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 $\sigma$ used for generating the population. However, this is well known, if seldom remembered, and I sort of did know, or I would not have done a simulation. Here is a simulation.

Here is an example for predicting 95% confidence intervals of $N(0,1)$ using 100, $n=2$, estimates of $\text{SD}$, and $\text{E}(s_{n=2})=\sqrt\frac{\pi}{2}\text{SD}$.

 RAND()   RAND()    Calc    Calc    
 N(0,1)   N(0,1)    SD      E(s)    
-1.1171  -0.0627    0.7455  0.9344  
 1.7278  -0.8016    1.7886  2.2417  
 1.3705  -1.3710    1.9385  2.4295  
 1.5648  -0.7156    1.6125  2.0209  
 1.2379   0.4896    0.5291  0.6632  
-1.8354   1.0531    2.0425  2.5599  
 1.0320  -0.3531    0.9794  1.2275  
 1.2021  -0.3631    1.1067  1.3871  
 1.3201  -1.1058    1.7154  2.1499  
-0.4946  -1.1428    0.4583  0.5744  
 0.9504  -1.0300    1.4003  1.7551  
-1.6001   0.5811    1.5423  1.9330  
-0.5153   0.8008    0.9306  1.1663  
-0.7106  -0.5577    0.1081  0.1354  
 0.1864   0.2581    0.0507  0.0635  
-0.8702  -0.1520    0.5078  0.6365  
-0.3862   0.4528    0.5933  0.7436  
-0.8531   0.1371    0.7002  0.8775  
-0.8786   0.2086    0.7687  0.9635  
 0.6431   0.7323    0.0631  0.0791  
 1.0368   0.3354    0.4959  0.6216  
-1.0619  -1.2663    0.1445  0.1811  
 0.0600  -0.2569    0.2241  0.2808  
-0.6840  -0.4787    0.1452  0.1820  
 0.2507   0.6593    0.2889  0.3620  
 0.1328  -0.1339    0.1886  0.2364  
-0.2118  -0.0100    0.1427  0.1788  
-0.7496  -1.1437    0.2786  0.3492  
 0.9017   0.0022    0.6361  0.7972  
 0.5560   0.8943    0.2393  0.2999  
-0.1483  -1.1324    0.6959  0.8721  
-1.3194  -0.3915    0.6562  0.8224  
-0.8098  -2.0478    0.8754  1.0971  
-0.3052  -1.1937    0.6282  0.7873  
 0.5170  -0.6323    0.8127  1.0186  
 0.6333  -1.3720    1.4180  1.7772  
-1.5503   0.7194    1.6049  2.0115  
 1.8986  -0.7427    1.8677  2.3408  
 2.3656  -0.3820    1.9428  2.4350  
-1.4987   0.4368    1.3686  1.7153  
-0.5064   1.3950    1.3444  1.6850  
 1.2508   0.6081    0.4545  0.5696  
-0.1696  -0.5459    0.2661  0.3335  
-0.3834  -0.8872    0.3562  0.4465  
 0.0300  -0.8531    0.6244  0.7826  
 0.4210   0.3356    0.0604  0.0757  
 0.0165   2.0690    1.4514  1.8190  
-0.2689   1.5595    1.2929  1.6204  
 1.3385   0.5087    0.5868  0.7354  
 1.1067   0.3987    0.5006  0.6275  
 2.0015  -0.6360    1.8650  2.3374  
-0.4504   0.6166    0.7545  0.9456  
 0.3197  -0.6227    0.6664  0.8352  
-1.2794  -0.9927    0.2027  0.2541  
 1.6603  -0.0543    1.2124  1.5195  
 0.9649  -1.2625    1.5750  1.9739  
-0.3380  -0.2459    0.0652  0.0817  
-0.8612   2.1456    2.1261  2.6647  
 0.4976  -1.0538    1.0970  1.3749  
-0.2007  -1.3870    0.8388  1.0513  
-0.9597   0.6327    1.1260  1.4112  
-2.6118  -0.1505    1.7404  2.1813  
 0.7155  -0.1909    0.6409  0.8033  
 0.0548  -0.2159    0.1914  0.2399  
-0.2775   0.4864    0.5402  0.6770  
-1.2364  -0.0736    0.8222  1.0305  
-0.8868  -0.6960    0.1349  0.1691  
 1.2804  -0.2276    1.0664  1.3365  
 0.5560  -0.9552    1.0686  1.3393  
 0.4643  -0.6173    0.7648  0.9585  
 0.4884  -0.6474    0.8031  1.0066  
 1.3860   0.5479    0.5926  0.7427  
-0.9313   0.5375    1.0386  1.3018  
-0.3466  -0.3809    0.0243  0.0304  
 0.7211  -0.1546    0.6192  0.7760  
-1.4551  -0.1350    0.9334  1.1699  
 0.0673   0.4291    0.2559  0.3207  
 0.3190  -0.1510    0.3323  0.4165  
-1.6514  -0.3824    0.8973  1.1246  
-1.0128  -1.5745    0.3972  0.4978  
-1.2337  -0.7164    0.3658  0.4585  
-1.7677  -1.9776    0.1484  0.1860  
-0.9519  -0.1155    0.5914  0.7412  
 1.1165  -0.6071    1.2188  1.5275  
-1.7772   0.7592    1.7935  2.2478  
 0.1343  -0.0458    0.1273  0.1596  
 0.2270   0.9698    0.5253  0.6583  
-0.1697  -0.5589    0.2752  0.3450  
 2.1011   0.2483    1.3101  1.6420  
-0.0374   0.2988    0.2377  0.2980  
-0.4209   0.5742    0.7037  0.8819  
 1.6728  -0.2046    1.3275  1.6638  
 1.4985  -1.6225    2.2069  2.7659  
 0.5342  -0.5074    0.7365  0.9231  
 0.7119   0.8128    0.0713  0.0894  
 1.0165  -1.2300    1.5885  1.9909  
-0.2646  -0.5301    0.1878  0.2353  
-1.1488  -0.2888    0.6081  0.7621  
-0.4225   0.8703    0.9141  1.1457  
 0.7990  -1.1515    1.3792  1.7286  

