Convergence in distribution of $(X_1 X_2+X_2 X_3+\ldots+X_n X_{n+1})/\sqrt n$

Question

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.

Answer 1

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.

Answer 2

In general, the Martingale CLT may be useful too, but here is a more concrete simple solution that exploits the properties of the Normal distribution.

Without loss of generality, by potentially dropping a term, we can suppose $n=2k+1$ is odd. By conditioning on $X_{2i+1},i=1,\cdots,k$, we obtain \begin{align*} S_n &\stackrel{d}{=} (X_1^2 +X_3^2)^{1/2}X_2 + (X_3^2 + X_5^2)^{1/2}X_4+\cdots\\ &\stackrel{d}{=} \Big( X_1^2 + \sum_{i=1}^{k-1} 2X_{2i+1}^2 + X_{2k+1}^2 \Big)^{1/2} Z\\ &=: (X_1^2 + 2 \chi_{k-1} + X_{2k+1}^2)^{1/2} Z, \end{align*} where $Z\sim {\cal N}(0,1)$ is independent from $X_1,X_{2k+1}$ and the chi-square random variable $\chi_{k-1}$. The Strong Law of Large Numbers implies that $$ \frac{1}{\sqrt{n}} (X_1^2 + 2 \chi_{k-1} + X_{2k+1}^2)^{1/2} \to 1, $$ almost surely, as $n\to\infty$. This in view of Slutsky’s theorem, implies the desired result.

Answer 3

Here's an elementary solution using characteristic functions. Let $\phi_0$ denote the characteristic function of the distribution $\mathcal N(0,1)$.

For $n\geq 1$, let $Y_n = \frac 1{\sqrt n} (X_1X_2+\ldots+X_n X_{n+1})$ and note that $$ \begin{align} \forall n\geq 3, \forall t\in \mathbb R, \; \phi_{Y_n}(t)&= \phi_{Y_{n-2}}\Big(t \sqrt{\frac{n-2}n}\Big) E\Big[E\big[\exp(i\frac{t}{\sqrt n})X_n(X_{n-1}+X_{n-2})\big|X_n\big]\Big] \\ &= \phi_{Y_{n-2}}\Big(t \sqrt{\frac{n-2}n}\Big) E\Big[\phi_0\big(\frac{t\sqrt 2X_n}{\sqrt n}\big)\Big] \\ &= \phi_{Y_{n-2}}\Big(t \sqrt{\frac{n-2}n}\Big)\Big( 1+\frac{2t^2}n\Big)^{-1/2}. \end{align} $$

Iterating, for $n\geq 3$ and $k\leq (n-1)/2$, $$\forall t\in \mathbb R,\; \phi_{Y_n}(t) = \phi_{Y_{n-2k}}\Big(t \sqrt{\frac{n-2k}n}\Big)\Big( 1+\frac{2t^2}n\Big)^{-k/2}.$$

Besides, similar conditioning shows that $\forall t\in \mathbb R$, $\phi_{Y_2}(t) = \phi_{Y_1}(t) = (1+t^2)^{-1/2}$, thus

$$\begin{align} \forall n\geq 2, \forall t\in \mathbb R,\; \phi_{Y_{2n}}(t) &= \phi_{Y_{2n-2(n-1)}}\Big(t \sqrt{\frac{2n-2(n-1)}{2n}}\Big)\Big( 1+\frac{2t^2}{2n}\Big)^{-(n-1)/2} \\ &=\Big(1+\frac{t^2}n\Big)^{-1/2}\Big( 1+\frac{t^2}{n}\Big)^{-(n-1)/2} \\ &=\Big( 1+\frac{t^2}{n}\Big)^{-n/2} \\&\xrightarrow[n\to \infty]{} \exp\Big(-\frac {t^2}2\Big) \end{align}$$ and similarly $$ \phi_{Y_{2n+1}}(t) = \Big(1+\frac{t^2}{2n+1}\Big)^{-1/2}\Big( 1+\frac{2t^2}{2n+1}\Big)^{-n/2} \xrightarrow[n\to \infty]{} \exp\Big(-\frac {t^2}2\Big). $$

By Levy's theorem, $Y_n$ converges in distribution to $\mathcal N(0,1)$.