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

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

4. Coronary Artery Disease

5. Coronary Artery Bypass Graft (CABG) Surgery

6. Percutaneous Coronary Intervention (PCI)

8. 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

9. 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.

10. 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)

11. 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.

12. 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 = 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.

13. How to estimate MCF?

14. (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,+)

15. Transmission data MCF and 95% confidence limits

16. 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.

17. 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

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

23. 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,