It is known that sample variance and sample mean are independent for normally distributed variables, which means knowing sample mean does not say anything about to estimate the variance of the underlying distribution.

However, intuitively range of a sample (max of sample-min of sample) should say something about the variance of the underlying distribution. So by knowing just the sample range, we should able to reach (I am not sure but feel in that way) an estimate of the variance of the underlying distribution.

Let say we have an unbiased estimator of variance of the related distribution by using range: var1

I wonder both knowing range and sample mean at the same time; can we have more efficient unbiased estimator than var1 .

I feel "no" as answer of this question for normal distribution but "yes" for asymmetric distributions. However, I can not figure it out. However, I am not clear about this subject. I will be very glad for any help.

  • $\begingroup$ Look into the Rao-blackwell theorem $\endgroup$ Commented May 14, 2017 at 19:10
  • $\begingroup$ I can't figure out the relation with Rao-Blackwell theorem. Can you please explain a bit? . Sample mean is not a sufficient estimator for variance. $\endgroup$
    – oercim
    Commented May 14, 2017 at 19:36
  • $\begingroup$ i don't know much about estimation theory, but i know that sample min/max can be used to estimate variance, see for exampleHozo, Djulbegovic, Hozo; Estimating the mean and variance from the median, range, and the size of a sample bmcmedresmethodol.biomedcentral.com/articles/10.1186/… $\endgroup$
    – rep_ho
    Commented Sep 25, 2020 at 18:16

1 Answer 1


In the case of independent identically distributed draws from the normal distribution, the sample mean and sample variance are sufficient statistics -- nothing further will convey additional information about the population mean and variance.

In other cases, the sample range may indeed add information. For example, consider a uniform distribution on $(\mu-\sqrt{3}\,\sigma,\mu+\sqrt{3}\,\sigma)$ -- which has mean $\mu$ an standard deviation $\sigma$. Then the sample range will contain all the information about $\sigma$ but the sample variance will add nothing to that. [Note that this is symmetric, not skew]

In many other cases each may contain some information that the other does not, but it will nearly always be the case that both together do not contain all the information about the population variance.

In the case where you know the distribution you're dealing with and you have some estimator of a parameter, the Rao-Blackwell theorem gives a way of improving that estimator, by conditioning it on a sufficient statistic. This can be a useful way to obtain efficient estimators, but this doesn't directly tell us the best way to use sample range with sample variance if they are the only two statistics we have -- unless the two sample statistics are between them sufficient, in which case start with any estimator (such as the variance) and attempt to apply the theorem.

The information - if any - that could be had from using both the sample variance and the sample range will depend on the precise distribution you're looking at. If you really don't know what you're dealing with you may not be able to say much. [Typically the range will contain information about variance when the distribution is quite short tailed but it may not be useful when the distribution is heavy-tailed.]

Using such a strategy, in the case of a symmetric triangular distribution for which we observe only the variance and range (at one particular sample size) I was able to identify (via a little simulation work) an estimator of population variance that itself performed better than either a scaled squared range or the sample variance (after noting that their logs were linearly related, one based on a suitably chosen weighted geometric mean of the two).


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