Suppose that Ito process $$X_t=\int_0^tK_sds+\int_0^tH_sdB_s=0, \,t\geq0.$$ Then $$K_t=H_t=0 \text{ a.s.},\,t\geq0.$$
To prove this, it is suggested to apply Ito formula to the process $Y_t=e^{-X^2_t}$, but I fail to see the connection.
Asked
Active
Viewed 22 times
0
-
One possibility is to argue using martingales: Since $X_t=0$ we know that $(X_t)_t$ is a martingale and therefore the drift must be zero. Applying Itô's formula then gives $H=0$. – saz Feb 15 '20 at 15:59
-
Why do you imply that $X_t$ is a martingale? – TheGrandDuke Feb 16 '20 at 13:28
-
Well, the constant process $X_t=0$ is a martingale, isn't it? – saz Feb 16 '20 at 13:51
-
In the exercise it is not said that the equality holds for all $t\geq0$. Or it doesn't make sense if it holds only for some certain $t$? – TheGrandDuke Feb 16 '20 at 13:58