# Why can we neglect the mean in the variance when the time step is very small?

Can anyone tell me why we can neglect the mean in the variance when the time step is very small? See the following picture: Usually, we choose a time step of one day. Is it small enough?

• 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. – Will Gu Feb 21 '17 at 18:23
• 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! – fni Mar 4 '17 at 11:24
• @fnic yes that makes sense. – A.Oreo Mar 5 '17 at 6:10

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 $N$.
• 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 $\sqrt{N}\gg N$. – Richard Hardy Feb 21 '17 at 18:16
• Yes, exactly. I just meant something like $\sqrt{0.0001}=0.01>0.0001$ – fni Feb 22 '17 at 0:30
• @RichardHardy but the time step of $\bar R$ is same as $R_j,$ if $\bar R$ is so small that neglected, why doesn't $R_j?$ – A.Oreo Feb 22 '17 at 1:39
• $\bar R$ is as small as $R_j,$ why not neglect $R_j?$ – A.Oreo Feb 22 '17 at 1:44