Intuition between Ito-Formula

Question

What is the intuition behind Ito formula ? It's looking comming from no where to me. I recall that if $f\in \mathcal C^2(\mathbb R)$, then, $$f(B_t)-f(B_0)=\int_0^t f'(B_s)dB_s+\frac{1}{2}\int_0^t f''(B_s)ds.$$

Answer 1

Let $\{t_i\}_{i=1}^{n}$ be a partition of $[0, t]$ and $\delta_n=\max\{t_{i+1}-t_i\}_{i=0}^{n-1}$. Clearly $$f(B_t)=f(B_0)+\sum_{i=0}^{n-1}(f(B_{t_{i+1}})-f(B_{t_{i}}))\tag1 $$ Apply Taylor’s formula to $f(B_{t_{i+1}})-f(B_{t_{i}})$ to obtain $$f(B_{t_{i+1}})-f(B_{t_{i}})=f'(B_{t_{i}})(B_{t_{i+1}}-B_{t_{i}})+\frac{1}{2}f''(\theta_i)(B_{t_{i+1}}-B_{t_{i}})^2\tag 2$$ where $\theta_i\in(B_{t_{i}},B_{t_{i+1}})$. Thus, $$f(B_t)=f(B_0)+\sum_{i=0}^{n-1}f'(B_{t_{i}})(B_{t_{i+1}}-B_{t_{i}})+\frac{1}{2}\sum_{i=0}^{n-1}f''(\theta_i)(B_{t_{i+1}}-B_{t_{i}})^2\tag 3$$ Taking limits as $\delta_n\to 0$, the first sum in $(3)$ converges to the Ito's integral $\int_{0}^{t}f'(B_s)dB_s$. On the other hand, if $g$ be a bounded continuous function ,then for any $\theta_i\in(B_{t_{i}},B_{t_{i+1}})$, the limit in probability $$\lim_{\delta_n\to 0}\sum_{i=0}^{n-1}g(\theta_i)(B_{t_{i+1}}-B_{t_{i}})^2=\int_{0}^{t}g(B_s)ds$$

Answer 2

From my poor experience in stochastic calculus, I would say that if a stochastic process depends on the Brownian $B_s$, the differential stochastic equation for such a process can be intuitively derived expanding its differential in terms of the Brownian, $\textit{i.e.}$ $$df(s,B_s) =\frac{\partial f}{\partial s}ds+ \frac{\partial f}{\partial B_{s}}dB_s + \frac{1}{2}\frac{\partial^2 f}{\partial B_s^2}(dB_s)^2+...$$ I know that this is not very rigorous, but it works.

From the definition of the Brownian one knows that $dB_s^2 = ds$, hence keeping terms of the first order $$df(s,B_s) =\left(\frac{\partial f}{\partial s}+\frac{1}{2}\frac{\partial^2 f}{\partial B_s^2}\right)ds+ \frac{\partial f}{\partial B_{s}}dB_s\tag1$$

Your expresion is the analogous of $(1)$ but in integral form.

Hope this helps