Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have a sequence of independent but not identically-distributed random variables $X_1, X_2, \ldots, X_n$ where $X_i\sim A_i$, with each $A_i$ having a support over $\mathbb{R}$ and subject to the following properties with known parameters:

  • $\mathbf{E}[X_i]=0$ (zero mean)
  • $\mathbf{var}[X_i]=\sigma^2_i<\infty$ (finite variance)
  • $\mu_3(X_i)=0$ (symmetry)
  • $\mu_4(X_i)<\infty$

I am interested in the convergence properties of the sum of squares of $X_i$: $\sum_{i=1}^n X_i^2$

Specifically, I am interested in the properties of $A_i$ such that

$$\frac{\sum_{i=1}^n (X_i^2-\sigma^2_i)}{\sqrt{\sum_{i=1}^n (\mu_4(i)-\sigma_i^4)}}\xrightarrow{d}\mathcal{N}(0,1)$$

That is, I am interested what additional conditions would $A_i$ have to obey in order for the normalised sum of squares to converge to a standard Gaussian distribution.

I am aware that the Lendeberg-Feller condition specifies if that most of the mass of $X^2_i$ is in the small interval around the mean, then CLT applies. This basically excludes heavy-tailed $A_i$. Unfortunately, it's unwieldy to apply.

I also know about the Lyapunov Condition: one can easily show that if $A_i$'s are Gaussian, then the normalised sum of their squares converges to a standard Gaussian.

I have three specific and one general questions:

  1. If $A_i$'s are platykurtic, does its square obey either Lyapunov and/or Lendeberg-Feller conditions (or any other conditions that ensure convergence to CLT)?
  2. If $A_i$'s belong to the exponential family, what are the least restrictive conditions that one can impose on their parameters so that the sum of squares of random variables drawn from them converges to a Gaussian? (I am looking for a looser condition than parameters yielding Gaussian $A_i$'s)
  3. If I impose constraints on higher central moments (as opposed to absolute moments) of $A_i$'s (but not absolute moments), can that ensure that the CLT applies to the sum of squares? If so, how loose can the constraints be (i.e. can I get away by saying that moments are finite)?

General question: the above three questions relate to specific distribution properties I thought of (and that are relevant to my project). Are there other properties of $A_i$'s that I should look into? Are there classes of $A_i$ for whose sum of squares CLT always applies?

Background: I am working on a variance detection problem and need to analytically characterise the asymptotic performance of a detector.

share|improve this question
1  
I'm somewhat confused by your notation. What is the difference between $A_i$ and $X_i$? Also, I'm not sure you actually do know the Lyapunov condition, which only requires a $2+\epsilon$ order absolute moment (and which follows from the Lindeberg-Feller condition using Markov/Chebychev). In particular, if for some $\epsilon > 0$, $\sup_i \mathbb{E} |X_i|^{4+\epsilon} < \infty$, then the CLT applies. –  Mark Meckes Nov 9 '11 at 16:54
    
I apologize for the notation. $X_i$ is a random variable, and $A_i$ is its distribution. Hmm, so are you saying that if I bound the 6th central moment of $X_i$ (which would be equivalent to bounding the 6th absolute moment), CLT would apply? –  Bullmoose Nov 9 '11 at 19:18
1  
Yes, then you would have a bound on the 3rd absolute moment of $X_i^2$, and you could apply the Berry-Esseen theorem. –  Mark Meckes Nov 9 '11 at 20:16
    
Beautiful! Thanks! I somehow missed the Berry-Essen theorem for non-identically distributed summands. If you write your comments in an answer, I'll accept it. –  Bullmoose Nov 10 '11 at 1:26
add comment

1 Answer

up vote 2 down vote accepted

The easiest condition would be a bound on $\sup_i \mathbb{E} X_i^6$, which would allow you to apply the Berry–Esseen theorem. More generally, if for some $0<\varepsilon < 2$ you have a uniform bound on $\mathbb{E} |X_i|^{4+\varepsilon}$, then you can apply the Lyapunov condition (which, as I noted above, follows from Lindeberg's condition by the Markov/Chebychev inequality).

share|improve this answer
add comment

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.