= Solution
For a <submartingale> $(Z_n)$ and $a<b$, let $U_N(a,b)$ count the completed <upcrossings> of $[a,b]$ by time $N$. The <Doob upcrossing inequality> gives
$$
(b-a)\mathbb E U_N(a,b)\leq\mathbb E(Z_N-a)^+.
$$
For a <martingale> $(X_n)$, a useful sharper form is
$$
\boxed{(b-a)\mathbb E U_N(a,b)\leq\mathbb E(X_N-a)^-.}
$$
Here $x^+=\max(x,0)$ and $x^-=\max(-x,0)$. To see the martingale form, use a bounded <predictable process> $H_k\in\{0,1\}$ that holds one unit after an observation at or below $a$ and releases it on the next observation at or above $b$. Its gain $G_N=\sum_{k=1}^NH_k(X_k-X_{k-1})$ is at least $(b-a)U_N-(X_N-a)^-$: completed trades earn at least $b-a$, and any unfinished trade loses at most $(X_N-a)^-$. Since $\mathbb E G_N=0$, the bound follows. For the submartingale version, apply the same strategy to $(Z_n-a)^+$, itself a <submartingale>. Each purchase occurs at value zero, so its gain is at least $(b-a)U_N$. The complementary predictable strategy has nonnegative expected gain, bounding this strategy's expected gain by $\mathbb E(Z_N-a)^+-\mathbb E(Z_0-a)^+$, and in particular by the stated right side.
The almost-sure <martingale convergence theorem> says that if $C=\sup_n\mathbb E|X_n|<\infty$, then $X_n$ converges almost surely to a finite, integrable <random variable> $Y$, with $\mathbb E|Y|\leq C$. Indeed, the <Doob upcrossing inequality> bounds the expected total number of <upcrossings> of every rational interval by $(C+|a|)/(b-a)$. By <monotone convergence>, each such total count is finite almost surely; intersect these probability-one events over all rational $a<b$. If the lower and upper limits of a <sample path> differed, some rational interval strictly between them would be crossed infinitely often. Thus each remaining path has a limit in the extended real line. The <Fatou lemma> gives
$$
\mathbb E\bigl[\liminf_n|X_n|\bigr]\leq C,
$$
so the limit is finite almost surely and integrable. Almost-sure convergence by itself neither preserves <expectations> nor supplies this uniform first-moment bound.
For the loss of <expectation>, let $(\eta_k)$ be independent variables with the fair <Bernoulli distribution> and set
$$
X_0=1,\qquad X_n=2^n\mathbf1_{\{\eta_1=\cdots=\eta_n=1\}}.
$$
Conditional on the past, survival doubles the value with <probability> $1/2$ and otherwise sets it to zero, so this is a <nonnegative martingale>. An infinite string of successes has <probability> zero. The process therefore eventually vanishes almost surely, giving \b[yes: $X_0=1$ and $Y=0$ almost surely are possible]. Its <expectations> remain one at every finite time; the process is not <uniformly integrable>.
A nonintegrable limit is also possible for a signed <martingale>. Take independent $\xi_k$ with
$$
\mathbb P(\xi_k=4^k)=\mathbb P(\xi_k=-4^k)=2^{-k-1},\qquad
\mathbb P(\xi_k=0)=1-2^{-k},\qquad k\geq1,
$$
and let $M_0=1$, $M_n=1+\sum_{k=1}^n\xi_k$. Every increment has mean zero and a finite absolute <expectation>, so each $M_n$ is integrable and $(M_n)$ is a <martingale>. Since $\sum_k\mathbb P(\xi_k\ne0)<\infty$, the <Borel-Cantelli lemma> makes only finitely many increments nonzero almost surely; hence the finite limit $Y=1+\sum_k\xi_k$ exists almost surely. To compute its absolute <expectation>, let $A_k$ be the event that only the $k$th increment is nonzero. <Independence> gives
$$
\mathbb P(A_k)=2^{-k}\prod_{j\ne k}(1-2^{-j})\geq c2^{-k},\qquad
c=\prod_{j\geq1}(1-2^{-j})>0.
$$
The <infinite product> is positive because its tail logarithms are bounded in absolute value by a constant times the summable series $\sum_j2^{-j}$. The events $A_k$ are disjoint, and on $A_k$, $|Y|\geq4^k-1$. Therefore
$$
\mathbb E|Y|\geq c\sum_{k\geq1}2^{-k}(4^k-1)=\infty.
$$
Thus \b[yes: a finite almost-sure martingale limit can have infinite absolute <expectation>], even with deterministic $M_0=1$. This is <almost-sure martingale convergence to a nonintegrable limit>; the missing hypothesis is the uniform first-moment bound. For a <nonnegative martingale>, in contrast, the <Fatou lemma> would force the limit to be integrable.
Back to article page
