Annualizing Volatility with the Square-Root-of-Time Rule

Summary

The document raises a practical question about scaling volatility across time horizons with the square-root-of-time rule. It notes that this relationship is commonly used and can be derived under a log-normal assumption, then contrasts volatility scaling with mean scaling, which the author sees as less sensitive to the horizon.

The author reports that, in their sample, changing the horizon produced roughly a twofold difference in volatility estimates, using either simple or log returns. They also assert that the rate of change in scaled volatility exceeds that of scaled mean for positive horizons. No answer, derivation, dataset description, or practical validation is provided, so the document does not establish whether this sensitivity is expected or whether the rule is suitable in a particular application. Its value is in identifying a scaling assumption that deserves scrutiny when comparing or annualizing estimates across horizons.

Key ideas

• The square-root-of-time rule is presented as a common method for scaling volatility across horizons.
• The author connects the rule to a log-normal assumption.
• The document claims volatility scaling is more sensitive to the horizon than mean scaling.
• A roughly twofold volatility difference across horizons is reported for the author's sample.
• The question receives no answer, and the sample and practical suitability are not established.

Tags

# Is $\sigma_{1} = \frac{\sigma_{\tau}}{\sqrt{\tau}}$ suitable for volatility scaling?


# Is $\sigma_{1} = \frac{\sigma_{\tau}}{\sqrt{\tau}}$ suitable for volatility scaling?












It seems to be the de-facto method; and I see how we get it from log-normal assumption.

However volatility scaling seems to be way more sensitive to $\tau$ than mean scaling -- as in two ~ 2 times (be it using simple or log returns) in my sample.

And it indeed it is: \begin{equation} \frac{d}{d\tau}\big(\sigma_{1}(\tau)\big) > \frac{d}{d\tau}\big(\mu_{1}(\tau) \big)\hspace{2.5 mm} \mid \hspace{2.5 mm} \tau > 0 \end{equation}

But is this good for practical use?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.