Estimating Annualized Volatility from Absolute Returns

Summary

The document asks whether daily, weekly, or monthly returns can stand in for standard deviation when annualizing volatility. It contrasts square-root-of-time scaling of standard deviation with the relationship between expected absolute returns and standard deviation under a normal-return assumption. Under that assumption, expected absolute return is a constant fraction of standard deviation, so an estimate based on mean absolute returns needs that conversion before annualization.

The discussion is a question rather than a resolved treatment. It does not establish that an observed return equals expected absolute return or volatility, and it gives no data or empirical validation. The conversion depends on the assumed return distribution and concerns an expectation; realized returns vary. Annualization by the square root of time also relies on assumptions about how returns aggregate, including the absence of material dependence or changing volatility.

Key ideas

• Square-root-of-time scaling applies to standard deviation under appropriate aggregation assumptions.
• Under normally distributed returns, expected absolute return is a fixed fraction of standard deviation.
• Mean absolute returns require a distribution-based conversion before using them as a volatility estimate.
• A single observed return is not the same quantity as expected absolute return or standard deviation.

Tags

# returns, standard deviation and mean absolute deviation


# returns, standard deviation and mean absolute deviation












I'm trying to understand the relationships between return, standard deviation and mean absolute deviation.

I saw someone mention:

$daily return * 16 \approx annualized volatility$

$weekly return * 7.2 \approx annualized volatility$

$monthly return * 3.5 \approx annualized volatility$

So he assumes the returns are equal to the standard deviation, because:

$\large \sigma_d\sqrt{252} = \sigma_a$

$\large \sigma_w\sqrt{52} = \sigma_a$

$\large \sigma_m\sqrt{12} = \sigma_a$

Is this correct? Because:

$\large E(|R_t|) = \sqrt{\frac{2}{\pi}}\sigma_t$

Which would result in:

$\large\frac{E(|R_d|)}{\sqrt{\frac{2}{\pi}}}\sqrt{252} = \sigma_a$

$\large\frac{E(|R_w|)}{\sqrt{\frac{2}{\pi}}}\sqrt{52} = \sigma_a$

$\large\frac{E(|R_m|)}{\sqrt{\frac{2}{\pi}}}\sqrt{12} = \sigma_a$

Am I wrong in considering the above mentioned returns as $\large E(|Rt|)$ instead of $\large \sigma_t$ ?

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