Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 32 / 1

Put $C={\sup }_p\mathbb{E}|X_p|<\infty$. Since $x\mapsto x^{+}$ is an increasing convex function, the conditional Jensen inequality and the submartingale property giveFor fixed $n$ and $p\ge n$, the tower property of conditional expectation therefore impliesChoose versions for this countable family so that all these inequalities hold outside one null set. Its nonnegative increasing limit $M_n$ is ${\mathcal{F}}_n$-measurable. The monotone convergence theorem shows thatIn particular the limit is finite almost surely. Thus the increasing conditional means converge to an integrable nonnegative $M_n$, rather than merely to a possibly infinite extended value. This is the construction of the positive martingale majorant of an L1-bounded submartingale.

Use conditional monotone convergence and the tower property of conditional expectation in the preceding construction:This proves the martingale property. The construction gives $M_n\ge 0$ and ${\sup }_n\mathbb{E}{M}_n\le C$, so $M$ is bounded in $L^1$. Taking $p=n$ in the increasing family also gives $M_n\ge X_n^{+}\ge X_n$. Define $Y_n=M_n-X_n$. Then $Y$ is nonnegative and adapted, andThus $Y$ is a supermartingale, with $\mathbb{E}|Y_n|\le \mathbb{E}{M}_n+\mathbb{E}|X_n|\le 2C$. The requested conclusion isThis positive martingale majorant of an L1-bounded submartingale is not asserted to have uniform integrability; boundedness in $L^1$ alone does not imply that stronger property.

The almost sure supermartingale convergence theorem applies to the $L^1$-bounded supermartingale $X$. Hence $X_n\to X_{\infty }$ almost surely, and the Fatou lemma gives $\mathbb{E}|X_{\infty }|\le {\sup }_n\mathbb{E}|X_n|<\infty$. Take the version of $X_{\infty }$ obtained from the sequence, so it is measurable for ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. SetThe tower property of conditional expectation makes $M$ a martingale, and the uniform integrability of conditional expectations makes it uniformly integrable. Explicitly, with $Z=X_{\infty }$, $C=\mathbb{E}|Z|$, and $R,K>0$, we have $|M_n|\le R+\mathbb{E}[|Z|1_{\{|Z|>R\}}\mid {\mathcal{F}}_n]$, soFirst choose $R$ large, then $K$ large. This proves uniform integrability uniformly in $n$.

The uniformly integrable martingale convergence theorem gives a limit $M_{\infty }$ both almost surely and in $L^1$. For $A\in {\mathcal{F}}_j$ and $n\ge j$, $\mathbb{E}[M_n{1}_A]=\mathbb{E}[X_{\infty }{1}_A]$. Passing to the $L^1$ limit and then extending from the algebra ${\bigcup }_j{\mathcal{F}}_j$ to ${\mathcal{F}}_{\infty }$ identifies $M_{\infty }=X_{\infty }$ almost surely. Finally,Thus $X=M+Y$ with $M$ uniformly integrable and $Y$ a supermartingale tending almost surely to zero. The terminal decomposition of an L1-bounded supermartingale does not require $Y\ge 0$ or $Y_n\to 0$ in $L^1$. Indeed, for $X$ equal to the negative of the fair-coin doubling martingale, $X_{\infty }=0$, $M=0$, and $Y=X$ is negative and has constant absolute mean one.