Mean Cumulative Function (MCF) For Recurrent Events

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Mean Cumulative Function (MCF) For Recurrent
Events

• (Only what I learned so far.)

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Outline

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

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Coronary Artery Disease

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Coronary Artery Bypass Graft (CABG) Surgery

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Percutaneous Coronary Intervention (PCI)
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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

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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.

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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)

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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.

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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.

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How to estimate MCF?
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How to estimate MCF?
 
 
Calculation of MCF for Artificial Data
System Repair Histories for Artificial data
Unit (Age in Months, Cost in 100) (Age in Months, Cost in 100) (Age in Months, Cost in 100) (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 1/60.17 0.17
2 (5,3) 6 3/60.50 0.67
3 (8,2) 6 2/60.33 1
4 (8,1) 6 1/60.17 1.17
5 (12,1) 6 1/60.17 1.33
6 (12,) 5    
7 (14,1) 5 1/50.20 1.53
8 (16,2) 5 2/50.40 1.93
9 (16,) 4    
10 (18,3) 4 3/40.75 2.68
11 (19,2) 4 2/40.50 3.18
12 (20,) 3    
13 (26,1) 3 1/30.33 3.52
14 (29,) 2    
15 (33,) 1    
16 (39,2) 1 2/12.00 5.52
17 (42,) 0
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Transmission data MCF and 95 confidence limits
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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.

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Using SAS for MCF estimation
Obs id days value
1 251 761 -1
2 252 759 -1
3 327 98 1
4 327 667 -1
5 328 326 1
6 328 653 1
     
89 422 582 -1

• symbol cblue vplus
• proc reliability datavalve
• unitid id
• mcfplot daysvalue(-1) / cframe ligr ccensor
  megr
• inset / cfill ywh
• run

Repair Data Analysis Repair Data Analysis Repair Data Analysis Repair Data Analysis Repair Data Analysis Repair Data Analysis
Age Sample MCF Standard Error 95 Confidence Limits 95 Confidence Limits Unit ID
Age Sample MCF Standard Error Lower Upper Unit ID
61 0.024 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
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S-plus for MCF

• SPLIDA (S-PLUS Life Data Analysis)
• By W. Q. Meeker

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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.