Showing that Ito integrals of functions form martingales.

Question

My textbook says that $X_t = (\int_{0}^{t} b(s)dB_s)^2 - \int_{0}^{t} b(s)^2ds )$ $\forall t\in [0,1]$ is a martingale. Where $b \in H^2(0,1)$ and for all $t\in [0,1], (\int_{0}^{t} b(s)dB_s)^2\in L^2(\Omega)$

my proof: \begin{align}E(X_{t+1}|F_t) &= E((\int_{0}^{t+1} b(s)dB_s)^2 - \int_{0}^{t+1} b(s)^2ds )|F_t)\\ &=(\int_{0}^{t} b(s)dB_s)^2 - \int_{0}^{t} b(s)^2ds )+ E((\int_{0}^{1} b(s)dB_s)^2 - \int_{0}^{1} b(s)^2ds )|F_t)\\ &=X_t +E(\int_{0}^{1} b(s)dB_s)^2 - \int_{0}^{1} b(s)^2ds ) \end{align} Then since $H^2$ functions can be approximated by simple functions $g\in H_0^2$, we can write:

$$=X_t + +E(\int_{0}^{1} g(s)dB_s)^2 - \int_{0}^{1} g(s)^2ds)$$ By Iso Isometry this is equal to $$ = X_t +\int_0^1Eg(s)^2 ds - E(\int_{0}^{1} g(s)^2ds)$$

Can I then take the expectation inside the integral by Fubini's theorem and write: $$ = X_t +\int_0^1Eg(s)^2 ds - \int_{0}^{1} Eg(s)^2ds$$

$$=X_t$$

Is this correct?

Answer 1

Three things:

• Don't change the bounds on the integration, when you split the integral up use $\int_t^{t+1}$ because in general $b(0) \neq b(t)$
• Apply Ito Isometry directly to the second argument before you approximate with simple functions
• You shouldn't need to approximate with simple functions or use Fubini