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Say I have a $K$ arm experiment that generates survival (time-to-event) endpoints. There are $K-1$ experimental arms and a single control arm.

Say I compute a log rank test statistic comparing the hazards for each arm to control. There will be $K-1$ standardized test statistics, $Z_1, \dots, Z_{k-1}$. What is the correlation between test statistics?

I've found course-notes that say when $K=2$ the correlation is 0.5. Under the null they would be distributed as $(Z_1, Z_2) \sim \mathcal{N}_2(0, \Sigma)$, where $$\Sigma = \begin{pmatrix} 1 & 0.5 \\ 0.5 & 1 \end{pmatrix}.$$ This seems plausible but I don’t see why it’s true. If it is true does it expand to multiple log rank test statistics?

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  • $\begingroup$ Do you have a source for your course-notes? $\endgroup$ Commented Oct 6, 2019 at 12:33
  • $\begingroup$ It's actually a talk, not course notes. Here's the link, look at page 22: people.bath.ac.uk/mascj/talks_2019/cj_novartis_mar_2019.pdf $\endgroup$
    – Eli
    Commented Oct 7, 2019 at 14:19
  • $\begingroup$ Correlation in this context seems misplaced? For survival endpoints, I don't think of multiple tests as having between group correlation. The fixed denominator is not a form of correlation, it is an experimental condition set by the experimental design. The only correlation involves the baseline disease under observation which is controlled for by the design, not the analysis. $\endgroup$
    – Todd D
    Commented Feb 22, 2020 at 18:24

1 Answer 1

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The (signed, two-group) logrank test statistic is $$Q_{AB}=\int_0^T W_{AB}(t) \left(\frac{dN_A(t)}{Y_A(t)}- \frac{dN_B(t)}{Y_B(t)}\right)$$ where $N_A$ counts the number of events in treatment $A$ and $Y_A$ counts the number at risk in treatment $A$, and $W_{AB}(t)$ is a weight function.

Under the strong null that the treatments don't differ at all, and for large sample sizes, $W_{AB}(t)$ will be approximately the same for any pair of treatments and approximately deterministic (call it $W(t)$). And again, under the strong null, the processes $dN/Y$ will have the same distribution for each treatment.

This means $$\mathrm{cov}[Q_{AB},Q_{AC}]\approx \int W^2(t)\,\mathrm{cov}\left[\frac{dN_A(t)}{Y_A(t)}- \frac{dN_B(t)}{Y_B(t)}, \frac{dN_A(t)}{Y_A(t)}- \frac{dN_C(t)}{Y_C(t)}\right]$$ Now, different treatment arms are independent under the null so this reduces to $$\int W^2(t)\,\mathrm{var}\left[\frac{dN_A(t)}{Y_A(t)}\right]$$ and (since different arms have the same distribution under the null), $$\mathrm{var}[Q_{AB}]=2\int W^2(t)\,\mathrm{var}\left[\frac{dN_A(t)}{Y_A(t)}\right]$$

Generalising this, $\mathrm{cor}[Q_{AB},Q_{AC}]=1/2$, $\mathrm{cor}[Q_{AB},Q_{BC}]=-1/2$,and $\mathrm{cor}[Q_{AB},Q_{CD}]=0$, giving us all the correlations.

None of this necessarily holds under the weak null where $E[Q_{AB}]=0$ but the treatments are different -- eg, if A is better early on and B is better later on in a way that exactly cancels for the logrank test. It also doesn't hold if censoring depends on treatment assignment, but in that case you have worse problems.

Another way to deduce this is from the fact that the logrank statistics are asymptotically multivariate Normal. Think of a graph whose vertices are arms and whose edges are logrank statistics. We must have $Q_{AB}+Q_{BC}+Q_{CA}=0$ and similarly for other sums around closed loops. Under the null, the variances of all the logrank statistics must be equal, the covariances of overlapping pairs such as $\mathrm{cov}[Q_{AB},Q_{BC}]$ and $\mathrm{cov}[Q_{BC},Q_{CA}]$ must all be equal (by symmetry), and $Q_{AB}=-Q_{BA}$, and the covariances of non-overlapping pairs must be zero. That's enough to determine the correlation matrix. For example, $\mathrm{var}[Q_{AB}+Q_{BC}+Q_{CA}]=0$ implies $3+6\rho=0$ where $\rho$ is the common correlation, which gives $\rho=-1/2$.

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