Let $X_1,X_2,\ldots$ be independent random variables with $X_k$ distributed as $\mathcal{N}(0,1)$ and $S_n=X_1 X_2+X_2 X_3+\ldots+X_n X_{n+1}.$
Show that $\frac{S_n}{\sqrt{n}}$ converges in distribution to the standard normal distribution.
Attempt: I tried to use the classical CLT but the variables $X_iX_{i+1}$ and $X_{i+1}X_{i+2}$ are dependent.
Best Answer
A (rather heavy) hammer to crack this nut is to use a central limit theorem for functionals of Markov chains, à la Jeffrey Rosenthal for example.
To do so, note that $Y_k=(X_k,X_{k+1})$ defines a stationary ergodic Markov chain $(Y_k)$ hence, for every suitable measurable function $h$, $\frac1{\sqrt{n}}\sum\limits_{k=1}^nh(Y_k)$ converges in distribution to $\sigma$ times a standard normal random variable, where $\sigma^2=\gamma_0+2\sum\limits_{k=0}^{+\infty}\gamma_k$ and $\gamma_k=E(h(Y_0)h(Y_k))$ for every $k$.
Here, $h(Y_k)=X_kX_{k+1}$ hence $\gamma_0=1$ and $\gamma_k=0$ for every $k\geqslant1$. Thus, $\sigma^2=1$ and the result holds.