In "Volatility Trading" Euan Sinclair defines Yang-Zhang volatility estimator as

$$ \sigma = \sqrt{\sigma^2_o + k\sigma^2_c + (1-k)\sigma^2_{rs}} $$

where $$ \sigma^2_o \propto Variance\left(ln\left(\frac{o_i}{o_{i-1}}\right)\right) $$ $$ \sigma^2_c \propto Variance\left(ln\left(\frac{c_i}{c_{i-1}}\right)\right) $$ $$ \sigma^2_{rs} = \frac{1}{N} \sum_{i=1}^N \left( \left(ln \frac{h_i}{c_i}\right) \left(ln \frac{h_i}{o_i}\right) + \left(ln \frac{l_i}{c_i}\right) \left(ln \frac{l_i}{o_i}\right) \right) $$

/* I'm using $\propto$ symbol as "proportional to" to avoid unbiasing the $Variance$ via multiplying $Variance$ by $\frac{N}{N-1}$. See the actual formulas on the screenshot below in the References. */

However, TTR package 1 uses different formulas for $\sigma_o^2$, $\sigma_c^2$:

$$ \sigma^2_o \propto Variance\left(ln\left(\frac{o_i}{c_{i-1}}\right)\right) $$ $$ \sigma^2_c \propto Variance\left(ln\left(\frac{c_i}{o_{i}}\right)\right) $$

I plotted Garman-Klass, Parkinson, Yang-Zhang (TTR and Sinclair's) estimators on a chart:

chart with various volatility estimators

It shows how Sinclair's Yang-Zhang definition systematically deviates (and overestimates?) the volatility compared to the rest of the estimators.


Does Sinclair's formula have a typo?


TTR Yang-Zhang volatility estimator

  • Yang-Zhang volatility estimator from Sinclair's book: screenshot yang-zhang volatility from Sinclair's book

1 Answer 1


When in doubt, consult the original paper:

In the beginning of the paper, the authors describe the following definitions for the normalized open and close (p. 479):

\begin{align*} o&=\ln(O_1) - \ln(C_0) = \ln\left(\frac{O_1}{C_0}\right), \quad \text{normalized open;}\\ c&=\ln(C_1) - \ln(O_1)= \ln\left(\frac{C_1}{O_1}\right), \quad \text{normalized close.} \end{align*}

Furthermore in Section II they define their volatility estimator as (pp. 482 - 488):

\begin{align*} V &= V_O + k \cdot V_C + (1-k) \cdot V_{RS}\\ V_O&= \frac{1}{n-1}\sum_{i=1}^n (o_i - \bar{o})^2\\ V_C&= \frac{1}{n-1}\sum_{i=1}^n (c_i - \bar{c})^2\\ \bar{o}&=\frac{1}{n}\sum_{i=1}^n o_i\\ \bar{c}&=\frac{1}{n}\sum_{i=1}^n c_i,\\ \end{align*} where $V_{RS}$ is derived later in the paper. Writing out one of the variance measures — using the notation from the original paper — gives you a clear indication that the TTR package has defined the Yang-Zhang estimator as originally intended: \begin{align*} V_O&= \frac{1}{n-1}\sum_{i=1}^n (o_i - \bar{o})^2\\ &=\frac{1}{n-1}\sum_{i=1}^n \left(o_i - \frac{1}{n}\sum_{i=1}^n o_i\right)^2\\ &=\frac{1}{n-1}\sum_{i=1}^n \left(\ln\left(\frac{O_i}{C_{i-1}}\right) - \frac{1}{n}\sum_{i=1}^n \ln\left(\frac{O_i}{C_{i-1}}\right)\right)^2\\ &=V_O^{\text{TTR}}. \end{align*} You can do the same derivation for $V_C$. Yes, I believe Sinclair has a few mistakes in his book.

  • 1
    $\begingroup$ The quality of questions and answers here is unmatched. Thank you Pleb! $\endgroup$ Commented Aug 2, 2022 at 18:57

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