 0.0344  -0.1892    0.8188  1.0263  mean E(.)
                    SD pred E(s) pred   
-1.9600  -1.9600   -1.6049 -2.0114    2.5%  theor, est
 1.9600   1.9600    1.6049  2.0114   97.5%  theor, est
                    0.3551 -0.0515    2.5% err
                   -0.3551  0.0515   97.5% err

Drag the slider down to see the grand totals. Now, I used the ordinary SD estimator to calculate 95% confidence intervals around a mean of zero, and they are off by 0.3551 standard deviation units. The E(s) estimator is off by only 0.0515 standard deviation units. If one estimates standard deviation, standard error of the mean, or t-statistics, there may be a problem.

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.

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 decrement 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. 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 $E(s)=\sqrt{\frac{\pi }{2}}\frac{d}{\sqrt 2}=\frac{\sqrt\pi }{2}d\approx0.886d$. For the usual calculation, for $n<10$, $\text{SD}$s suffer from very significant underestimation called small number bias, which only approaches 1% underestimation of $\sigma$ when $n$ is approximately $25$. Since many biological experiments have $n<25$, this is indeed an issue. For $n=1000$, the error is approximately 25 parts in 100,000. In general, small number bias correction implies that the unbiased estimator of population standard deviation of a normal distribution is

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

From Wikipedia under creative commons licensing one has a plot of SD underestimation of $\sigma$ <a title="By Rb88guy (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0) or GFDL (http://www.gnu.org/copyleft/fdl.html)], via Wikimedia Commons" href="https://commons.wikimedia.org/wiki/File%3AStddevc4factor.jpg"><img width="512" alt="Stddevc4factor" src="https://upload.wikimedia.org/wikipedia/commons/thumb/e/ee/Stddevc4factor.jpg/512px-Stddevc4factor.jpg"/></a>

Since SD is a biased estimator of population standard deviation, it cannot be the minimum variance unbiased estimator MVUE of population standard deviation unless we are happy with saying that it is MVUE as $n\rightarrow \infty$, which I, for one, am not.

Concerning non-normal distributions and approximately unbiased $SD$ read this.

Now comes the question Q1

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 How can I find the standard deviation of the sample standard deviation from a normal distribution?.

Next question, Q2

Would someone please explain to me why we are using $\text{SD}$ anyway as it is clearly biased and misleading? That is, why not use $\text{E}(s)$ for most everything? Supplementary, it has become clear in the answers below that variance is unbiased, but its square root is biased. I would request that answers address the question of when unbiased standard deviation should be used.

As it turns out, a partial answer is that to avoid bias in the simulation above, the variances could have been averaged rather than the SD-values. To see the effect of this, if we square the SD column above, and average those values we get 0.9994, the square root of which is an estimate of the standard deviation 0.9996915 and the error for which is only 0.0006 for the 2.5% tail and -0.0006 for the 95% tail. Note that this is because variances are additive, so averaging them is a low error procedure. However, standard deviations are biased, and in those cases where we do not have the luxury of using variances as an intermediary, we still need small number correction. Even if we can use variance as an intermediary, in this case for $n=100$, the small sample correction suggests multiplying the square root of unbiased variance 0.9996915 by 1.002528401 to give 1.002219148 as an unbiased estimate of standard deviation. So, yes, we can delay using small number correction but should we therefore ignore it entirely?

The question here is when should we be using small number correction, as opposed to ignoring its use, and predominantly, we have avoided its use.

