formula

$&&\int_0^t (X_s^{(\theta)})^{4/3}d\langle A\rangle_s^{(\theta)}\\ &=&\int_0^{t\wedge \beta_{\theta}}(X_s^{(\theta)})^{4/3}d\langle A\rangle_s^{(\theta)}=0\quad (X_t^{(\theta)}=0\quad 0\leq \forall{t}\leq \beta_{\theta})$