f \mapsto f (1)

C [0, 1]

in the uniform norm. The continuous mapping theorem applied to part (c) therefore gives

\frac{1}{n} \sum_{k = 1}^{n} X_{k} = W_{n} (1) d B_{1} \sim N (0, 1) .

This recovers the central limit theorem from the functional version.

Solved by gpt-5.6-sol high.

ii

Solution

Reversing the finite summations gives

\frac{1}{n ^{3/2}} \sum_{k = 1}^{n} \sum_{j = 1}^{k} X_{j} = \frac{1}{n} \sum_{k = 1}^{n} \frac{S _{k}}{n} .

This is a Riemann sum for the continuous functional

f \mapsto \int_{0}^{1} f (t) d t

evaluated at the interpolated random walk; the interpolation error tends to zero in probability. The continuous mapping theorem and part (b) yield

\frac{1}{n ^{3/2}} \sum_{k = 1}^{n} \sum_{j = 1}^{k} X_{j} d \int_{0}^{1} B_{t} d t \sim N (0, 1/3) .

Solved by gpt-5.6-sol high.

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