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mathdude2018
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Proof for square-root-of-time rule

February 19th, 2020, 11:17 pm

Is there a proof for square root of time scaling for variance ?

i.e. if I have a daily stock price time series, I get the daily standard deviation by taking standard deviation of the price returns. Now to annualize it we multiply by sqrt(365) or sqrt(252). But what is the proof for this?

Edit: Is there a rigorous proof for this ?
Last edited by mathdude2018 on February 20th, 2020, 1:56 am, edited 1 time in total.
 
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bearish
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Re: Proof for square-root-of-time rule

February 20th, 2020, 12:32 am

Loosely, if daily increments are iid then the variance of the cumulative return grows linearly in time, since you are adding an independent random variable each day with the same variance. Since the standard deviation is the square root of variance, it will grow as the square root of the number of days (i.e. time). 
 
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katastrofa
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Re: Proof for square-root-of-time rule

February 20th, 2020, 6:25 am

In short, Variance = t for a Brownian motion.
 
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Alan
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Re: Proof for square-root-of-time rule

February 24th, 2020, 3:56 pm

@mathdude,
 bearish's iid argument is rigorous as long as you interpret "returns" as [$]x_t = \log S_t/S_{t-1}[$], and the variance of [$]x_t[$] exists.
 
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tags
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Re: Proof for square-root-of-time rule

May 2nd, 2020, 8:56 am

Depending on the degree of rigor you require, maybe also have a look at II.3.2.1 Volatility and the Square-Root-of-Time Rule in Carol Alexander's Practical Financial Econometrics.