Mean Cumulative Function (MCF) For Recurrent Events (Only what I learned so far.)

Outline Background What is MCF? How to estimate MCF? Plot of MCF Compare two MCFs

Coronary Artery Disease

Coronary Artery Bypass Graft (CABG) Surgery

Percutaneous Coronary Intervention (PCI)

Limitations associated with PCI: Restenosis The treated vessel becomes blocked again. Usually occurs within 6 months after the initial procedure Balloon angioplasty alone: 40% Stenting: 25% In-stent restenosis Scar tissue overgrow and obstruct the blood flow Typically within 3 to 6 months Brachytherapy Drug elute stent

Recurrent events A sample units can undergo repeated events, such as repairs of products, recurrences of tumors, and restenosis of coronary artery in our case.

Analysis of recurrent events Time-to-first event simple and easy to interpret. conventional survival analysis method ignores information hence inefficient Wei, Lin Weissfeld (WLW) marginal model event number is used as a stratification variable; separate model per stratum Prentice, Williams and Peterson (PWP) conditional method: ‘at-risk process’ for jth event only becomes 1 after the (j - 1)th event Andersen and Gill (AG) method: ‘at-risk process’ remains at 1 until unit is censored Wayne Nelson: Mean cumulative function (MCF)

What is MCF? Product reliability analysis When a repairable system fails, it is repaired and placed back in service. As a repairable system ages, it accumulates a history of repairs and costs of repairs. At a particular age t, there is a population distribution of cumulative cost (or number) of repairs; the distribution has a mean M(t), called the Mean Cumulative Function (MCF) for the cost (or number) of repairs.

How to estimate MCF? Calculate the nonparametric estimate of the population MCF M(t) for the number of repairs of N units. List all repair and censoring ages in order from smallest to largest as in column (1) of Table 2. Denote each censoring age with a +. If a repair age of a unit equals its censoring age, put the repair age first. If two or more units have a common age, list them in a suitable order, possibly random. For each sample age, write the number I of units that passed through that age ("at risk") in column (2) as follows. If the earliest age is a censoring age, then write           I = N - 1; otherwise, write I = N. Proceed down column (2) writing the same I value for each successive repair age. At each censoring age, reduce the I value by one. For the last age, I = 0.  For each repair, calculate its observed mean number of repairs at that age as 1/I. For example, for the repair at 28 miles, 1 / 34 = 0.03, which appears in column (3). For a censoring age, the observed mean number is zero, corresponding to a blank in column (3). However, the censoring ages determine the I values of the repairs and thus are properly taken into account. In column (4), calculate the sample mean cumulative function M*(t) for each repair as follows. For the earliest repair age this is the corresponding mean number of repairs, namely 0.03 in Table 2. For each successive repair age this is the corresponding mean number of repairs (column (3)) plus the preceding mean cumulative number (column (4)). For example, at 19,250 miles this is 0.04 + 0.27 = 0.31. Censoring ages have no mean cumulative number. For each repair, plot on graph paper its mean cumulative number (column (4)) against its age (column (1)) as in Figure 2. This plot displays the nonparametric estimate M*(t), also called the sample MCF, as a red staircase function. Censoring times are not plotted.

How to estimate MCF?

(Age in Months, Cost in $100) How to estimate MCF?     Calculation of MCF for Artificial Data System Repair Histories for Artificial data Unit (Age in Months, Cost in $100) 6 (5,$3) (12,$1) (12,+)   5 (16,+) 4 (2,$1) (8,$1) (16,$2) (20,+) 3 (18,$3) (29,+) 2 (8,$2) (14,$1) (26,$1) (33,+) 1 (19,$2) (39,$2) (42,+) Event (Age,Cost) Number in Service Mean Cost MCF 1 (2,$1) 6 $1/6=0.17 0.17 2 (5,$3) $3/6=0.50 0.67 3 (8,$2) $2/6=0.33 4 (8,$1) 1.17 5 (12,$1) 1.33 (12,+)   7 (14,$1) $1/5=0.20 1.53 8 (16,$2) $2/5=0.40 1.93 9 (16,+) 10 (18,$3) $3/4=0.75 2.68 11 (19,$2) $2/4=0.50 3.18 12 (20,+) 13 (26,$1) $1/3=0.33 3.52 14 (29,+) 15 (33,+) 16 (39,$2) $2/1=2.00 5.52 17 (42,+)

Transmission data MCF and 95% confidence limits

Using SAS for MCF estimation RELIABILITY procedure nonparametric estimates of population MCF and its 95% confidence interval plot the estimated MCF for the number of repairs or the cost of repairs estimates of the difference of two MCFs and confidence intervals plot the difference of two MCFs and confidence intervals.

Using SAS for MCF estimation Obs id days value 1 251 761 -1 2 252 759 3 327 98 4 667 5 328 326 6 653 …   89 422 582 symbol c=blue v=plus; proc reliability data=valve; unitid id; mcfplot days*value(-1) / cframe = ligr ccensor = megr; inset / cfill = ywh ; run; Repair Data Analysis Age Sample MCF Standard Error 95% Confidence Limits Unit ID Lower Upper 61 0.024 -0.023 0.072 393 76 0.049 0.034 -0.018 0.116 395 84 0.073 0.041 -0.008 0.154 330 87 0.098 0.047 0.006 0.19 331 92 0.122 0.052 0.021 0.223 390 98 0.146 0.056 0.037 0.256 327 …   761 . 251

S-plus for MCF SPLIDA (S-PLUS Life Data Analysis) By W. Q. Meeker

References Nelson, Wayne (2003), Recurrent-Events Data Analysis for Repairs, Disease Episodes, and Other Applications, ASA SIAM Series on Statistics and Applied Probability, SIAM, Philadelphia, PA. Nelson, W. (1995), "Confidence Limits for Recurrence Data--Applied to Cost or Number of Product Repairs," Technometrics, 37, 147 -157.