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I am very confused dealing $dW(t)$, what is it? $W(t)$ is nowhere differentiable, we cannot write $W'(t)~dt$, but $dW(t)$ is a notation often used in my professor's notes.

Edward Wang
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    Note that $W(t)$ has quadratic variance meaning the process does not deviate much from $y=\sqrt{t}$. So we can write $(dW(t))^2 = dt $ – Btzzzz Nov 26 '16 at 20:56
  • The usual interpretation of $dW(t)dW(t)=dt$ is the quadratic variation $W,W=t$. Does anybody know what they could mean with $t_1$ and $t_2$? – parsiad Nov 26 '16 at 21:44
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    Contrarily to what is sometimes asserted, it is not recommended to use notations such as $(dW(t))^2=dt$ if one wants to get any understanding of the subject. Re the notation $dW(t)$ itself, the whole object of Itô's calculus is to give a rigorous meaning to so-called stochastic integrals $$\int_0^tX_sdW(s)$$ for suitable stochastic processes $(X_t)$ eventhough, as you note, the paths $t\mapsto W(t)$ are almost surely non differentiable. – Did Nov 26 '16 at 21:44
  • @EdwardWang Indeed, would you have examples of sources using the notation $dW(t_1)dW(t_2)$? – Did Nov 26 '16 at 21:45
  • @parsiad No idea why your answer was downvoted. – Did Nov 26 '16 at 21:46
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    @Did: I would guess it was because I did not explicitly discuss $dW(t_1)dW(t_2)$. I will undelete it (in case it is useful to OP) and bear the brunt of the downvote :-) – parsiad Nov 26 '16 at 21:50
  • @Did $dW_1(t)dW_2(t) = 0$ oh sorry, my professor means the cross variation, or quadratic covariance – Edward Wang Mar 15 '17 at 20:11

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Whenever you see an SDE of the form $$ dX_{s}=a(s,X_{s})ds+b(s,X_{s})dW(s), $$ just remember that this is simply a "short form" for $$ X_{s}=x+\int_{t}^{s}a(u,X_{u})du+\int_{t}^{s}b(u,X_{u})dW(u) $$ where $x$ is the initial condition at time $t$.

That is, the only place $dW(u)$ appears is in an Ito integral, whose definition you might already be familiar with. No derivatives are considered.

Addendum: While $dW(t)dW(t)=dt$ is just short form to describe the quadratic variation of the process $[W,W](t)=t$, I am not sure what the notation $dW(t_1)dW(t_2)$ could mean.

parsiad
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  • @downvoter Why the downvote? The answer addresses the part of the question that one can understand. – Did Nov 26 '16 at 21:56