Let . Prove that and have the same distribution.

Stochastic Processes Solution – Problem 5: Prove that and have the same distribution

Question

Let $M_{t}=\operatorname*{max}_{0\leq s\leq t}B_{s}$ . Prove that $M(t)-B(t)$ and $M(t)$ have the same distribution.

Step-by-step solution

Step 1. By the reflection principle and the distributional properties of the Brownian motion maximum, for a standard Brownian motion $B_t$ , the maximum $M_t = \max_{0 \le s \le t} B_s$ has the same distribution as $|B_t|$ . That is, for any $a > 0$ , $P(M_t > a) = P(|B_t| > a) = 2P(B_t > a)$ .

Step 2. Consider the process $M_t - B_t$ . By Levy's theorem, the process $Y_t = M_t - B_t$ is equal in distribution to the reflected Brownian motion $|B_t|$ as a stochastic process. That is, $\{M_t - B_t\}_{t \ge 0}$ and $\{|B_t|\}_{t \ge 0}$ share the same finite-dimensional distributions. This conclusion implies not only that they are identically distributed at each fixed time $t$ , but also that they are equal in law as processes.

Step 3. By the conclusion of Levy's theorem above, for a fixed time $t$ , the random variable $M_t - B_t$ has the same distribution as $|B_t|$ .

Step 4. Returning to the distribution of $M_t$ : by Step 1, $M_t$ and $|B_t|$ are identically distributed. Since $M_t - B_t$ and $|B_t|$ are identically distributed, and $M_t$ is also identically distributed with $|B_t|$ , by transitivity we conclude that $M_t - B_t$ and $M_t$ have the same distribution.

Step 5. Direct verification via the joint density. The joint probability density function of $(M_t, B_t)$ is: $f(m, x) = \frac{2(2m-x)}{\sqrt{2\pi t^3}} \exp\left(-\frac{(2m-x)^2}{2t}\right), \quad m > 0, x \le m$ Let $Y = M_t - B_t$ . Since $M_t$ has the same distribution as $|B_t|$ , the density of $M_t$ is $f_{M_t}(m) = \sqrt{\frac{2}{\pi t}} e^{-m^2/2t}, m > 0$ . Levy's theorem directly shows that $\{M_t - B_t\}$ is equal in distribution to $\{|B_t|\}$ , and $\{M_t\}$ is equal in distribution to $\{|B_t|\}$ . Therefore $M_t - B_t \stackrel{d}{=} |B_t| \stackrel{d}{=} M_t$ .

Final answer

QED.

Marking scheme

The following rubric is designed for an undergraduate-level mathematics grading of this problem.

1. Checkpoints (max 7 pts total)

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

Chain A: Theoretical Citation Path (Based on Levy's Theorem or Reflection Principle)

[2 pts] Establishing the distribution of $M_t$ : Explicitly state that $M_t$ and $|B_t|$ are identically distributed, or write out the density/distribution function of $M_t$ (citing the reflection principle or maximum distribution properties).
[4 pts] Core argument ( $M_t - B_t$ distribution): Explicitly cite Levy's Theorem stating that the process $\{M_t - B_t\}_{t \ge 0}$ (or the random variable $M_t - B_t$ ) is equal in distribution to $\{|B_t|\}_{t \ge 0}$ (or $|B_t|$ ). Alternative: If the theorem name is not cited but a detailed probabilistic analysis, diagram, or reflection-transformation argument correctly establishes the distributional relationship between $M_t - B_t$ and reflected Brownian motion, award full 4 pts. If only asserting $M_t - B_t \stackrel{d}{=} |B_t|$ without any justification, theorem name, or derivation, this core step receives 0 pts.
[1 pt] Conclusion: Combine the above two points ( $M_t \stackrel{d}{=} |B_t|$ and $M_t - B_t \stackrel{d}{=} |B_t|$ ) and use transitivity to conclude $M_t - B_t \stackrel{d}{=} M_t$ .

Chain B: Analytical Computation Path (Based on Joint Density Function)

[2 pts] Write out the joint density: Correctly write the joint probability density function of $(M_t, B_t)$ : $f(m, x) = \frac{2(2m-x)}{\sqrt{2\pi t^3}} e^{-(2m-x)^2/2t}$ (with implicit or explicit $m>0, x \le m$ ).
[2 pts] Set up the integral/change of variables: Let $Y = M_t - B_t$ and correctly set up the integral expression for the marginal density of $Y$ . If the integration limits are clearly wrong (e.g., not accounting for $m \ge 0$ ), this item receives 0 pts.
[2 pts] Execute the integral computation: Correctly compute the above integral, deriving the half-normal density $\sqrt{\frac{2}{\pi t}} e^{-y^2/2t}$ (or equivalent form). If the main steps are correct but there is a minor coefficient error, deduct 1 pt; if the computation logic is confused leading to a patched-together result, this item receives 0 pts.
[1 pt] Comparison and conclusion: Explicitly state that the computed density matches that of $M_t$ (or $|B_t|$ ), thereby completing the proof.

Total (max 7)

2. Zero-credit items

Merely recopying the problem conditions or the definitions of $M_t, B_t$ with no substantive derivation.
Only verifying that first or second moments are equal (e.g., $E[M_t - B_t] = E[M_t]$ ) without proving distributional equality.
Incorrectly assuming $M_t$ and $B_t$ are independent and attempting to derive the result by directly subtracting marginal distributions.
Merely asserting by intuition that "by symmetry..." without any concrete support from the reflection principle, Levy's theorem, or integral computation.

3. Deductions

(-1) Imprecise logical statements: Confusing "almost sure equality" with "equality in distribution ( $\stackrel{d}{=}$ )" (e.g., claiming $M_t - B_t = M_t$ ), or omitting necessary domain specifications ( $m>0, y>0$ ) in differential equations/integrals.
(-2) Integration limit errors in the computation path: In Chain B, if the integration variable bounds violate the constraints $M_t \ge B_t$ or $M_t \ge 0$ .
(-1) Notational confusion: Confusing random variables (uppercase) with realized values (lowercase), causing logical reading difficulties.

Stochastic Processes – Problem 5: Prove that and have the same distribution