In the given Image you can see a plot with error bars of different length on each upper and lower side. The bars were generated by calculating the standard error once for all points above the mean value and once for those below.


As information on this "method" is very sparse I ask myself wether it is correct to use them. Or is it just the wrong term (lower/upper standard error) I used? And finally what is the meaning of them?

  • $\begingroup$ I think this needs some more context, what is "this method"? Are errors assumed to be Normal? Symmetric? If not, the confidence interval doesn't need to be symmetric. As far as the meaning of them, do you mean standard error in general, or in this context? $\endgroup$
    – Gschneider
    May 5, 2012 at 14:21
  • $\begingroup$ "Standard errors" denote the (estimated) sampling variability of parameter estimates. What you have here are "prediction errors". $\endgroup$
    – StasK
    May 5, 2012 at 16:20
  • $\begingroup$ Also are these pointwise bands or simultaneous? $\endgroup$ May 8, 2012 at 15:47

2 Answers 2


According to your updated question, the claim of @onestop is still valid: it's not ok to call them standard errors. Furthermore, the method seems strange and non-standard at all. What really was done in your case is to divide the population in two (values upper and lower than the mean) and calculate the standard error of THAT population, not of your real population and therefore, I find it personally strange to assign the length of the error bars in that way. Apparently the idea that was done here was taken from here. However, IMHO, the idea of dividing the sample and calculating an "upper and lower" standard deviation doesn't make much sense (or at least it botters me).

In physics (my area and apparently yours), however, it has been somewhat standard to show 68% confidence intervals for the sample median or the mean (depending on your choice of a location statistic; let's call this statistic $\bar{X}$ for the moment) in the following way for non-symmetric distributions (apparently emulating what would be a central credible interval): with your data points, you calculate $\bar{X}$ and then report an upper error bar of length $L_u$, where $L_u$ is calculated in order to satisfy $P(\bar{X}<\mu<\bar{X}+L_u)= 0.34$, where $\mu$ is the real (unknown) parameter. Then, for your lower error bar of length $L_l$, you repeat the same procedure but now downwards of the location statistic $\bar{X}$, i.e., $P(\bar{X}-L_l<\mu<\bar{X})= 0.34$. Of course, because the distribution of $\bar{X}$ is usually not known this is usually done with non-parametric methods (such as the Bootstrap or some variant of it).

As was also pointed out by @onestop, you can also obtain bayesian credible intervals, where you actually calculate the probability (density, in the continuous case) of your parameter given your data. Let's call this probability $p(x|D)$. The length of the lower error bar is now calculated in a more "natural way" (at least for me), in order to satisfy $P(\hat{x}-L_l<x<\hat{x}|D)=0.34$, and the length of the upper error bar is now calculated in order to satisfy $P(\hat{x}<x<\hat{x}+L_u|D)=0.34$, where $\hat{x}$ is your point estimate of the parameter (usually the median or even the mode).

All of the above, of course, makes sense only if your parameter is unimodal.

  • $\begingroup$ Thank you Néstor, I think your 34% method is what I was looking for. Is it correct to calculate e.g. $L_u$ by ordering all $N$ points by value and take the one $n = 0.34 \cdot N$? $\endgroup$
    – severin
    May 9, 2012 at 11:22
  • $\begingroup$ Unless your sample size is high enough as to be representative of your population I wouldn't. I would always go with a bootstrap to be sure :-). $\endgroup$
    – Néstor
    May 9, 2012 at 15:53
  • $\begingroup$ sample sizes vary from 30 to 250. So I use bootstrap to create e.g. 100 samples out of the original sample. Then I sort each sample by size and pick the one with the index=round(0.341 * sampleSize). Unfortunately the result needs a very large number of bootstrappedsamples to be stable due to the rounding. Is it therefore valid to sum all samples and then pick the 34.1% ? $\endgroup$
    – severin
    May 9, 2012 at 18:08
  • $\begingroup$ I may be wrong, but you can't just use percentiles in order to get confidence intervals. I think that what you are trying to do is to create the so-called distribution-free confidence intervals (although somewhat modified this time). I think you'll like this blogpost: probabilityandstats.wordpress.com/2010/02/22/… as it has references and a very nice explaination of what you are trying to do ;-). $\endgroup$
    – Néstor
    May 9, 2012 at 19:08
  • $\begingroup$ Thanks a lot for this answer. What happens with values equal to the mean? $\endgroup$
    – danijar
    Aug 10, 2016 at 5:59

It's OK to call them error bars, but as they're asymmetric they are not representing the standard error so it's not correct to talk about 'lower/upper standard error'.

I assume the error bars here represent confidence intervals, though they might also be credible intervals if they were constructed using Bayesian methods.

Hard to be sure if you're not going to tell us where you got this graph from.

  • $\begingroup$ @severin, let me see if I understood correctly: you have, say, $n_{U_{\infty}}$ points for each value of $U_{\infty}$. For the sake of notation, let's fix $U_{\infty}=i$ (e.g., $U_{\infty}=i=2\ m/s$). From there, you calculated the mean value of the $n_i$ points, say, $\hat{\Omega}_{R}(i)$, and, in order to calculate the "upper standard error", from the total $n_i$ points you used only the points that where above $\hat{\Omega}_{R}(i)$, and repeated the same procedure for the "lower standard error" but taking only the points that where under $\hat{\Omega}_{R}(i)$? $\endgroup$
    – Néstor
    May 7, 2012 at 16:41
  • $\begingroup$ @Nesp, correct. Someone else calculated the arithmetic mean $\bar{\Omega}(i)$ for each $i = U_\infty$, then took all $\Omega(i) > \bar{\Omega}(i)$ to calculate the upper error, using the formula for the standard error. The same for the points below. $\endgroup$
    – severin
    May 8, 2012 at 11:27

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