Median Statistics Estimate of the Distance to the Galactic Center

Median statistics benefits from ignoring the measurements' individual errors, at the expense of having a larger uncertainty about the median than that of a method that utilizes the error information. For a sufficiently large number of statistically independent values, it is expected that there exists a true median with half of the data points lying above and below it. Each individual measurement has a 50% probability of being greater or less than the true median. Gott et al. (2001) explains that for $i=1,2,\,\ldots .,\,N$ independent measurements ${M}_{i}$, the probability of the median falling between ${M}_{i}$ and ${M}_{i+1}$ follows the binomial distribution

$$\begin{eqnarray}&&P=\displaystyle \frac{{2}^{-N}N!}{i!(N-1)!}.\end{eqnarray} \tag{ 1 }$$

The one (two) standard deviation error associated with the median is defined in Gott et al. (2001) as the range about the median including 68.27% (95.45%) of the probability.⁵ The one standard deviation error given by a Gott et al. (2001) 68.27% confidence range is smaller than that obtained by binning the measurements and integrating outwards to 68.27% of the total area around the median (Crandall & Ratra 2014). We call the error determined from the probability distribution of Equation (1) ${\sigma }_{\mathrm{Gott}}$, while we refer to the result from the integration of the binned measurements' method as ${\sigma }_{\mathrm{med}}$.

Utilizing the idea that "better" measurements should have more weight, weighted mean statistics yields the benefit of a smaller error about the central estimate and takes the risk of under-weighting values with inaccurate uncertainties (see, e.g., Podariu et al. 2001). The weighted mean is defined as

$$\begin{eqnarray}&&{M}_{\mathrm{wm}}=\displaystyle \frac{{\displaystyle \sum }_{i=1}^{N}{M}_{i}{/\sigma }_{i}^{2}}{{\displaystyle \sum }_{i=1}^{N}1/{\sigma }_{i}^{2}},\end{eqnarray} \tag{ 2 }$$

where ${\sigma }_{i}$ are the one standard deviation errors. The weighted mean standard deviation is

$$\begin{eqnarray}&&{\sigma }_{\mathrm{wm}}=\displaystyle \frac{1}{\sqrt{{\displaystyle \sum }_{i=1}^{N}1/{\sigma }_{i}^{2}}}.\end{eqnarray} \tag{ 3 }$$

In our weighted mean analysis, and other analyses that use the errors, ${\sigma }_{i}$ is the quadrature sum of the (symmetrized) statistical and systematic (if quoted) errors.

It may also be of value to consider the arithmetic mean,

$$\begin{eqnarray}&&{M}_{{\rm{m}}}=\displaystyle \frac{1}{N}\displaystyle \sum _{i=1}^{N}{M}_{i}.\end{eqnarray} \tag{ 4 }$$

The underlying assumptions here are that each of the measurements have roughly the same uncertainty, and that the data come from a normally distributed set. The standard error of the mean is

$$\begin{eqnarray}&&{\sigma }_{{\rm{m}}}=\sqrt{\displaystyle \frac{1}{{N}^{2}}\displaystyle \sum _{i=1}^{N}{({M}_{i}-{M}_{{\rm{m}}})}^{2}}.\end{eqnarray} \tag{ 5 }$$

Note that the standard deviation of the data set, σ, and the standard error of the mean, ${\sigma }_{m}$, differ by the square root of the amount of measurements: ${\sigma }_{m}=\sigma /\sqrt{N}$.

The central estimates and associated errors are recorded in Table 2 for each of the data sets of Table 1. From column 2 of Table 2, we see our median, weighted mean, and arithmetic mean central estimates of 8 kpc coincide with those of Vallée (2017) (at the bottom of his Table 1). However, we are unable to reproduce his weighted mean and arithmetic mean error bars of ±0.2 kpc (he does not quote a median error bar); our weighted (arithmetic) mean error bar is ±0.04 (0.4) kpc.

