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I have easy access to the the mean, standard deviation and min/max values for a variable. I also have the number of elements used to compute those values. The data is real numbers with an absolute zero.

Given this information, is it possible to estimate the median value?

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    $\begingroup$ No. Unless you are willing to assume more about the data than what you have described. But doing this would also most probably go counter the reasons you have to want to estimate the median in the first place. $\endgroup$
    – user603
    Jun 18, 2014 at 21:30
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    $\begingroup$ Do you have (or can you get) any good information on the shape of the distribution? If it is symmetric, then the median should equal the mean. Are the min and max values about equally distant from the mean? That would be some (small) indication that the distribution is symmetric. How much import do you plan to place on the estimated median? $\endgroup$
    – Joel W.
    Jun 18, 2014 at 22:11

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It's certainly possible to place some bounds on the median, but without further assumptions they might potentially be pretty weak bounds. The problem is that the only gauge you have on how skew it might be (particularly, in the sense of the second Pearson skewness) is the relative positions of the extrema to the mean, and they're typically a very weak indicator of that. Adding in the fact that the variable is nonnegative gives a second very weak indicator of skewness (the relative size of the standard deviation and mean).

But the second Pearson skewness does give us a bound: for a distribution, the median cannot be more than one standard deviation from the mean. (For a sample, because of the effect of the usual Bessel correction on standard deviation, it must lie somewhat inside those limits.)

If the standard deviation is small, that may be adequate for some purposes.

If we denote the median as $\stackrel{\sim}{x}$, the mean as $\bar{x}$, the usual sample standard deviation as $s_{n-1}$ (and let $s_n=\sqrt{\frac{n-1}{n}}s_{n-1}$ be the uncorrected s.d.), the minimum as $x_{(1)}$ and the maximum as $x_{(n)}$ then naively, we can immediately say that

$$\max(x_{(1)},\bar{x}-s_n)\leq\,\,\stackrel{\sim}{x}\,\,\leq \min(x_{(n)},\bar{x}+s_n)\,.$$

By more careful consideration of all the information, knowing the minimum and maximum might bound the result still further, but my guess is not necessarily by very much (it may help more in some cases than others). Knowing the sample size, $n$, may also add some important information, particularly if $n$ is small.

The fact that the variable is non-negative might help. Markov's inequality suggests that the median cannot be more than twice the mean, perhaps that may sometimes improve the bound from the mean plus a standard deviation (though, if the s.d. were greater than the mean, you'd usually expect the median to be lower than the mean; again it may be possible to get better bounds still).

Anyway, adding that bound to our previous naive bounds, we have:

$$\max(x_{(1)},\bar{x}-s_n)\leq\,\,\stackrel{\sim}{x}\,\,\leq \min(x_{(n)},\bar{x}+s_n,2\bar{x})\,.$$

(In that situation we also know that the median is above $0$, but given we know $x_{(1)}$, that knowledge doesn't ever improve the lower bound.)


Edit: I simulated a few data sets from different distributions (partly to see how the bounds behaved and partly as a double check that I hadn't made any egregious errors). One of the examples did have the property that $2\bar x$ was a bit less than $\bar x +s_n$ (thus reducing the upper bound on the median, so adding that third component does sometimes help), but as I expected might often be the case, the actual median was less than the mean (so it didn't make the upper bound very close).

Still, the intervals did actually enclose the median for every example I did.


If you assumed some distributional form (like, say, normality), then of course you can get much better estimates (/intervals).

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  • $\begingroup$ Thanks. That was really interesting. I'll test for the normality of the data, but I doubt it is normal. $\endgroup$
    – ahoffer
    Jun 19, 2014 at 3:39
  • $\begingroup$ It doesn't have to be normal, that was just an example; if gamma or Weibull or lognormal or whatever models are typical for your problem, that may help you identify a more narrow interval (but your inference would still be conditional on such assumptions). $\endgroup$
    – Glen_b
    Jun 19, 2014 at 3:44
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    $\begingroup$ If you can evaluate the distributional shape, is there a reason you can't compute the median, even approximately? $\endgroup$
    – Glen_b
    Jun 19, 2014 at 3:50
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If you want potentially practical estimates, that can be subjected to bound analysis, per prior provided suggestions, it is perhaps feasible to assume, especially for noisy data, that an applicable distribution is lognormal for your data or can be used as an approximation thereof.

In this proposed parametric case, per Wikipedia on the lognormal distribution, an examination of the provided formula for the quantile, mean, median and variance, suggests by taking appropriate ratios, one can derive an expression to arrive at an estimate for the median employing interestingly all provided data (that is, employing all information content) that can further be subject to bounds, as derived particularly for the median.

For example, the ratio of the expression for the quantile (employing the high or low or both data points) to the reported mean results in a quadratic equation based estimate(s) for sigma (which represents the standard deviation of natural log transform applied to your underlying data, albeit, unknown). Similarly, the ratio of the reported variance to the reported mean produces an expression leading to an estimate for sigma squared. Combining, based on some simulations for guidance, said estimates for a more accurate value for said sigma leads to a factor correction (using the formula for the associated median) to your reported mean for a possible median estimate that you are interested in providing.

I would try simulation exercises, for tuning the methodology, based on the lognormal and also for various distributions that may potentially fit your data, or can justifiably be applicable to your data, to see if there is associated value in the proposed approximate parametric suggested methodology.

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  • $\begingroup$ Prior comment by Glen_b appears agreeable to some form of approximate analysis for a median estimate. $\endgroup$
    – AJKOER
    Aug 21, 2022 at 1:10

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