Limiting Distributions and expectation of sequence of RVs

Question

We're given a sequence $X_i$ of i.i.d non-negative random variables, with $\mu = 4$ and $\sigma^2 = 16$. We're asked to calculate: $$ \lim_{n \to \infty} E[cos (\sqrt{S_n} - 2\sqrt{n})],$$ where $S_n = \sum_{i=1}^n X_i$.

I have no clue where to start. Any ideas are appreciated.

Answer 1

First, note that $$ \sqrt{n}(S_n/n-\mu)\xrightarrow{d}\mathcal{N}(0,\sigma^2), $$ and, therefore, $$ \sqrt{S_n}-\sqrt{\mu n}=\sqrt{n}\left(\sqrt{S_n/n}-\sqrt{\mu}\right)\xrightarrow{d}\mathcal{N}\!\left(0,\sigma^2/(2\sqrt{\mu})^2\right)\overset{d}{=}N. $$ Thus, since $x\mapsto\cos(x)$ is bounded and continuous, $$ \mathsf{E}\cos(\sqrt{S_n}-\sqrt{\mu n})\to \mathsf{E}\cos(N). $$

The answer is $e^{-1/2}$.

Answer 2

alternative procedure.

$$\frac{\left(\sqrt{S_n}-2\sqrt{n}\right)\cdot\left(\sqrt{S_n}+2\sqrt{n}\right))}{4\sqrt{n}}=\frac{S_n-4n}{4\sqrt{n}}$$

that is

$$\left(\sqrt{S_n}-2\sqrt{n}\right)=\frac{4\sqrt{n}}{\sqrt{S_n}+2\sqrt{n}}\cdot\frac{S_n-4n}{4\sqrt{n}}=\frac{4}{\sqrt{\overline{X}_n}+2}\cdot\frac{S_n-4n}{4\sqrt{n}}=A\cdot B$$

Now:

$$A\xrightarrow{\mathcal{P}}1$$

$$B\xrightarrow{d}N(0;1)$$

thus using Slutsky Theorem you have that

$$A\cdot B\xrightarrow{d}N(0;1)$$