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If I have $N$ repeated occurrences of a measurement $x$ with uncorrelated errors and identical uncertainties $u_x$, and take the mean $\langle x\rangle$, the uncertainty on the mean becomes:

$$u_{\langle x\rangle} = \frac{u_x}{\sqrt{N}}$$

where $N$ is the number of measurements I have taken. This is derived from the law of propagation of uncertainties (for example, see this answer).

If I take the median instead of the mean, I'm mathematically only propagating information from one or two data points, which would mean $N=1$ or $N=2$. But that seems unfair, for surely the principle should hold that if I repeat the measurement many times and take the median, the resulting uncertainty goes down.

Is there any established way of propagating the uncertainty when taking the median of a series of measurements?

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  • $\begingroup$ Related but different question: stats.stackexchange.com/q/47738/12615 $\endgroup$ – gerrit Nov 15 '16 at 18:05
  • $\begingroup$ There is a significant section about this on wikipedia. Did you look there? en.wikipedia.org/wiki/Median#The_sample_median $\endgroup$ – Rob Jeffries Nov 15 '16 at 18:16
  • $\begingroup$ @RobJeffries That appears to address on how to estimate an uncertainty based on the median, not how an already-estimated uncertainty propagates upon the calculation of the median, unless I misunderstand it. I'm not asking about how to estimate anything based on the spread of my data. $\endgroup$ – gerrit Nov 15 '16 at 18:20
  • $\begingroup$ I don't follow. Your question asks how to calculate an uncertainty in a median value given a set of measurements that define the median. The straightforward answer is that it depends on the underlying properties of the population and there is no trivial solution. $\endgroup$ – Rob Jeffries Nov 15 '16 at 18:22
  • $\begingroup$ @RobJeffries No, I'm not asking how to estimate/calculate the uncertainty, I'm asking how to propagate an uncertainty when calculating the median, assuming uncorrelated errors and identical uncertainties for each measurement (which is of course a calculation, but a different one). I already have an estimate for the uncertainty on each of my $N$ measurements. When I take the mean, this simply propagates as $\sqrt{N}$ from the law of propagation of uncertainties. The question is what happens in case of the median. $\endgroup$ – gerrit Nov 15 '16 at 18:25
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I do not think you are estimating the uncertainty of your median correctly.

The ratio of the variance of the mean to the variance of the median is given by $4n/[\pi(2n+1)]$, where $N=(2n+1)$ is the total number of data points in the sample you have used to construct the median. I think this approximation is only true if your outliers are symmetric.

Thus the uncertainty in the median is given by $$ \Delta x_{Med} = \Delta \bar{x} \sqrt{\pi(2n+1)/4n},$$

In the limit of large $N$ (and hence large $n$), this tends to $$ \Delta x_{Med} = \Delta \bar{x} \sqrt{\pi/2},$$

The uncertainty in the mean $\Delta \bar{x}$ is the standard deviation of the points divided by $\sqrt{N}$ and therefore the precision of your estimate will improve as $\sqrt{N}$.

This information is given at http://mathworld.wolfram.com/StatisticalMedian.html

If you have outliers, then the formula that you begin your question with is incorrect (see the first sentence of the answer you refer to). The data points will not be normally distributed according to your estimate of their uncertainties and the standard error of the mean that I quote above will be larger than $u_x/\sqrt{N}$ because the standard deviation of the data will be larger than $u_x$ (I refer you also to the last paragraph of my answer to that question).

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  • $\begingroup$ I did have an incorrect estimate of my uncertainty (but for a different reason). $\endgroup$ – gerrit Nov 15 '16 at 19:31

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