 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-38/1/a/solution

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

Write

B = β

for the driving Brownian motion throughout; the two symbols in the displayed formulas refer to the same process. For

t < 1

the deterministic integrand is square-integrable on

[0, t]

, so

I_{t} = \int_{0}^{t} \frac{d B _{s}}{1 - s}, X_{t} = (1 - t) I_{t}

is well defined. The Itô product rule, with zero quadratic covariation against the deterministic factor, gives

d X_{t} = - I_{t} d t + (1 - t) \frac{d B _{t}}{1 - t} = d B_{t} - \frac{X _{t}}{1 - t} d t, X_{0} = 0.

For the other form, apply the same Itô product rule to

B_{t} / (1 - t)

\frac{B _{t}}{1 - t} = \int_{0}^{t} \frac{d B _{s}}{1 - s} + \int_{0}^{t} \frac{B _{s}}{( 1 - s ) ^{2}} d s .

Multiplying by

1 - t

X_{t} = B_{t} - (1 - t) \int_{0}^{t} \frac{B _{s}}{( 1 - s ) ^{2}} d s, t < 1.

The drift coefficient is singular only at the terminal endpoint, so this calculation is performed on compact subintervals of

[0, 1)

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