I am interested in median-to-median propagation of standard error: The values in my variable (say, $Y$) are all medians (of something, say, $X$), and are each associated with a standard error. Next, I want to summarise this current variable further using median. I am unable to figure out how to propagate the existing median SEs in this last step.

$$ Y_{i} = median(X_{j}) $$ $$ Z = median(Y_{i}) $$

Many threads on Stack Exchange I saw discuss mean-to-mean propagation, which isn't what I want. For mean, the partial derivative required for the propagation formula is only concerned with the summing of individual values, and hence the formula is fairly straightforward (equation 2 here). However, since there is no summing involved in median, I don't know how to proceed with this propagation.

This post on Physics SE was the closest I could get, but I am confused as the answer only shows the relationship between uncertainties in mean and median, and doesn't explicitly talk about propagation. According to the linked answer, for large samples:

$$ SE_{median} = SE_{mean} \sqrt{\pi/2} $$

Am I right in interpreting from this that for my purpose, I need to first calculate $ Z^{'} = mean(Y_{i}) $ then find $ SE_{Z^{'}} $ (using the propagation formula for mean), which I will substitute in the formula to get:

$$ SE_{Z} = SE_{Z^{'}} \sqrt{\pi/2} $$

$$ SE_{Z} = \frac{1}{N} \sqrt{\sum (SE_{Y_{i}})^2} \sqrt{\pi/2} $$

Surely, there's a more straightforward way to achieve this propagation, directly from the error values of the medians?

  • $\begingroup$ Check the sampling distribution of the Wikipedia article referred to en.m.wikipedia.org/wiki/Median#The_sample_median $\endgroup$
    – seanv507
    Jan 11, 2022 at 8:01
  • $\begingroup$ @seanv507 I had seen this article, but haven't made much progress on a second read. I may be missing something but I don't think it throws much light on propagating, again. To clarify though, I am not looking for the theoretical background to this, but rather for a way to do the calculations with my current variables. $\endgroup$ Jan 11, 2022 at 12:38
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    $\begingroup$ So I don't really understand what you want to do, but statisticians tend to use the mean/variance precisely because errors propagate nicely etc (as opposed to eg median/range). There is no equivalent for median. There will be just different approximate formulas. $\endgroup$
    – seanv507
    Jan 11, 2022 at 12:59
  • $\begingroup$ @seanv507 Yes, that makes sense. Is the last equation in my question statistically correct? If so, I could propagate the error in a roundabout way by first propagating SE of mean. $\endgroup$ Jan 11, 2022 at 13:04
  • $\begingroup$ I don't have the time to look at this right now, but typically people use the bootstrap to do things like this. this is likely to work better than these normal approximations. stat.ucla.edu/~rgould/252w02/bsintro $\endgroup$
    – seanv507
    Jan 11, 2022 at 14:42

1 Answer 1


One of the main uses of the bootstrap is to calculate statistics of quantities with no analytical formula.

So if you want the standard error of the median of medians Then

  1. generate a realisation of your raw data
  2. calculate your Yis
  3. calculate a realisation of Z

Repeat this 1000 times and you have 1000 realizations of Z Now calculate the standard deviation of these Zs ( or the percentiles etc)

  • $\begingroup$ Does this mean that in the case of median, there is no way to propagate the SE calculated for $Y$ (via bootstrapping) to $Z$, and that the SE needs to be calculated via bootstrapping separately for $Z$? $\endgroup$ Jan 12, 2022 at 1:48
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    $\begingroup$ What I am suggesting is that the standard way of getting the Variance etc of Z is to do a bootstrap of your Xs and make all the calculations to create the Z. You do not do 2 bootstraps. No there is no exact way of propagating the variances. There are various approximations which might depend on how close your distribution is to normal etc, but the best way of checking the accuracy of the approximations is in any case to use a bootstrap. $\endgroup$
    – seanv507
    Jan 12, 2022 at 8:14

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