I was reading Chapter 4 of Jean-Philippe Bouchaud's book "Theory of Financial Risk and Derivative Pricing: From Statistical Physics to Risk Management" and in section 4.2.2 author was comparing standard errors of standard deviation and mean absolute deviation estimators. Given $N$ samples $X_1, ..., X_N$, let $\sigma_e^2$ be an estimator of the variance, i.e. $$\sigma_e^2 = \frac{1}{N} \sum_{i=1}^N (X_i - m_e)^2$$ where $m_e$ is a mean estimator, i.e. $m_e = \frac{1}{N} \sum_{i=1}^N X_i$.

Then the author procedes (where I think he made a typo) to approximate (when $N$ is large) the standard error of $\sigma_e^2$ as follows:

\begin{aligned} [\Delta (\sigma_e^2)] [\Delta (\sigma_e^2)]^2 &= \text{Var}\left(\frac{1}{N} \sum_{i=1}^N (X_i - m_e)^2\right) \\ &\approx \text{Var}\left(\frac{1}{N} \sum_{i=1}^N (X_i - m)^2\right) \\ &= \frac{\langle (X_1-m)^4 \rangle - \langle (X_1-m)^2 \rangle^2}{N} \\ &= \frac{\sigma^4}{N}(2 + \kappa) \end{aligned}

Then comes the confusing part: he then says that $$ \frac{\Delta \sigma_e}{\sigma} = \frac{1}{2 \sqrt{N}}\sqrt{2 + \kappa} \approx \frac{1}{\sqrt{2N}} \left(1 + \frac{1}{4}\kappa\right) $$ Although I agree with the second approximation when $\kappa$ is small, why does the first equation hold? We do have $$ \frac{\Delta\sigma_e^2}{\sigma^2} = \frac{\sqrt{2 + \kappa}}{\sqrt{N}} $$ from above, but where does an extra $\frac{1}{2}$ come from when you look at $\Delta{\sigma_e}$ instead of $\Delta{\sigma_e^2}$? Is it simply a typo or am I overlooking an obvious approximation?

Remark: if you also found the notation here confusing, from my understanding, $\Delta \sigma_e^2$ is standard error for the ("typical") estimator of $\sigma^2$, which is $\sqrt{Var(\frac{1}{N} \sum_i (X_i - m_e)^2)}$, while $\Delta \sigma_e$ here is standard error for the estimator of $\sigma$, which is $\sqrt{Var(\sqrt{\frac{1}{N} \sum_i (X_i - m_e)^2)}}$.

  • $\begingroup$ Any idea why there is a subscript $_e$ (when working with $X$)? $\endgroup$ Mar 22 at 11:39
  • 1
    $\begingroup$ subscript e is used for estimators $\endgroup$
    – pierce
    Mar 22 at 12:09


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