Submartingale

An integrable adapted process $(X_n)$ is a submartingale when

For a submartingale with ${\sup }_n\mathbb{E}|X_n|=C<\infty$, its positive part is a submartingale by the conditional Jensen inequality. Thus the displayed conditional expectations increase in $p$. Their limits are finite almost surely and have expectations at most $C$, by the monotone convergence theorem. Conditional monotone convergence and the tower property of conditional expectation give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Moreover $M_n\ge X_n^{+}$. Consequently $Y_n=M_n-X_n$ is a nonnegative supermartingale, ${\sup }_n\mathbb{E}{Y}_n\le 2C$, and $X=M-Y$. The resulting martingale need not have uniform integrability.

Under the usual filtration conditions, a càdlàg submartingale of Class D on a finite horizon has a unique decomposition into a uniformly integrable martingale and a predictable integrable nondecreasing process starting at zero. A general continuous semimartingale decomposition instead allows an arbitrary continuous finite-variation drift. A convex function of a semimartingale need not be a submartingale when that drift has unrestricted sign.

If $(X_m{)}_{m\le n}$ is a nonnegative submartingale, then for every $a>0$,Stop at the first crossing of $a$ and use the submartingale property on that event.

For a nonnegative submartingale with maximum $M_t$, the first-moment maximal estimate is $\lambda \mathbb{P}(M_t\ge \lambda )\le \mathbb{E}[X_t{1}_{\{M_t\ge \lambda \}}]$. Integrate $2\lambda$ times the tail probability only up to $R$, obtaining $\mathbb{E}(M_t\wedge R{)}^2\le 2\mathbb{E}[X_t(M_t\wedge R)]$. The Cauchy-Schwarz inequality gives $\Vert M_t\wedge R{\Vert }_2\le 2\Vert X_t{\Vert }_2$, and monotone convergence gives the result as $R\to \infty$. Truncation avoids assuming beforehand that the maximum has a finite second moment.

// Target: brownian-motion.bigb

For a square-integrable martingale with $M_0=0$ and $V=\mathbb{E}{M}_n^2$, apply the Doob maximal inequality to $(M_k+c{)}^2$ for $c\ge 0$. An upper crossing of $x$ implies a square crossing of $(x+c{)}^2$, giving $(V+c^2)/(x+c{)}^2$. Minimization at $c=V/x$ gives the displayed bound. A one-step martingale taking $x$ with probability $V/(V+x^2)$ and $-V/x$ otherwise attains equality when $V>0$.

For independent zero-mean square-integrable increments, their partial sums form a martingale, and their squares form a nonnegative submartingale. Apply the Doob maximal inequality to the squared process at threshold $x^2$. Independence and zero means make its terminal second moment the sum of the increment variances. Identical distributions are not required.

For a centered square-integrable martingale with $v=\mathbb{E}{M}_n^2$ and $\lambda >0$, apply the Doob maximal inequality for a nonnegative submartingale to $(M_k+c{)}^2$ for $c\ge 0$. Conditional Jensen inequality gives the submartingale property, and crossing $\lambda$ forces its square above $(\lambda +c{)}^2$. The bound $(v+c^2)/(\lambda +c{)}^2$ is minimized by $c=v/\lambda$. This extends the Cantelli inequality from one random variable to a martingale maximum, without requiring independent increments.

For a nonnegative martingale, $T_R=\inf \{n:M_n\ge R\}$ satisfies $\mathbb{P}(T_R<\infty )\le \mathbb{E}{M}_0/R$. Stop at $T_R\wedge n$ and use nonnegativity before passing to the increasing event. A limit of thresholds from below also gives the bound for $\{{\sup }_n{M}_n\ge R\}$ even when the supremum is not attained.

For independent centered random variables and $S_m={\sum }_{j\le m}{X}_j$,Apply the Doob maximal inequality for a nonnegative submartingale to $(S_m+c{)}^2$ and optimize at $c=\mathrm{Var}(S_n)/x$.