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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.

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What do you mean by every book? Could you list a couple? – cardinal Nov 14 '11 at 2:58
My mistake. I meant every online source I've come across from googling. This was stated in a Mathematical Finance class without justification, and I've been spending hours trying to figure out how this comes about. – j.diddland Nov 14 '11 at 6:13
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Sorry. The point of my question, which may not have been clear, was to get a feel for the level at which you were expecting an answer. What textbook does the course use? Do you know about quadratic variation? At the level of say, S. Shreve, Stochastic Calculus for Finance or, maybe, Karatzas & Shreve, Brownian Motion and Stochastic Calculus? Or, maybe, the course is more at the level of J. C. Hull, Options, Futures, and Other Derivatives? Providing this kind of info will help me or someone else provide an answer at the appropriate level. Cheers. :) – cardinal Nov 14 '11 at 9:43

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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 $$ \mathbb{E} \left[ dB_t^n \right] = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi dt}} \exp\left(-\frac{x^2}{2 dt}\right) x^n dx \, .$$

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Sorry but the nonrigorous shorthand $dB_t^2=dt$ refers to a much deeper result than the (true) fact that $B_{t+s}-B_t$ is centered with variance $s$, which your answer reduces to. As such, one may find said answer rather misleading. Or, since the question is badly formulated and the OP never answered @cardinal's fully justified demand for background, one can also consider all this rather moot... – Did Dec 11 '12 at 6:13
Certainly not. First, because neither $E(dB_t^4)$ nor $dt^2$ are well defined objects. Second, and even more importantly, because what the shorthand $dB_t^2=dt$ refers to is the whole class of Doob's semimartingale decompositions which Itô's formula provides (for example, to stay at the level of a toy example, the fact that $t\mapsto B_t^2-t$ is a martingale, which is not reducible to the fact that $E(B_t^2)=t$). – Did Dec 11 '12 at 6:44
I am aware of a fairly rigorous analysis that shows a term containing $dB_t^2$ in Ito's lemma, for example, converges to one that contains $dt$ almost surely, using the Borel-Cantelli lemma. See pages 4 to 6 in the lecture math.nyu.edu/faculty/goodman/teaching/StochCalc2012/notes/… . – William S. Wong Dec 11 '12 at 6:47
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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.

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Sorry but the nonrigorous shorthand $dB_t^2=dt$ refers to a much deeper result than the (true) fact that $B_{t+s}-B_t$ is centered with variance $s$, which your answer reduces to. As such, one may find said answer rather misleading. Or, since the question is badly formulated and the OP never answered @cardinal's fully justified demand for background, one can also consider all this rather moot... – Did Dec 11 '12 at 6:13

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