Testing equality of censored distributions

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 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 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 $$\sum \limits_{j = 1}^d \Delta N(t_j) \cdot \dfrac{\overline{Y}_1(t_j) \overline{Y}_2(t_j)}{\left(\overline{Y}(t_j)\right)^2} \cdot \dfrac{\overline{Y}(t_j) - \Delta N(t_j)}{\overline{Y}(t_j) - 1} \cdot \left(W(t_j)\right)^2.$$