Suppose is a standard Brownian motion, and suppose is a continuous function with . Prove that for any ,

Stochastic Processes Solution – Problem 12: Prove that for any ,

Question

Suppose $(B_{t})$ is a standard Brownian motion, and suppose $f$ is a continuous function with $f(0)=0$ . Prove that for any $\epsilon>0$ ,

$\mathbb{P}\left(\sup_{0\leq t\leq1}|B_{t}-f(t)|\leq\epsilon\right)>0.$

Step-by-step solution

Step 1. Since $f$ is continuous on the closed interval $[0,1]$ , it is uniformly continuous. Given $\epsilon > 0$ , there exists $\delta > 0$ such that whenever $t_1, t_2 \in [0,1]$ with $|t_1 - t_2| \leq \delta$ , we have $|f(t_1) - f(t_2)| \leq \frac{\epsilon}{2}$ . Choose a positive integer $n$ such that $\frac{1}{n} \leq \delta$ , and partition $[0,1]$ into $n$ equal subintervals with partition points $t_0 = 0, t_1 = \frac{1}{n}, t_2 = \frac{2}{n}, \dots, t_n = 1$ . Then for each $k = 1, 2, \dots, n$ , we have $|f(t_k) - f(t_{k-1})| \leq \frac{\epsilon}{2}$ .

Step 2. Define the event $A$ as the event that $|B_{t_k} - f(t_k)| \leq \frac{\epsilon}{2}$ for all $k = 0, 1, \dots, n$ . Note that $B_{t_0} = B_0 = 0$ and $f(0) = 0$ , so the inequality at $k=0$ holds automatically. The finite-dimensional distributions of Brownian motion are multivariate normal, and such distributions have strictly positive joint density on all of $\mathbb{R}^n$ . Therefore, the probability $\mathbb{P}(A)$ is strictly positive.

Step 3. For any point $t \in [0,1]$ , there exists some $k = 1, 2, \dots, n$ such that $t$ lies in $[t_{k-1}, t_k]$ . By the triangle inequality: $|B_t - f(t)| \leq |B_t - B_{t_k}| + |B_{t_k} - f(t_k)| + |f(t_k) - f(t)|$ Since $t \in [t_{k-1}, t_k]$ , the distance between $t$ and $t_k$ is at most $\frac{1}{n} \leq \delta$ , so by uniform continuity of $f$ , $|f(t_k) - f(t)| \leq \frac{\epsilon}{2}$ . Since event $A$ includes $|B_{t_k} - f(t_k)| \leq \frac{\epsilon}{2}$ , combined with the continuity of Brownian motion paths and the boundedness of $B_{t_{k-1}}, B_{t_k}$ , we can deduce $|B_t - B_{t_k}| \leq \frac{\epsilon}{2}$ . Substituting into the inequality: $|B_t - f(t)| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Step 4. From Step 3, whenever event $A$ occurs, $\sup_{0 \leq t \leq 1} |B_t - f(t)| \leq \epsilon$ necessarily holds, meaning $A$ is a subset of $\left\{ \sup_{0 \leq t \leq 1} |B_t - f(t)| \leq \epsilon \right\}$ . By monotonicity of probability, since $\mathbb{P}(A) > 0$ : $\mathbb{P}\left( \sup_{0 \leq t \leq 1} |B_t - f(t)| \leq \epsilon \right) \geq \mathbb{P}(A) > 0$

Final answer

QED.

Marking scheme

This rubric is designed for an undergraduate/graduate level mathematics grader, targeting the Support Theorem for Brownian motion. The rubric is based on the official discretization proof path while also accommodating the Girsanov transform approach from advanced probability theory.

1. Checkpoints (max 7 pts total)

Score exactly one chain; take the maximum subtotal among chains; do not add points across chains.

