Why do we need Ito's integrals? Why couldnt we just use Riemann integral when integrating with respect to Brownian motion?

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2either a stupid question or a very deep one! – Trajan Sep 17 '20 at 19:34
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And what exactly is a "Riemann integral with respect to...(Brownian motion or whatever)"? It's the former case, I'm afraid. – Sep 17 '20 at 19:43
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1See also this question – saz Sep 18 '20 at 13:05
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Thanks that's excellent – Trajan Sep 18 '20 at 14:26
1 Answers
Because the paths of the Brownian motion cannot be differentiated with respect to time. This means that the Brownian motion has non zero quadratic variation (indeed, for $T \geq 0$, the quadratic variation $P\{[W, W](T)=T\}=1$). Informally:
$$ dW_t dW_t = dt. $$
Consequently, when characterizing the differential of a function of a Brownian motion in terms of its Taylor expansion, one has a nonzero second degree differential associated to the double derivative of the function with respect to the Brownian motion, that is
$$ df(t, x) = f_t dt + f_{x}dx +\frac{1}{2}f_{xx}dt $$
where $x=W(t)$ and the term $\frac{1}{2}f_{xx}dt$ is the term that is due to the quadratic variation of the Brownian motion $W(t)$. Such is the famous Ito-Doeblin formula. Integrating, you have the answer to your question:
$$ \begin{align} f(T, W(T)) & = f(0, W(0)) + \int_0^T f_t(t, W(t))dt + \int_0^T f_x(t, W(t))dW(t) \\\\ & + \frac{1}{2}\int_0^T f_{xx}(t, W(t))dt. \end{align} $$

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