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

Stochastic Processes Solution – Problem 28: prove that

Question

Let $B_t$ be a one-dimensional (standard) Brownian motion with $B_0 = 0$ . Given $a>0, b>0$ , prove that $P(B(s)\neq 0, 0<s<t \mid B(0)=a, B(t)=b) = 1 - e^{-2ab/t}$ .

Step-by-step solution

Step 1. By the reflection principle, for a Brownian motion starting at $B(0)=a$ and ending at $B(t)=b$ , if the path touches the origin $0$ during the time interval $(0, t)$ , then there is a one-to-one correspondence between such paths and Brownian motion paths starting from $-a$ and ending at $b$ . The unconditional transition probability density from $a$ to $b$ is: $p_t(a, b) = \dfrac{1}{\sqrt{2\pi t}} \exp\left( -\dfrac{(b-a)^2}{2t} \right).$

Step 2. Correspondingly, the transition probability density from the reflected point $-a$ to $b$ is: $p_t(-a, b) = \dfrac{1}{\sqrt{2\pi t}} \exp\left( -\dfrac{(b+a)^2}{2t} \right).$ This density $p_t(-a, b)$ corresponds to the contribution from paths starting at $a$ , reaching $b$ , and touching $0$ along the way.

Step 3. Under the condition $B(0)=a, B(t)=b$ , the probability that the path touches $0$ during $(0, t)$ is the ratio of the two densities: $P(\exists s \in (0, t), B(s)=0 \mid B(0)=a, B(t)=b) = \dfrac{p_t(-a, b)}{p_t(a, b)}.$ Substituting the density formulas yields: $\dfrac{p_t(-a, b)}{p_t(a, b)} = \dfrac{\exp\left( -\dfrac{(b+a)^2}{2t} \right)}{\exp\left( -\dfrac{(b-a)^2}{2t} \right)} = \exp\left( \dfrac{(b-a)^2 - (b+a)^2}{2t} \right).$

Step 4. Computing the difference in the exponent: $(b-a)^2 - (b+a)^2 = (b^2 - 2ab + a^2) - (b^2 + 2ab + a^2) = -4ab.$ Therefore the probability of touching $0$ is: $\exp\left( \dfrac{-4ab}{2t} \right) = e^{-2ab/t}.$ The desired event $B(s) \neq 0$ means the path does not touch $0$ , so its probability is: $P(B(s) \neq 0, 0 < s < t \mid B(0)=a, B(t)=b) = 1 - e^{-2ab/t}.$

Final answer

QED.

Marking scheme

The following is the grading rubric for this problem. Please grade according to the three parts below, for a total of 7 points.

1. Checkpoints (Key Scoring Points, Total 7 Points)

Note: If the student solves the problem via the "method of images" for the heat equation, this is logically equivalent to using the reflection principle, and the corresponding scoring points below apply.

Unconditional transition density (denominator) [1 point]
Correctly writes (or implicitly uses in the computation) the unconditional transition probability density from $B(0)=a$ to $B(t)=b$ :

$p_t(a, b) = \frac{1}{\sqrt{2\pi t}} \exp\left( -\frac{(b-a)^2}{2t} \right)$

(Note: If the student directly writes a joint probability expression containing this term as a normalizing factor or denominator, credit is also given.)

Core application of the reflection principle [3 points] [Additive]
Statement/logical invocation of the principle [2 points]: Clearly states that paths from $a$ to $b$ that touch the origin $0$ are in one-to-one correspondence with paths from $-a$ to $b$ (or from $a$ to $-b$ ); or directly invokes the "reflection principle"/"method of images" to construct the probability of touching $0$ .
Reflection term expression [1 point]: Correctly writes the density function, probability measure, or heat kernel term corresponding to the reflected path:

$p_t(-a, b) = \frac{1}{\sqrt{2\pi t}} \exp\left( -\frac{(b+a)^2}{2t} \right)$

Constructing the conditional probability [2 points]
Establishes the correct conditional probability expression. Identifies that the "probability of touching the origin" equals the ratio of the "reflected path density" to the "direct path density":

$P(\text{hit } 0) = \frac{p_t(-a, b)}{p_t(a, b)}$

Alternatively: if using the method of images, directly writes the joint density of not touching the origin as $p_t(a,b) - p_t(-a,b)$ , and divides by $p_t(a,b)$ for normalization.
Algebraic simplification and final conclusion [1 point]
Correctly computes the difference in the exponent ( $(b-a)^2 - (b+a)^2 = -4ab$ ), obtains $e^{-2ab/t}$ , and uses the complement idea ( $1 - P(\text{hit})$ ) to arrive at the final result.
(Note: Simply writing the last line $1-e^{-2ab/t}$ without showing the exponent simplification does not earn this point.)

Total (max 7)

2. Zero-credit items

Only copying the given conditions from the problem (e.g., $B_0=0, a>0, b>0$ ).
Only listing the standard definition of Brownian motion (e.g., $B_t \sim N(0,t)$ ) without applying it to the endpoints $a, b$ .
Directly writing down the extreme value distribution formula for the Brownian bridge without any derivation based on the reflection principle or density integration.
Claiming "obviously" or "by a theorem in the textbook" and directly stating the conclusion.

3. Deductions

Failure to take the complement [-1]: Correctly derives $e^{-2ab/t}$ but mistakenly presents it as the final answer (i.e., computes the probability of touching the origin rather than not touching it), without further explanation.
Algebraic/sign error [-1]: A sign error occurs during the expansion or simplification of the exponent (e.g., computing $(b-a)^2 - (b+a)^2$ incorrectly), causing the final result to have an incorrect form.
Logical gap [Cap at 4/7]: Mentions the reflection principle but does not write out the specific density functions or ratio formula, jumping directly to the final result.

Stochastic Processes – Problem 28: prove that