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Can anyone tell me why we can neglect the mean in the variance when the time step is very small? See the following picture:

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Usually, we choose a time step of one day. Is it small enough?

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  • $\begingroup$ I personally think that it only makes sense in practice. Mathematically, by the law of large numbers, $\bar R_n \rightarrow \mu$ when $n \rightarrow \infty$. Is the daily return $\mu = 0$? Not necessarily. $\endgroup$ – Will Gu Feb 21 '17 at 18:23
  • $\begingroup$ Think about it terms of a diffusion process $Return=dS/S = \mu dt + \sigma dW$, then you realize that $E[dS/S]=\mu dt$, i.e. the average return is proportional to the time interval $dt$, while $Var(dS/S)=E[(dS/S)^2] - E[dS/S]^2= \sigma dt + (\mu dt)^2$. You see immediately that as $dt \to 0$ the second term goes to zero much faster than the first! $\endgroup$ – fni Mar 4 '17 at 11:24
  • $\begingroup$ @fnic yes that makes sense. $\endgroup$ – A.Oreo Mar 5 '17 at 6:10
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The average return scales linearly with the time period, i.e. $R_N = N R_1$, while the standard deviation scales with the square root, i.e. $\sigma_N = \sqrt{N}\sigma_1$. As the period becomes really small, $\sqrt{N}$ becomes much bigger than $N$.

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  • $\begingroup$ Perhaps an even better exposition would involve using $\frac{1}{N}$ in place of $N$ (as if dividing a longer period into many short periods), because now you need $N\ll 1$ to achieve $\sqrt{N}\gg N$. $\endgroup$ – Richard Hardy Feb 21 '17 at 18:16
  • $\begingroup$ Yes, exactly. I just meant something like $\sqrt{0.0001}=0.01>0.0001$ $\endgroup$ – fni Feb 22 '17 at 0:30
  • $\begingroup$ @RichardHardy but the time step of $\bar R$ is same as $R_j,$ if $\bar R$ is so small that neglected, why doesn't $R_j?$ $\endgroup$ – A.Oreo Feb 22 '17 at 1:39
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it is because "R-bar" over 1 day is a very tiny number , virtually zero

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  • $\begingroup$ $\bar R$ is as small as $R_j,$ why not neglect $R_j?$ $\endgroup$ – A.Oreo Feb 22 '17 at 1:44

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