It came as a bit of a shock to me the first time I did a normal distribution Monte Carlo simulation and discovered that the mean of $100$ standard deviations from $100$ samples, all having a sample size of only $n=2$, proved to be much less than, i.e., averaging $ \sqrt{\frac{2}{\pi }}$ times, the population generating $\sigma$. However, this is well known, if seldom remembered, and I sort of did know, or I would not have done a simulation. 

My reasoning was as follows, the population mean, $\mu$, of two values can be anywhere with respect to a $x_1$ and is definitely not located at $\frac{x_1+x_2}{2}$, which latter makes for an absolute minimum possible sum squared so that we are underestimating $\sigma$ substantially, as follows

w.l.o.g. let $x_2-x_1=d$, then $\Sigma_{i=1}^{n}(x_i-\bar{x})^2$ is $2 (\frac{d}{2})^2=\frac{d^2}{2}$, the least possible result. 

For the usual calculation, for $n<10$, $\text{SD}$s suffer from very significant underestimation called [small number bias](https://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation), which only approaches 1% underestimation of $\sigma$ when $n$ is approximately $100$. That means that standard deviation calculated as 

$\text{SD}=\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}$   ,

is a biased estimator of the population standard deviation ($\sigma$). Note, in that formula we decrementing the degrees of freedom of $n$ by 1 and dividing by $n-1$, i.e., we do some correction, but it is only asymptotically correct, and $n-3/2$ would be a better [rule of thumb](https://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Rule_of_thumb_for_the_normal_distribution). For our $x_2-x_1=d$ example the $\text{SD}$ formula would give us $SD=\frac{d}{\sqrt 2}\approx 0.707d$, a statistically implausible minimum value as $\mu\neq \bar{x}$, where a better expected value ($s$) would be $E(s)=\sqrt{\frac{\pi }{2}}\frac{d}{\sqrt 2}=\frac{\sqrt\pi }{2}d\approx0.886d$. 
In general, [small number bias correction](http://stats.stackexchange.com/questions/11707/why-is-sample-standard-deviation-a-biased-estimator-of-sigma?rq=1) implies that the unbiased estimator of population standard deviation of a normal distribution is 

$\text{E}(s)\,=\,\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{2}}>\text{SD}=\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}\; .$

Since SD is a biased estimator of population standard deviation, it cannot be the minimum variance unbiased estimator [MVUE](https://en.wikipedia.org/wiki/Minimum-variance_unbiased_estimator#Definition) of population standard deviation unless we are happy with saying that it is MVUE as $n\rightarrow \infty$, which I, for one, am not. 

Now comes the question

**Can it be proven that the $\text{E}(s)$ above is MVUE for $\sigma$ of a normal distribution of sample-size $n$, where $n$ is a positive integer greater than one?**

Hint: (But not the answer) see http://stats.stackexchange.com/questions/27976/how-can-i-find-the-standard-deviation-of-the-sample-standard-deviation-from-a-no