Table 2.  R₀ (in kpc) Central Estimates and Errors

	Vallée	Vallée:	Vallée:	Independent
		Updated	Independent	from 2011
Median, Integrala	${{8.00}_{-0.34}^{+0.36}}_{-1.26}^{+0.54}$	${{8.03}_{-0.32}^{+0.31}}_{-1.27}^{+0.83}$	${{8.02}_{-0.55}^{+0.26}}_{-1.24}^{+0.86}$	${{7.96}_{-0.50}^{+0.29}}_{-1.20}^{+0.90}$
1σ Range	7.66–8.36	7.71–8.34	7.47–8.28	7.46–8.25
2σ Range	6.74–8.54	6.76–8.86	6.78–8.88	6.76–8.86
Median, Gottb	${{8.00}_{-0.00}^{+0.20}}_{-0.30}^{+0.30}$	${{8.03}_{-0.05}^{+0.17}}_{-0.33}^{+0.24}$	${{8.02}_{-0.16}^{+0.18}}_{-0.38}^{+0.25}$	${{7.96}_{-0.23}^{+0.11}}_{-0.30}^{+0.24}$
1σ Range	8.00–8.20	7.98–8.20	7.86–8.20	7.73–8.07
2σ Range	7.70–8.30	7.70–8.27	7.64–8.27	7.66–8.20
Weighted Mean	8.02 ± 0.04	7.99 ± 0.03	7.93 ± 0.03	7.93 ± 0.03
1σ Range	7.99–8.06	7.95–8.02	7.90–7.97	7.89–7.96
Arithmetic Mean	8.00 ± 0.08	7.99 ± 0.08	7.97 ± 0.11	7.92 ± 0.09
1σ Range	7.91–8.08	7.91–8.07	7.86–8.08	7.84–8.01

Notes.

^aErrors are estimated by binning the measurements to 0.1 kpc and integrating outwards until reaching 68.27% and 95.45% of the area under the distribution. ^bErrors are estimated from the median statistics probability distribution of Equation (1).

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The last column of Table 2 summarizes our main result. As discussed below, we find the error distribution for our chosen 28 measurements are somewhat non-Gaussian, but not excessively so.⁶ Consequently we recommend that the median central value and the symmetrized errors for the 68.27% and 95.45% confidence ranges as defined in Gott et al. (2001) be used to describe the value of and errors on R₀. This gives ${R}_{0}=7.96\pm 0.17$ (±0.27) kpc, with symmetrized 1σ (2σ) error, though it might be preferable to use the unsymmetrized result of ${R}_{0}={7.96}_{-0.23}^{+0.11}$ (${}_{-0.30}^{+0.24}$) kpc to take into account the slightly asymmetric nature of the set of measurements. For most practical purposes, ${R}_{0}=8.0\pm 0.3$ (2σ error) serves as an appropriate summary estimate to one decimal place.

After determining our central estimates, we construct our error distributions by using

$$\begin{eqnarray}&&{N}_{{\sigma }_{i}}=\displaystyle \frac{{R}_{i}-{R}_{\mathrm{CE}}}{\sqrt{{\sigma }_{i}^{2}+{\sigma }_{\mathrm{CE}}^{2}}}.\end{eqnarray} \tag{ 6 }$$

Here, ${R}_{\mathrm{CE}}$ is the central estimate of R_i and ${\sigma }_{\mathrm{CE}}$ is the error of the central estimate of R_i. ${N}_{{\sigma }_{i}}$ represents how much R_i deviates from the central estimate, taking into account both the error associated with the measurement and the error associated with the central estimate. In this paper we do not symmetrize ${\sigma }_{\mathrm{CE}}$ for the median statistics cases (the data are not symmetric enough to justify it). Thus, when applicable, we use the upper/right-side error ${\sigma }_{\mathrm{CE}}^{u}$ for when ${R}_{i}\geqslant {R}_{\mathrm{CE}}$ and the lower/left-side error ${\sigma }_{\mathrm{CE}}^{l}$ for when ${R}_{i}\leqslant {R}_{\mathrm{CE}}$.

We label our error distributions ${N}_{\sigma }^{\mathrm{med}}$, ${N}_{\sigma }^{\mathrm{Gott}}$, ${N}_{\sigma }^{\mathrm{wm}+}$, and ${N}_{\sigma }^{\mathrm{mean}}$. These represent differing combinations of central estimates and errors, defined as

$$\begin{eqnarray}&&{N}_{{\sigma }_{i}}^{\mathrm{med}}=\displaystyle \frac{{R}_{i}-{R}_{\mathrm{med}}}{\sqrt{{\sigma }_{i}^{2}+{\sigma }_{\mathrm{med}}^{2}}},\end{eqnarray} \tag{ 7 }$$

