# Unbiased estimator of standard deviation

I'm reading "Properties of range-based volatility estimators" where the authors talk about using the range of a distribution ($$h$$ - $$l$$) to estimate its volatility. Specifically, they say,

Daily return $$c$$ is a random variable drawn from a normal distribution with zero mean and variance (volatility) $$\sigma^{2}$$:

$$c ∼ N(0, σ^{2})$$

Our goal is to estimate (unobservable) volatility $$\sigma^{2}$$ from observed variables $$c$$, $$h$$ and $$l$$. Since we know that $$c^{2}$$ is an unbiased estimator of $$\sigma^{2}$$,

$$E[c^{2}] = \sigma^{2}$$

we have the first volatility estimator (subscript $$_{s}$$ stands for ”simple”) $$\widehat{\sigma^{2}_{s}} = c^{2}$$

They then go on to say: As can be easily proved, an unbiased estimator $$\hat{\sigma_{s}}$$ of the standard deviation $$\sigma$$ based on $$\sqrt{\widehat{\sigma^{2}_{s}}}$$ is:

$$\hat{\sigma_{s}} = \sqrt{\widehat{\sigma_{s}^{2}}} \times \sqrt{\pi/2} = |c| \times \sqrt{\pi/2}$$

My question is where does the $$\sqrt{\pi / 2}$$ come from?

• The unbiased estimator of $\sigma$ is not the square root of the unbiased estimator of $\sigma^2$. Ask Jensen. Apr 15, 2020 at 14:46
• – whuber
Apr 15, 2020 at 15:38

While $$\mathbb E_\sigma[c^2]=\sigma^2$$, \begin{align} \mathbb E_\sigma[|c|] &= \int_0^\infty \sqrt{2/\pi}\, \sigma^{-1} x\, \exp\{-x^2/2\sigma^2\}\,\text{d}x\tag{symmetry}\\ &= \sigma\int_0^\infty \sqrt{2/\pi}\, y\, \exp\{-y^2/2\}\,\text{d}y\tag{y=\sigma x}\\ &= \sigma\int_0^\infty \sqrt{2/\pi}\, z^{1/2}\, \exp\{-z/2\}\,\frac{z^{-1/2}}{2}\text{d}z\tag{z=y^2}\\ &= \sigma\sqrt{1/2\pi}\int_0^\infty \exp\{-z/2\}\,\text{d}z\\ &= \sigma\sqrt{1/2\pi}\,2\,[-\exp\{-z/2\}]_0^\infty = \sigma\sqrt{2/\pi} \end{align}