 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-33/4/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

For

x > 0

, reflect a Brownian path after its first hitting of

x

. The Strong Markov property and symmetry preserve its law and exchange the events

{S_{1} \geq x, B_{1} < x}

and

{B_{1} > x}

. The terminal equality event has probability zero, so the Brownian reflection principle gives

P (S_{1} \geq x) = 2 P (B_{1} \geq x), P (S_{1} \leq x) = 2Φ (x) - 1.

Continuity of this distribution also gives no atom at zero. Therefore

x ↓ 0 lim \frac{P ( S _{1} \leq x )}{x} = 2 Φ^{'} (0) = \frac{2}{π} .

A useful global bound follows from the maximum of the standard normal distribution density:

P (S_{1} \leq x) = 2 \int_{0}^{x} \frac{e ^{- u^{2} /2}}{2 π} d u \leq \frac{2}{π} x (x \geq 0) .

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