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?


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.