Here is another example, the minimum number of points in space to establish a linear trend that has an error is three. If we fit these points with ordinary least squares the result for many such fits is a folded normal residual pattern if there is non-linearity and half normal if there is linearity. In the half-normal case our distribution mean requires small number correction. If we try the same trick with 4 or more points, the distribution will not generally be normal related or easy to characterize. Can we use variance to somehow combine those 3-point results? Perhaps, perhaps not. However, it is easier to conceive of problems in terms of distances and vectors.

  • 3
    $\begingroup$ Q1: See the Lehmann-Scheffe theorem. $\endgroup$ Commented Dec 8, 2016 at 15:57
  • 1
    $\begingroup$ Nonzero bias of an estimator is not necessarily a drawback. For example, if we wish to have an accurate estimator under square loss, we are willing to induce bias as long as it reduces the variance by a sufficiently large amount. That is why (biased) regularized estimators may perform better than the (unbiased) OLS estimator in a linear regression model, for example. $\endgroup$ Commented Dec 14, 2016 at 20:20
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    $\begingroup$ @Carl many terms are used differently in different application areas. If you're posting to a stats group and you use a jargon term like "bias", you would naturally be assumed to be using the specific meaning(s) of the term particular to statistics. If you mean anything else, it's essential to either use a different term or to define clearly what you do mean by the term right at the first use. $\endgroup$
    – Glen_b
    Commented Dec 15, 2016 at 3:35
  • 3
    $\begingroup$ "bias" is certainly a term of jargon -- special words or expressions used by a profession or group that are difficult for others to understand seems pretty much what "bias" is. It's because such terms have precise, specialized definitions in their application areas (including mathematical definitions) that makes them jargon terms. $\endgroup$
    – Glen_b
    Commented Dec 15, 2016 at 3:50
  • 2
    $\begingroup$ @Carl, agree with Glen_b. Honestly, for me discussions with you are often more difficult than usual precisely because you are using well-established statistical terms to denote something quite different than they denote in statistics (which may probably make sense in physics). So purely for convenience, I would recommend sticking to the classical statistical definitions and terms when discussing with statisticians (e.g. here at Cross Validated). $\endgroup$ Commented Dec 15, 2016 at 6:51

5 Answers 5


The sample standard deviation $S=\sqrt{\frac{\sum (X - \bar{X})^2}{n-1}}$ is complete and sufficient for $\sigma$ so the set of unbiased estimators of $\sigma^k$ given by

$$ \frac{(n-1)^\frac{k}{2}}{2^\frac{k}{2}} \cdot \frac{\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n+k-1}{2}\right)} \cdot S^k = \frac{S^k}{c_k} $$

(See Why is sample standard deviation a biased estimator of $\sigma$?) are, by the Lehmann–Scheffé theorem, UMVUE. Consistent, though biased, estimators of $\sigma^k$ can also be formed as

$$ \tilde{\sigma}^k_j= \left(\frac{S^j}{c_j}\right)^\frac{k}{j} $$

(the unbiased estimators being specified when $j=k$). The bias of each is given by

$$\operatorname{E}\tilde{\sigma}^k_j - \sigma^k =\left( \frac{c_k}{c_j^\frac{k}{j}} -1 \right) \sigma^k$$

& its variance by

$$\operatorname{Var}\tilde{\sigma}^{k}_j=\operatorname{E}\tilde{\sigma}^{2k}_j - \left(\operatorname{E}\tilde{\sigma}^k_j\right)^2=\frac{c_{2k}-c_k^2}{c_j^\frac{2k}{j}} \sigma^{2k}$$

For the two estimators of $\sigma$ you've considered, $\tilde{\sigma}^1_1=\frac{S}{c_1}$ & $\tilde{\sigma}^1_2=S$, the lack of bias of $\tilde{\sigma}_1$ is more than offset by its larger variance when compared to $\tilde{\sigma}_2$:

$$\begin{align} \operatorname{E}\tilde{\sigma}_1 - \sigma &= 0 \\ \operatorname{E}\tilde{\sigma}_2 - \sigma &=(c_1 -1) \sigma \\ \operatorname{Var}\tilde{\sigma}_1 =\operatorname{E}\tilde{\sigma}^{2}_1 - \left(\operatorname{E}\tilde{\sigma}^1_1\right)^2 &=\frac{c_{2}-c_1^2}{c_1^2} \sigma^{2} = \left(\frac{1}{c_1^2}-1\right) \sigma^2 \\ \operatorname{Var}\tilde{\sigma}_2 =\operatorname{E}\tilde{\sigma}^{2}_1 - \left(\operatorname{E}\tilde{\sigma}_2\right)^2 &=\frac{c_{2}-c_1^2}{c_2} \sigma^{2}=(1-c_1^2)\sigma^2 \end{align}$$ (Note that $c_2=1$, as $S^2$ is already an unbiased estimator of $\sigma^2$.)

