Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 26K (Probability and Measure)

The strong law of large numbers states that if $(X_i{)}_{i\ge 1}$ are independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, thenalmost surely. Under the stronger fourth-moment hypothesis, it has a short proof. Put $\mu =\mathbb{E}{X}_1$, $Y_i=X_i-\mu$, and $S_n={\sum }_{i=1}^n{Y}_i$. Independence and centering make every term in the expansion of $S_n^4$ vanish unless each index occurs at least twice, soFor every $\epsilon >0$, the Markov inequality givesThe probabilities are summable, so the first Borel-Cantelli lemma says that $|S_n|>\epsilon n$ occurs only finitely often almost surely. Apply this simultaneously to $\epsilon =1,1/2,1/3,\dots$ to obtain $S_n/n\to 0$ almost surely, which is the claimed law.

For $n>m$, independence and $\mathbb{E}{X}_k=0$ giveThus $(S_n)$ is a Cauchy sequence in the complete L2 space and converges in $L^2$ to some random variable $S$. The Cauchy-Schwarz inequality gives $\Vert Z{\Vert }_1\le \Vert Z{\Vert }_2$ on a probability space, so the convergence also holds in $L^1$. Finally, convergence in $L^2$ implies convergence in probability, which implies convergence in distribution.

For every finite Rademacher sum, independence and the vanishing of odd moments giveThe same calculation applied to $S_n-S_m$ shows that $(S_n)$ is Cauchy in $L^4$, because the tail of $\sum a_k^2$ tends to zero. Its $L^4$ limit agrees almost surely with its $L^2$ limit $S$, since both modes imply convergence in probability. Passing to the limit givesand therefore

A finite linear combination of independent Gaussian random variables is Gaussian, soSince $v_n\to v:={\sum }_{k\ge 1}{a}_k^2<\infty$, the characteristic functions converge pointwise:The Lévy continuity theorem yields convergence in distribution toOn the given common probability space, the same variance calculation actually proves convergence in $L^2$ as well.