By the ,
$$
\frac{T_n}{n}\longrightarrow\sigma^2
$$
almost surely. Brownian scaling and a maximal inequality show that changing Brownian time by $o(n)$ changes its value by $o_{\mathbb P}(\sqrt n)$; explicitly, first restrict to $|T_n-n\sigma^2|\leq\delta n$, bound the Brownian maximum over a time interval of length $2\delta n$, and then let $\delta\downarrow0$. Consequently
$$
\frac{B_{T_n}-B_{n\sigma^2}}{\sqrt n}\longrightarrow0
$$
in probability.

But
$$
\frac{B_{n\sigma^2}}{\sqrt n}\sim N(0,\sigma^2)
$$
for every $n$. Since $B_{T_n}$ has the law of $S_n$, <Slutsky theorem> proves the <Central limit theorem from the Skorokhod embedding>:
$$
\boxed{\frac{S_n}{\sqrt n}\ \xrightarrow{d}\ N(0,\sigma^2)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 5 / b / Solution

{c}

The law of large numbers gives $T_n/n\to\sigma^2$. Brownian maximal estimates then show
$$
\frac{B_{T_n}-B_{n\sigma^2}}{\sqrt n}\longrightarrow0
$$
in probability. Since $B_{n\sigma^2}/\sqrt n\sim N(0,\sigma^2)$, the embedded random walk satisfies the central limit theorem.


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 Incoming links: Central limit theorem from the Skorokhod embedding