Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 28 / 3

Use the natural filtration ${\mathcal{G}}_n=\sigma ({\xi }_1,\dots ,{\xi }_n)$. Define the deterministic tail $r_n={\sum }_{j=n}^{\infty }{\delta }_j$ and put $Z_n={\xi }_n+r_n$. This is nonnegative and integrable. Since $r_n={\delta }_n+r_{n+1}$,Thus the deterministic summable drift correction produces a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives a finite limit $Z_{\infty }$ almost surely. Its finiteness is also reflected in the bound $\mathbb{E}{Z}_n\le \mathbb{E}{Z}_1$ and Fatou's lemma, which gives $\mathbb{E}{Z}_{\infty }\le \mathbb{E}{Z}_1<\infty$. The convergence theorem follows from the upcrossing argument: bounded expected negative parts of $-Z$ rule out infinitely many crossings of rational intervals, and the expectation bound excludes an infinite limit.

Since $r_n\to 0$, the required conclusion is $\boxed{{\xi }_n⟶Z_{\infty }\text{ almost surely, with }{Z}_{\infty }<\infty }$. The original sequence need not itself be a supermartingale; subtracting the vanishing deterministic tail recovers it from one that is.

Set $\inf \varnothing =\infty$. Since each $Y_j$ is ${\mathcal{F}}_j$-measurable, the eventis in ${\mathcal{F}}_n$, so ${\tau }_a$ is a stopping time. The stopped predictable budget for adapted increments is $A_n=a-U_{n\wedge {\tau }_a}$, for $n\ge 1$.

If ${\tau }_a=k<\infty$, the sum before the overshooting increment is $U_k={\sum }_{j<k}{Y}_j\le a$. If ${\tau }_a>n$, the sum through time $n$ is at most $a$, and in particular $U_n\le a$. ThusThe process is adapted, and boundedness gives integrability even when the increments have no finite expectations. Directly from the one-step-lagged sum,This decrement is ${\mathcal{F}}_n$-measurable, and $0\le Y_n{1}_{\{{\tau }_a>n\}}\le a$. ThereforeHence $A$ is a nonnegative supermartingale; in fact its paths are nonincreasing. At time zero one may set $U_0=0$ and $A_0=a$; then $A_1=a$ too. Stopping the sum including $Y_{{\tau }_a}$ would lose nonnegativity, which is why the shifted definition of $U$ matters.

Work from time one; a finite initial prefix does not affect convergence. Fix a positive integer $a$, and use the stopping time ${\tau }_a$ and budget $A_n$ from part (b). DefineIt is nonnegative. For each finite $n$, the stopped variable $X_{n\wedge {\tau }_a}$ is integrable: it is bounded by the finite sum $X_1+\cdots +X_n$, and the budget is bounded by $a$.

The stopped increment of $X$ isTaking conditional expectations, the assumed drift bound givesThe last term is bounded by $a$; no integrability assumption on the unlocalized $Y_n$ has been used. Its contribution is exactly cancelled by the budget decrement from part (b), givingThus $Z^{(a)}$ is an integrable nonnegative supermartingale. By the almost sure supermartingale convergence theorem it has a finite almost-sure limit. The bounded decreasing budget $A_n$ has a finite pathwise limit as well, so $X_{n\wedge {\tau }_a}=Z_n^{(a)}-A_n$ converges to a finite limit almost surely.

Take the intersection of these probability-one convergence events over all positive integers $a$. On the further probability-one event ${\sum }_{j\ge 1}{Y}_j<\infty$, choose an integer $a$ strictly larger than that total. Then ${\tau }_a=\infty$, so the stopped sequence is the original one. ConsequentlyThis proves supermartingale convergence with summable adapted drift under almost-sure summability alone; it does not require $\mathbb{E}{\sum }_j{Y}_j<\infty$.