# Propagation of asymmetric uncertainties

I have two (fully independent) measurements of the same quantity X. Each of them reports a measurement $$X_{-\sigma_L}^{+\sigma_R}$$ where $$\sigma_L$$ and $$\sigma_R$$ are the left and right uncertainties (asymmetric error bars). In other words, if we call the measurements $$A$$ and $$B$$, and the subscript $$A$$ and $$B$$ stands for the measurements, we have

$$A_{-\sigma_{L,A}}^{+\sigma_{R,A}}$$

$$B_{-\sigma_{L,B}}^{+\sigma_{R,B}}$$

Now I need to calculate the different of these measurements, $$\Delta = A-B$$. What will be $$\sigma_{L,\Delta}$$, $$\sigma_{R,\Delta}$$? In other words, how do I propagate independent asymmetric error bars?

• Could you explain more precisely and quantitatively what these "left" and "right" uncertainties mean or how they are computed?
– whuber
Commented Dec 23, 2018 at 15:29
• It comes from the fact that the $\chi^2$ distribution is not symmetric around its minimum, therefore taking the 68% confidence interval leads to an asymmetric interval Commented Dec 23, 2018 at 15:38
• So, $A$ and $B$ are assumed to be $\chi^2$-distributed and the $\sigma$'s are confidence bounds? One can then probably try to find the actual distribution of $\Delta$ and compute new confidence bounds. A very simple approximate estimate would be $\sigma_{L,\Delta} = \sigma_{L,A} - \sigma_{R,B}$ and $\sigma_{R,\Delta} = \sigma_{R,A} - \sigma_{L,B}$. They are correct, if the $\sigma$'s are strict bounds. Commented Dec 23, 2018 at 21:35
• @Jonas I also thought about something similar, which makes sense to me, however I was wondering if it respects the rules for the propagation of uncertainties. After all, when we sum two Gaussian distributed symmetric errors we do NOT just sum their uncertainties, but use the square root of the sum of the squares Commented Dec 23, 2018 at 21:48
• Depends on how the „error bars“ are defined. If they are standard deviations, you are right. If they are variances, you would just sum. Commented Dec 23, 2018 at 22:15

The sum of $$\chi^2$$-distributed independent values is another $$\chi^2$$-distributed value; see https://math.stackexchange.com/questions/1096298/question-about-sum-of-chi-squared-distribution