 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/29k/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

B_{u} = u W_{1/ u}

. For

s > 0

u = 1/ s

. Then

W_{s} - a s \leq b ⟺ \frac{W _{s}}{s} - a \leq \frac{b}{s} ⟺ B_{u} - b u \leq a .

0 < s \leq t

corresponds exactly to

u \geq 1/ t

, and the inequality at

s = 0

is automatic because

b > 0

, this is the pathwise identity

{sup_{0 \leq s \leq t} (W_{s} - a s) \leq b} = {sup_{u \geq 1/ t} (B_{u} - b u) \leq a} .

Since

B

and

W

P (0 \leq s \leq t sup (W_{s} - a s) \leq b) = P (u \geq 1/ t sup (W_{u} - b u) \leq a) .

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