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

Question

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.

Answer 1

@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:

Answer 2

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 $\sigma^2$.

Answer 3

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 $$ 0 < \mathrm{Var}[S_n] = \mathrm{E}[S_n^2] - \mathrm{E}^2[S_n] \;\;\Leftrightarrow\;\; \mathrm{E}^2[S_n] < \mathrm{E}[S_n^2] \;\;\Leftrightarrow\;\; \mathrm{E}[S_n] < \sqrt{\mathrm{E}[S_n^2]} =\sigma. $$

Answer 4

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.