Enlarge the space by an independent Brownian motion $W$ and define
$$
B_t=\int_0^t\mathbf1_{\{A_s>0\}}A_s^{-1/2}\,dX_s
+\int_0^t\mathbf1_{\{A_s=0\}}\,dW_s.
$$
The two terms have zero cross-variation and
$$
[B]_t=\int_0^t\mathbf1_{\{A_s>0\}}ds
+\int_0^t\mathbf1_{\{A_s=0\}}ds=t.
$$
The <Lévy characterization of Brownian motion> makes $B$ a Brownian motion. The residual $\int\mathbf1_{\{A=0\}}dX$ has zero quadratic variation and is therefore constant, so
$$
X_t-X_0=\int_0^tA_s^{1/2}\,dB_s.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 202 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-202/4/b/solution