Plot showing contributions of bias & variance to MSE at sample sizes from one to 20 for the two estimators

The mean square error of $a_k S^k$ as an estimator of $\sigma^2$ is given by

$$ \begin{align} (\operatorname{E} a_k S^k - \sigma^k)^2 + \operatorname{E} (a_k S^k)^2 - (\operatorname{E} a_k S^k)^2 &= [ (a_k c_k -1)^2 + a_k^2 c_{2k} - a_k^2 c_k^2 ] \sigma^{2k}\\ &= ( a_k^2 c_{2k} -2 a_k c_k + 1 ) \sigma^{2k} \end{align} $$

& therefore minimized when

$$a_k = \frac{c_k}{c_{2k}}$$

, allowing the definition of another set of estimators of potential interest:

$$ \hat{\sigma}^k_j= \left(\frac{c_j S^j}{c_{2j}}\right)^\frac{k}{j} $$

Curiously, $\hat{\sigma}^1_1=c_1S$, so the same constant that divides $S$ to remove bias multiplies $S$ to reduce MSE. Anyway, when $j=k$ these are the uniformly minimum-square-error location-invariant & scale-equivariant estimators of $\sigma^k$ (you don't want your estimate to change at all if you measure in kelvins rather than degrees Celsius, & you want it to change by a factor of $\left(\frac{9}{5}\right)^k$ if you measure in Fahrenheit).

None of the above has any bearing on the construction of hypothesis tests or confidence intervals (see e.g. Why does this excerpt say that unbiased estimation of standard deviation usually isn't relevant?). And $\tilde{\sigma}^k_j$ & $\hat{\sigma}^k_j$ exhaust neither estimators nor parameter scales of potential interest—consider the maximum-likelihood estimator $\sqrt{\frac{n-1}{n}}S$, or the median-unbiased estimator $\sqrt{\frac{n-1}{\chi^2_{n-1}(0.5)}}S$; or the geometric standard deviation of a lognormal distribution $\mathrm{e}^\sigma$. It may be worth showing a few more-or-less popular estimates made from a small sample ($n=2$) together with the upper & lower bounds, $\sqrt{\frac{(n-1)s^2}{\chi^2_{n-1}(\alpha)}}$ & $\sqrt{\frac{(n-1)s^2}{\chi^2_{n-1}(1-\alpha)}}$, of the equal-tailed confidence interval having coverage $1-\alpha$:

confidence distribution for sigma showing estimates

The span between the most divergent estimates is negligible in comparison with the width of any confidence interval having decent coverage. (The 95% C.I., for instance, is $(0.45s,31.9s)$.) There's no sense in being finicky about the properties of a point estimator unless you're prepared to be fairly explicit about what you want you want to use it for—most explicitly you can define a custom loss function for a particular application. A reason you might prefer an exactly (or almost) unbiased estimator is that you're going to use it in subsequent calculations during which you don't want bias to accumulate: your illustration of averaging biased estimates of standard deviation is a simple example of such (a more complex example might be using them as a response in a linear regression). In principle an all-encompassing model should obviate the need for unbiased estimates as an intermediate step, but might be considerably more tricky to specify & fit.

† The value of $\sigma$ that makes the observed data most probable has an appeal as an estimate independent of consideration of its sampling distribution.


For the more restricted question

Why is a biased standard deviation formula typically used?

the simple answer

Because the associated variance estimator is unbiased. There is no real mathematical/statistical justification.

may be accurate in many cases.

However, this is not necessarily always the case. There are at least two important aspects of these issues that should be understood.

First, the sample variance $s^2$ is not just unbiased for Gaussian random variables. It is unbiased for any distribution with finite variance $\sigma^2$ (as discussed below, in my original answer). The question notes that $s$ is not unbiased for $\sigma$, and suggests an alternative which is unbiased for a Gaussian random variable. However it is important to note that unlike the variance, for the standard deviation it is not possible to have a "distribution free" unbiased estimator (*see note below).

Second, as mentioned in the comment by whuber the fact that $s$ is biased does not impact the standard "t test". First note that, for a Gaussian variable $x$, if we estimate z-scores from a sample $\{x_i\}$ as $$z_i=\frac{x_i-\mu}{\sigma}\approx\frac{x_i-\bar{x}}{s}$$ then these will be biased.

However the t statistic is usually used in the context of the sampling distribution of $\bar{x}$. In this case the z-score would be $$z_{\bar{x}}=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\approx\frac{\bar{x}-\mu}{s/\sqrt{n}}=t$$ though we can compute neither $z$ nor $t$, as we do not know $\mu$. Nonetheless, if the $z_{\bar{x}}$ statistic would be normal, then the $t$ statistic will follow a Student-t distribution. This is not a large-$n$ approximation. The only assumption is that the $x$ samples are i.i.d. Gaussian.

