Why do we need Ito's integrals?

Question

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

Answer 1

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:

$$ \begin{align} f(T, W(T)) & = f(0, W(0)) + \int_0^T f_t(t, W(t))dt + \int_0^T f_x(t, W(t))dW(t) \\\\ & + \frac{1}{2}\int_0^T f_{xx}(t, W(t))dt. \end{align} $$