Let be a one-dimensional standard Brownian motion with . Given , prove that .

Stochastic Processes Solution – Problem 17: prove that

Question

Let $B_t$ be a one-dimensional standard Brownian motion with $B_0 = 0$ . Given $C>0$ , prove that $P(\sup_{0\leq\mu\leq t}|B(\mu)|>C)\leq t/C^{2}$ .

Step-by-step solution

Step 1. For standard Brownian motion $B_s$ , which is a continuous martingale, Doob's $L^p$ maximal inequality with $p = 2$ gives: $E\left[ \sup_{0 \le s \le t} |B_s|^2 \right] \le 4 \, E[ |B_t|^2 ].$ Since $E[|B_t|^2] = t$ , we obtain $E\left[ \sup_{0 \le s \le t} |B_s|^2 \right] \le 4t.$

Step 2. For the nonneg. r.v. $X = \sup_{0 \le s \le t} |B_s|^2$ , Markov's inequality gives: $P( X > C^2 ) \le \frac{E[X]}{C^2} \le \frac{4t}{C^2}.$ This bound $4t/C^2$ is weaker than the required $t/C^2$ , so a more refined method is needed.

Step 3. Define $\tau = \inf\{ s \ge 0 : |B_s| > C \}$ . Then $\tau \wedge t$ is a bounded stopping time. Since $B_t^2 - t$ is a martingale, by the Optional Stopping Theorem: $E[ B_{\tau \wedge t}^2 - (\tau \wedge t) ] = 0$ , so $E[ B_{\tau \wedge t}^2 ] = E[ \tau \wedge t ]$ .

Step 4. On the event $\{ \tau \le t \}$ , we have $|B_{\tau \wedge t}| = |B_\tau| \ge C$ , hence $B_{\tau \wedge t}^2 \ge C^2$ . On the event $\{ \tau > t \}$ , $B_{\tau \wedge t}^2 = B_t^2 \ge 0$ . Therefore: $E[ B_{\tau \wedge t}^2 ] \ge C^2 P(\tau \le t)$ . Since $E[ \tau \wedge t ] \le t$ : $C^2 P(\tau \le t) \le E[ B_{\tau \wedge t}^2 ] = E[ \tau \wedge t ] \le t.$ Hence: $P(\tau \le t) \le \frac{t}{C^2}$ , and so $P\left( \sup_{0 \le s \le t} |B_s| > C \right) = P(\tau \le t) \le \frac{t}{C^2}.$

Final answer

QED.

Marking scheme

This rubric is based on the official solution (Optional Stopping Theorem) and the mathematically equivalent standard approach (Doob's inequality).

1. Checkpoints (max 7 pts total)

Scoring rule: The following contains two parallel solution paths (Chain A / Chain B). Score only the path yielding the highest marks; do not combine points across paths.

Chain A: Optional Stopping Theorem Path (Official Solution, Steps 3--4)

Define the stopping time $\tau = \inf\{s \ge 0 : |B_s| > C\}$ and explicitly consider the truncated stopping time $\tau \wedge t$ (or explain the need for boundedness). [1 pt]
Construct or identify $M_s = B_s^2 - s$ as a martingale. [1 pt]
Apply the Optional Stopping Theorem to the bounded stopping time $\tau \wedge t$ , obtaining $E[B_{\tau \wedge t}^2] = E[\tau \wedge t]$ or $E[B_{\tau \wedge t}^2 - (\tau \wedge t)] = 0$ . [2 pts]
Key estimate: Use the definition of the stopping time to show that on $\{\tau \le t\}$ , $|B_{\tau \wedge t}| \ge C$ , thereby establishing the lower bound $E[B_{\tau \wedge t}^2] \ge C^2 P(\tau \le t)$ . [2 pts]
Combine with $E[\tau \wedge t] \le t$ to complete the algebraic derivation. [1 pt]

Chain B: Doob / Kolmogorov Inequality Path (Direct Method)

Identify $X_s = B_s^2$ as a nonneg. submartingale, or directly cite Kolmogorov's inequality for continuous martingales. [2 pts]
Correctly state Doob's maximal probability inequality: $P(\sup_{0\le s \le t} X_s \ge \lambda) \le \frac{1}{\lambda} E[X_t]$ . [3 pts]
Substitute $\lambda = C^2$ and $E[B_t^2] = t$ to obtain the final result. [2 pts]

Total (max 7)

2. Zero-credit items

Merely copying the problem statement, the definition of Brownian motion, or the standard variance formula $E[B_t^2]=t$ with no subsequent derivation.
Attempting to bound the probability via the fixed-time tail $P(|B_t| > C)$ (i.e., normal tail probability) instead of the supremum probability; this is a conceptual error.
Asserting the conclusion holds by intuition without any martingale theory or inequality.

3. Deductions

*(Apply the most severe single deduction; total score cannot go below 0.)*

Inequality citation error (Cap at 3/7): Using the $L^p$ -norm inequality (as in Steps 1--2, yielding $4t/C^2$ ) without further refinement via OST: total capped at 3 pts.
Stopping time rigor (-1): In Chain A, applying OST directly to the unbounded stopping time $\tau$ without mentioning truncation $\tau \wedge t$ or justifying the limit.
Logical gap (-1): In Chain A, not explaining why $E[B_{\tau \wedge t}^2] \ge C^2 P(\tau \le t)$ (i.e., omitting the case analysis on $\{\tau \le t\}$ vs. $\{\tau > t\}$ ).
Notation confusion (-1): Confusing random variables with constants, or failing to distinguish $B_t$ from $|B_t|$ .

Stochastic Processes – Problem 17: prove that