Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am looking for a "counterexample" to a central limit type setup. Here is my question: Is there an example of a sequence of identically distributed random variables $(X_n)_{n\in\mathbb{N}}$, with mean zero and variance 1, which is only UNCORRELATED, but not independent (so we do not have an i.i.d. sequence), that violates the statement of the central limit theorem, i.e., we do NOT have that $$T^{-1/2}\sum_{t=1}^TX_t \Rightarrow N(0,1), \quad \text{ as }\,\,T \rightarrow \infty.$$ (In particular, is there an example where the above expression does not converge in distribution at all?) This would show that independence in the CLT cannot be weakened to mere uncorrelatedness. I can't figure out an example for the above.

Many thanks for any thoughts!

share|improve this question
3  
take $X_i = X\epsilon_i$ where $X$ is a fixed normal and the $\epsilon_i$ are i.i.d $\pm 1$ with prob $\frac 12$ each, also independent of X. –  mike Jan 10 '13 at 19:37
    
@mike: you'd consider promote that comment to an answer –  leonbloy Jan 10 '13 at 20:08
    
ok, many thanks for the example - very nice. but is there also an example where the above sum, scaled by root T, does not converge at all (I will update the question above)? –  s_2 Jan 10 '13 at 21:39

1 Answer 1

up vote 2 down vote accepted

Adapting @mike's indication, try $X_n=Y_kZ_n$ for every $\varphi(k)\leqslant n\lt\varphi(k+1)$ and every $k\geqslant1$, where:

  • $(\varphi(k))_{k\geqslant1}$ is an increasing sequence such that $\varphi(k)/\varphi(k+1)\to0$ (say, $\varphi(k)=k!$),
  • $(Z_n)_{n\geqslant1}$ is i.i.d. centered with variance $1$,
  • $(Y_k)_{k\geqslant1}$ is independent of $(Z_n)_{n\geqslant1}$.

The sequence $(X_n)_{n\geqslant1}$ is uncorrelated and, for each $n=\varphi(k+1)-1$, the normalized sum $T_n=\frac1{\sqrt{n}}\sum\limits_{i=1}^nX_i$ is approximately $Y_kZ$ in distribution, where $Z$ is standard normal.

Thus, if $(Y_k)_{k\geqslant1}$ does not converge in distribution, neither does $(T_n)_{n\geqslant1}$. A simple example is $Z_n$ standard normal, $Y_{2k}=2Z$ and $Y_{2k-1}=3Z$ for every $k\geqslant1$. Then the set of limit points of $(T_n)_{n\geqslant1}$ in distribution is the set of centered normal distributions with variance in $[2,3]$.

share|improve this answer
    
hi did, many thanks for your great answer! –  s_2 Jan 15 '13 at 15:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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