Let $\{W_t : t \geq 0\}$ and $\{\tilde{W_t} : t \geq 0\}$ be two independent Brownian motions. It is not clear to me how to apply Ito formula to represent a process as an Ito process. For instance, I would like to represent the following two processes
$$W(t)^2\tilde{W}(t) - t\quad \text{and}\quad exp[W(t)\tilde{W}(t) - \frac{1}{2}(\tilde{W}(s)^2 + W(s)^2)ds]$$
as Ito processes.
Can someone please show me how to proceed ?
Thanks in advance.
Let $w,b$ be Wiener processes and let $t$ be the time variable. Then a function $f$ (with certain nice conditions) that is a function of $(w,b,t)$, will be equal to (using the Ito rule):
$ d f(w_t,b_t,t) = \partial_w [f] dw_t + \frac{1}{2} \partial_{ww} [f] d \langle w,w \rangle_t + \partial_b [f] db_t + \frac{1}{2} \partial_{bb} [f] d \langle b,b \rangle_t + \partial_{bw} [f] d \langle b,w \rangle_t + \partial_t [f] dt $
Let $z_t = w_t^2 b_t -t$. The above formula applied to $z$ gives us
$ dz_t=d[w_t^2b_t-t] = 2w_tb_t dw_t+w_t^2 db_t+(b_t-1)dt. $
You have made a mistake in defining the second process - the $ds$ term cannot exist unless it is under an integral - which is missing.