$B(t)$ is brownian motion. I want Find $d(M(t))^2$,where $M(t)=e^{B(t)-\frac{t}{2}}$,

Question

let $B(t)$ is brownian motion.
Find $d(M(t))^2$,where $M(t)=e^{B(t)-\frac{t}{2}}$,

Answer 1

Hint By Itô's formula,

$$f(t,B_t) -f(0,B_0) = \int_0^t \partial_x f(s,B_s) \, dB_s + \int_0^t \left( \frac{1}{2} \partial_x^2 f(s,B_s) + \partial_t f(s,B_s) \right) \, ds \tag{1}$$

for any (nice) function $f=f(t,x)$. For $$f(t,x) := \exp(2x-t)$$ we have $M_t^2 = f(t,B_t)$. Calculate the derivatives $\partial_x f$, $\partial_x^2 f$, $\partial_t f$ and plug them into $(1)$ in order to obtain an expression for $M_t^2$.