Question

(Estimating the survival function) Assume that \(n\) i.i.d. replications from the proportional excess model is observed over the time-interval \([0, \tau]\). (a) Suggest an estimator of the survival function, and derive an optional variation estimator of the variance for \(n^{1 / 2}\left(\hat{S}_{0}(t)-S_{0}(t)\right)\). (b) Make an i.i.d. decomposition formula for \(n^{1 / 2}\left(\hat{S}_{0}(t)-S_{0}(t)\right)\), and give the recipe for making a \(95 \%\) confidence band.

Short answer

For a proportional excess model with \(n\) independent and identically distributed replications over a time-interval \([0, \tau]\), the Kaplan-Meier estimator is proposed to estimate the survival function. The Greenwood variance estimator is then calculated for the Kaplan-Meier estimator. The i.i.d. decomposition formula for \(n^{1/2}\left(\hat{S}_{0}(t)-S_{0}(t)\right)\) is given as a summation of i.i.d. terms with a specified distribution. To construct a 95% confidence band, critical values are obtained from the distribution of these terms, and the confidence band is formed as \(\hat{S}_{0}(t) \pm z_{0.975}\cdot \frac{1}{\sqrt{n}}\cdot \hat{\sigma}(\hat{S}_{0}(t))\). If the distribution is unknown, nonparametric distribution-free methods such as the log-rank test can be employed.

Step-by-step solution

Propose an estimator

One popular estimator for the survival function is the Kaplan-Meier estimator, also known as the product limit estimator. It is given by the following formula: \(\hat{S}_{0}(t) = \prod_{i:t_{i}\leq t} (1 - \frac{d_i}{n_i})\) where \(d_i\) represents the number of failures (or deaths) at time \(t_{i}\) and \(n_i\) is the total number of individuals at risk just before time \(t_{i}\).

Find the variance

We can calculate the Greenwood variance estimator for the Kaplan-Meier estimator, which is given by: \(\hat{\sigma}^2(\hat{S}_{0}(t)) = \hat{S}_{0}^2(t) \sum_{i: t_i\leq t} \frac{d_i}{n_i(n_i-d_i)}\) Now we can find the optimal variation estimator for the variance of \(n^{1/2}\left(\hat{S}_{0}(t)-S_{0}(t)\right)\): \(\frac{1}{n}\hat{\sigma}^2(\sqrt{n}\left(\hat{S}_{0}(t)-S_{0}(t)\right)) = \hat{S}_{0}^2(t) \sum_{i: t_i\leq t} \frac{d_i}{n_i(n_i-d_i)}\) #b) i.i.d. decomposition and 95% confidence band#

i.i.d. decomposition formula

The i.i.d. decomposition formula for \(n^{1/2}\left(\hat{S}_{0}(t)-S_{0}(t)\right)\) can be written as: \(n^{1/2}\left(\hat{S}_{0}(t)-S_{0}(t)\right) = \sum_{i=1}^{n}Z_i\) where \(Z_i\) are i.i.d. terms with a specified distribution.

Recipe for 95% confidence band

To construct a 95% confidence band, we need to find the critical value for the distribution of \(n^{1/2}\left(\hat{S}_{0}(t)-S_{0}(t)\right)\). If the distribution of the \(Z_i\) terms is known, we can use the quantiles of this distribution. For example, if the distribution is assumed to be normal, we can use the standard normal quantiles. Let \(z_{0.975}\) be the 97.5th quantile of the \(Z_i\) distribution. The 95% confidence band will then be between \(\hat{S}_{0}(t) \pm z_{0.975}\cdot \frac{1}{\sqrt{n}}\cdot \hat{\sigma}(\hat{S}_{0}(t))\). If the distribution of the \(Z_i\) terms is not known, one can use a nonparametric distribution-free method such as the log-rank test to construct the confidence band.

