If $\mathcal G\subseteq\mathcal H$, then $\mathbb E[X\mid\mathcal G]$ is already $\mathcal H$-measurable, so
$$
\mathbb E[\mathbb E[X\mid\mathcal G]\mid\mathcal H]
=\mathbb E[X\mid\mathcal G]
=\mathbb E[X\mid\mathcal G\cap\mathcal H].
$$
If $\mathcal H\subseteq\mathcal G$, the <tower property of conditional expectation> gives
$$
\mathbb E[\mathbb E[X\mid\mathcal G]\mid\mathcal H]
=\mathbb E[X\mid\mathcal H]
=\mathbb E[X\mid\mathcal G\cap\mathcal H].
$$
Finally, if $\mathcal G$ and $\mathcal H$ are independent, the $\mathcal G$-measurable variable $\mathbb E[X\mid\mathcal G]$ is independent of $\mathcal H$. Its conditional expectation given $\mathcal H$ is its mean $\mathbb E[X]$. Part c shows that the right side is also $\mathbb E[X]$.

Solution

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

 Solution

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