Prove that $S_n/\sqrt{n}-S_{4n}/\sqrt{4n} \overset w \longrightarrow \mathcal{N}(0,\sigma^2)$

Question

For $X_1, X_2,\ldots$ i.i.d., $\mathbb{E}[X_1]=0$ and $\mathbb{E}[X_1^2]=\sigma^2$ it should be possible to prove $$\frac{S_n}{\sqrt{n}}-\frac{S_{4n}}{\sqrt{4n}} \overset w \longrightarrow \mathcal{N}(0,\sigma^2)$$ with the Lindeberg theorem.

But I can't see how, in particular which elements to use so you can apply the theorem. I'd appreciate any hints.

Answer 1

Hint: $S_n - S_{4n}/2$ is the sum of $4n$ independent random variables, of which $n$ have the distribution of $X_1/2$ and $3n$ have the distribution of $-X_1/2$. Apply Lindeberg to the sequence $Y_j$ where $Y_j$ has the distribution of $-X_1/2$ if $j \equiv 1 \bmod 4$ and $X_1/2$ otherwise.

On second thought, you don't really need Lindeberg if you consider these in groups of four...

Answer 2

You don't need to use Lindeberg. Note that $$S_{4n} = S_n + \sum_{i=n+1}^{4n} X_i,$$ and that the two terms are independent. So you can separate $X_n \equiv S_n/\sqrt{n}-S_{4n}/\sqrt{4n}$ into two independent parts and apply central limit theorem to both. This shows that $$ X_n \Rightarrow \frac 1 2 Z_1 + \frac {\sqrt{3}}{2} Z_2 \sim N(0, \sigma^2), $$ where $Z_1$ and $Z_2$ are independent $\mathcal N(0, \sigma^2)$.