My understanding is that the confidence interval for a hazard ratio should be symmetrical about the mean (the distance between the lower limit and the mean is the same as the distance between the mean and the upper limit). But often when I see a hazard ratio in the published literature on clinical trials this is not the case.

For example, a study reporting HR, 0.69; 95% CI, 0.54 to 0.89 in mCRC for cetuximab plus FOLFOX-4 vs FOLFOX-4 alone found here: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7044820/pdf/bmjopen-2019-030738.pdf.

The distance between the lower limit and the mean is 0.69 - 0.54 = 0.15, while the distance between the mean and the upper limit is 0.89 - 0.69 = 0.20. Shouldnt the distance for both be 0.15 to be symmetric?

  • 2
    $\begingroup$ “ The distance between the lower limit and the mean is 0.69 - 0.54 = 0.15, while the distance between the mean and the upper limit is 0.89 - 0.69 = 0.20.” Have you tried taking the ratios rather than the differences? You’ll find they are both ≈ 1.28 $\endgroup$ Oct 2, 2021 at 23:38
  • 1
    $\begingroup$ Thank you @MichaelMacAskill but why are the ratios symmetric? Also, they are not exactly symmetric: 0.89/0.69 = 1.289855 (1.29), while 0.69/0.54 = 1.277778 (1.28) does this present a problem? Will the ratios always be symmetric for hazard ratios? $\endgroup$ Oct 3, 2021 at 13:08

3 Answers 3


The log (partial) likelihood maximization methods used to fit survival models work in the log-hazard scale. The regression coefficient estimates are in that scale, and in that scale the covariance of the estimates is assumed to be multivariate normal. In the log-hazard scale, confidence intervals (CI) are thus symmetric.

When you exponentiate those coefficients and corresponding CI limits to get results in terms of hazard ratios (HR), the CI are necessarily asymmetric. That's even true for the usual null-hypothesis assumption of log-hazard-ratio of 0 or HR of 1. For example, if the log-hazard-ratio estimate has CI of (-0.2, +0.2) around a point estimate of 0, in the (exponentiated) HR scale the CI are (0.8187, 1.2214), asymmetric about the point estimate of HR = 1.

More broadly, as other answers nicely demonstrate, there is no need for CI to be symmetric at all. They often are taken to be symmetric when coefficient estimates can be assumed to have symmetric distributions, but even then there is no rule requiring such a choice. Any interval containing 95% of the probability distribution can be taken as 95% CI.

  • 1
    $\begingroup$ Thank you. This presents 2 questions, in journal articles reporting hazard ratios, i.e. of some new intervention assumed to improve survival in cancer patients, I assume the hazard ratios not on the log scale unless they are explicitly called "log-hazard ratios"? Second, consider a hazard ratio of 1.9 (95% CI, 1.6 2.3). When I take the log of the confidence intervals and hazard ratios: log(1.9) = 0.6418539, log(1.6) = 0.4700036, log(2.3) = -0.8329091 The confidence intervals are still not symmetric about the hazard ratio 0.6418539 - 0.4700036 = 0.1718503, 0.8329091 - 0.6418539 = 0.1910552. $\endgroup$ Oct 3, 2021 at 12:51
  • $\begingroup$ @JamesMoore Clinical papers typically report hazard ratios, not their logs. If authors call values "hazard ratios," take them at their word. The apparent lack of symmetry in transforming to log-HR comes from rounding. Your values shows the upper CI farther from the point estimate than the lower CI in the log scale. Yet a HR of 1.9 might be 1.949 before rounding; an upper CI of 2.3 might be as low as 2.251; a lower CI of 1.6 might be as high as 1.649. Then the direction of "asymmetry" in log scale reverses. Somewhere in the middle of the rounding ranges there is symmetry in log scale. $\endgroup$
    – EdM
    Oct 3, 2021 at 13:55
  • 1
    $\begingroup$ @JamesMoore one example of symmetry in log scale compatible with rounding to one decimal in HR scale: HR point estimate of 1.90 (log, 0.6419); upper CI, 2.275 (log, 0.822), rounding up to 2.3; lower CI, 1.587 (log, 0.4618), rounding up to 1.6. In the log scale, both upper and lower CI are 0.1801 away from the point estimate. Exponentiation can lead to surprisingly big differences for small changes in values. $\endgroup$
    – EdM
    Oct 3, 2021 at 14:11

My understanding is that the confidence interval for a hazard ratio should be symmetrical about the mean (the distance between the lower limit and the mean is the same as the distance between the mean and the upper limit).

