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.
Wiener Process $dB^2=dt$
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stochastic-processes
stochastic-calculus
brownian-motion
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2What do you mean by *every book*? Could you list a couple? – 2011-11-14
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0My 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. – 2011-11-14
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1Sorry. 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. :) – 2011-11-14
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0I think the answer to my bounty clarification request is that a simple calculation shows that the standard deviation of $dB^2$ is actually of the order of $dt^{3/2}$, while its expectation is of the order of $dt$. So the randomness can be ignored. – 2013-09-26