# Why is sample standard deviation a biased estimator of $\sigma$?

According to the Wikipedia article on unbiased estimation of standard deviation the sample SD

$$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2}$$

is a biased estimator of the SD of the population. It states that $E(\sqrt{s^2}) \neq \sqrt{E(s^2)}$.

NB. Random variables are independent and each $x_{i} \sim N(\mu,\sigma^{2})$

My question is two-fold:

• What is the proof of the biasedness?
• How does one compute the expectation of the sample standard deviation

My knowledge of maths/stats is only intermediate.

• You will find both questions are answered in the Wikipedia article on the Chi distribution.
– whuber
Jun 8, 2011 at 12:58
• You might also be interested in reading about Bessel's correction. Jul 19, 2021 at 21:36

@NRH's answer to this question gives a nice, simple proof of the biasedness of the sample standard deviation. Here I will explicitly calculate the expectation of the sample standard deviation (the original poster's second question) from a normally distributed sample, at which point the bias is clear.

The unbiased sample variance of a set of points $x_1, ..., x_n$ is

$$s^{2} = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \overline{x})^2$$

If the $x_i$'s are normally distributed, it is a fact that

$$\frac{(n-1)s^2}{\sigma^2} \sim \chi^{2}_{n-1}$$

where $\sigma^2$ is the true variance. The $\chi^2_{k}$ distribution has probability density

$$p(x) = \frac{(1/2)^{k/2}}{\Gamma(k/2)} x^{k/2 - 1}e^{-x/2}$$

using this we can derive the expected value of $s$;

\begin{align} E(s) &= \sqrt{\frac{\sigma^2}{n-1}} E \left( \sqrt{\frac{s^2(n-1)}{\sigma^2}} \right) \\ &= \sqrt{\frac{\sigma^2}{n-1}} \int_{0}^{\infty} \sqrt{x} \frac{(1/2)^{(n-1)/2}}{\Gamma((n-1)/2)} x^{((n-1)/2) - 1}e^{-x/2} \ dx \end{align}

which follows from the definition of expected value and fact that $\sqrt{\frac{s^2(n-1)}{\sigma^2}}$ is the square root of a $\chi^2$ distributed variable. The trick now is to rearrange terms so that the integrand becomes another $\chi^2$ density:

\begin{align} E(s) &= \sqrt{\frac{\sigma^2}{n-1}} \int_{0}^{\infty} \frac{(1/2)^{(n-1)/2}}{\Gamma(\frac{n-1}{2})} x^{(n/2) - 1}e^{-x/2} \ dx \\ &= \sqrt{\frac{\sigma^2}{n-1}} \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \int_{0}^{\infty} \frac{(1/2)^{(n-1)/2}}{\Gamma(n/2)} x^{(n/2) - 1}e^{-x/2} \ dx \\ &= \sqrt{\frac{\sigma^2}{n-1}} \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \cdot \frac{ (1/2)^{(n-1)/2} }{ (1/2)^{n/2} } \underbrace{ \int_{0}^{\infty} \frac{(1/2)^{n/2}}{\Gamma(n/2)} x^{(n/2) - 1}e^{-x/2} \ dx}_{\chi^2_n \ {\rm density} } \end{align}

now we know the integrand the last line is equal to 1, since it is a $\chi^2_{n}$ density. Simplifying constants a bit gives

$$E(s) = \sigma \cdot \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) }$$

Therefore the bias of $s$ is

$$\sigma - E(s) = \sigma \bigg(1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \bigg) \sim \frac{\sigma}{4 n} \>$$ as $n \to \infty$.

It's not hard to see that this bias is not 0 for any finite $n$, thus proving the sample standard deviation is biased. Below the bias is plot as a function of $n$ for $\sigma=1$ in red along with $1/4n$ in blue: • (+1) Nice answer. I hope you don't mind, I tweaked a couple of very minor things and added an asymptotic result regarding the bias. I suppose you could superimpose the curve $(4n)^{-1}$ onto your plot, but it's probably unnecessary. Cheers. :) May 8, 2012 at 1:49
• You really went to a lot of pains to do this Macro. When I first saw the post about a minute ago I was thinking of showing the bias using Jensen's rule but someone already did it. May 8, 2012 at 2:23
• of course this is a round-a-bout way to show that the standard deviation is biased - I was mainly answering the original poster's second question: "How does one compute the expectation of the standard deviation?". May 8, 2012 at 2:24
• Another point perhaps worth mentioning is that this calculation allows one to read off immediately what the UMVU estimator of the standard deviation is in the Gaussian case: One simply multiplies $s$ by the reciprocal of the scale factor that appears in the proof. This generalizes to UMVU estimators of $\sigma^k$ fairly readily. May 8, 2012 at 11:42
• Sorry, Macro. The same basic integral approach you've used will work, you'll just end up with a different scaling factor of $s^k$, with the gamma arguments you get being functions of $k$. That's what I meant, but it came out a bit too terse. :) May 8, 2012 at 15:13

