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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.

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    $\begingroup$ Welcome to Math.SE. Please add two things: 1) some context (for example, is this a homework question?) and 2) an explanation of what you have tried and where you are getting stuck. You will find us much more willing to help if you include this information! $\endgroup$ – Nick Peterson Jun 27 '13 at 16:06
  • $\begingroup$ Look up the Central Limit Theorem. Do the terms of the sum meet the criteria for the CLT to apply? What is the effect of summing these instead of averaging? $\endgroup$ – Dale M Jun 27 '13 at 23:24
  • $\begingroup$ I tried to use classical CLT but variables $X_i X_{i+1}, X_{i+1} X_{i+2}$ are dependent. $\endgroup$ – Rafał Jun 28 '13 at 12:05
  • $\begingroup$ Is there a way to do it by writing the series as the sum of two separate series, both having independent terms? i.e. $X_1 X_2 + X_3 X_4 + ...$ and $X_2 X_3 + X_4 X_5 + ...$? I only thought about it briefly, but perhaps something might come of it... $\endgroup$ – gogurt Jun 28 '13 at 16:29
  • $\begingroup$ @gogurt The beginning is straightforward, each of these two sums converges to N(0,1/2). Unfortunately, they are not independent... $\endgroup$ – Did Jun 28 '13 at 16:30
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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.

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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 $\mathcal N(0,1)$.

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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.

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