Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 27K (Probability and Measure)

The space $L^2(X,\mathcal{F},\nu )$ consists of equivalence classes, modulo equality almost everywhere, of measurable real functions satisfyingIt has inner product and induced normThe Cauchy-Schwarz inequality makes the inner product finite and verifies the norm axioms.

To prove completeness, let $(f_n)$ be Cauchy and choose a subsequence such thatBy the triangle inequality, the partial sumshave uniformly bounded $L^2$ norm. The monotone convergence theorem shows that their pointwise limit $g$ belongs to $L^2$ and is finite almost everywhere. The telescoping series therefore converges almost everywhere to a measurable function $f$, andThe original Cauchy sequence then converges to $f$ in $L^2$. Thus the result that an L2 space is a Hilbert space gives

Let $U$ have the uniform distribution on $[0,1]$, and setThe variables have the same law. They are not independent random variables, since

Put ${\mu }_f={\int }_0^{1}fd\lambda$. Subtracting this constant does not affect variance, and pairwise independent random variables have zero pairwise covariance. ThereforeOnly pairwise independence is needed for the off-diagonal covariance terms to vanish.

Fix an integer $k>n$ and partition $[0,1]$, up to endpoints of measure zero, intoFor $p\ne q$, defineThese functions have integral zero and $L^2$ norm one, so they are admissible.

Letbe the interval occupancy counts. Since each $Z_i$ is uniform, $\mathbb{E}{N}_p=n/k$, and henceFor every outcome,because the nonnegative integer counts satisfy $N_p^2\ge N_p$ and ${\sum }_p{N}_p=n$. No independence assumption was used.

Averaging over all ordered pairs givesThe diagonal terms vanish, so some admissible pair $p\ne q$ satisfies the same lower bound. ThereforeLetting $k\to \infty$ proves the dependent uniform sampling variance lower bound