Central Limit Theorem with $\sqrt{S_{n}}$

Question

Let $X_{1}, X_{2}, \ldots$ be i.i.d. with $X_{i} \geq 0, E X_{i}=1, \operatorname{Var} X_{i}=1.$ Roughly, by the central limit theorem we might expect that $2(\sqrt{S_{n}}-\sqrt{n})=\int_{n}^{S_{n}} \frac{d x}{x^{1 / 2}} \approx \frac{S_{n}-n}{\sqrt{n}} \Rightarrow Z$ with $Z \sim N(0,1)$ Give a complete proof that $2 (\sqrt{S_{n}}-\sqrt{n}) \Rightarrow Z$

$S_n = X_1+X_2+...+X_n$.

I am not sure how to deal with the $\sqrt{S_{n}}$. I cannot even compute the characteristic function. Any hint will be appreciated! Thank you!

Answer 1

Write

$$ 2(\sqrt{S_n} - \sqrt{n}) = \frac{2}{\sqrt{S_n/n} + 1} \cdot \frac{ S_n - n}{\sqrt{n}} $$

and note that

• $S_n/n \to \mathbb{E}[X_1] = 1$ in distribution by either version of LLN.

• $(S_n - n)/\sqrt{n} \to \mathcal{N}(0, \sigma^2)$ in distribution by CLT.

So by the converging together lemma (a.k.a. Slutsky's Theorem), it follows that

$$ 2(\sqrt{S_n} - \sqrt{n}) \xrightarrow[n\to\infty]{\text{law}} \frac{2}{\sqrt{1} + 1} \cdot \mathcal{N}(0, \sigma^2) = \mathcal{N}(0, \sigma^2). $$

Answer 2

You didn't write it but I am assuming you mean that $S_n=X_1+\cdots +X_n$ (some people use other conventions, for instance the the wikipedia page uses a different convention than yours).

Applying the central limit theorem yields that $$ \frac{S_n-n}{\sqrt{n}} $$ converges in law to Gaussian random variable with mean $0$ and variance $\sigma^2$, let's call such a random variable $G$ to avoid conflicting notation with your $Z$.

Rearranging the convergence, we can equivalently say that $S_n=n+\sqrt{n}G+o(\sqrt{n})$, using the asymptotic notation $o(f(n))$ to denote a (potentially random) quantity whose limit as $n\to\infty$ when divided by $f(n)$ equals $0$.

Equivalently $$ \frac{S_n}{n}=1+\frac{G}{\sqrt{n}}+o(n^{-1/2}), $$ so we can take the square root of both sides (using the fractional binomial theorem) to conclude that $$ \frac{\sqrt{S_n}}{\sqrt n}=1+\frac{G}{2\sqrt{n}}+o(n^{-1/2}). $$ Multiplying through by $\sqrt{n}$ and rearranging yields the result that $ \sqrt{S_n}-\sqrt{n} $ converges in law to $G/2$, i.e. to a Gaussian random variable with mean $0$ and variance $\sigma^2/4$.

Answer 3

Let $N(\mu,\sigma^2)$ denote a normal random variable of mean $\mu$ and variance $\sigma^2$. We can proceed by using Taylor's theorem and the strong law of large numbers (SLLN). By the SLLN, there is an event $E$ with $\mathbf P(E) = 1$ such that $(S_n-n)/n\longrightarrow 0$ on $E$. On $E$ we write: \begin{align*} 2(\sqrt{S_n} - \sqrt n) &= 2(\sqrt{n + S_n-n} - \sqrt{n})\\ &= 2\sqrt{n}\bigg(\sqrt{1+\frac{S_n-n}{n}}-1\bigg)\\ &= 2\sqrt{n}\bigg(1 + \frac{1}{2}\cdot\frac{S_n-n}{n} + O\bigg(\frac{S_n-n}{n}\bigg)^2 - 1\bigg) \\ &= \frac{S_n-n}{\sqrt n} + \sqrt{n}\cdot O\bigg(\frac{S_n-n}{n}\bigg)^2 \\ &= \frac{S_n-n}{\sqrt n} + O\bigg(\frac{S_n-n}{n^{3/4}}\bigg)^2. \end{align*} By Theorem 2.5.11 of Durrett's Probability Theory and Examples, 5th ed., $$ \frac{S_n-n}{n^{3/4}} \longrightarrow 0 \quad\text{a.s.} $$ Convergence a.s. implies convergence in distribution. Since $X_i-1$ has mean $0$ and variance $\sigma^2$ and $0$ is a constant, we apply the converging together lemma to conclude that $$ 2(\sqrt{S_n}-\sqrt n) \Rightarrow N(0,\sigma^2) + 0 = \sigma N(0,1), $$ as desired.

Answer 4

Let $S_n':=S_n/n$. Then $\sqrt{n}(S_n'-1)\xrightarrow{d}\mathcal{N}(0,\sigma^2)$. Applying the delta method, $$ \sqrt{S_n}-\sqrt{n}=\sqrt{n}\!\left(\sqrt{S_n'}-1\right)\xrightarrow{d}\mathcal{N}\!\left(0,\sigma^2/(2\sqrt{1})^2\right). $$