$$\begin{eqnarray}&&{N}_{{\sigma }_{i}}^{\mathrm{Gott}}=\displaystyle \frac{{R}_{i}-{R}_{\mathrm{med}}}{\sqrt{{\sigma }_{i}^{2}+{\sigma }_{\mathrm{Gott}}^{2}}},\end{eqnarray} \tag{ 8 }$$

$$\begin{eqnarray}&&{N}_{{\sigma }_{i}}^{\mathrm{wm}+}=\displaystyle \frac{{R}_{i}-{R}_{\mathrm{wm}}}{\sqrt{{\sigma }_{i}^{2}+{\sigma }_{\mathrm{wm}}^{2}}},\end{eqnarray} \tag{ 9 }$$

$$\begin{eqnarray}&&{N}_{{\sigma }_{i}}^{\mathrm{mean}}=\displaystyle \frac{{R}_{i}-{R}_{{\rm{m}}}}{\sqrt{{\sigma }_{i}^{2}+{\sigma }_{{\rm{m}}}^{2}}}.\end{eqnarray} \tag{ 10 }$$

Since the central estimates are calculated from the data, they must to some degree be correlated with the error measurements. If the errors are Gaussian and the weighted mean has been determined from the measurements, then it is correlated with the measurements and a more appropriate error distribution is then⁷

$$\begin{eqnarray}&&{N}_{{\sigma }_{i}}^{\mathrm{wm}-}=\displaystyle \frac{{R}_{i}-{R}_{\mathrm{wm}}}{\sqrt{{\sigma }_{i}^{2}-{\sigma }_{\mathrm{wm}}^{2}}}.\end{eqnarray} \tag{ 11 }$$

The derivation of an equivalent error distribution that accounts for the correlation is nontrivial for a median central estimate, however Equation (6) provides a valuable limiting case.⁸

We choose to use the above five error distributions to attempt to gain some insight into the R₀ measurements' error distribution.⁹

We numerically study our error distributions using the one-sample K–S test (Feigelson & Babu 2012). This non-parametric, distribution-free test determines the probability that the given sample distribution comes from a well-defined probability density function (PDF), at a chosen significance level α. In this paper we use Gaussian, Student's t, Cauchy, and Laplace (Double Exponential) distributions. The qualitative returns of a K–S test are a D statistic and a P value. The D statistic is the supremum of, or the largest distance between, the cumulative sample distribution and the cumulative PDF. The closer this value is to zero, the better the sample distribution is well described by the PDF. For a sample distribution of N measurements there is a critical value ${D}_{\mathrm{crit}}(N)$ that must be less than the test result, D, in order to not reject the null hypothesis at the specified significance level (which is conventionally set at $\alpha =0.05$ for a confidence level of 95%). For N = 28 measurements ${D}_{\mathrm{crit}}=0.24993$.¹⁰ The P value follows from the D statistic and represents not the probability that the sample set is from the proposed PDF, but rather the probability that we cannot reject the null hypothesis that the distributions are the same. It is for this reason that the probabilities of the K–S test should be used as qualitative indicators of distribution fitting. It is of interest to study the K–S test results for as many PDF's as possible. We choose the PDF with the lowest D statistic and the highest P value as the best representation of the error distribution under study.

We define our PDF's as functions of $| {\boldsymbol{N}}| =| {N}_{\sigma }/S|$, where S is a scale factor. When S = 1 and $| {\boldsymbol{N}}| =| {N}_{\sigma }|$, $P(| {\boldsymbol{N}}| )$ is the standard form of the PDF. When $S\gt 1$, the distribution is broader than the standard form, while $S\lt 1$ corresponds to a narrower distribution. While ${N}_{{\sigma }_{i}}$ is computed with unsymmetrized errors, the distribution of ${N}_{\sigma }$ is symmetrized for the K–S test.

