Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 29 / 1

Write $Y=\mathbb{E}[X\mid \mathcal{G}]$. The conditional expectation is an $L^2$ contraction by the conditional Jensen inequality, so $X,Y$ are square-integrable. The defining integral identity for conditional expectation, first with bounded measurable truncations of $Y$ and then with their $L^2$ limit, givesFor clarity, if $Y^{(K)}=\max (-K,\min (Y,K))$, then $\mathbb{E}[X{Y}^{(K)}]=\mathbb{E}[Y{Y}^{(K)}]$, and the Cauchy-Schwarz inequality permits the limit $K\to \infty$. Expanding the square now proves the equality case for conditional second moments:A nonnegative random variable of zero expected value vanishes almost surely. Therefore $\boxed{X=Y\text{ almost surely}.}$

Equality of the probability distributions implies equality of all integrals for which they are finite, in particular $\mathbb{E}[(\mathbb{E}[X\mid \mathcal{G}]{)}^2]=\mathbb{E}[X^2]<\infty$. Part (a), applied to this conditional expectation, therefore gives $\boxed{X=\mathbb{E}[X\mid \mathcal{G}]\text{ almost surely}.}$ The second-moment assumption justifies the squared-norm argument; the next part removes it.

Again put $Y=\mathbb{E}[X\mid \mathcal{G}]$. For any fixed real $c$, the random variableis nonnegative and integrable. The sign function of $Y-c$ is bounded and $\mathcal{G}$-measurable, so conditioning and equality of the probability distributions giveThus $D_c=0$ almost surely. On $\{Y>c\}$ this forces $X\ge c$; on $\{Y<c\}$ it forces $X\le c$; on $\{Y=c\}$ it forces $X=c$. In particular $\{Y=c\}\subseteq \{X=c\}$ up to a null set. These two events have equal probability, again by equality of the probability distributions, so they agree up to a null set. ConsequentlyTaking $c=0$ establishes the suggested sign assertion, but this alone would not exclude unequal positive values. Apply the same argument simultaneously to the countable set of rational $c$. Outside the union of their null sets, unequal $X$ and $Y$ would have a rational strictly between them, giving opposite signs. This proves the conditional expectation preserving a distribution result:Only integrability was used; no finite second moment or hidden square-integrability assumption is needed.