Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 201 / 3

The conditional expectation $\mathbb{E}[X\mid \mathcal{G}]$ is the almost surely unique integrable, $\mathcal{G}$-measurable random variable $Z$ such that ${\int }_{A}Zd\mathbb{P}={\int }_{A}Xd\mathbb{P}$ for every $A\in \mathcal{G}$.

Conditional expectation is the orthogonal projection from $L^2(\mathcal{F})$ onto $L^2(\mathcal{G})$. The difference $X-\mathbb{E}[X\mid \mathcal{G}]$ is orthogonal to every $\mathcal{G}$-measurable square-integrable variable, including $\mathbb{E}[X\mid \mathcal{G}]-\mathbb{E}[X\mid \mathcal{H}]$. The Pythagorean theorem in an inner-product space applied togives the identity.

Put $M=\mathbb{E}[Y\mid \mathcal{G}]$. Expanding and using $\mathbb{E}[YM]=\mathbb{E}[M^2]$ givesIf $M\stackrel{d}{=}Y$, the right side is zero. Therefore $Y=M$ almost surely.

The conditional Jensen inequality states that for an integrable $X$ and convex $\varphi$, whenever the terms are integrable,For differentiable $\varphi$, the supporting-line inequality $\varphi (x)\ge \varphi (m)+{\varphi }^{\prime }(m)(x-m)$ with $m=\mathbb{E}[X\mid \mathcal{G}]$ gives the result after conditional expectation. Approximation by supporting affine functions proves the general convex case.

Let $M=\mathbb{E}[Y\mid \mathcal{G}]$ and choose the stated strictly convex differentiable $\varphi$ with $|\varphi (x)|\le |x|$. Conditional Jensen inequality gives $\varphi (M)\le \mathbb{E}[\varphi (Y)\mid \mathcal{G}]$. Since $M\stackrel{d}{=}Y$, both sides have the same expectation, so equality holds almost surely. Strictness of the supporting-line inequality on $\{Y\ne M\}$ then forces $Y=M$ almost surely.