Asymptotic Standard Errors for Volatility Estimators
Summary
The document examines the large-sample uncertainty of a variance estimator based on squared deviations from the sample mean. It presents an approximation in which the variance of that estimator depends on the fourth central moment, summarized through kurtosis, and asks how this translates into the standard error of the standard deviation estimate. The central issue is the factor of one half that appears when moving from variance to standard deviation.
The author distinguishes the standard error of the variance estimate from that of its square root and questions whether the cited result follows by a first-order approximation or reflects a typo. No resolution is included, so the post does not establish which interpretation is correct. Its useful point is that uncertainty for a nonlinear transformation of an estimator generally requires a separate calculation; the displayed approximation also depends on a large sample and a finite fourth moment.
Key ideas
- The variance estimator’s asymptotic uncertainty depends on the fourth central moment of the data.
- Kurtosis affects the estimated variance’s standard error.
- The standard error of a standard deviation estimate is distinct from that of a variance estimate.
- A first-order approximation is relevant when transforming variance uncertainty into standard deviation uncertainty.
- The post raises the factor-of-one-half issue but does not resolve it.
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# Comparing standard error asymptotics of standard deviation and mean absolute deviation estimators
# Comparing standard error asymptotics of standard deviation and mean absolute deviation estimators
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 $\sqrt{Var(\sqrt{\frac{1}{N} \sum_i (X_i - m_e)^2)}}$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.