This is not correct. Hazard ratios have no reason to be symmetric. (They could in principle be symmetric in specific cases, e.g. if the point estimate is $1.0$ - but they are not symmetric in general.)

Here is an argument. Wikipedia says:

In its simplest form, the hazard ratio can be interpreted as the chance of an event occurring in the treatment arm divided by the chance of the event occurring in the control arm, or vice versa, of a study.

Now, note that whether one arm is considered "treatment" and the other one "control" is absolutely up to interpretation. We could switch the labels at will. And such switching would turn the hazard ratio into its reciprocal, $\text{HR}'=\frac{1}{\text{HR}}$.

If, now, hazard ratio CIs were always symmetrical, then this inversion would need to respect this symmetry. Thus, a CI of $(r-\epsilon,r+\epsilon)$ for $\text{HR}$ would need to turn into a CI $(\frac{1}{r+\epsilon},\frac{1}{r-\epsilon})$ for $\text{HR}'$ that is also symmetric about $\frac{1}{r}$, or $$\frac{1}{r+\epsilon} = \frac{1}{r}-\delta, \qquad\frac{1}{r-\epsilon}=\frac{1}{r}+\delta. $$ But this is not possible mathematically in general. Taking reciprocals is not a symmetric operation.

  • 1
    $\begingroup$ Ah! I think I know how I got confused! I read the following: tinyurl.com/vk7t46zb showing that a HR's CI is calculated by taking the natural log of the HR and then the CI is found on the natural log scale. It said the log transformed CI will be symmetrical around the point estimate so I thought that HRs and their CI were always reported on the log scale, and thus should be symmetrical. But maybe HRs are calculated on a linear scale, transformed to the log scale to calculate the CI, and then the HR and CI are exponentiated to give the HR and 95% CI we see reported in journal articles? $\endgroup$ Oct 3, 2021 at 13:41
  • 4
    $\begingroup$ @JamesMoore the analysis usually proceeds in the log-hazard space and then at the last moment we exponentiate to give a more meaningful interpretation. $\endgroup$
    – mdewey
    Oct 3, 2021 at 13:47

It is just that we are used to symmetric confidence intervals, because this is the most typical/common situation: the case of some sort of linear regression with Gaussian distributed errors.

But... there is not a single unique confidence interval. Any region with 95% confidence will do.

Often one chooses the region that is smallest in some sense. For instance by using a hypothesis test (the confidence interval can be seen a region of hypothesis tests) that computes the p-values based on the region with highest probability density.

Example 1

Below is an example from this question that shows the estimation of the rate parameter $\lambda$ of a population that is exponential distributed. The estimate is based on the observed sample mean $\bar{x}$. You can see that the confidence interval boundaries (thick black lines) are not symmetrical around the estimate (dotted line). This has two reasons.

  • The estimate for the rate parameter $\hat\lambda$ is not a linear function of the observed $\bar{x}$.
  • The sample distribution of $\bar{x}$ for a given parameter $\lambda$ is not symmetric.

Example 2

A very clear example is the estimation of the parameter $p$ in a binomial distribution.

Say you estimate the number of red and blue balls in an urn. If in some sample you observe zero red balls then the estimate for the number of red balls will be zero, but the confidence interval won't be symmetric (or it shouldn't be) or otherwise you include negative values.

Example 3

The answer of Christoph Hanck in the previous linked question shows an example of the Gaussian distribution and how the confidence interval will turn out to be symmetric.

But if you transform the parameter for which you compute the interval then it will again become non-symmetric. Possibly that is the case for the hazard ratio.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.