(Commonly the t-test is applied more broadly for possibly non-Gaussian $x$. This does rely on large-$n$, which by the central limit theorem ensures that $\bar{x}$ will still be Gaussian.)

*Clarification on "distribution-free unbiased estimator"

By "distribution free", I mean that the estimator cannot depend on any information about the population $x$ aside from the sample $\{x_1,\ldots,x_n\}$. By "unbiased" I mean that the expected error $\mathbb{E}[\hat{\theta}_n]-\theta$ is uniformly zero, independent of the sample size $n$. (As opposed to an estimator that is merely asymptotically unbiased, a.k.a. "consistent", for which the bias vanishes as $n\to\infty$.)

In the comments this was given as a possible example of a "distribution-free unbiased estimator". Abstracting a bit, this estimator is of the form $\hat{\sigma}=f[s,n,\kappa_x]$, where $\kappa_x$ is the excess kurtosis of $x$. This estimator is not "distribution free", as $\kappa_x$ depends on the distribution of $x$. The estimator is said to satisfy $\mathbb{E}[\hat{\sigma}]-\sigma_x=\mathrm{O}[\frac{1}{n}]$, where $\sigma_x^2$ is the variance of $x$. Hence the estimator is consistent, but not (absolutely) "unbiased", as $\mathrm{O}[\frac{1}{n}]$ can be arbitrarily large for small $n$.

Note: Below is my original "answer". From here on, the comments are about the standard "sample" mean and variance, which are "distribution-free" unbiased estimators (i.e. the population is not assumed to be Gaussian).

This is not a complete answer, but rather a clarification on why the sample variance formula is commonly used.

Given a random sample $\{x_1,\ldots,x_n\}$, so long as the variables have a common mean, the estimator $\bar{x}=\frac{1}{n}\sum_ix_i$ will be unbiased, i.e. $$\mathbb{E}[x_i]=\mu \implies \mathbb{E}[\bar{x}]=\mu$$

If the variables also have a common finite variance, and they are uncorrelated, then the estimator $s^2=\frac{1}{n-1}\sum_i(x_i-\bar{x})^2$ will also be unbiased, i.e. $$\mathbb{E}[x_ix_j]-\mu^2=\begin{cases}\sigma^2&i=j\\0&i\neq{j}\end{cases} \implies \mathbb{E}[s^2]=\sigma^2$$ Note that the unbiasedness of these estimators depends only on the above assumptions (and the linearity of expectation; the proof is just algebra). The result does not depend on any particular distribution, such as Gaussian. The variables $x_i$ do not have to have a common distribution, and they do not even have to be independent (i.e. the sample does not have to be i.i.d.).

The "sample standard deviation" $s$ is not an unbiased estimator, $\mathbb{s}\neq\sigma$, but nonetheless it is commonly used. My guess is that this is simply because it is the square root of the unbiased sample variance. (With no more sophisticated justification.)

In the case of an i.i.d. Gaussian sample, the maximum likelihood estimates (MLE) of the parameters are $\hat{\mu}_\mathrm{MLE}=\bar{x}$ and $(\hat{\sigma}^2)_\mathrm{MLE}=\frac{n-1}{n}s^2$, i.e. the variance divides by $n$ rather than $n^2$. Moreover, in the i.i.d. Gaussian case the standard deviation MLE is just the square root of the MLE variance. However these formulas, as well as the one hinted at in your question, depend on the Gaussian i.i.d. assumption.

Update: Additional clarification on "biased" vs. "unbiased".

Consider an $n$-element sample as above, $X=\{x_1,\ldots,x_n\}$, with sum-square-deviation $$\delta^2_n=\sum_i(x_i-\bar{x})^2$$ Given the assumptions outlined in the first part above, we necessarily have $$\mathbb{E}[\delta^2_n]=(n-1)\sigma^2$$ so the (Gaussian-)MLE estimator is biased $$\widehat{\sigma^2_n}=\tfrac{1}{n}\delta^2_n \implies \mathbb{E}[\widehat{\sigma^2_n}]=\tfrac{n-1}{n}\sigma^2 $$ while the "sample variance" estimator is unbiased $$s^2_n=\tfrac{1}{n-1}\delta^2_n \implies \mathbb{E}[s^2_n]=\sigma^2$$

Now it is true that $\widehat{\sigma^2_n}$ becomes less biased as the sample size $n$ increases. However $s^2_n$ has zero bias no matter the sample size (so long as $n>1$). For both estimators, the variance of their sampling distribution will be non-zero, and depend on $n$.

As an example, the below Matlab code considers an experiment with $n=2$ samples from a standard-normal population $z$. To estimate the sampling distributions for $\bar{x},\widehat{\sigma^2},s^2$, the experiment is repeated $N=10^6$ times. (You can cut & paste the code here to try it out yourself.)

