Past exam of the martingales and financial mathematics course of the University of Oxford / 2011 / A / 2

Convergence in probability meansThe probability concerns each index separately; it does not require eventual closeness along almost every sample path.

For $1\le p<\infty$, convergence in Lp means the variables belong to $L^p$ andEquivalently the $p$th absolute moment of the difference tends to zero. The analogous definition for $p=\infty$ uses the essential-supremum norm.

Almost sure convergence meansOutside a single null set, the numerical sequence converges pointwise.

For events $E_n$, the Borel-Cantelli first lemma states that ${\sum }_n\mathbb{P}(E_n)<\infty$ implies $\mathbb{P}(E_n\text{ infinitely often})=0$. Indeed,These unions decrease to $\mathrm{limsup}{E}_n$, so continuity from above proves the assertion.

The Borel-Cantelli second lemma states that independent events with ${\sum }_n\mathbb{P}(E_n)=\infty$ satisfy $\mathbb{P}(E_n\text{ infinitely often})=1$. For every fixed $m$, independence and $1-x\le e^{-x}$ giveContinuity from above shows there is probability zero of no event after $m$. A countable union over $m$ gives probability zero of only finitely many occurrences. Independence is needed for this second conclusion, but not for the first.

For fixed $\epsilon >0$ and $n>\epsilon$,so there is convergence in probability to one. The absolute difference is exactly $n$ on the exceptional event, henceFor every $p>1$ this tends to infinity, so convergence in Lp fails. In contrast its first moment is $1/\log n\to 0$, so convergence actually holds in $L^1$.

There are exactly $n$ summands in $S_n$, and the triangle inequality givesThe last step is convergence of arithmetic means of a sequence tending to zero. Thus $S_n/n\to 1$ in $L^1$, and Markov's inequality gives

Let $E_n=\{|X_n-1|=n\}$. These events are independent and ${\sum }_{n\ge 2}\mathbb{P}(E_n)={\sum }_{n\ge 2}1/(n\log n)=\infty$. By the Borel-Cantelli second lemma, they occur infinitely often almost surely.

If $S_n/n$ had any finite sample-path limit, thenBut on $E_{n+1}$, $|X_{n+1}|/n$ is either $1$ or $(n+2)/n$, contradicting this convergence infinitely often. Therefore the averages have no finite almost sure limit. This is a rare-jump counterexample to a strong law with convergent first-moment averages.