5
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.