Writing $c_i(t)=\langle\mathbf1_{[0,t]},f_i\rangle_{L^2}$ gives $X_n(t)=\sum_{i=1}^n\alpha_ic_i(t)$, hence
$$
X_n(t)\sim N\left(0,\sum_{i=1}^nc_i(t)^2\right).
$$
By <Parseval identity>, the variance tends to $\lVert\mathbf1_{[0,t]}\rVert_2^2=t$. Thus $X_n(t)\xrightarrow dN(0,t)$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-201/1/c/i/solution