TITLE: Wiener Process $dB^2=dt$
QUESTION [21 upvotes]: Why is $dB^2=dt$? Every online source I've come across lists this as an exercise or just states it, but why isn't this ever explicitly proved? I know that $dB=\sqrt{dt}Z$, but I don't know what squaring a Gaussian random variable means.

REPLY [2 votes]: For convenience we will informally regard Brownian motion as a random walk over infinitesimal time intervals of length $\Delta t$, whose increments
$$
\Delta B_{t}:=B_{t+\Delta t}-B_{t} \simeq \mathcal{N}(0, \Delta t)
$$
over the time interval $[t, t+\Delta t]$ will be approximated by the Bernoulli random variable
$$
\Delta B_{t}=\pm \sqrt{\Delta t}
$$
with equal probabilities $(1 / 2,1 / 2)$.
The choice of the square root in $(4.2)$ is in fact not fortuitous. Indeed, any choice of $\pm(\Delta t)^{\alpha}$ with a power $\alpha>1 / 2$ would lead to explosion of the process as $d t$ tends to zero, whereas a power $\alpha \in(0,1 / 2)$ would lead to a vanishing process.
Note that we have
$$
\mathbb{E}\left[\Delta B_{t}\right]=\frac{1}{2} \sqrt{\Delta t}-\frac{1}{2} \sqrt{\Delta t}=0
$$
and
$$
\operatorname{Var}\left[\Delta B_{t}\right]=\mathbb{E}\left[\left(\Delta B_{t}\right)^{2}\right]=\frac{1}{2} \Delta t+\frac{1}{2} \Delta t=\Delta t
$$
According to this representation, the paths of Brownian motion are not differentiable, although they are continuous by Property 2, as we have
$$
\frac{d B_{t}}{d t} \simeq \frac{\pm \sqrt{d t}}{d t}=\pm \frac{1}{\sqrt{d t}} \simeq \pm \infty
$$
After splitting the interval $[0, T]$ into $N$ intervals

Notes from: http://www.ntu.edu.sg/home/nprivault/MA5182/brownian-motion-stochastic-calculus.pdf