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Following this question, proving the equivalence between equation $(1)$ and $(2)$, I deduced that $$d\langle X \rangle_t = d X_t dX_t$$ (where $X_t$ was an Ito's process, hence a semimartingale).

I would like to discover why the equation above holds.

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  • How do you define (rigorously) the right-hand side? See also this question. – saz Jan 17 '16 at 07:38
  • @saz I'm not a mathematician (as you may have noticed), so I don't know how rigorous this is, but I took the definition of $dX_t$ as an Ito's drift-diffusion process ($dX_t = \mu(t, X_t)dt + \sigma(t, X_t) , dW_t $) and I squared it. From there I proceeded as shown in the question in your link. But even if the equation in my question seems intuitively right, I can't understand the mathematical steps to prove it – Puzzle Jan 17 '16 at 11:35
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    The point is that you can't simply multiply/square differentials. As discussed in the linked question, statements like $$dB_t , dB_t = dt$$ are just rules to simplify calculations. – saz Jan 17 '16 at 11:53
  • @saz Can I write the following? $\langle X \rangle_t $ is by definition $\int_0^t \sigma^2_s d_s = \sigma^2 dt$ and $d X^2_t = dX_td X_t=(\mu dt + \sigma d W_t)^2= \sigma^2 dt$. Then, $ d \langle X \rangle_t = d X_t d X_t $ – Puzzle Jan 17 '16 at 12:49
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    Obviously you can write it, but it is not mathematically rigorous (because you are using rules which are not rigorous, see my previous comment). – saz Jan 17 '16 at 13:31

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