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This is math finance question. I'm trying to figure out the formula for converting the standard deviation of % changes in prices for a sample of days into the standard deviation of % changes in prices annualized.

Here is an excerpt: "Historical statistical volatility is calculated as follows: Daily returns are calculated as return = natural log of (p2/p1). The standard deviation of this return is then calculated for the last 21, 42, 63, 126, 252, 504, and 756 days, corresponding to an average trading month, two months, three months, six months, one year, two years, and three years. The standard deviation is then annualized by multiplying by the square root of (252/number of trading days)"

Why do we multiply by the square root of (252/number of trading days)? I don't know how to move on from here?

I know that standard deviation = sqrt(variance). But why do we multiply by the sqrt(252/number of trading days) and not just by 252/number of trading days.

Jwan622
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1 Answers1

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If return of day $i$ is $r_i$, $r_i$ is independently and identically distributed (a common assumption) across $i$ and standard deviation of $r_i$ is $\sigma$, then variance of $n$ day return, $r_1+r_2+\cdots+r_n$, equals expected value of $\{(r_1+r_2+\cdots+r_n)-E[r_1+r_2+\cdots+r_n]\}^2$. The expression can be expanded to $n$ square terms of the form $(r_i-E[r_i])^2$ each of which has expected value of $\sigma^2$ and $n(n-1)/2$ cross-terms of the form $(r_i-E[r_i])(r_j-E[r_j])$, each of which has expected value of $0$ because of independence. So variance of $r_1+r_2+\cdots+r_n$ equals $n\sigma^2$ and standard deviation equals $\sigma\sqrt{n}$. So the ratio of standard deviation of annual return to standard deviation of n-day return is $\dfrac{\sigma\sqrt{252}}{\sigma\sqrt{n}}=\sqrt{252/n}$.

Michael Hardy
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user96614
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