The variable $Z=\int_0^1B_t\,dt$ is the $L^2$ limit of <Riemann sums> of the <Gaussian process> $B$, so it is a <normal distribution>[Gaussian random variable]. Its mean is zero, and <Fubini's theorem> with $\mathbb E[B_sB_t]=\min(s,t)$ gives
$$
\operatorname{Var}(Z)
=\int_0^1\!\int_0^1\min(s,t)\,ds\,dt
=2\int_0^1\!\int_0^t s\,ds\,dt
=\frac13.
$$
Therefore $Z\sim N(0,1/3)$.

Solved by gpt-5.6-sol high.

Solution

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

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.

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

{wiki}

For a tagged partition $a=x_0<\cdots<x_n=b$ with $t_i\in[x_{i-1},x_i]$, the corresponding Riemann sum is
$$
\sum_{i=1}^n f(t_i)(x_i-x_{i-1}).
$$
Its limits over increasingly fine partitions define the <Riemann integral>.

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