3.1 Estimating the survivor function: non-parametric estimation

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  typically the first step in the presence of censored data is to obtain the Kaplan-Meier estimate of the survivor function

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  some values may be right censored and there may also be individuals with the same observed survival time

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  example data: $1,3,7^{\ast },9,9,{10}^{\ast }$ ($\ast$ means censored)

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  suppose that there are $r$ observed failures with $r\le n$

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  further suppose the event times are in ascending order the $j$th of which is denoted $t_{(j)}$ for $j=1,\mathrm{\dots },r$

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  the ordered event times are thus:

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  the number in the sample who are at risk just prior to time $t_{(j)}$ are given by $n_j$ for $j=1,2,\mathrm{\dots },r$ and $d_j$ are the number of who fail at this time

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  the estimated probability of survival through interval $j$ is then:

  $$p_j=\frac{n_j-d_j}{n_j}$$

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  assuming that the event times occur independently of one another leads to the Kaplan-Meier estimator:

  $$\widehat{S}(t)=\prod _{j=1}^k{p}_j=\prod _{j=1}^k\left(\frac{n_j-d_j}{n_j}\right)$$

  for $k=r+1$ sub-intervals and with $\widehat{S}(t)=1$ for

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  $\widehat{S}(t)=\widehat{p}(T>t)=p_1\times p_2\times \mathrm{\dots }\times p_k$

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  Kaplan-Meier also known as the product limit estimator

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  note if no censoring then this is simply the empirical survivor function

Unnumbered Figure: Link

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  we also need to address the question of uncertainty in our estimate

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  various formula exist for the estimated variance

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  Greenwood’s formula

  $$\widehat{\text{Var}}(\widehat{S}(t))=\widehat{S}{(t)}^2\sum _{j=1}^k\left(\frac{d_j}{n_j\{n_j-d_j\}}\right)$$

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    can give confidence bands

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    can give confidence bands $>1$

Unnumbered Figure: Link

Unnumbered Figure: Link

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  note how the plot appears smoother with more data and more distinct failure times

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  the survivor function commences at 1 and the estimated survival close to 10% at about 2.5 years