We define a Gaussian distribution of ${N}_{\sigma }$ with an expected 68.27% and 95.45% of the values falling within $| {N}_{\sigma }| \leqslant 1$ and $| {N}_{\sigma }| \leqslant 2$, respectively, as

$$\begin{eqnarray}&&P(| {\boldsymbol{N}}| )=\displaystyle \frac{1}{\sqrt{2\pi }}\exp (-| {\boldsymbol{N}}{| }^{2}/2).\end{eqnarray} \tag{ 12 }$$

The second distribution that we consider is a Laplace (Double Exponential), given by

$$\begin{eqnarray}&&P(| {\boldsymbol{N}}| )=\displaystyle \frac{1}{2}\exp (-| {\boldsymbol{N}}| ).\end{eqnarray} \tag{ 13 }$$

The Laplace PDF is sharply peaked, with longer (smaller) tails than a Gaussian (Cauchy) distribution. For this distribution, 68.27% and 95.45% of the values correspond to $| {N}_{\sigma }| \leqslant 1.2$ and $| {N}_{\sigma }| \leqslant 3.1$, respectively. The Cauchy (Lorentz) distribution

$$\begin{eqnarray}&&P(| {\boldsymbol{N}}| )=\displaystyle \frac{1}{\pi }\displaystyle \frac{1}{1+| {\boldsymbol{N}}{| }^{2}}\end{eqnarray} \tag{ 14 }$$

has much higher probability in the tails, with an expected 68.27% and 95.45% of the values falling within $| {N}_{\sigma }| \leqslant 1.8$ and $| {N}_{\sigma }| \leqslant 14$, respectively. The Student's t distribution is defined by

$$\begin{eqnarray}&&P(| {\boldsymbol{N}}| )=\displaystyle \frac{{\rm{\Gamma }}[(n+1)/2]}{\sqrt{\pi n}\ {\rm{\Gamma }}(n/2)}\displaystyle \frac{1}{{(1+| {\boldsymbol{N}}{| }^{2}/n)}^{(n+1)/2}},\end{eqnarray} \tag{ 15 }$$

where n is a positive non-zero parameter and Γ is the gamma function. When n = 1 this is the Cauchy distribution, and when $n\to \infty$ it becomes the Gaussian distribution. Thus, for $n\gt 1$, it is a function with slightly less extended tails than a Cauchy, that decrease as n increases. In this case, the limits corresponding to 68.27% and 95.45% of the values depend on the value of n.

Our K–S test results, for the 28 independent R₀ values listed in column 5 of Table 1, are shown in Table 3. While some S = 1 entries have low probabilities, and $P=11.7 \%$ for the S = 1 Gaussian case of the weighted mean central estimate and the 1σ error distribution of Equation (11), overall, allowing S to vary a little away from unity, it is fair to conclude that the errors of the 28 measurement data set are not very non-Gaussian, although they are slightly so.¹¹ Tables 4 and 5, which show the probabilities corresponding to $| {N}_{\sigma }| \leqslant 1$ and $| {N}_{\sigma }| \leqslant 2$ and the $| {N}_{\sigma }|$ values corresponding to 68.27% and 95.45% of the probability for these favored distributions, reinforce this conclusion.

Table 3.  K–S Test Probabilities

	${N}_{\sigma }^{\mathrm{med}}$		${N}_{\sigma }^{\mathrm{Gott}}$ c		${N}_{\sigma }^{\mathrm{wm}+}$		${N}_{\sigma }^{\mathrm{wm}-}$		${N}_{\sigma }^{\mathrm{mean}}$
PDF	Sa	P(%)b	Sa	P(%)b	Sa	P(%)b	Sa	P(%)b	Sa	P(%)b
Gaussian	1	69.4	1	53.4	1	11.9	1	11.7	1	17.8
Gaussian	0.85	99.5	1.24	99.6	1.68	99.9	1.73	99.8	1.56	99.9
Laplace	1	39.0	1	82.6	1	47.9	1	45.3	1	57.3
Laplace	0.77	93.6	1.13	97.7	1.40	99.8	1.52	99.9	1.28	99.0
Cauchy	1	4.1	1	32.8	1	64.6	1	88.7	1	50.2
Cauchy	0.51	84.6	0.70	84.8	0.77	90.2	0.83	97.2	0.75	88.1
	n = 100		n = 3		n = 2		⋯e		n = 2
Student's td	1	67.7	1	97.5	1	81.1	⋯	⋯	1	88.8
	n = 100		n = 4		n = 5		n = 2		n = 34
Student's td	0.85	99.4	1.11	99.7	1.50	99.9	1.28	99.9	1.53	99.9

Notes.