Key concepts

Kaplan-Meier Estimator

The Kaplan-Meier estimator is a non-parametric statistic used to estimate the survival function from life-time data. In simpler terms, it helps to estimate the probability that a subject will survive a certain amount of time. The estimator is particularly useful in medical research to measure the fraction of patients living for a certain amount of time after treatment.

The Kaplan-Meier estimator is advantageous because it can handle different types of data, such as right-censored data which occurs when a patient's survival time is unknown after a certain date. The formula for the Kaplan-Meier estimator for the survival function at time t is given by:
\[\begin{equation}\hat{S}_{0}(t) = \prod_{i:t_{i}\leq t} \left(1 - \frac{d_i}{n_i}\right)\end{equation}\]
The product runs over all observed time points up to time t, where \(d_i\) is the number of 'failures' (like deaths or disease recurrences) and \(n_i\) is the number at risk just before time \(t_i\). Intuitively, the estimator multiplies the probabilities of surviving each time interval up to time t.

Greenwood's Formula

When using the Kaplan-Meier estimator, it's also important to estimate the variance of the estimated survival probabilities. Greenwood's formula gives us a method to do this. The estimated variance of the Kaplan-Meier estimator using Greenwood's formula is:
\[\begin{equation}\hat{\sigma}^2(\hat{S}_{0}(t)) = \hat{S}_{0}^2(t) \sum_{i: t_i\leq t} \frac{d_i}{n_i(n_i-d_i)}\end{equation}\]

Why Is Variance Important?

Variance lets us understand the spread or uncertainty around the estimate. In the context of survival analysis, it tells us how much the survival probabilities might fluctuate due to chance alone.

By applying Greenwood's formula, we can calculate confidence intervals around the Kaplan-Meier estimator to determine with some level of certainty whether the observed variation in survival time is real or due to random fluctuations.

Confidence Intervals

Confidence intervals give a range within which we can expect the true survival function to lie, with a certain degree of confidence. For example, a 95% confidence interval around the Kaplan-Meier estimate means that there's a 95% probability that the interval contains the true survival function at time t.

Using the variance estimated by Greenwood's formula, the 95% confidence interval for the survival function can be constructed. It is commonly expressed as:
\[\begin{equation}\hat{S}_{0}(t) \pm z_{0.975}\cdot \frac{1}{\sqrt{n}}\cdot \hat{\sigma}(\hat{S}_{0}(t))\end{equation}\]
where \(z_{0.975}\) is the 97.5th percentile of the standard normal distribution, corresponding to the upper tail probability of 0.025. The \(\hat{\sigma}(\hat{S}_{0}(t))\) is the standard deviation estimated using Greenwood's formula, and \(n\) is the sample size. This interval provides a statistical assurance regarding the range in which the true survival function is likely to lie.

i.i.d. Decomposition

The i.i.d. decomposition stands for independent and identically distributed decomposition. This principle allows us to express a complex, aggregate random variable as a sum of simpler random variables that are independent and identically distributed.

In the context of the Kaplan-Meier estimator, we can use i.i.d decomposition to express the variance of the survival estimates. The central limit theorem plays a crucial role here; it tells us that under certain conditions, the distribution of the sum (or average) of a large number of i.i.d. random variables will approach a normal distribution, regardless of the underlying individual distributions.

For the Kaplan-Meier estimator, the i.i.d. decomposition can be quantified as:
\[\begin{equation}\sqrt{n}\left(\hat{S}_{0}(t)-S_{0}(t)\right) = \sum_{i=1}^{n}Z_i\end{equation}\]
Each \(Z_i\) term in the sum represents the contribution of individual data points to the aggregate variance of the Kaplan-Meier estimate. This decomposition is essential in statistical theory because it underpins the construction of confidence intervals and hypothesis testing for the survival function.

The i.i.d. decomposition offers an insightful statistical perspective and simplifies the complex analysis associated with survival data. It also provides the foundation for constructing confidence bands, as each \(Z_i\) term is used to find the critical values for a distribution, which is pivotal in determining the width of the confidence bands.