Why volatility scales with the square root of time

Question

Daily returns X1, \dots, Xn are independent with mean \mu and standard deviation \sigma = 1\% per day. Over a year of n = 252 trading days, consider the total return S = \sum{i=1}^n Xi (not the average). What are the variance and standard deviation of the total? What annual volatility does that imply?

Accepted Answer

Variance of a sum. For independent variables, variances add: \operatorname{Var}(S) = \sum{i=1}^n \operatorname{Var}(Xi) = n\sigma^2. So the standard deviation of the total scales with \sqrt{n}: 	ext{SD}(S) = \sigma\sqrt{n}. Annual number. With \sigma = 1\% and n = 252, 	ext{SD}(S) = 1\% 	imes \sqrt{252} \approx 1\% 	imes 15.87 = 15.9\%. So a 1\% daily volatility implies roughly 16\% annual volatility. Notice the mean of the sum grows linearly (n\mu) while the standard deviation grows only as \sqrt{n}, which is why a Sharpe-like signal-to-noise ratio improves with \sqrt{n}. Contrast with t…

Why volatility scales with the square root of time

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