Reversing the finite summations gives
$$
\frac1{n^{3/2}}\sum_{k=1}^n\sum_{j=1}^kX_j
=\frac1n\sum_{k=1}^n\frac{S_k}{\sqrt n}.
$$
This is a <Riemann sum> for the continuous functional $f\mapsto\int_0^1f(t)\,dt$ evaluated at the interpolated random walk; the interpolation error tends to zero in probability. The <continuous mapping theorem> and part (b) yield
$$
\frac1{n^{3/2}}\sum_{k=1}^n\sum_{j=1}^kX_j
\xrightarrow d\int_0^1B_t\,dt
\sim N(0,1/3).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 201 / 4 / d / ii / Solution

The evaluation map $f\mapsto f(1)$ is <continuous function>[continuous] on $C[0,1]$ in the <uniform norm>. The <continuous mapping theorem> applied to part (c) therefore gives
$$
\frac1{\sqrt n}\sum_{k=1}^nX_k=W_n(1)
\xrightarrow d B_1\sim N(0,1).
$$
This recovers the <central limit theorem> from the functional version.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 201 / 4 / d / i / Solution

{wiki}

If $X_n\xrightarrow dX$ and a measurable function $g$ is continuous at $X$ almost surely, then $g(X_n)\xrightarrow d g(X)$. The same principle holds for random elements of metric spaces.


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 Incoming links: Continuous mapping theorem