How do you convert an uncertainty quoted in dex to a 1-sigma uncertainty?

Question

The black hole mass measurement uncertainties from this paper https://arxiv.org/pdf/2006.15150.pdf are quoted in "dex" and are said to be equal to 0.4 dex. I have seen from certain papers that 1 dex equals to one order of magnitude meaning $0.4$ dex $= 10^{0.4}$. But I'm not exactly sure what this means and how it translates to the scale on which the measurement was made (i.e. $log_{10}(M_{BH}/M_{\odot})$ ).

As an example: how would I get the uncertainty in $log_{10}(M_{BH}/M_{\odot})$ units of measure on a measurement that is 5.80 $log_{10}(M_{BH}/M_{\odot})$.

Also, would this be a 1, 2 or 3 sigma uncertainty?

I hope my question is clear, thank you!

Answer 1

Uncertainties span a confidence interval

Adding and subtracting the uncertainty from the central value gives you, respectively, the upper and the lower values of the confidence interval (CI) — whatever it represents.

For instance, a value $x=5\pm2$ means that your CI is $[x_\mathrm{lo},x_\mathrm{hi}] = [3,7]$. The same is true for asymmetric uncertainties, so that the CI of $5_{-2}^{+3}$ is $[3,8]$.

The same approach is used for logarithmically distributed values. If you have, say, a black hole with a mass of $\log(M/M_\odot) = 5.80$, with a uncertainty of $\sigma_{\log\!M}=0.4\,\mathrm{dex}$, this means that the CI is $$ \begin{array}{rcl} [\log\!M_\mathrm{lo},\log\!M_\mathrm{hi}] & = & [\log\!M-\sigma_{\log\!M},\,\log\!M+\sigma_{\log\!M}]\\ & = & [5.4,\,6.2]. \end{array} $$

Symmetric uncertainties in logspace become asymmetric in linspace

In linspace, the mass is $$ M/M_\odot = 10^{5.8}\simeq6.3\times10^5, $$ and the CI is $$ \begin{array}{rcl} [M_\mathrm{lo},\,M_\mathrm{hi}] & = & 10^{[5.4,\,6.2]}M_\odot\\ & = & [2.5\times10^5 M_\odot,\,1.6\times10^6 M_\odot]. \end{array} $$

In linspace, the uncertainties now become asymmetric. Writing the mass as $M_\mathrm{-\sigma_{M-}}^{+\sigma_{M+}}$, the lower and upper uncertainties are $$ \begin{array}{rcl} \sigma_{M-} & = & M-M_\mathrm{lo} = \Big(6.3\times10^5 M_\odot - 2.5\times10^5 M_\odot \Big)= 3.8\times10^5 M_\odot;\\ \sigma_{M+} & = & M_\mathrm{hi}-M = \Big(1.6\times10^6 M_\odot - 6.3\times10^5 M_\odot \Big) = 9.5\times10^5 M_\odot. \end{array} $$

In other words you can write $$ \begin{array}{rcl} M = \big(6.3_{-3.8}^{+9.5}\big) \times10^5 M_\odot. \end{array} $$

What do the uncertainties and the confidence interval represent?

There is no consensus. It should be clearly stated in the paper, but often it's not. In astronomy, symmetric uncertainties often represent one standard deviation ($1\sigma$). If they represent $3\sigma$, the text will sometimes say so.

In the case of asymmetric uncertainties, there is even less consensus. Asymmetric uncertainties are, by nature, not normal distributed, and hence the term "standard deviation" loses its meaning. There are several options for the meaning of $x_0$, $\sigma_{-}$, and $\sigma_{+}$ in the notation "${x_0}_{-\sigma_{-}}^{+\sigma_{+}}$", including the following common cases:

• Sometimes the central value $x_0$ represents the mean, and the upper and lower uncertainties $\sigma_{-}$ and $\sigma_{+}$ represent some end result of an error propagation.
• Other times $x_0$ represents the median, and $\sigma_{-}$ and $\sigma_{+}$ are the 84th and the 16th percentiles, respectively. I personally like that, because
  1. the total probability inside the CI is 68%, like for normal distributions,
  2. you can be certain that $X_0$ lies inside the CI (which you can't for the mean), and
  3. it doesn't matter whether you consider $x_0$ or $\log x_0$.
• Yet other times (especially in 2D distributions), $x_0$ represents the mode.

I expanded a bit on this issue in this answer about propagation of asymmetric errors.