Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 25J / Solution

Let $M$ be the linear subspace of $L^2(P)$ consisting of classes admitting $\mathcal{G}$-measurable representatives. It is closed. Indeed, if $Y_n\in M$ converges in $L^2$ to $Y$, choose a subsequence converging almost surely: take $\Vert Y_{n_j}-Y{\Vert }_2^2\le 2^{-3j}$ and apply Markov inequality and the Borel-Cantelli lemma to errors exceeding $2^{-j}$. The pointwise limit of its $\mathcal{G}$-measurable representatives, set to zero wherever the finite limit does not exist, is $\mathcal{G}$-measurable and equals $Y$ almost surely. This also works when $\mathcal{G}$ is not complete.

Here is a proof of the orthogonal projection construction using only completeness. Put $d={\inf }_{W\in M}\Vert X-W{\Vert }_2$ and take $W_n$ approaching this infimum. The parallelogram law givesCompleteness and closedness give a minimizer $Y\in M$. For every $Z\in M$, expand $\Vert X-Y-tZ{\Vert }_2^2$ for real $t$. Its minimum at zero impliesThis $Y$ is unique almost surely, because the difference of two possible choices belongs to $M$ and is orthogonal to itself. It is the conditional expectation $E(X\mid \mathcal{G})$, as also follows by testing indicators of events in $\mathcal{G}$.

For the jointly Gaussian pair, if $u>0$, setThe residual $R=X-Y$ has mean zero and covariance $\mathrm{Cov}(R,G)=v-(v/u)u=0$. Since $(R,G)$ is jointly Gaussian, its characteristic function factors when this covariance is zero, so $R$ and $G$ are independent. Hence $E(RZ)=E(R)E(Z)=0$ for every square-integrable function $Z$ of $G$; independence first proves this for bounded functions, and truncation plus Cauchy-Schwarz inequality gives the general case. This proves the Gaussian conditional expectation formula, including a zero residual variance. If $u=0$, $G=\mu$ almost surely and the covariance inequality gives $v=0$; $Y=\nu$ is the answer.