How to compute ito's calculus without knowing its solution already?

Question

I'm having trouble computing Itô's calculus. Take $\int_0^tB_sdB_s$ for example, can I solve it base solely on Itô-Doeblin Formula, instead of assuming $f(x)=\frac{x^2}{2}$? Please help!

Answer 1

Itô's formula states that

$$f(B_t) -f(B_0) = \int_0^t f'(B_s) \, dB_s + \frac{1}{2} \int_0^t f''(B_s) \, ds $$

for any twice differentiable function $f$. Equivalently,

$$\int_0^t f'(B_s) \, dB_s = f(B_t)-f(B_0) - \frac{1}{2} \int_0^t f''(B_s) \, ds \tag{1}.$$

Now if we are interested in the stochastic integral

$$\int_0^t B_s \, dB_s$$

then we can write

$$\int_0^t B_s \, dB_s = \int_0^t f'(B_s) \, dB_s$$

for any function $f$ which satisfies

$$f'(x)=x. \tag{2}$$

Clearly, the function $f(x)=x^2/2$ satisfies $(2)$. Using $(1)$ for $f(x)=x^2/2$ we get

$$\begin{align*}\int_0^t B_s \, dB_s &= \int_0^t f'(B_s) \, dB_s \\ &= f(B_t)-f(B_0)- \frac{1}{2} \int_0^t f''(B_s) \, ds \\ &= \frac{1}{2} B_t^2 - \frac{1} {2} t.\end{align*}$$