 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-36/3/c/solution

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

The initial values are

β_{0} (t) = 1

and

β_{1} (t) = 0

. Repeatedly applying the Brownian moment recursion to odd indices reduces them to

β_{1}

, so every odd moment vanishes. For even indices, if the formula holds at

2 k - 2

, then

β_{2 k} (t) = k (2 k - 1) \int_{0}^{t} \frac{( 2 k - 2 )! s ^{k - 1}}{2 ^{k - 1} ( k - 1 )!} d s = \frac{( 2 k )!}{2 ^{k} k !} t^{k} .

The base case

k = 0

β_{0} = 1

. Thus

β_{2 k + 1} (t) = 0, β_{2 k} (t) = \frac{( 2 k )! t ^{k}}{2 ^{k} k !} (k \geq 0) .

Equivalently the even moments are

(2 k - 1)!! t^{k}

, the familiar Gaussian moment formula with variance

t

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