from F (t). Observe n units until time t c. Failure times are recorded for the r> units that fail in (,t c]; n r unfailed at t c. Prediction problem: Find an upper bound for the number of future failures,, intheinterval(t c,t w], t c <t w. n Units at Start r Failures tc? tw 23 24

Distribution of and Naive Prediction Bound Conditional on DATA, the number of failures in (t c,t w] is distributed as where ρ = BIN(n r, ρ) Pr(tc <T tw) Pr(T >t c) Obtain ˆρ by evaluating at θ. = F (tw; θ) F (tc; θ). 1 F (t c; θ) The naive 1(1 α)% upper prediction bound for is (1 α) = ˆ 1 α, the estimate of the 1 α quantile of the distribution of. This is computed as the smallest integer such that BINCDF(, n r, ˆρ) > 1 α. 25 Example 3: Prediction of the Number of Future Failures n = 1, units put into service; 8 failures in 48 months. Want an upper prediction bound on the number of the remaining n r = 1 8 = 992 units that will fail between 48 and 6 months. Weibull time to failure distribution assumed; ML estimates: α = 1152, β =1.518 giving ρ = ˆF (6) ˆF (48) 1 ˆF (48) =.3233. Point prediction for the number failing between 48 and 6 months is (n r) ˆρ = 992.3233 = 32.7. 26 Calibration of the Naive -Prediction Bound for the Number of Field Failures Example 3. Calibration functions for upper and lower prediction bounds on the number of future field failures Find α c such that CP[PI(α c); θ] =Pr [ (1 α c) ] =1 α A Monte Carlo evaluation of the unconditional coverage probability is CP[PI(α c); θ] = 1 B P j B j=1 where ] P j =BINCDF (1 α c) [ j ; n r j, ρ is the conditional coverage probability for the jth simulated interval evaluated at ρ. Similar for the lower prediction bound. CP[PI(alpha_cal);thetahat]..2.4.6.8 1...2.4.6.8 1. 27 28 Example 3. Calibration functions for upper and lower prediction bounds on the number of future field failures CP[PI(1-alpha_cal);thetahat].88.9.92.94.96.98 1. number of simulated samples= 1 1-alpha=.95 1-alpha_cal_upper=.986 1-alpha_cal_lower=.981.88.9.92.94.96.98 1. Example 3 Computations The naive 95% upper prediction bound on is ˆ.95 = 42, the smallest integer such that BINCDF(, 992,.3233) >.95. From simulation CP[PI(.9863); θ].95. Thus the calibrated 95% upper prediction bound on is = ˆ.9863 = 45, the smallest integer such that BINCDF(, 992,.3233).9863. 32 Naive.95 42 Calibrated.986 45 Number Failing 29 3

Bearing-Cage Field-Failure Data (from Abernethy et al. 1983) Staggered Entry Prediction Problem n1 Failures Observed in the Field Number of Failures: r 1 r 2 n2 r 3 n3 t c1 t c2 t c3 Future 1? 2? 3? t w1 t w2 t w3 A total of 173 units failed introduced into service over a period of eight years (about 16 in the past three years). Time measured in hours of service. Six out of 173 units failed. ns r s t cs s? t ws Unexpected failures early in life mandated a design change. Total Number of Future Failures:? How many failures in the next year (point prediction and upper prediction bound requested), assuming 3 hours of service. 31 32 Within-Sample Prediction With Staggered Entry Bearing Cage Data and Future-Failure Risk Analysis The objective it to predict future events in a process based on several sets of early data from the process. Units enter the field in groups over time. Need to predict the total number of new failures (in all groups) when unfailed units are observed for an additional period of length t. For group i, n i units are followed for a period of length t cj and r i failures were observed, i =1,...,s. DATA i for set i (i =1,...,s) are the first r i of n i failure times, say t (i1) < <t (iri ). 33 Group Hours in Failed At Risk i Service n i r i n i r i ρ i (n i r i ) ρ i 1 5 288 288.763.2196 2 15 148 148.1158.1714 3 25 125 1 124.1558.1932 4 35 112 1 111.1962.2178 5 45 17 1 16.2369.2511 6 55 99 99.2778.275..................... 17 165 6 6.7368.442 18 175.7791 19 185 1 1.8214.82 2 195.8638 21 25 2 2.962.181 Total 173 6 5.57 34 Distribution of the Number of Future Failures Conditional on DATA i, the number of additional failures i in group i during interval (t cj,t wi ](wheret wi = t cj + t) is distributed as i BIN(n i r i,ρ i )with ρ i = Pr(t cj <T t wi ) Pr(T >t cj ) = F (t wi; θ) F (t cj ; θ). 1 F (t cj ; θ) Obtain ρ i by evaluating ρ =(ρ i,...,ρ s)at θ. Let = s i=1 i be the total number of additional failures over t. Conditional on the DATA (and the fixed censoring times) SBINCDF(k; n r, ρ) a sum of s independent binomials; n r =(n 1 r 1,...,n s r s)and ρ =(ρ 1,...,ρ s). A naive 1(1 α)% upper prediction bound (1 α) is computed as the smallest integer k such that SBINCDF(k, n r, ρ ) 1 α. 35 Calibration of the Naive Prediction Bound for the Staggered Entry Number of Field Failures Find α c such that CP[PI(α c); θ] =Pr [ (1 α c) ] =1 α A Monte Carlo evaluation of the unconditional coverage probability is CP[PI(α c); θ] = 1 B P j B j=1 where [ ] P j =SBINCDF (1 α c) j ; n r, ρ is the conditional coverage probability for the jth simulated interval evaluated at ρ. Similar for the lower prediction bound. 36

Example 4: Calibration functions for upper and lower prediction bounds on the number of future field failures with staggered entry Example 4: Calibration functions for upper and lower prediction bounds on the number of future field failures with staggered entry CP[PI(alpha_cal);thetahat]..2.4.6.8 1...2.4.6.8 1. 37 CP[PI(1-alpha_cal);thetahat].88.9.92.94.96.98 1. number of simulated samples= 1 1-alpha=.95 1-alpha_cal_upper=.991 1-alpha_cal_lower=.959.88.9.92.94.96.98 1. 38 Example 4 Computations Concluding Remarks and Future Work The naive 95% upper prediction bound on is ˆ.95 =9, the smallest integer such that SBINCDF(, n r, ρ) >.95. From simulation CP[PI(.9916); θ].95. Thus the calibrated 95% upper prediction bound on is = ˆ.9916 = 11, the smallest integer such that SBINCDF(, n r, ρ) >.9916. Methodology can be extended to: Staggered entry with differences among cohort distributions. Staggered entry with differences in remaining warranty period. Modeling of spatial and temporal variability in environmental factors like UV radiation, acid rain, temperature, and humidity. Naive.95 Calibrated.991 Today, the computational price is small; general-purpose software needed. 5.57 9 12 Number Failing 39 Asymptotic theory promises good approximation when not exact; use simulation to verify and compare with other approximate methods. 4