Chain A: Discretization and Uniform Continuity Method (Official Solution Approach)

Uniform continuity and grid setup (2 pts)
State that $f$ is continuous on the closed interval $[0,1]$ and hence uniformly continuous, or directly argue that a sufficiently fine partition can be chosen. [1 pt]
Establish specific partition points $t_0, t_1, \dots, t_n$ (or choose $n$ ) so that the oscillation of $f$ on each subinterval $|f(t) - f(s)|$ is less than a specific threshold (such as $\epsilon/2$ or $\epsilon/3$ ). [1 pt]
Positive probability of finite-dimensional distributions (2 pts)
Define the skeleton event $A$ , requiring $B_t$ at partition points $t_k$ to fall within a neighborhood of $f(t_k)$ (e.g., $|B_{t_k} - f(t_k)| \leq \delta$ ). [1 pt]
Cite that the finite-dimensional distributions of Brownian motion (multivariate normal) have strictly positive density on $\mathbb{R}^n$ , thereby concluding $\mathbb{P}(A) > 0$ . [1 pt]
Path estimate and conclusion (3 pts)
Use the triangle inequality to decompose the path error: for any $t$ , $|B_t - f(t)| \leq |B_t - B_{t_k}| + |B_{t_k} - f(t_k)| + |f(t_k) - f(t)|$ (or similar form). [1 pt]
Use the path continuity of Brownian motion to control the interpolation term $|B_t - B_{t_k}|$ (i.e., show there exist paths making this term sufficiently small, or as in the official solution, derive the bound using continuity). [1 pt]
Use the set inclusion relation and monotonicity of probability to conclude: $\mathbb{P}(\sup |B_t - f(t)| \leq \epsilon) \geq \mathbb{P}(A) > 0$ . [1 pt]

Chain B: Girsanov Transform Method (Advanced Probability Approach)

Approximation and space setup (2 pts)
State that if $f$ is not smooth enough (not in the Cameron-Martin space), one needs to approximate $f$ by smooth functions (e.g., polynomials) $g$ , reducing the problem to proving $B_t$ has positive probability in a tubular neighborhood of $g$ . [1 pt]
Define the transformed process $\tilde{B}_t = B_t - g(t)$ (or $B_t - f(t)$ if $f$ is assumed smooth) and prepare to use Girsanov's theorem. [1 pt]
Measure change construction (2 pts)
Correctly write the Radon-Nikodym derivative (likelihood ratio) $Z = \exp(\int_0^1 \dot{g}(s)dB_s - \frac{1}{2}\int_0^1 \dot{g}(s)^2 ds)$ . [1 pt]
State that under the new measure $\mathbb{Q}$ , $\tilde{B}_t$ is a standard Brownian motion. [1 pt]
Positivity of probability and conclusion (3 pts)
State that $Z > 0$ almost surely (i.e., the two measures are equivalent). [1 pt]
Cite that the tubular neighborhood probability of standard Brownian motion near $0$ is positive ( $\mathbb{P}(\sup_{t \in [0,1]} |B_t| \leq \epsilon) > 0$ ). [1 pt]
Combine to conclude the original probability is positive. [1 pt]

Total (max 7)

2. Zero-credit items

Only copying the given conditions from the problem ( $B_t$ is Brownian motion, $f$ is continuous, etc.).
Only asserting by intuition that "Brownian motion can reach any point in space, so the probability is positive" without any mathematical derivation or analysis of continuity/measures.
Listing the definition of Brownian motion (e.g., independent increments, normality) without applying it to prove the probability near $f$ .

3. Deductions

Logic gap (-2 pts): In Chain A, if the logical connection "discrete points close implies entire path close" is not established via the triangle inequality or other means, and only the finite-point probability is computed before directly claiming the proof is complete.
Measure confusion (-1 pt): In Chain B, if the $P$ -measure and $Q$ -measure expectation/probability symbols are confused, causing logical inconsistency.
Constant handling imprecision (no deduction): If only the fitting of coefficients $\epsilon/2, \epsilon/3$ , etc., is imprecise but does not affect the core logic, no deduction.

Stochastic Processes – Problem 12: Prove that for any ,