<h4>Uncertainties span a confidence interval</h4>
<p>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.</p>
<p>For instance, a value <span class="math-container">$x=5\pm2$</span> means that your CI is <span class="math-container">$[x_\mathrm{lo},x_\mathrm{hi}] = [3,7]$</span>. The same is true for asymmetric uncertainties, so that the CI of <span class="math-container">$5_{-2}^{+3}$</span> is <span class="math-container">$[3,8]$</span>.</p>
<p>The same approach is used for logarithmically distributed values. If you have, say, a black hole with a mass of <span class="math-container">$\log(M/M_\odot) = 5.80$</span>, with a uncertainty of <span class="math-container">$\sigma_{\log\!M}=0.4\,\mathrm{dex}$</span>, this means that the CI is
<span class="math-container">$$
\begin{array}{rcl}
[\log\!M_\mathrm{lo},\log\!M_\mathrm{hi}] &amp; = &amp; [\log\!M-\sigma_{\log\!M},\,\log\!M+\sigma_{\log\!M}]\\
&amp; = &amp; [5.4,\,6.2].
\end{array}
$$</span></p>
<h4>Symmetric uncertainties in logspace become asymmetric in linspace</h4>
<p>In linspace, the mass is
<span class="math-container">$$
M/M_\odot = 10^{5.8}\simeq6.3\times10^5,
$$</span>
and the CI is
<span class="math-container">$$
\begin{array}{rcl}
[M_\mathrm{lo},\,M_\mathrm{hi}] &amp; = &amp; 10^{[5.4,\,6.2]}M_\odot\\
 &amp; = &amp; [2.5\times10^5 M_\odot,\,1.6\times10^6 M_\odot].
\end{array}
$$</span></p>
<p>In linspace, the uncertainties now become asymmetric. Writing the mass as <span class="math-container">$M_\mathrm{-\sigma_{M-}}^{+\sigma_{M+}}$</span>, the lower and upper uncertainties are
<span class="math-container">$$
\begin{array}{rcl}
\sigma_{M-} &amp; = &amp; M-M_\mathrm{lo} = \Big(6.3\times10^5 M_\odot - 2.5\times10^5 M_\odot \Big)= 3.8\times10^5 M_\odot;\\
\sigma_{M+} &amp; = &amp; M_\mathrm{hi}-M = \Big(1.6\times10^6 M_\odot - 6.3\times10^5 M_\odot \Big) = 9.5\times10^5 M_\odot.
\end{array}
$$</span></p>
<p>In other words you can write
<span class="math-container">$$
\begin{array}{rcl}
M = \big(6.3_{-3.8}^{+9.5}\big) \times10^5 M_\odot.
\end{array}
$$</span></p>
<h4>What do the uncertainties and the confidence interval represent?</h4>
<p>There is no consensus. It should be clearly stated in the paper, but often it's not. In astronomy, symmetric uncertainties <em>often</em> represent one standard deviation (<span class="math-container">$1\sigma$</span>). If they represent <span class="math-container">$3\sigma$</span>, the text will <em>sometimes</em> say so.</p>
<p>In the case of asymmetric uncertainties, there is even less consensus. Asymmetric uncertainties are, by nature, not normal distributed, and hence the term &quot;standard deviation&quot; loses its meaning. There are several options for the meaning of <span class="math-container">$x_0$</span>, <span class="math-container">$\sigma_{-}$</span>, and <span class="math-container">$\sigma_{+}$</span> in the notation &quot;<span class="math-container">${x_0}_{-\sigma_{-}}^{+\sigma_{+}}$</span>&quot;, including the following common cases:</p>
<ul>
<li>Sometimes the central value <span class="math-container">$x_0$</span> represents the <em>mean</em>, and the upper and lower uncertainties <span class="math-container">$\sigma_{-}$</span> and <span class="math-container">$\sigma_{+}$</span> represent some end result of an error propagation.</li>
<li>Other times <span class="math-container">$x_0$</span> represents the <em>median</em>, and <span class="math-container">$\sigma_{-}$</span> and <span class="math-container">$\sigma_{+}$</span> are the 84th and the 16th percentiles, respectively. I personally like that, because
<ol>
<li>the total probability inside the CI is 68%, like for normal distributions,</li>
<li>you can be certain that <span class="math-container">$X_0$</span> lies <em>inside</em> the CI (which you can't for the mean), and</li>
<li>it doesn't matter whether you consider <span class="math-container">$x_0$</span> or <span class="math-container">$\log x_0$</span>.</li>
</ol>
</li>
<li>Yet other times (especially in 2D distributions), <span class="math-container">$x_0$</span> represents the <em>mode</em>.</li>
</ul>
<p>I expanded a bit on this issue in <a href="https://astronomy.stackexchange.com/questions/47539/how-do-you-propagate-asymmetric-errors-on-the-practical-way-to/47541#47541">this answer about propagation of asymmetric errors</a>.</p>

CC BY-SA 4.0
2022-07-15T22:42:54.250Z
2022-07-16T07:40:22.710Z
2022-07-16T07:40:22.710Z
pela
<p>The black hole mass measurement uncertainties from this paper <a href="https://arxiv.org/pdf/2006.15150.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/2006.15150.pdf</a> are quoted in &quot;dex&quot; 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 <span class="math-container">$0.4$</span> dex <span class="math-container">$= 10^{0.4}$</span>. But I'm not exactly sure what this means and how it translates to the scale on which the measurement was made (i.e. <span class="math-container">$log_{10}(M_{BH}/M_{\odot})$</span> ).</p>
<p>As an example: how would I get the uncertainty in <span class="math-container">$log_{10}(M_{BH}/M_{\odot})$</span> units of measure on a measurement that is 5.80 <span class="math-container">$log_{10}(M_{BH}/M_{\odot})$</span>.</p>
<p>Also, would this be a 1, 2 or 3 sigma uncertainty?</p>
<p>I hope my question is clear, thank you!</p>

<black-hole><mass>
How do you convert an uncertainty quoted in dex to a 1-sigma uncertainty?
Os GS
# How do you convert an uncertainty quoted in dex to a 1-sigma uncertainty?

(Asked by: username_0 on 2022-07-15)

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!

**Tags:** black-hole, mass


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## 1 answers

### Answer by username_1 on 2022-07-15. Score: 2 | Accepted answer

#### 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](https://astronomy.stackexchange.com/questions/47539/how-do-you-propagate-asymmetric-errors-on-the-practical-way-to/47541#47541).


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