Why Brownian Motion Steps Scale with the Square Root of Time
Summary
The discussion explains why a normal random-walk increment over a time step of length t/n is scaled by the square root of t/n. Each increment can be written as a standard normal random variable multiplied by that scale, so its variance is t/n. For independent increments, the variances add across n steps, giving total variance t over the full interval, while the expected value remains zero. This matches the defining variance behavior of Brownian motion as the step size shrinks. A second intuition invokes quadratic variation: Brownian paths have unbounded total variation, but their accumulated squared increments converge to elapsed time. The answers offer a concise conceptual derivation rather than a full proof of convergence or discussion of assumptions such as independence. The explanation is useful for understanding diffusion scaling in continuous-time models, but does not address drift terms or applications to asset pricing directly.
Key ideas
- A normal increment over a time step of length t/n is scaled by the square root of that duration.
- The scaling makes the variance of each increment equal to t/n.
- For independent increments, variances sum to t across n steps, consistent with Brownian motion.
- Brownian motion has unbounded total variation but quadratic variation equal to elapsed time.
- The explanation omits a formal convergence proof and does not cover drift.
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# 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 $\sqrt{(t/n)}$?
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 $t$.
## Answer by mark leeds (score 5)
https://quant.stackexchange.com/a/43224
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 $t$.
## Answer by alexbougias (score 1)
https://quant.stackexchange.com/a/43200
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 $QV \to t$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.