% n=sample size, N=number of samples
n=2; N=1e6;
% generate standard-normal random #'s
z=randn(n,N); % i.e. mu=0, sigma=1
% compute sample stats (Gaussian MLE)
zbar=sum(z)/n; zvar_mle=sum((z-zbar).^2)/n;
% compute ensemble stats (sampling-pdf means)
zbar_avg=sum(zbar)/N, zvar_mle_avg=sum(zvar_mle)/N
% compute unbiased variance

Typical output is like

zbar_avg     =  1.4442e-04
zvar_mle_avg =  0.49988
zvar_avg     =  0.99977

confirming that \begin{align} \mathbb{E}[\bar{z}]&\approx\overline{(\bar{z})}\approx\mu=0 \\ \mathbb{E}[s^2]&\approx\overline{(s^2)}\approx\sigma^2=1 \\ \mathbb{E}[\widehat{\sigma^2}]&\approx\overline{(\widehat{\sigma^2})}\approx\frac{n-1}{n}\sigma^2=\frac{1}{2} \end{align}

Update 2: Note on fundamentally "algebraic" nature of unbiased-ness.

In the above numerical demonstration, the code approximates the true expectation $\mathbb{E}[\,]$ using an ensemble average with $N=10^6$ replications of the experiment (i.e. each is a sample of size $n=2$). Even with this large number, the typical results quoted above are far from exact.

To numerically demonstrate that the estimators are really unbiased, we can use a simple trick to approximate the $N\to\infty$ case: simply add the following line to the code

% optional: "whiten" data (ensure exact ensemble stats)
[U,S,V]=svd(z-mean(z,2),'econ'); z=sqrt(N)*U*V';

(placing after "generate standard-normal random #'s" and before "compute sample stats")

With this simple change, even running the code with $N=10$ gives results like

zbar_avg     =  1.1102e-17
zvar_mle_avg =  0.50000
zvar_avg     =  1.00000
  • 3
    $\begingroup$ @amoeba Well, I'll eat my hat. I squared the SD-values in each line then averaged them and they come out unbiased (0.9994), whereas the SD-values themselves do not. Meaning that you and GeoMatt22 are correct, and I am wrong. $\endgroup$
    – Carl
    Commented Dec 8, 2016 at 7:27
  • 3
    $\begingroup$ @Carl: It's generally true that transforming an unbiased estimator of a parameter doesn't give an unbiased estimate of the transformed parameter except when the transformation is affine, following from the linearity of expectation. So on what scale is unbiasedness important to you? $\endgroup$ Commented Dec 8, 2016 at 8:29
  • 4
    $\begingroup$ Carl: I apologize if you feel my answer was orthogonal to your question. It was intended to provide a plausible explanation of Q:"why a biased standard deviation formula is typically used?" A:"simply because the associated variance estimator is unbiased, vs. any real mathematical/statistical justification". As for your comment, typically "unbiased" describes an estimator whose expected value is correct independent of sample size. If it is unbiased only in the limit of infinite sample size, typically it would be called "consistent". $\endgroup$
    – GeoMatt22
    Commented Dec 9, 2016 at 6:38
  • 4
    $\begingroup$ (+1) Nice answer. Small caveat: That Wikipedia passage on consistency quoted in this answer is a bit of a mess and the parenthetical statement made related to it is potentially misleading. "Consistency" and "asymptotic unbiasedness" are in some sense orthogonal properties of an estimator. For a little more on that point, see the comment thread to this answer. $\endgroup$
    – cardinal
    Commented Dec 10, 2016 at 21:45
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    $\begingroup$ +1 but I think @Scortchi makes a really important point in his answer that is not mentioned in yours: namely, that even for Gaussian population, the unbiased estimate of $\sigma$ has higher expected error than the standard biased estimate of $\sigma$ (due to the high variance of the former). This is a strong argument in favour of not using an unbiased estimator even if one knows that the underlying distribution is Gaussian. $\endgroup$
    – amoeba
    Commented Dec 13, 2016 at 14:52

Q2: Would someone please explain to me why we are using SD anyway as it is clearly biased and misleading?

This came up as an aside in comments, but I think it bears repeating because it's the crux of the answer:

The sample variance formula is unbiased, and variances are additive. So if you expect to do any (affine) transformations, this is a serious statistical reason why you should insist on a "nice" variance estimator over a "nice" SD estimator.

In an ideal world, they'd be equivalent. But that's not true in this universe. You have to choose one, so you might as well choose the one that lets you combine information down the road.

Comparing two sample means? The variance of their difference is sum of their variances.
Doing a linear contrast with several terms? Get its variance by taking a linear combination of their variances.
Looking at regression line fits? Get their variance using the variance-covariance matrix of your estimated beta coefficients.
Using F-tests, or t-tests, or t-based confidence intervals? The F-test calls for variances directly; and the t-test is exactly equivalent to the square root of an F-test.

