Theorem 1.6.6: Donsker | Brownian Motion and Stochastic Calculus

Theorem 1.6.6 (Donsker).label Let $\seq{S_n}$ be the symmetric random walk on $\integer$. For each $n \in \nat$, let $S^{n}_{t} = L_{n}P_{n}$ be the $n$-adic linearisation of

\[P_{n}: \bracs{k/n|k \in \nat_0}\to \real \quad \frac{k}{n}\mapsto \frac{S_{k}}{\sqrt{n}}\]

Let $B$ be the standard Brownian motion, then $S^{n} \to B$ in distribution as $n \to \infty$.

Proof. For any $0 \le s < t \le <\infty$,

\[S^{n}_{t} - S^{n}_{s} = \frac{S_{n\fl{t}_n}- S_{n\fl{s}_n}}{\sqrt{n}}+ \frac{R_{n, s, t}}{\sqrt{n}}\]

where $\sup_{n \in \nat}\norm{R_{n, s, t}}_{L^\infty}< \infty$^[1], so $R_{n, s, t}/\sqrt{n}\to 0$ in $L^{\infty}$ as $n \to \infty$.

Suppose that $S^{n}$ converges in distribution. By the above calculation, it is sufficient to consider the convergence in distribution of $(S_{n\floor{t}_n}- S_{n\floor{s}_n})/\sqrt{n}$.

Since $\sqrt{n}(\fl{t}_{n} - \fl{s}_{n})/\sqrt{n}\to (t - s)$ as $n \to \infty$, for any $0 \le t_{1} < \cdots < t_{k} < \infty$,

\[\frac{1}{\sqrt{n}}\paren{S_{n\fl{t_2}_n} - S_{n\fl{t_1}_n}, \cdots, S_{n\fl{t_k}_n} - S_{n\fl{t_{k-1}}_n}}\to (X_{1}, \cdots, X_{k})\]

in distribution by the central limit theorem, where $\bracs{X_j}_{1}^{k}$ are independent random variables with $X_{j} \sim \gamma_{0, t_{j+1} - t_{j}}$. Therefore $S^{n} \to B$ in distribution.

To see that $S^{n}$ does converge in distribution, it is sufficient to show that it is tight. Given that $S^{n}_{0} = 0$ for all $n \in \nat$, it is sufficient to check (2) of Proposition 1.6.3.

Let $\lambda > 0$. To achieve the appropriate order of bound, apply Doob’s maximal inequality to $S_{n}^{4}$.

\begin{align*}\bp\bracs{\max_{1 \le k \le n}|S_k| > \lambda \sqrt{n}}&= \bp\bracs{\max_{1 \le k \le n}|S_k^4| > \lambda^4n^2}\le \frac{1}{\lambda^{4}n^{2}}\ev\braks{|S_n^4|}\\&= \frac{1}{\lambda^{4} n^{2}}\braks{n + {n \choose 2}{4 \choose 2}}\\&= \frac{3n^{2} - 2n}{\lambda^{4}n^{2}}\le \frac{3}{\lambda^{4}}\end{align*}

For any $\delta > 0$ and $\eps > 0$, if $3/\sqrt{N}< \eps$, then for any $0 \le s < t$ with $t - s < \delta$, and $n \ge N$,

\begin{align*}\bp\bracs{|S_t^n - S^n_s| \ge 2\eps}&\le \bp\bracs{|S_{n\floor{t}_n} - S_{n\floor{s}_n}| \ge \frac{\eps}{\sqrt{\delta}}\sqrt{n \delta}}\\&\le \frac{3 \delta^{2}}{\eps^{4}}\end{align*}

In particular, for any $\eta > 0$, a choice of $\delta = \eta \eps^{4}/3$ yields

\[\bp\bracs{|S_t^n - S^n_s| \ge 2\eps}\le \frac{3 \delta^{2}}{\eps^{4}}= \frac{\eta^{2}\eps^{4}}{3}= \eta \delta\]

Partition the interval $[0, N]$ into $\lceil N/\delta \rceil$ pieces of of intervals of length $\delta$, then

\[\bp\bracs{\sup_{\substack{0 \le s < t \\ t - s < \delta}}|S_t^n - S_s^n| \ge 2\eps}\le (1 + N/\delta)\eta \delta = \eta \delta + N\eta\]

$\square$

I believe that 3 is an easy bound, but I cannot be bothered to check.keyboard_return

Direct References

circleProposition 1.6.3