Template:Mean cumulative function for recurrence data

The Mean Cumulative Function (MCF) for Recurrence Data

In non-parametric analysis of recurrent events data, each population unit can be described by a cumulative history function for the cumulative number of recurrences. It is a staircase function that depicts the cumulative number of recurrences of a particular event, such as repairs, over time. Figure CHM depicts a unit's cumulative history function.

Cumulative number of failures.

The nonparametric model for a population of units is described as the population of cumulative history functions (curves). It is the population of all staircase functions of every unit in the population. At age t, the units have a distribution of their cumulative number of events. That is, a fraction of the population has accumulated 0 recurrences, another fraction has accumulated 1 recurrence, another fraction has accumulated 2 recurrences, etc. This distribution differs at different ages, [math]\displaystyle{ t }[/math] , and has a mean [math]\displaystyle{ M(t) }[/math] called the mean cumulative function (MCF) . The [math]\displaystyle{ M(t) }[/math] is the pointwise average of all population cumulative history functions(see Fig.MCFDemo).

Illustration of MCF and population distribution at age t.

For the case of uncensored data, the mean cumulative function [math]\displaystyle{ M{{(t)}_{i}}\ }[/math] values at different recurrence ages [math]\displaystyle{ {{t}_{i}} }[/math] are estimated by calculating the average of the cumulative number of recurrences of events for each unit in the population at [math]\displaystyle{ {{t}_{i}} }[/math] . When the histories are censored, the following steps are applied.

1st Step - Order All Ages:

Order all recurrence and censoring ages from smallest to largest. If a recurrence age for a unit is the same as its censoring (suspension) age, the recurrence age goes first. If multiple units have a common recurrence or censoring age, then these units could be put in a certain order or be sorted randomly.

2nd Step - Calculate the Number, [math]\displaystyle{ {{r}_{i}} }[/math] , of Units that Passed Through Age [math]\displaystyle{ {{t}_{i}} }[/math] :

[math]\displaystyle{ \begin{align} & {{r}_{i}}= & {{r}_{i-1}}\quad \quad \text{if }{{t}_{i}}\text{ is a recurrence age} \\ & {{r}_{i}}= & {{r}_{i-1}}-1\text{ if }{{t}_{i}}\text{ is a censoring age} \end{align} }[/math]

[math]\displaystyle{ N }[/math] is the total number of units and [math]\displaystyle{ {{r}_{1}}=N }[/math] at the first observed age which could be a recurrence or suspension.

3rd Step - Calculate [math]\displaystyle{ MCF }[/math] Estimate, [math]\displaystyle{ {{M}^{*}}(t) }[/math]: For each sample recurrence age [math]\displaystyle{ {{t}_{i}}, }[/math] calculate the mean cumulative function estimate as follows:

[math]\displaystyle{ {{M}^{*}}({{t}_{i}})=\frac{1}{{{r}_{i}}}+{{M}^{*}}({{t}_{i-1}}) }[/math]

where [math]\displaystyle{ {{M}^{*}}(t)=\tfrac{1}{{{r}_{1}}} }[/math] at the earliest observed recurrence age, [math]\displaystyle{ {{t}_{1}} }[/math] .

Example 1

A health care company maintains five identical pieces of equipment used by a hospital. When a piece of equipment fails, the company sends a crew to repair it. The following table gives the failure and censoring ages for each machine, where the + sign indicates a censoring age.

[math]\displaystyle{ \begin{matrix} Equipment ID & Months \\ \text{1} & \text{5, 10 , 15, 17+} \\ \text{2} & \text{6, 13, 17, 19+} \\ \text{3} & \text{12, 20, 25, 26+} \\ \text{4} & \text{13, 15, 24+} \\ \text{5} & \text{16, 22, 25, 28+} \\ \end{matrix} }[/math]

[math]\displaystyle{ }[/math] Estimate the MCF values and the 95% confidence limits ignoring the repair duration.

Solution to Example 1

The MCF estimate is obtained as follows:

[math]\displaystyle{ \begin{matrix} ID & Months ({{t}_{i}}) & State & {{r}_{i}} & 1/{{r}_{i}} & {{M}^{*}}({{t}_{i}}) \\ \text{1} & \text{5} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.20} \\ \text{2} & \text{6} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.20 + 0}\text{.20 = 0}\text{.40} \\ \text{1} & \text{10} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.40 + 0}\text{.20 = 0}\text{.60} \\ \text{3} & \text{12} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.60 + 0}\text{.20 = 0}\text{.80} \\ \text{2} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.80+0}\text{.20 =1}\text{.00} \\ \text{4} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.00 + 0}\text{.20 = 1}\text{.20} \\ \text{1} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.20 + 0}\text{.20 =1}\text{.40} \\ \text{4} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.40 + 0}\text{.20 = 1}\text{.60} \\ \text{5} & \text{16} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.60 + 0}\text{.20 = 1}\text{.80} \\ \text{2} & \text{17} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.80 + 0}\text{.20 = 2}\text{.0} \\ \text{1} & \text{17} & \text{S} & \text{4} & {} & {} \\ \text{2} & \text{19} & \text{S} & \text{3} & {} & {} \\ \text{3} & \text{20} & \text{F} & \text{3} & \text{0}\text{.33} & \text{2}\text{.00 + 0}\text{.33 = 2}\text{.33} \\ \text{5} & \text{22} & \text{F} & \text{3} & \text{0}\text{.33} & \text{2}\text{.33 + 0}\text{.33 = 2}\text{.66} \\ \text{4} & \text{24} & \text{S} & \text{2} & {} & {} \\ \text{3} & \text{25} & \text{F} & \text{2} & \text{0}\text{.50} & \text{2}\text{.66 + 0}\text{.50 = 3}\text{.16} \\ \text{5} & \text{25} & \text{F} & \text{2} & \text{0}\text{.50} & \text{3}\text{.16 + 0}\text{.50 = 3}\text{.66} \\ \text{3} & \text{26} & \text{S} & \text{1} & {} & {} \\ \text{5} & \text{28} & \text{S} & \text{0} & {} & {} \\ \end{matrix} }[/math]