Theory

The two-sample problem in censored data analysis

We consider two independent censored samples of sizes \(n_1\) and \(n_2\), one sample with survival function \(S_1(t)\) and cumulative hazard \(\Lambda_1(t)\), the other sample with survival function \(S_2(t)\) and cumulative hazard \(\Lambda_2(t)\). The censoring distribution need not be the same in both groups.

Consider the hypothesis \[ \begin{aligned} \mathrm{H}_0: &\, S_1(t)=S_2(t) \quad \forall t>0, \\ \mathrm{H}_1: &\, S_1(t)\neq S_2(t) \quad \text{ for some } t>0, \end{aligned} \quad \Longleftrightarrow \quad \begin{aligned} \mathrm{H}_0: &\, \Lambda_1(t)=\Lambda_2(t) \quad \forall t>0, \\ \mathrm{H}_1: &\, \Lambda_1(t)\neq \Lambda_2(t) \quad \text{ for some } t>0. \end{aligned} \]

Weighted logrank statistics

The class of weighted logrank test statistics for testing \(H_0\) against \(H_1\) is \[ W_K=\int_0^\infty K(s)\,\mathrm{d}(\widehat{\Lambda}_1-\widehat{\Lambda}_2)(s), \] where \[ K(s)=\sqrt{\frac{n_1n_2}{n_1+n_2}} \frac{\overline{Y}_1(s)}{n_1}\, \frac{\overline{Y}_2(s)}{n_2}\,\frac{n_1+n_2}{\overline{Y}(s)}W(s) \] and \(W(s)\) is a weight function – it governs the power of the test against different alternatives.

  1. \(W(s)=1\) is the logrank test. It is most powerful against proportional hazards alternatives, i.e., \(\lambda_1(t)=c\lambda_2(t)\) for some positive constant \(c\neq 1\).

  2. \(W(s)=\overline{Y}(s)/(n+1)\) is the Gehan-Wilcoxon test which in the uncensored case is equivalent to two-sample Wilcoxon test. This test puts more weight on early differences in hazard functions than on differences that occur later.

  3. \(W(s)=\widehat{S}(s-)\) is the Prentice-Wilcoxon test (also known as Peto-Prentice or Peto and Peto test). It is especially powerful against alternatives with decreasing hazard ratios – strong early effects that weaken over time. Preferred compared to Gehan-Wilcoxon test.

  4. \(W(s)=\left[\widehat{S}(s-)\right]^\rho\left[1-\widehat{S}(s-)\right]^\gamma\) is the \(G(\rho,\gamma)\) class of tests of Fleming-Harrington. The logrank test is a special case for \(\rho=\gamma=0\), the Prentice-Wilcoxon test is a special case for \(\rho=1,\ \gamma=0\). The \(G(0,1)\) test is especially powerful against alternatives with increasing hazard ratios – little initial difference in hazards that gets stronger over time.

Let \(\widehat\sigma^2_K\) be the estimated variance of \(W_K\) under \(H_0\), which is of the form \[ \widehat{\sigma}_K^2 = \int \limits_{0}^{\infty} \dfrac{K^2(s)}{\overline{Y}_1(s) \overline{Y}_2(s)} \left(1 - \dfrac{\Delta \overline{N}(s)-1}{\overline{Y}(s) - 1}\right) \,\mathrm{d} \overline{N}(s). \] It can be shown that, under \(H_0\), \[ \frac{W_K}{\widehat\sigma_K}\stackrel{D}{\longrightarrow}\mathsf{N}(0,1) \qquad\text{ and }\qquad \frac{W^2_K}{\widehat\sigma^2_K}\stackrel{D}{\longrightarrow}\chi^2_1. \]

Calculation of weighted logrank statistics

Let \(t_1<t_2<\cdots<t_d\) be the ordered distinct failure times in both samples. The weighted logrank test statistic (without the normalizing constant depending on \(n_1\) and \(n_2\)) can be written as \[ \sum_{j=1}^d \biggl(\Delta\overline{N}_1(t_j)-\Delta\overline{N}(t_j) \frac{\overline{Y_1}(t_j)}{\overline{Y}(t_j)}\biggr) W(t_j). \] The variance estimator \(\widehat{\sigma}_K^2\) (again without the normalizing constant depending on \(n_1\) and \(n_2\)) can be calculated using the following formula \[ \sum \limits_{j = 1}^d \Delta N(t_j) \cdot \dfrac{\overline{Y}_1(t_j) \overline{Y}_2(t_j)}{\left(\overline{Y}(t_j)\right)^2} \cdot \dfrac{\overline{Y}(t_j) - \Delta N(t_j)}{\overline{Y}(t_j) - 1} \cdot \left(W(t_j)\right)^2. \]

Task 1

Download the dataset km_all.RData.

The dataframe inside is called all. It includes 101 observations and three variables. The observations are acute lymphatic leukemia [ALL] patients who had undergone bone marrow transplant. The variable time contains time (in months) since transplantation to either death/relapse or end of follow up, whichever occured first. The outcome of interest is time to death or relapse of ALL (relapse-free survival). The variable delta includes the event indicator (1 = death or relapse, 0 = censoring). The variable type distinguishes two different types of transplant (1 = allogeneic, 2 = autologous).

Using ordinary R functions (not the survival library), calculate and print the following table:

\(j\) \(t_j\) \(d_{1j}\) \(d_{j}\) \(y_{1j}\) \(y_j\) \(e_j=d_j\frac{y_{1j}}{y_j}\) \(d_{1j}-e_j\) \(v_j=d_j\dfrac{y_{1j} (y_j-y_{1j}) (y_j-d_j)}{y_j^2 (y_j - 1)}\)
1 … … … … … … … …
… … … … … … … … …
d … … … … … … … …

where \(j\) is the order of the failures, \(t_j\) is the ordered failure time, \(d_{1j}=\Delta\overline{N}_1(t_j)\), \(d_j=\Delta\overline{N}(t_j)\), \(y_{1j}=\overline{Y}_1(t_j)\), \(y_j=\overline{Y}(t_j)\).

