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?

  • 1
    $\begingroup$ The unbiased estimator of $\sigma$ is not the square root of the unbiased estimator of $\sigma^2$. Ask Jensen. $\endgroup$
    – Xi'an
    Apr 15, 2020 at 14:46
  • 1
    $\begingroup$ See stats.stackexchange.com/a/27984/919. $\endgroup$
    – whuber
    Apr 15, 2020 at 15:38

1 Answer 1


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}


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