# 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 $$R_i = \sqrt{\frac{t}{n}}$$, 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 $$t$$.

• 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. – mark leeds Dec 24 '18 at 2:17

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 $$t$$.
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 $$QV \to t$$