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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?

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    $\begingroup$ Could you explain more precisely and quantitatively what these "left" and "right" uncertainties mean or how they are computed? $\endgroup$
    – whuber ♦
    Commented Dec 23, 2018 at 15:29
  • $\begingroup$ 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 $\endgroup$
    – johnhenry
    Commented Dec 23, 2018 at 15:38
  • $\begingroup$ 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. $\endgroup$
    – Jonas
    Commented Dec 23, 2018 at 21:35
  • $\begingroup$ @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 $\endgroup$
    – johnhenry
    Commented Dec 23, 2018 at 21:48
  • $\begingroup$ Depends on how the „error bars“ are defined. If they are standard deviations, you are right. If they are variances, you would just sum. $\endgroup$
    – Jonas
    Commented Dec 23, 2018 at 22:15

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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

How the values and uncertainties propagate depend on how you did the values in the first place; are they mean, mode, or median values?

The asymmetric uncertainties also depend on how they were set up. With asymmetric distributions, the confidence intervals can be set up in three different ways: (i) half of the area either side of the quoted value (eg 34% left and right of median), resulting in an asymmetric interval; (ii) a symmetric interval (eg about the median) such that the total area between the upper and lower limits is 68% - results in a symmetric interval; and (iii) upper and lower limits such that the area between them is 68% and the distance between the upper and lower is as small as possible, resulting in an asymmetric interval.

Once you now that, you can apply that to the final distribution to get the correct bounds for your estimate.

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