Why are confidence intervals of hazard ratios not symmetric? 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?
 A: 
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.
A: 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.

A: 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.
