MathIsimple

Stochastic Processes – Problem 43: Prove that

Question

Let BtB_t be a one-dimensional (standard) Brownian motion with B0=0B_0 = 0. Let x>0x>0 and A[0,)A\subset[0,\infty) be a measurable set. Prove that

Px(B(s)0,0st,B(t)A)=Px(B(t)A)Px(B(t)A).P_{x}(B(s)\geq0,0\leq s\leq t,B(t)\in A)=P_{x}(B(t)\in A)-P_{-x}(B(t)\in A).

Step-by-step solution

Step 1. Let τ0=inf{s0:B(s)=0}\tau_0 = \inf\{s \ge 0: B(s) = 0\} denote the first hitting time of 00 by the Brownian motion. Assume the Brownian motion starts from x>0x > 0, and the target set is A[0,)A \subset [0, \infty). By the law of total probability, partitioning the event {B(t)A}\{B(t) \in A\} according to whether the path hits the origin before time tt, we obtain Px(B(t)A)=Px(τ0t,B(t)A)+Px(τ0>t,B(t)A)P_x(B(t) \in A) = P_x(\tau_0 \le t, B(t) \in A) + P_x(\tau_0 > t, B(t) \in A).

Step 2. For the first term Px(τ0t,B(t)A)P_x(\tau_0 \le t, B(t) \in A) on the right-hand side, apply the reflection principle of Brownian motion. The reflection principle states that for paths starting from x>0x > 0 that hit 00 before time tt, the post-τ0\tau_0 behavior is symmetric with that of a path starting from x-x. Specifically, the probability of starting from xx, hitting 00, and eventually reaching AA equals the probability of starting from x-x and eventually reaching AA. Since x<0-x < 0 and A[0,)A \subset [0, \infty), any path from x-x reaching AA must pass through 00, so Px(τ0t,B(t)A)=Px(B(t)A)P_x(\tau_0 \le t, B(t) \in A) = P_{-x}(B(t) \in A).

Step 3. For the second term Px(τ0>t,B(t)A)P_x(\tau_0 > t, B(t) \in A), since the starting point x>0x > 0 and the path is continuous, if the first hitting time τ0\tau_0 of 00 is greater than tt, then the path never touches 00 during [0,t][0, t], i.e., B(s)>0B(s) > 0 for all 0st0 \le s \le t (hence B(s)0B(s) \ge 0). Therefore this term equals the probability that the path remains non-negative on [0,t][0, t] and lands in AA at time tt, i.e., Px(τ0>t,B(t)A)=Px(B(s)0,0st,B(t)A)P_x(\tau_0 > t, B(t) \in A) = P_x(B(s) \ge 0, 0 \le s \le t, B(t) \in A).

Step 4. Substituting the results of Steps 2 and 3 into the decomposition from Step 1, we get Px(B(t)A)=Px(B(t)A)+Px(B(s)0,0st,B(t)A)P_x(B(t) \in A) = P_{-x}(B(t) \in A) + P_x(B(s) \ge 0, 0 \le s \le t, B(t) \in A). Rearranging this equation yields the desired conclusion.

Final answer

QED.

Marking scheme

This rubric strictly follows the official solution approach, with a total of 7 points. Please score according to the proof path used by the student, choosing Chain A or Chain B; do not mix chains.


1. Checkpoints (max 7 pts)

Select exactly one scoring path | If both are partially addressed, take the single-chain maximum; do not accumulate across chains.

Chain A: Event Decomposition and Reflection Principle (Official Approach)

