Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 21 / 5

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.

For $t>0$, put $M_t={\sup }_{0\le s\le t}{B}_s$. For $\lambda >0$, reflect the Brownian path after its first hit of $\lambda$. The Strong Markov property and symmetry of subsequent Brownian increments make this reflection preserve the path distribution. It exchanges the events $\{M_t\ge \lambda ,B_t<\lambda \}$ and $\{B_t>\lambda \}$. Since $B_t$ has no atom at $\lambda$, the Brownian reflection principle givesThe case $\lambda =0$ follows by decreasing positive levels to zero: the probability is one. At $t=0$, $M_0=0$ and the probability is zero for every $\lambda \ge 0$.

The displayed tail probabilities are exactly those of $|B_t|$, so $M_t$ has the half-normal distribution of $\sqrt{t}|Z|$ with $Z\sim N(0,1)$. ThusFor $A_t={\sup }_{s\le t}|B_s|$, the pointwise inequality $A_t\ge M_t$ gives the lower non-strict bound. To prove strictness when $t>0$, fix $\lambda >0$. Reflection at $\lambda$ also givesSubtracting this from $\mathbb{P}(B_t<-\lambda )=\mathbb{P}(B_t>\lambda )$ givesOn this event $A_t\ge |B_t|>\lambda >M_t$, so $A_t^2-M_t^2$ is strictly positive with positive probability. Finally, $|B_t|$ is a continuous nonnegative submartingale by the conditional Jensen inequality, and part (a) gives $\Vert A_t{\Vert }_2\le 2\Vert B_t{\Vert }_2=2\sqrt{t}$. Its square is therefore integrable, and the strict pointwise comparison on a positive-probability event givesThis is the strict comparison of one-sided and absolute Brownian maxima. The strict lower inequality printed with $t\ge 0$ needs the qualification $t>0$: at $t=0$ both maxima vanish and all three quantities are zero.