You don't need normality. All you need is that $$s^2 = \frac{1}{n-1} \sum_{i=1}^n(x_i - \bar{x})^2$$ is an unbiased estimator of the variance $\sigma^2$. Then use that the square root function is strictly concave such that (by a strong form of Jensen's inequality)
$$E(\sqrt{s^2}) < \sqrt{E(s^2)} = \sigma$$ unless the distribution of $s^2$ is degenerate at $\sigma^2$.

• Does this assume that $s^2 \neq 0$ and therefore $s^2 > 0$? The strong form of Jensen's inequality would only hold if square root was strictly convex but we lose the strictness if $s^2 = 0$. I assume that's an adequate assumption since $s^2 = 0$ only if we're dealing with a constant, in which case it is obvious that $s = \sigma$? May 5, 2021 at 0:49
• @David you are right that we need $\sigma > 0$. I believe I avoided $\sigma = 0$ by stating that $s^2$ should not have a degenerate distribution. If $\sigma = 0$, the distribution of $s^2$ is degenerate in 0. I guess that with independent variables, this is the only case where $s^2$ is degenerate. It's not really a problem if $s^2 = 0$ occurs with positive probability.
– NRH
May 11, 2021 at 9:59

Complementing NRH's answer, if someone is teaching this to a group of students who haven't studied Jensen's inequality yet, one way to go is to define the sample standard deviation $$S_n = \sqrt{\sum_{i=1}^n\frac{(X_i-\bar{X}_n)^2}{n-1}} ,$$ suppose that $S_n$ is non degenerate (therefore, $\mathrm{Var}[S_n]\ne0$), and notice the equivalences $$0 < \mathrm{Var}[S_n] = \mathrm{E}[S_n^2] - \mathrm{E}^2[S_n] \;\;\Leftrightarrow\;\; \mathrm{E}^2[S_n] < \mathrm{E}[S_n^2] \;\;\Leftrightarrow\;\; \mathrm{E}[S_n] < \sqrt{\mathrm{E}[S_n^2]} =\sigma.$$

This is a more general result without assuming of Normal distribution. The proof goes along the lines of this paper by David E. Giles.

First, we consider Taylor's expanding $$g(x) = \sqrt{x}$$ about $$x=\sigma^2$$, we have $$g(x) = \sigma + \frac{1}{2 \sigma}(x-\sigma^2) - \frac{1}{8 \sigma^3}(x-\sigma^2)^2 + R(x),$$ where $$R(x) =- \left(\frac{1}{8 \tilde \sigma^3} - \frac{1}{8 \sigma^3}\right)(x-\sigma^2)^2$$ for some $$\tilde \sigma$$ between $$\sqrt{x}$$ and $$\sigma$$.

Let $$\kappa = E(X - \mu)^4 / \sigma^4$$ be the kurtosis. It could be shown that $$E\left[\sqrt{n}(S_n^2 - \sigma^2)\right]^2 \rightarrow \sigma^4(\kappa-1)$$ and $$n ER(S_n^2) \rightarrow 0$$ (and the proofs are beyond the discussion of this thread. See for example CLT that states that $$\sqrt{n}(S_n^2 - \sigma^2)$$ converges to $$N(0, \sigma^4(\kappa-1))$$).

Thus, $$E(S_n) = Eg(S_n^2) =\sigma + \frac{1}{2 \sigma} E(S_n^2 - \sigma^2) - \frac{1}{8\sigma^3} E(S_n^2 - \sigma^2)^2 + o(n^{-1}).$$ $$= \sigma - \frac{\sigma}{8}\left[ \frac{\kappa - 1}{n}\right] + o(n^{-1}).$$

For normal distribution, setting $$\kappa = 3$$ gives the first order bias $$-\frac{\sigma}{4n}$$ as shown above.