  • Total probability decomposition [additive]
  • Use the first hitting time τ0\tau_0 to decompose Px(B(t)A)P_x(B(t)\in A) into two parts: {τ0t}\{ \tau_0 \le t \} (hit the origin) and {τ0>t}\{ \tau_0 > t \} (did not hit the origin).
  • +1 pt: Write a decomposition similar to Px(B(t)A)=Px(τ0t,B(t)A)+Px(τ0>t,B(t)A)P_x(B(t) \in A) = P_x(\tau_0 \le t, B(t) \in A) + P_x(\tau_0 > t, B(t) \in A).
  • Identify the target term (meaning of not hitting the origin) [additive]
  • +1 pt: State that for x>0x>0, the event {τ0>t}\{ \tau_0 > t \} is equivalent to the path staying strictly positive (or non-negative) on [0,t][0,t], thereby confirming this term is the left-hand side of the identity to be proved.
  • Apply the reflection principle (core step) [additive]
  • +3 pts: Apply the reflection principle to establish symmetry, stating that the probability of starting from xx, hitting the origin, and eventually reaching AA equals the probability of starting from x-x (hitting the origin) and reaching AA.
  • *Note: If the student writes Px(τ0t,B(t)A)=Px(τ0t,B(t)A)P_x(\tau_0 \le t, B(t) \in A) = P_{-x}(\tau_0 \le t, B(t) \in A) or directly writes =Px(B(t)A)= P_{-x}(B(t) \in A), full 3 pts are awarded.*
  • Argue "must pass through the origin" (geometric property) [additive]
  • +1 pt: Explicitly explain that since x<0-x < 0 and A[0,)A \subset [0, \infty), a continuous path from x-x to AA must pass through the origin, so the condition τ0t\tau_0 \le t is redundant (has probability 1) for the event {B(0)=x,B(t)A}\{B(0)=-x, B(t) \in A\}.
  • *Note: If the student writes the result directly without mentioning this geometric intuition or implying this logic, this 1 pt is not awarded.*
  • Algebraic rearrangement and conclusion [additive]
  • +1 pt: Substitute all terms back into the decomposition and rearrange to obtain the final conclusion.

Chain B: Density Function Computation (Integral Method)

  • Cite the reflection principle density formula [additive]
  • +4 pts: Directly write (or derive) the transition probability density for starting from x>0x>0, not touching 0 on [0,t][0,t], and B(t)=yB(t)=y (i.e., the absorbing-barrier Brownian motion density):

p(t,x,y)=12πt(e(yx)22te(y+x)22t)p^*(t, x, y) = \frac{1}{\sqrt{2\pi t}} \left( e^{-\frac{(y-x)^2}{2t}} - e^{-\frac{(y+x)^2}{2t}} \right).

  • *Note: This step contains the core application of the reflection principle and carries higher weight.*
  • Identify the first integral [additive]
  • +1 pt: Identify and state that A12πte(yx)22tdy\int_A \frac{1}{\sqrt{2\pi t}} e^{-\frac{(y-x)^2}{2t}} dy equals the unconditional probability Px(B(t)A)P_x(B(t) \in A).
  • Identify the second integral [additive]
  • +1 pt: Identify and state that A12πte(y+x)22tdy\int_A \frac{1}{\sqrt{2\pi t}} e^{-\frac{(y+x)^2}{2t}} dy equals Px(B(t)A)P_{-x}(B(t) \in A) (or Px(B(t)A)P_x(B(t) \in -A)).
  • Conclusion [additive]
  • +1 pt: Combine the integral terms and explain that this equals the required probability difference, completing the proof.

Total (max 7)


2. Zero-credit items

  • Merely copying the problem: Copying the identity to be proved verbatim with no intermediate steps.
  • Circular reasoning: Starting from the identity to be proved, performing algebraic manipulations to arrive at 0=00=0 or a tautology, without establishing the reversibility of the logic or with incorrect reasoning direction.
  • Empty citation: Writing only "by the reflection principle" without any specific probability decomposition, density formula, or geometric correspondence.
  • Incorrect premise: Assuming AA must be symmetric (e.g., [a,a][-a, a]) or assuming AA contains a negative part (the problem specifies A[0,)A \subset [0, \infty)), causing the argument to deviate entirely from the problem.

3. Deductions

Apply at most the single largest deduction (score cannot go below 0).

  • Logical gap (missing geometric argument): [-1]
  • In Chain A, the student correctly applies the reflection principle to obtain the x-x term, but completely fails to explain why Px(τ0t,)P_{-x}(\tau_0 \le t, \dots) simplifies to Px()P_{-x}(\dots) (i.e., does not mention the fact that "from negative to positive one must cross zero").
  • Notation confusion: [-1]
  • Seriously confusing probability P()P(\cdot) with probability density p()p(\cdot), or confusing the random variable B(t)B(t) with specific values x,yx, y, rendering expressions mathematically meaningless (but if the core idea is clear, deduct only 1 pt).
  • Missing measure-theoretic details: [no deduction]
  • If the student does not explicitly mention "AA is a measurable set" or use measure-theoretic language to extend from density integrals to general sets, no deduction; this is assumed understood at the undergraduate level.
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