2
$\begingroup$

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?

$\endgroup$
  • 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 at 14:46
  • 1
    $\begingroup$ See stats.stackexchange.com/a/27984/919. $\endgroup$ – whuber Apr 15 at 15:38
4
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.