Wiener Process $dB^2=dt$

Question

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.

Answer 1

For independent random variables, the variance of the sum equals the sum of the variances. So $\mathbb{E}((\Delta B)^2)=\Delta t$, i.e. if you increment $t$ a little bit, then the variance of the value of $B$ before that increment plus the variance of the increment equals the variance of the value of $B$ after the increment.

Or you could say $$ \frac{\mathbb{E}((\Delta B)^2)}{\Delta t} = 1. $$ That much follows easily from the first things you hear about the Wiener process. I could then say "take limits", but that might be sarcastic, so instead I'll say that for a fully rigorous answer, I'd have to do somewhat more work.

Answer 2

Obviously $dB_t^2 \neq dt$, since $dB_t \sim \mathcal{N} (0, dt)$ is a random variable, while $dt$ is deterministic.

As Michael Hardy said, they really meant to say $\mathbb{E} \left[ dB_t^2 \right] = dt$. To convince yourself, compute $$ \mathbb{E} \left[ dB_t^n \right] = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi dt}} \exp\left(-\frac{x^2}{2 dt}\right) x^n dx \, .$$