Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 26K (Probability and Measure)

The product sigma-algebra isThe product measure ${\mu }_1\otimes {\mu }_2$ is the probability measure on this sigma-algebra satisfyingThe measurable rectangles form a pi-system generating ${\mathcal{A}}_1\otimes {\mathcal{A}}_2$. Any two probability measures agreeing on them therefore agree on the generated sigma-algebra by the pi-lambda theorem, proving uniqueness.

Saying that $U$ has law $\mu$ means that $\mu$ is the pushforward measureFor a real random variable, $\mu$ is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

The pair $U,V$ consists of independent random variables whenfor every pair of Borel sets $A,B$; equivalently, their joint law is the product of their marginal laws.

Let $U$ be uniformly distributed on $X$ and let $\eta$ be the law of $(U,U)$. Both marginals are $\mu$, becauseHowever, $\eta$ is concentrated on the diagonal $\{(x,x):x\in X\}$, which has two-dimensional Lebesgue measure zero. Thus $\eta$ is not Lebesgue measure on $X^2$.

On the probability space $(X^2,\eta )$, put $A=I\times X$ and $B=X\times J$. The marginal conditions give $\eta (A)=x$ and $\eta (B)=y$, while $A\cap B=I\times J$. The elementary Fréchet inequalities yieldHence

Take independent random variables $U_1,\dots ,U_d$, each uniform on $[0,1/2]$, and independent Rademacher random variables ${\epsilon }_1,\dots ,{\epsilon }_{d-1}$. Defineand let $\eta$ be the joint law of $(X_1,\dots ,X_d)$.

Each $X_i$ is uniform on $[-1/2,1/2]$. If the $d$th coordinate is omitted, the remaining signs are the independent ${\epsilon }_i$. If coordinate $j<d$ is omitted, the remaining signs arewhich are again uniformly distributed over all $2^{d-1}$ sign vectors because this map from $({\epsilon }_1,\dots ,{\epsilon }_{d-1})$ is a bijection. The independent magnitudes then show that every $(d-1)$-coordinate marginal is Lebesgue measure.

The full law is not $d$-dimensional Lebesgue measure, becausealmost surely, whereas this product is equally likely to have either sign under independent Lebesgue coordinates.