^aScale factor S is first set at S = 1 (representing the case when $| {N}_{\sigma }| =1$ corresponds to 1 standard deviation for a Gaussian distribution) and is then allowed to vary with the width of the function as D is minimized. ^bThis is the P value described in Section 3.3. It is the probability that we cannot reject the hypothesis that the sample distribution ${N}_{\sigma }$ came from a distribution created from the probability density function. ^cWe use the errors corresponding to 68.27% confidence in ${N}_{\sigma }^{\mathrm{Gott}}$ because we use 1 standard deviation for ${N}_{\sigma }^{\mathrm{med}}$. ^dWe allow n to vary between 1 and 100 for the Student's t distribution. ^eThe K–S test using a Student's t PDF on ${N}_{\sigma }^{\mathrm{wm}-}$ for S = 1 yielded a best fit of n = 1 which is the Cauchy distribution.

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Table 4.  $| {N}_{\sigma }|$ Expected Fractions

	${N}_{\sigma }^{\mathrm{med}}$			${N}_{\sigma }^{\mathrm{Gott}}$			${N}_{\sigma }^{\mathrm{wm}+}$			${N}_{\sigma }^{\mathrm{wm}-}$			${N}_{\sigma }^{\mathrm{mean}}$
PDF	Sa	$| {N}_{\sigma }| \leqslant 1$ b	$| {N}_{\sigma }| \leqslant 2$ b	Sa	$| {N}_{\sigma }| \leqslant 1$ b	$| {N}_{\sigma }| \leqslant 2$ b	Sa	$| {N}_{\sigma }| \leqslant 1$ b	$| {N}_{\sigma }| \leqslant 2$ b	Sa	$| {N}_{\sigma }| \leqslant 1$ b	$| {N}_{\sigma }| \leqslant 2$ b	Sa	$| {N}_{\sigma }| \leqslant 1$ b	$| {N}_{\sigma }| \leqslant 2$ b
Gaussian	1	0.68	0.95	1	0.68	0.95	1	0.68	0.95	1	0.68	0.95	1	0.68	0.95
Gaussian	0.85	0.76	0.98	1.24	0.58	0.89	1.68	0.45	0.77	1.73	0.44	0.75	1.56	0.48	0.80
Laplace	1	0.63	0.87	1	0.63	0.87	1	0.63	0.87	1	0.63	0.87	1	0.63	0.87
Laplace	0.78	0.73	0.92	1.13	0.59	0.83	1.40	0.51	0.76	1.52	0.48	0.73	1.28	0.54	0.79
Cauchy	1	0.50	0.71	1	0.50	0.71	1	0.50	0.71	1	0.50	0.71	1	0.50	0.71
Cauchy	0.51	0.70	0.84	0.70	0.61	0.79	0.77	0.58	0.77	0.83	0.56	0.75	0.75	0.59	0.77
	n = 100			n = 3			n = 2			⋯c			n = 2
Student's t	1	0.58	0.82	1	0.61	0.86	1	0.58	0.82	⋯	⋯	⋯	1	0.58	0.82
	n = 100			n = 4			n = 5			n = 2			n = 34
Student's t	0.85	0.76	0.98	1.11	0.58	0.85	1.50	0.47	0.76	1.28	0.48	0.74	1.53	0.48	0.80
Observed	⋯	0.86	1.00	⋯	0.54	0.93	⋯	0.50	0.71	⋯	0.50	0.68	⋯	0.50	0.71

Notes.

^aScale factor S is first set at S = 1 (representing the case when $| {N}_{\sigma }| =1$ corresponds to 1 standard deviation for a Gaussian distribution) and is then allowed to vary with the width of the function as D is minimized. ^bThe fraction of data points that lie within $| {N}_{\sigma }| \leqslant 1$ or $| {N}_{\sigma }| \leqslant 2$. ^cThe Student's t test on ${N}_{{\sigma }_{\mathrm{wm}-}}$ for S = 1 yielded a best fit of n = 1, which is the Cauchy distribution.

