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

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  • $\begingroup$ hamilton's text had a nice description of going from discrete random walk to brownian motion. So, check that out. It's essentially for the reason you mentioned : In the discrete case ( RW case), you need that for the RW to be consistent with brownian motion in the limit as the timestep goes towards zero. $\endgroup$ – mark leeds Dec 24 '18 at 2:17
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I don't think I was clear in my comment so I'm putting it in an answer to have more space. The variance of a brownian motion, z, is $t$. (i.e: $E(z^{2}) = t$ ). Notice that $R_{i}$ really equals $\sqrt{\frac{t}{n}} \times \epsilon$ where $\epsilon \sim N(0,1)$. I think they leave the $\epsilon$ out because the variance is 1 but showing the consistency is clearer if we define it that way.

By definition, brownian motion is defined as the sum of a bunch of random walks as the step size goes to zero. So, given the definition of $R_{i}$, you end up with the variance of the sum, being $\sum_{i = 1}^{n} \frac{t}{n} = n \times \frac{t}{n} = t$. Therefore, in the limit, the sum of the n random walks is consistent with brownian motion as the step size goes to zero because the expected value is still zero and the variance is $t$.

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My intuition is that the term $\sqrt{dt}$, is a result of Brownian Motion's properties. Total variation of the Brownian Motion is unbounded and therefore $TV \to \infty$. However the Quadratic Variation is bounded and $QV \to t$

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  • $\begingroup$ Can you explain maybe further what you mean by a result of the Brownian Motion's properties? Is there are proof that the increments are sqrt(dt)? Or what exactly is the link to the quadratic variation? $\endgroup$ – 303 Waters Please Dec 24 '18 at 23:19
  • $\begingroup$ I think you were asking Coxswaiiin but I changed my comment to an answer and that may answer the first part of your question. As far as the second part,.I don't think the quadratic variation plays a role in the OP's question. $\endgroup$ – mark leeds Dec 27 '18 at 15:53

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