Use these values to perform logrank test on your own.

BONUS: Try to implement \(G(\rho,\gamma)\) test, remember, you need left-continuous version of Kaplan-Meier estimator of survival function.


Implementation in R

The function survdiff in the library survival can calculate \(G(\rho,0)\) statistics. The default is \(\rho=0\), that is, the logrank test. Output is a list with:

library(survival)
survdiff(Surv(time,delta)~type,data=all)
survdiff(Surv(time,delta)~type,data=all,rho=2)
## Call:
## survdiff(formula = Surv(time, delta) ~ type, data = all)
## 
##         N Observed Expected (O-E)^2/E (O-E)^2/V
## type=1 50       22     24.2     0.195     0.382
## type=2 51       28     25.8     0.182     0.382
## 
##  Chisq= 0.4  on 1 degrees of freedom, p= 0.5

The function FHtestrcc in the library FHtest can calculate \(G(\rho,\gamma)\) statistics for non-zero \(\gamma\). The default is \(\rho=0,\ \gamma=0\), that is, the logrank test. The parameter \(\gamma\) is entered as the argument lambda. For the proper choice of \(\rho\) and \(\gamma\) read Details of help(FHtestrcc). Output is a list with:

library(FHtest)
FHtestrcc(Surv(time,delta)~type,data=all)
FHtestrcc(Surv(time,delta)~type,data=all,rho=0.5,lambda=2)
## 
##  Two-sample test for right-censored data
## 
## Parameters: rho=0, lambda=0
## Distribution: counting process approach
## 
## Data: Surv(time, delta) by type
## 
##         N Observed Expected   O-E (O-E)^2/E (O-E)^2/V
## type=1 50       22     24.2 -2.17     0.195     0.382
## type=2 51       28     25.8  2.17     0.182     0.382
## 
## Statistic Z= 0.6, p-value= 0.537
## Alternative hypothesis: survival functions not equal

Caution

The library FHtest requires library Icens which has been removed from R and cannot be installed by standard methods. It is available from the Bioconductor repository. It can be downloaded manually and installed by a command such as:

install.packages("http://www.karlin.mff.cuni.cz/~vavraj/cda/data/Icens_1.38.0.zip", repos =NULL)
Task 1 (comparison with survdiff and FHtestrcc output)

Compare your calculation of logrank test with the output of the functions survdiff and FHtestrcc in the case of dataset all.


Task 2

In the all data, investigate the difference in survival and hazard functions according to the type of transplant. Remember that the outcome is relapse-free survival.

  1. Calculate and plot estimated survival functions by the type of transplant (1 = allogeneic, 2 = autologous). Distinguish the groups by color. Add a legend.

  2. Calculate and plot Nelson-Aalen estimators of cumulative hazard functions by the type of transplant (1 = allogeneic, 2 = autologous). Distinguish the groups by color. Add a legend.

  3. Calculate and plot smoothed hazard functions by the type of transplant (1 = allogeneic, 2 = autologous). Distinguish the groups by color. Add a legend.

  4. Perform logrank, Prentice-Wilcoxon and \(G(0,1)\) tests using functions survdiff or FHtestrcc. Interpret the results. Which test is more suitable in this case?

  • Tasks 1 and 2: Recommended to be finished till next exercise class (9th December)


Hint for obtaining smoothed hazard functions:

library(muhaz)
haz=muhaz(all$time,all$delta)
par(mar = c(4,4,1,1))
plot(haz, col = "black", lwd = 2)

Task 3

Activate the dataset veteran by calling data(veteran). It includes 137 observations and eight variables. The observations are lung cancer patients. The variable time contains survival time (in days) since the start of treatment. The variable status includes the event indicator (1 = death, 0 = censoring). The variable trt distinguishes two different treatments (1 = standard, 2 = experimental). The treatment was randomly assigned to the patients.

Compare survival distributions between standard and experimental treatments. Conduct the analysis according to the steps 1–4 of the previous task.

Compare survival distributions between patients with Karnofsky scores up to 60 and over 60 (regardless of treatment assignment). These scores are stored in variable karno. Conduct the analysis according to the steps 1–4 of the previous task.


Task 4 (simulation study)

Generate \(n_1=100\) censored observations from the Weibull distribution with shape parameter \(\alpha=0.7\) and scale parameter \(1/\lambda=2\). Its expectation is \(\Gamma(1+1/\alpha)/\lambda=2\Gamma(17/7)\doteq 2.53\). Generate another sample of \(n_2=100\) censored observations from the exponential distribution with the same mean. Let the censoring distribution in both samples be exponential with the rate \(\lambda=0.2\) (the expectation is \(1/\lambda=5\)), independent of survival. Put both samples in a single dataframe, introducing a variable distingushing which sample each observation comes from.

Compare survival distributions between the two groups using the generated data. Conduct the analysis according to the steps 1–4 of Task 2 (i.e., perform a descriptive analysis before you do the test). Display the true survival and hazard functions in the plots generated at steps 1–3 of the analysis.

  • Deadline for report (all Tasks 1-4): exercise class in two weeks (!Tuesday 17th December!)


Theoretical Task (voluntary)

Show that with uncensored data, the Gehan-Wilcoxon weighted logrank test with \(W(s)=\frac{\overline{Y}(s)}{n_1+n_2+1}\) (the weight is proportional to the total number of subjects at risk) is equivalent to the standard Wilcoxon test.