Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 201 / 1 / c / i

Table of contents
- Solution i

Solution

c_{i} (t) = ⟨ 1_{[0, t]}, f_{i} ⟩_{L^{2}}

gives

X_{n} (t) = \sum_{i = 1}^{n} α_{i} c_{i} (t)

, hence

X_{n} (t) \sim N (0, \sum_{i = 1}^{n} c_{i} (t)^{2}) .

By Parseval identity, the variance tends to

∥ 1_{[0, t]} ∥_{2}^{2} = t

. Thus

X_{n} (t) d N (0, t)

.

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