Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 26K (Probability and Measure)

The Lp space is the vector space of equivalence classes, modulo equality almost everywhere, of measurable functions $f$ with $\int |f{|}^{p}d\mu <\infty$, equipped with $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. To prove it is a Banach space, let $(f_n)$ be Cauchy in this norm. Choose a subsequence $(f_{n_j})$ such thatTake measurable representatives and put $g_j=|f_{n_{j+1}}-f_{n_j}|$, $G_M={\sum }_{j=1}^M{g}_j$. The Minkowski inequality gives $\Vert G_M{\Vert }_p\le {\sum }_j{2}^{-j}\le 1$. By the monotone convergence theorem, $G={\sum }_{j\ge 1}{g}_j$ satisfies $\int G^{p}d\mu \le 1$. Consequently $G$ is finite almost everywhere, and the telescoping seriesconverges absolutely almost everywhere. Define it arbitrarily on the null exceptional set. The bound $|f|\le |f_{n_1}|+G$ puts $f$ in $L^p$.

For the tails, monotone convergence and Minkowski again giveSince the original sequence is Cauchy, the triangle inequality then gives $f_n\to f$ in norm, not merely along the subsequence. HenceThis proof needs no finiteness or sigma-finiteness assumption on the measure space.

Suppose $X_n\to X$ in probability. For a fixed real $t$ and any $\delta >0$, split according to $|X_n-X|\le \delta$. The supplied exponential inequality and the bound $|e^{it{X}_n}-e^{itX}|\le 2$ giveLet $n\to \infty$, then $\delta \downarrow 0$. Thus the characteristic functions converge pointwise to ${\phi }_X$. The Lévy continuity theorem states that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies convergence in distribution. Every characteristic function is continuous at zero, so

The independent standard Gaussian random variables have mean zero and variance one. For $m>n$, the mixed products have expectation zero, givinguniformly over $m>n$ as $n\to \infty$. Thus $(Y_n)$ is Cauchy in the L2 space. Its completeness, proved in (i), suppliesMoreover $L^2$ convergence implies convergence of the means and second moments: $\mathbb{E}Y=0$ and $\mathbb{E}{Y}^2={\lim }_n{\sum }_{j\le n}{\alpha }_j^2={\sigma }^2$.

Finite independent Gaussian sums are Gaussian; directly, multiplication of their characteristic functions givesBy Markov inequality, $\mathbb{P}(|Y_n-Y|>ϵ)\le {ϵ}^{-2}\Vert Y_n-Y{\Vert }_2^2\to 0$. Part (ii) makes the same sequence converge in distribution to $Y$, and the characteristic-function estimate there makes ${\phi }_{Y_n}\to {\phi }_Y$. Uniqueness of distributions determined by characteristic functions therefore yieldsIf ${\sigma }^2=0$, all coefficients vanish and $Y=0$ almost surely; $N(0,0)$ here denotes that degenerate distribution.