Past exam of the martingales and financial mathematics course of the University of Oxford / 2009 / A / 1 / b

Independent random variables satisfy, for every finite selection of distinct indices $n_1,\dots ,n_m$ and Borel sets $C_1,\dots ,C_m$,The Second Borel-Cantelli lemma says that if events $E_n$ are independent and ${\sum }_n\mathbb{P}(E_n)=\infty$, thenHere ${\mathrm{limsup}}_n{E}_n={\bigcap }_N{\bigcup }_{n\ge N}{E}_n$ is the event that infinitely many of the $E_n$ occur.

By convergence in probability, choose recursively $n_k>n_{k-1}$ so thatThe corresponding events have a summable sequence of probabilities. The First Borel-Cantelli lemma, which requires no independence, therefore makes the probability of infinitely many of them zero. On the complementary event, $|X_{n_k}-X|\le 2^{-k}$ for all sufficiently large $k$, provingThe recursively chosen indices are strictly increasing and hence tend to infinity.

Suppose first that complete convergence holds. For each positive integer $j$, apply the First Borel-Cantelli lemma to $\{|X_n-X|>1/j\}$. Outside a countable union of null events, only finitely many of these exceedances occur for every $j$. This is almost sure convergence, with no independence assumption.

For the converse, an important point is that independence of $X_n$ does not immediately imply independence of $X_n-X$. Instead use the fact that an almost-sure limit of independent random variables is constant. Indeed, $Z={\mathrm{limsup}}_n{X}_n$ is measurable for the tail sigma-algebra. The Kolmogorov zero-one law makes each event $\{Z\le q\}$, $q\in \mathbb{Q}$, have probability zero or one. Since $Z=X$ almost surely and $X$ is finite, these probabilities form the distribution function of a point mass: there is a finite constant $c$ with $X=c$ almost surely.

Now $\{|X_n-c|>ϵ\}$ are independent events. Almost sure convergence says that their infinitely-often event has probability zero. The contrapositive of the Second Borel-Cantelli lemma givesReplacing $c$ by the almost surely equal $X$ leaves each probability unchanged. Thus complete convergence and almost sure convergence are equivalent for independent sequences.

Let $U$ have the uniform distribution on $(0,1)$, set $Y=1/U$, and take $X_n=Y/n$. The variable $Y$ is finite almost surely, so $X_n\to 0$ almost surely. For any $ϵ>0$ and all sufficiently large $n$,The harmonic series diverges, soThis dependent sequence fails complete convergence. All its members are scalar multiples of the same nonconstant random variable, which explains why the independence hypothesis cannot be dropped.