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 $t$.

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 $t$.