The logrank test statistic is equivalent to the score of a Cox regression. Is there an advantage of using a logrank test over a Cox regression?

Question

I have understood the logrank test as a "safe" or "conservative" way to check for a difference between two survival curves. It is "safe" in the sense that it is a nonparametric test of $h_1(t) = h_2(t)$, where $h_i( \cdot)$ is the hazard function of the $i^{th}$ group. I can't find sources that support this claim of "safety", so I've either invented this interpretation myself or observed it from others over the years.

I have seen sources that say that the logrank test statistic is equivalent to the score test (also called the Legrange multiplier test). I am pulling definitions from these course notes (pg. 14) for anyone who wants to look. The score of a Cox model is $$\sum_{i=1}^{n} \delta_{i}\left\{Z_{i}-E\left(Z ; T_{i}\right)\right\},$$ where $\delta_i$ is an indicator of whether the subject $i$ has an event, $T_i$ is the possibly-censored survival time, $Z_i$ is a covariate like treatment assignment, and $E\left(Z ; T_{i}\right)$ is the expectation of $Z_i$. This is equivalent (or analogous??) to the "Observed - Expected" form of the logrank test.

I have two related questions:

1. Since the logrank test and Cox regression have this equivalence, is my perception incorrect that the logrank test is "safer" than Cox regression?
2. A cox model assumes that the hazard functions are proportional with the same baseline hazard: $h(t | Z_i) = h_0(t) * \exp(Z_i \times \beta)$. I am not aware of any assumptions for the logrank test. My perception has always been that a logrank test statistics must be better than Cox regression in terms of power or asymptotic efficiency or something since it is not making an assumption of proportional hazards. Is this actually true?

Answer 1

There is no advantage to using the log-rank statistic and several disadvantages:

• Unlike the Cox model, log-rank does not generalize to a Bayesian framework
• the log-rank test only works for mutually exclusive categories and does not handle a continuous exposure variable
• log-rank does not allow for general covariate adjustment

Since the log-rank test is a special case of the Cox model, it does not have fewer assumptions or more power. IMHO we no longer need to be using or teaching the log-rank test.

Answer 2

The relation between the log-rank test and the Cox PH model is very similar to that of the two-sample t-test and linear regression. That is, for a standard log-rank test, it is roughly equivalent to fitting a Cox-PH model with an indicator for group. There is a corner case in which the standard log-rank has some advantage: if one of the groups events are earlier than all of the other (or all censored), this causes perfect separation of the data and can lead to unbounded MLE's, causing numeric issues.

But similar to the two sample t-test with unequal variances and linear regression, log rank statistics can be generalized in ways that are not so simple (or at least, not readily implemented) for regression models. In particular, the weighted log-rank statistic allows us to upweight parts of the survival curve where we suspect the hazard ratios to differ the most and thus regain power in the case of non-proportional hazards.

For some discussion, see "Is it time for the weighted log-rank test to play a more important role in confirmatory trials?" by Zheng Su (2018).

Answer 3

In the simple, semi-parametric Cox model

$$h(t | Z_i) = h_0(t) \exp(Z_i \times \beta)$$

where

$$ Z_i= \begin{cases} 0 & \text{if subject $i$ is in the 1st group}\\ 1 & \text{if subject $i$ is in the 2nd group} \end{cases} $$

the null hypothesis that $\beta = 0$ is just the hypothesis that the two groups have the (unspecified) hazard function $h_0$ in common. So any valid test of $\beta=0$ is a valid, non-parametric test of $h_1(t) = h_2$(t) (where $h_1(t)=h_0(t)$ & $h_2(t)=h_0(t)\exp(\beta)$).

It's rather an apples-and-oranges comparison to say the log-rank test is safer than Cox regression, but it is true that it doesn't rely on an assumption of proportional hazards for validity. As the score test for the Cox model, it's the locally most powerful test (the most sensitive to $\beta$s approaching zero); it has good power against alternatives with more-or-less proportional hazards, & fairly decent power against alternatives where the hazard for one group is consistently higher than that of the other over time.