Solution: Why $\sqrt{n}$ Is the Right Rescaling for Donsker

Exercise: Why $\sqrt{n}$ Is the Right Rescaling for Donsker

Part 1 — Variance of the rescaled process

For $t \in [0, 1]$ , let $k_n = \lfloor nt \rfloor$ . Then $k_n / n \to t$ , and:

\operatorname{Var}\!\left(W_n^{(\alpha)}(t)\right) = \operatorname{Var}(n^{-\alpha} S_{k_n}) = n^{-2\alpha} \operatorname{Var}(S_{k_n}) = n^{-2\alpha} k_n \sim n^{1 - 2\alpha} t

In the limit $n \to \infty$ :

Case	$\lim_{n\to\infty} \operatorname{Var}(W_n^{(\alpha)}(t))$
$\alpha < \tfrac{1}{2}$	$+\infty$ (exponent $1 - 2\alpha > 0$ )
$\alpha = \tfrac{1}{2}$	$t$ (exponent equals zero)
$\alpha > \tfrac{1}{2}$	$0$ (exponent $1 - 2\alpha < 0$ )

Part 2 — Why the non- $\frac{1}{2}$ choices fail

$\alpha < 1/2$ (under-rescaling). The variance blows up: the process at any fixed $t$ becomes arbitrarily spread out. No meaningful probability distribution emerges in the limit — almost every path has infinite amplitude. Pictorially, the rescaled path explodes.
$\alpha > 1/2$ (over-rescaling). The variance collapses to zero: the process concentrates at the origin. By Chebyshev's inequality, $W_n^{(\alpha)}(t) \to 0$ in probability for every $t$ , so the limit is the trivial constant-zero process. The rescaling has washed out all the randomness.
$\alpha = 1/2$ (Goldilocks). The variance stabilises at $t$ — finite, non-zero, and dependent on $t$ in exactly the right way. This is the unique rescaling that preserves both the non-degeneracy and the time-dependence of the process.

The

\sqrt{n}

rescaling is the only power-law rescaling that produces a non-trivial scaling limit.

Part 3 — Covariance structure

For $0 \le s \le t \le 1$ , let $k_n = \lfloor ns \rfloor$ and $m_n = \lfloor nt \rfloor$ , so $k_n \le m_n$ . Using $\operatorname{Cov}(S_{k_n}, S_{m_n}) = \min(k_n, m_n) = k_n$ (from independent increments, exactly as in the parent lesson's covariance calculation):

\operatorname{Cov}\!\left(W_n^{(1/2)}(s), W_n^{(1/2)}(t)\right) = n^{-1}\operatorname{Cov}(S_{k_n}, S_{m_n}) = \frac{k_n}{n} \to s = \min(s, t)

This matches the Brownian-motion covariance identity $\operatorname{Cov}(W_s, W_t) = \min(s, t)$ exactly. Since the rescaled process is also Gaussian (by the Central Limit Theorem applied to each increment), and all finite-dimensional distributions match those of Brownian motion, we have the finite-dimensional convergence that Donsker's theorem strengthens to functional (path-space) convergence.

Part 4 — The dimensional argument

Because $\operatorname{Var}(S_n) = n$ regardless of the distribution of the $X_i$ (as long as $\operatorname{Var}(X_i) = 1$ ), the standard deviation scales as $\sqrt{n}$ ; dividing space by $\sqrt{n}$ is the only normalisation that keeps the process in an order-one range as $n \to \infty$ . Everything else — Gaussianity of the limit, continuity of the paths, exact covariance $\min(s, t)$ — is then forced by the Central Limit Theorem.

Takeaways

The $\sqrt{n}$ scaling is uniquely determined by variance. It is not a modelling choice — it is the only power-law rescaling that yields a non-trivial, finite-variance limit.
Covariance matches Brownian motion without any further calculation. The $\min(s, t)$ structure of the limit is a direct consequence of $\operatorname{Cov}(S_m, S_n) = \min(m, n)$ , which in turn follows from independent increments.
Donsker's theorem is the path-space CLT. The pointwise CLT says each finite slice of the rescaled walk is Gaussian; Donsker upgrades this to convergence of the entire path in the space of continuous functions. The $\sqrt{n}$ normalisation is the same in both.
Practical heuristic. When you see $\sqrt{\Delta t}$ show up anywhere in quant finance — volatility scaling, Brownian increment size, option Greeks' time-decay structure — it is a trace of this same $\sqrt{n}$ rescaling.