I recently began studying survival analysis and there is something I am curious about.

Why do we prefer to use the ln-ln survival curve rather than the survival curve in a proportional hazard test? As you know, the $\hat{S}$ we use in ln-ln formula is the survival function based on Kaplan-Meier curves, so why we don't use the untransformed KM survival function?

  • $\begingroup$ It looks to me like the two answers are interpreting the intent of "ln-ln" differently. Could you give the formula you mean when you say "in ln-ln formula" explicitly in your question? $\endgroup$
    – Glen_b
    Commented Dec 27, 2015 at 23:50
  • $\begingroup$ @Glen_b; I think we both interpreted "ln-ln" to be the function $\ln( -\ln(x))$ $\endgroup$
    – Cliff AB
    Commented Dec 28, 2015 at 1:44
  • $\begingroup$ @CliffAB The first two sentences of the third paragraph of the other answer clearly indicates otherwise. It interprets it to be taking logs of both axes (hence making a power law linear). [In any case, the more usual interpretation of log-log with reference to a plot is to take log of each variable, so it would still be necessary for the OP to clarify which is intended. Your interpretation makes perfect sense in this context, but it's still unclear whether that's what is being asked about. It may even be that the OP is misinterpreting something that someone else has written.] $\endgroup$
    – Glen_b
    Commented Dec 28, 2015 at 1:52
  • $\begingroup$ @Glen_b: oh I see. I read the other answer too quickly. $\endgroup$
    – Cliff AB
    Commented Dec 28, 2015 at 1:56

2 Answers 2


Let us define the linear predictor (i.e. log hazard ratio) $\eta = X^T \beta$. Then the proportional hazards model can be written as

$h(t | \eta) = h_o(t) \exp(\eta)$

This relation is equivalent to

$S(t | \eta) = S_o(t)^{\exp(\eta)}$

Therefore, if we plot the cloglog (complimentary log log, or ln -ln as stated in the question) of the survival functions, we get

$\text{cloglog} (S(t | \eta) ) = \text{cloglog} (S_o(t)) - \eta$

i.e. if the proportional hazards assumption is true, the curves should differ only by a constant. It's much easier to visually assess whether two curves differ by an additive constant than whether one differs by an exponential factor.

As an example, here is simulated data that does not follow the proportional hazards model (it follows a proportional odds model instead). Looking at the cloglog plots (with the average of the two cloglog functions removed for easier comparison), we can see that the difference between these two functions is not exactly constant.


However, looking at the two survival curves, you would have to have a much better eye than I will ever have to determine that these two curves do not differ by an exponential factor.

enter image description here


Note that the curve is the same. The log-log is only a way of representing the same data.

There are many reasons to why a log-log plot is preferred, not only in survival analysis: the main one is that when functions go to 0, applying the logarithm allows you to differentiate very low values, for example if the scale goes between 1 and $10^{-5}$, the difference between $10^{-3}$ and $10^{-4}$ spans 20% of the vertical axis.

In particular, by using a log-log scale you are able to visually differentiate if you have an exponential or a power-law decay. The first will look as a sharp drop in the function while the second like a straight line. Of course that this is not rigorous (specially if you compare power-law with other, more similar, distributions like lognormal), but it can give you a quick idea of what you are looking to.


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