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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 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 $c=\mathrm{Var}(S_n)/x$.

For the continuous adapted process $X$, the supremum on each compact time interval is attained. Let $D_t=(\mathbb{Q}\cap [0,t])\cup \{t\}$. Continuity gives ${\sup }_{s\le t}{X}_s={\sup }_{r\in D_t}{X}_r$, and henceEach event inside the union is in ${\mathcal{F}}_r\subseteq {\mathcal{F}}_t$, so this proves carefully that $T$ is a stopping time. Merely checking rational times for $X_r\ge \lambda$ would miss a path that touches the level at an isolated irrational time; the approximation from below avoids this issue. This is the closed-level hitting times of continuous adapted processes criterion. For $\lambda =0$, nonnegativity makes $T=0$ and the later inequality is immediate.

For $\lambda >0$, put $A=\{T\le t\}$ and $\tau =T\wedge t$. On $A$, continuity implies $X_{\tau }\ge \lambda$. To justify the optional sampling step without extra filtration assumptions, round $\tau$ upwards to a finite dyadic grid in $[0,t]$, obtaining stopping times ${\tau }_n\downarrow \tau$. The event $A$ belongs to ${\mathcal{F}}_{\tau }$, and hence to ${\mathcal{F}}_{{\tau }_n}$: directly, $A\cap \{\tau \le u\}=\{T\le u\}$ for $u<t$, while for $u\ge t$ it is $A$. Discrete optional sampling for the sampled submartingale givesAs $n\to \infty$, continuity gives $X_{{\tau }_n}\to X_{\tau }$. Nonnegativity and the Fatou lemma therefore yieldConsequently the Doob maximal inequality for a nonnegative submartingale takes the precise form

For the second-moment bound, suppose first that $\mathbb{E}{X}_t^2<\infty$; otherwise the inequality with an infinite right side is automatic. Set $M_R=X_t^{\ast }\wedge R$. The layer cake representation and the preceding tail estimate giveThe integrands are nonnegative, so the interchange is justified by the Tonelli theorem. Since $M_R$ is bounded, division is legitimate when its norm is nonzero, giving $\Vert M_R{\Vert }_2\le 2\Vert X_t{\Vert }_2$; the zero-norm case is immediate. Finally, monotone convergence as $R\to \infty$ givesThis truncated layer-cake proof of the L2 maximal inequality proves square integrability of the maximum rather than assuming it during the Cauchy-Schwarz inequality step.

The Doob maximal inequality for a nonnegative submartingale states that, for a nonnegative submartingale $(M_j{)}_{0\le j\le N}$ and $u>0$,For completeness, partition the crossing event according to its first crossing time $j$. On that event $M_j\ge u$, while $\mathbb{E}[M_N\mid {\mathcal{F}}_j]\ge M_j$. Summing the corresponding expectations over $j$ proves $u\mathbb{P}(\max M_j\ge u)\le \mathbb{E}[M_N{1}_{\{\max M_j\ge u\}}]\le \mathbb{E}{M}_N$.

If $q>p$, the exponential $r^{S_j}$, with $r=q/p>1$, is increasing as a function of $S_j$. Apply the inequality to the martingale from part (a), with $u=r^k$:Letting $N\to \infty$ and using continuity of probability measures on increasing events givesThe same event is obtained if the supremum excludes time zero, since $k\ge 1$. If $p\ge q$, the right side is at least one and the bound is automatic; the nontrivial Doob maximal inequality argument is for $q>p$.

Put $v=\mathbb{E}{X}_n^2$. For $c\ge 0$, conditional Jensen inequality applied to the convex function $x\mapsto (x+c{)}^2$ shows thatis a nonnegative submartingale. Its integrability follows from the square integrability of $X_k$. If $X_k\ge \lambda$, then $Z_k\ge (\lambda +c{)}^2$, since $c\ge 0$. The Doob maximal inequality for a nonnegative submartingale yieldswhere zero mean removes the cross term. For completeness, the maximal inequality follows by stopping at the first crossing: on the event of a crossing at $k$, the submartingale property gives $\mathbb{E}[Z_n{1}_{\{T=k\}}]\ge \mathbb{E}[Z_k{1}_{\{T=k\}}]$. Sum over $k\le n$, and use nonnegativity on the event of no crossing.

The derivative of the last ratio isFor $v>0$, the minimum over $c\ge 0$ is attained at $c=v/\lambda$. Substitution gives the one-sided maximal inequality for a centered square-integrable martingale:If $v=0$, $X_n=0$ almost surely and $X_k=\mathbb{E}(X_n\mid {\mathcal{F}}_k)=0$ almost surely for every $k\le n$, so the bound also holds. The optimization is the same one underlying the Cantelli inequality, but the submartingale argument controls the entire finite-time maximum.

Put $V=\mathrm{Var}(S_n)={\sum }_{m=1}^n{v}_m$. If ${\max }_{m\le n}{S}_m\ge x$, thenThe Doob maximal inequality for a nonnegative submartingale therefore givesThe right side is minimized at $c=V/x$, and substitution gives

For $\theta >0$, the Exponential martingale for Brownian motion and the Doob maximal inequality for a nonnegative submartingale giveTaking $\theta =a/s$ gives $e^{-a^2/(2s)}$. Apply the same argument to $-B$ and use the union bound: