Random Walk with normal increments and n time periods why is the increment sqrt$\sqrt{(t/n)}$?

Question is basically in the title. I have found several sourcessources stating that Ri = sqrt(t/n)$R_i = \sqrt{\frac{t}{n}}$, but I couldn't find the intuition behind taking the square root. And it seems to be crucial since E(Ri^2)= t/n$\operatorname{E}\left[{R_i^2}\right]= \frac{t}{n}$ and from there derive the variance of the Brownian motion as being t$t$.

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asked Dec 23, 2018 at 23:14

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Random Walk with normal increments and n time periods why is the increment sqrt(t/n)?

Question is basically in the title. I have found several sources stating that Ri = sqrt(t/n), but I couldn't find the intuition behind taking the square root. And it seems to be crucial since E(Ri^2)= t/n and from there derive the variance of the Brownian motion as being t.

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Random Walk with normal increments and n time periods why is the increment sqrt$\sqrt{(t/n)}$?

Random Walk with normal increments and n time periods why is the increment sqrt(t/n)?

Random Walk with normal increments and n time periods why is the increment $\sqrt{(t/n)}$?

Random Walk with normal increments and n time periods why is the increment sqrt(t/n)?