 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-27/3/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

The nonnegative series in part (b) has independent squared-standard-normal terms. If

G

is standard normal, direct Gaussian integration gives

E e^{- a G^{2}} = (1 + 2 a)^{- 1/2}

for

a \geq 0

. Consequently, by independence and dominated convergence applied to the exponentials of increasing partial sums,

E exp (- λ \int_{0}^{1} W_{t}^{2} d t) = \prod_{n = 1}^{\infty} (1 + \frac{2 λ}{( n - \frac{1}{2} ) ^{2} π ^{2}})^{- 1/2} .

Use the permitted product identity with

x = 2 λ

. The answer is

E exp (- λ \int_{0}^{1} W_{t}^{2} d t) = \frac{1}{cosh 2 λ} (λ \geq 0) .

1

at zero. Its first derivative there gives mean

1/2

, in agreement with

\int_{0}^{1} E W_{t}^{2} d t

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