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Columns 4 and 5 of Table 3 show the probabilities are as high as 99.9% for a Gaussian distribution with S = 1.68 and a Laplacian distribution with S = 1.52, respectively. The non-Gaussianity associated with using the error bars from the R₀ measurements in weighted mean analyses can be substantiated from columns 4 and 5 of Tables 4 and 5: for the S = 1.68 Gaussian in ${N}_{\sigma }^{\mathrm{wm}+}$, only 45% (77%) of the probability lies within $| {N}_{\sigma }| \leqslant 1$ ($| {N}_{\sigma }| \leqslant 2$) and to attain the standard probability of 68.27% (95.45%) we must integrate out to $| {N}_{\sigma }| =1.7$ ($| {N}_{\sigma }| =3.4$); for the S = 1.52 Laplacian of ${N}_{\sigma }^{\mathrm{wm}-}$, only 48% (73%) of the probability lies within $| {N}_{\sigma }| \leqslant 1$ ($| {N}_{\sigma }| \leqslant 2$) and to attain the standard probability of 68.27% (95.45%) we must integrate out to $| {N}_{\sigma }| =1.7$ ($| {N}_{\sigma }| =4.7$). The Gaussian fits for ${N}_{\sigma }^{\mathrm{wm}+}$, ${N}_{\sigma }^{\mathrm{wm}-}$, and ${N}_{\sigma }^{\mathrm{mean}}$ require scale factors of S = 1.68, S = 1.73, and S = 1.56, respectively. For this reason, it is best to use median statistics to determine the error bars on R₀, which are looser than those from weighted mean statistics and arithmetic mean statistics. The probability distribution computed from Equation (1) then provides the best central estimate and errors bars for determining the somewhat non-Gaussian nature of the error distribution of the 28 independent R₀ measurements. The corresponding median-statistics error distribution of Equation (8) is best fit by an n = 4 Student's t PDF with an S = 1.1 scale factor, and is non-Gaussian to the degree that with a probability of $99.6 \%$, we cannot reject the hypothesis that it comes from a Gaussian distribution with an S = 1.24 scale. The slightly broader-than-expected Gaussian distributed error distribution could indicate some (slightly) improperly estimated systematic uncertainties. This is, however, perhaps a mild concern until we can compile and study a larger set of recent and statistically independent measurements of R₀.

Table 5.  $| {N}_{\sigma }|$ Limits

	${N}_{\sigma }^{\mathrm{med}}$			${N}_{\sigma }^{\mathrm{Gott}}$			${N}_{\sigma }^{\mathrm{wm}+}$			${N}_{\sigma }^{\mathrm{wm}-}$			${N}_{\sigma }^{\mathrm{mean}}$
PDF	Sa	68.27%b	95.45%b	Sa	68.27%b	95.45%b	Sa	68.27%b	95.45%b	Sa	68.27%b	95.45%b	Sa	68.27%b	95.45%b
Gaussian	1	1.0	2.0	1	1.0	2.0	1	1.0	2.0	1	1.0	2.0	1	1.0	2.0
Gaussian	0.85	0.9	1.7	1.24	1.2	2.5	1.68	1.7	3.4	1.73	1.7	3.5	1.56	1.6	3.1
Laplace	1	1.2	3.1	1	1.2	3.1	1	1.2	3.1	1	1.2	3.1	1	1.2	3.1
Laplace	0.78	0.9	2.4	1.13	1.3	3.5	1.40	1.6	4.3	1.52	1.7	4.7	1.28	1.5	4.0
Cauchy	1	1.8	14.0	1	1.8	14.0	1	1.8	14.0	1	1.8	14.0	1	1.8	14.0
Cauchy	0.51	0.9	7.0	0.70	1.3	9.7	0.77	1.4	10.6	0.83	1.5	11.5	0.75	1.4	10.6
	n = 100			n = 3			n = 2			⋯c			n = 2
Student's t	1	1.0	2.0	1	1.2	3.3	1	1.3	4.5	⋯	⋯	⋯	1	1.3	4.5
	n = 100			n = 4			n = 5			n = 2			n = 34
Student's t	0.85	0.9	1.7	1.11	1.3	3.2	1.50	1.7	4.0	1.28	1.7	5.8	1.53	1.5	3.2
Observed	⋯	0.8	1.9	⋯	1.3	2.3	⋯	1.9	3.1	⋯	2.1	3.5	⋯	1.7	2.5

Notes.

^aScale factor S is first set at S = 1 (representing the case when $| {N}_{\sigma }| =1$ corresponds to 1 standard deviation for a Gaussian distribution) and is then allowed to vary with the width of the function as D is minimized. ^bThe $| {N}_{\sigma }|$ limits containing 68.27% or 95.45% of the probability. For a Gaussian PDF with S = 1, 68.27% (95.45%) of the probability is contained within $| {N}_{\sigma }| =1$ ($| {N}_{\sigma }| =2$). ^cThe Student's t test on ${N}_{{\sigma }_{\mathrm{wm}-}}$ for S = 1 yielded a best fit of n = 1, which is the Cauchy distribution.

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