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If $(B_t)$ is a Brownian motion, we say formally that $dB_t=(dt)^{1/2}$. And then we use the formal notation $dt\cdot dB_t=dt\cdot dt=0$ and $dB_t\cdot dB_t=dt$.

Q1) Is there a rigorous reason for that or is it only a convention ? In general, if $\alpha >1$, why (dt)^\alpha =0$ ? Can we prove this ?

Q2) I know that the fact that $(B_t)$ has no bounded variation, but only quadratic variation gives us that $dB_t=(dt)^{1/2}$. But how can we justify this fact rigorously ?

This question is quite correlated to my other question here. But since the question is not the same, I prefered to open a new question.

user649261
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  • I think you should apply Ito's Formula to find out :) – Statistic Dean Mar 01 '19 at 10:09
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    A low-tech answer for Q1 would be that, for $\alpha > 1$, $$\lim_{|P|\to0}\sum_{\xi_i \in [x_{i-1},x_i]\in P} f(\xi_i) (x_i - x_{i-1})^{\alpha} \to 0$$ for any continuous function $f$ on $[a, b]$ and along partitions $P$ of $[a, b]$ whose mesh sizes go to $0$. – Sangchul Lee Mar 01 '19 at 10:15
  • Why the downvote ? – user649261 Mar 01 '19 at 10:19
  • For me, $df$ refer to the linear approximation of $f$. In other word, $d_xf=K(x)dt$ would be the linear approximation of $f$ arround $x$. In particular, if $\alpha >1$, then $f(x+dt)=f(x)+d_xf+o(dt)+J(x)dt^\alpha =f(x)+d_xf+o(dt).$ – Surb Mar 01 '19 at 10:23
  • @saz: the link doesn't work... – user649261 Mar 01 '19 at 13:23
  • For the very first part of your first question you might want to take a look at this question (and the linked questions as well). – saz Mar 01 '19 at 13:25

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