Put $S=X_1+X_2$. By symmetry,
$$
\mathbb E[X_1\mid S]=\mathbb E[X_2\mid S].
$$
Their sum is $S$, which is measurable with respect to $\mathcal G=\sigma(S)$, so
$$
2\mathbb E[X_1\mid\mathcal G]
=\mathbb E[X_1+X_2\mid\mathcal G]
=S.
$$
Therefore
$$
\boxed{\mathbb E[X_1\mid\mathcal G]=\frac{X_1+X_2}{2}}.
$$
This also follows from the <Gaussian conditional expectation> formula because $\operatorname{Cov}(X_1,S)/\operatorname{Var}(S)=1/2$.

Solution

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

 Solution

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