If $G\in\mathcal G\cap\mathcal H$, independence makes $G$ independent of itself, so
$$
\mathbb P(G)=\mathbb P(G)^2.
$$
Thus every event in $\mathcal G\cap\mathcal H$ has probability zero or one. The intersection is trivial modulo null sets, and the <independent sigma-algebras have trivial intersection> result gives
$$
\boxed{\mathbb E[X\mid\mathcal G\cap\mathcal H]=\mathbb E[X]}
$$
almost surely.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-201/1/c/solution