In each of these common scenarios, if you start with unbiased variances, you'll remain unbiased all the way (unless your final step converts to SDs for reporting).
Meanwhile, if you'd started with unbiased SDs, neither your intermediate steps nor the final outcome would be unbiased anyway.

  • $\begingroup$ Variance is not a distance measurement, and standard deviation is. Yes, vector distances add by squares, but the primary measurement is distance. The question was what would you use corrected distance for, and not why should we ignore distance as if it did not exist. $\endgroup$
    – Carl
    Commented Dec 11, 2016 at 3:39
  • $\begingroup$ Well, I guess I'm arguing that "the primary measurement is distance" isn't necessarily true. 1) Do you have a method to work with unbiased variances; combine them; take the final resulting variance; and rescale its sqrt to get an unbiased SD? Great, then do that. If not... 2) What are you going to do with a SD from a tiny sample? Report it on its own? Better to just plot the datapoints directly, not summarize their spread. And how will people interpret it, other than as an input to SEs and thus CIs? It's meaningful as an input to CIs, but then I'd prefer the t-based CI (with usual SD). $\endgroup$
    – civilstat
    Commented Dec 11, 2016 at 22:35
  • $\begingroup$ I do no think that many clinical studies or commercial software programs with $n<25$ would use standard error of the mean calculated from small sample corrected standard deviation leading to a false impression of how small those errors are. I think even that one issue, even if that is the only one, should be ignored. $\endgroup$
    – Carl
    Commented Dec 11, 2016 at 23:00
  • $\begingroup$ "so you might as well choose the one that lets you combine information down the road" and "the primary measurement is distance" isn't necessarily true. Farmer Jo's house is 640 acres down the road? One uses the appropriate measurement correctly for each and every situation, or one has a higher tolerance for false witness than I. My only question here is when to use what, and the answer to it is not "never." $\endgroup$
    – Carl
    Commented Dec 12, 2016 at 3:11
  • $\begingroup$ Well, +1 anyway, answer is not bad. $\endgroup$
    – Carl
    Commented Jul 17, 2020 at 6:47

This post is in outline form.

(1) Taking a square root is not an affine transformation (Credit @Scortchi.)

(2) ${\rm var}(s) = {\rm E} (s^2) - {\rm E}(s)^2$, thus ${\rm E}(s) = \sqrt{{\rm E}(s^2) -{\rm var}(s)}\neq{\sqrt{\rm var(s)}}$

(3) $ {\rm var}(s)=\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{n-1}$, whereas $\text{E}(s)\,=\,\,\frac{\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{2}}$$\neq\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}={\sqrt{\rm var(s)}}$

(4) Thus, we cannot substitute ${\sqrt{\rm var(s)}}$ for $\text{E}(s)$, for $n$ small, as square root is not affine.

(5) ${\rm var}(s)$ and $\text{E}(s)$ are unbiased (Credit @GeoMatt22 and @Macro, respectively).

(6) For non-normal distributions $\bar{x}$ is sometimes (a) undefined (e.g., Cauchy, Pareto with small $\alpha$) and (b) not UMVUE (e.g., Cauchy ($\rightarrow$ Student's-$t$ with $df=1$), Pareto, Uniform, beta). Even more commonly, variance may be undefined, e.g. Student's-$t$ with $1\leq df\leq2$. Then one can state that $\text{var}(s)$ is not UMVUE for the general case distribution. Thus, there is then no special onus to introducing an approximate small number correction for standard deviation, which likely has similar limitations to $\sqrt{\text{var}(s)}$, but is additionally less biased, $\hat\sigma = \sqrt{ \frac{1}{n - 1.5 - \tfrac14 \gamma_2} \sum_{i=1}^n (x_i - \bar{x})^2 }$ ,

where $\gamma_2$ is excess kurtosis. In a similar vein, when examining a normal squared distribution (a Chi-squared with $df=1$ transform), we might be tempted to take its square root and use the resulting normal distribution properties. That is, in general, the normal distribution can result from transformations of other distributions and it may be expedient to examine the properties of that normal distribution such that the limitation of small number correction to the normal case is not so severe a restriction as one might at first assume.

For the normal distribution case:

A1: By Lehmann-Scheffe theorem ${\rm var}(s)$ and $\text{E}(s)$ are UMVUE (Credit @Scortchi).

A2: (Edited to adjust for comments below.) For $n\leq 25$, we should use $\text{E}(s)$ for standard deviation, standard error, confidence intervals of the mean and of the distribution, and optionally for z-statistics. For $t$-testing we would not use the unbiased estimator as $\frac{ \bar X - \mu} {\sqrt{\text{var}(n)/n}}$ itself is Student's-$t$ distributed with $n-1$ degrees of freedom (Credit @whuber and @GeoMatt22). For z-statistics, $\sigma$ is usually approximated using $n$ large for which $\text{E}(s)-\sqrt{\text{var}(n)}$ is small, but for which $\text{E}(s)$ appears to be more mathematically appropriate (Credit @whuber and @GeoMatt22).

  • 2
    $\begingroup$ A2 is incorrect: following that prescription would produce demonstrably invalid tests. As I commented to the question, perhaps too subtly: consult any theoretical account of a classical test, such as the t-test, to see why a bias correction is irrelevant. $\endgroup$
    – whuber
    Commented Dec 9, 2016 at 21:24
  • 2
    $\begingroup$ There's a strong meta-argument showing why bias correction for statistical tests is a red herring: if it were incorrect not to include a bias-correction factor, then that factor would already be included in standard tables of the Student t distribution, F distribution, etc. To put it another way: if I'm wrong about this, then everybody has been wrong about statistical testing for the last century. $\endgroup$
    – whuber
    Commented Dec 9, 2016 at 21:30
  • 1
    $\begingroup$ Am I the only one who's baffled by the notation here? Why use $\operatorname{E}(s)$ to stand for $\frac{\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\sqrt{\frac{\Sigma_{i=1}^{n}(x_i-\bar{x})^2}{2}}$, the unbiased estimate of standard deviation? What's $s$? $\endgroup$ Commented Dec 9, 2016 at 21:58
  • 2
    $\begingroup$ @Scortchi the notation apparently came about as an attempt to inherit that used in the linked post. There $s$ is the sample variance, and $E(s)$ is the expected value of $s$ for a Gaussian sample. In this question, "$E(s)$" was co-opted to be a new estimator derived from the original post (i.e. something like $\hat{\sigma}\equiv s/\alpha$ where $\alpha\equiv\mathbb{E}[s]/\sigma$). If we arrive at a satisfactory answer for this question, probably a cleanup of the question & answer notation would be warranted :) $\endgroup$
    – GeoMatt22
    Commented Dec 9, 2016 at 22:20
  • 2
    $\begingroup$ The z-test assumes the denominator is an accurate estimate of $\sigma$. It's known to be an approximation that is only asymptotically correct. If you want to correct it, don't use the bias of the SD estimator--just use a t-test. That's what the t-test was invented for. $\endgroup$
    – whuber
    Commented Dec 9, 2016 at 22:58

I want to add the Bayesian answer to this discussion. Just because your assumption is that the data is generated according to some normal with unknown mean and variance, that doesn't mean that you should summarize your data using a mean and a variance. This whole problem can be avoided if you draw the model, which will have a posterior predictive that is a three parameter noncentral scaled student's T distribution. The three parameters are the total of the samples, total of the squared samples, and the number of samples. (Or any bijective map of these.)

Incidentally, I like civilstat's answer because it highlights our desire to combine information. The three sufficient statistics above are even better than the two given in the question (or by civilstat's answer). Two sets of these statistics can easily be combined, and they give the best posterior predictive given the assumption of normality.

  • $\begingroup$ How then does one calculate an unbiased standard error of the mean from those three sufficient statistics? $\endgroup$
    – Carl
    Commented Dec 14, 2016 at 17:44
  • $\begingroup$ @carl You can easily calculate it since you have the number of samples $n$, you can multiply the uncorrected sample variance by $\frac{n}{n-1}$. However, you really don't want to do that. That's tantamount to turning your three parameters into a best fit normal distribution to your limited data. It's a lot better to use your three parameters to fit the true posterior predictive: the noncentral scaled T distribution. All questions you might have (percentiles, etc.) are better answered by this T distribution. In fact, T tests are just common sense questions asked of this distribution. $\endgroup$
    – Neil G
    Commented Dec 15, 2016 at 0:30
  • $\begingroup$ How can one then generate a true normal distribution RV from Monte Carlo simulations(s) and recover that true distribution using only Student's-$t$ distribution parameters? Am I missing something here? $\endgroup$
    – Carl
    Commented Dec 15, 2016 at 2:57
  • $\begingroup$ @Carl The sufficient statistics I described were the mean, second moment, and number of samples. Your MLE of the original normal are the mean and variance (which is equal to the second moment minus the squared mean). The number of samples is useful when you want to make predictions about future observations (for which you need the posterior predictive distribution). $\endgroup$
    – Neil G
    Commented Dec 15, 2016 at 3:24
  • $\begingroup$ Though a Bayesian perspective is a welcome addition, I find this a little hard to follow: I'd have expected a discussion of constructing a point estimate from the posterior density of $\sigma$. It seems you're rather questioning the need for a point estimate: this is something well worth bringing up, but not uniquely Bayesian. (BTW you also need to explain the priors.) $\endgroup$ Commented Dec 16, 2016 at 14:22

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