# Does Feller-Levy Condition imply holospoudicity?

Given a double array of random variables $X_{nj}, j=1,\dots, k_n, n\in\mathbb{N}$ with $k_n \to \infty$ as $n \to \infty$, suppose

• for each $n$, $X_{nj}, j=1,\dots, k_n$ are independent,
• each $X_{nj}$ has finite mean $\mu_{nj}$ and finite variance $\sigma_{nj}$.

Define $S_n := \sum_{j=1}^{k_n} X_{nj}$ and $s_n := \sqrt{\sum_{j=1}^{k_n} \mathrm{var} X_{nj}}$. Also define the following statements:

• (Central limit, or CL): $(S_n-\mathrm{E}S_n)/s_n \to N(0,1)$ in distribution;

• (Lindeberg condition, or L): $\forall \epsilon >0, \quad \lim_{n \to \infty} \frac{\sum_{j=1}^{k_n} \mathrm{E} [(X_{nj} - \mu_{nj})^2 I_{\{|X_{nj} - \mu_{nj}| > \epsilon s_n\}} ] }{s_n^2} = 0;$

• (Holospoudicity, or H): $\forall \epsilon >0, \quad\lim_{n\to\infty} \max_{j=1,\dots,k_n} P(|X_{nj}| > \epsilon) = 0$;

• (Feller-Levy Condition, or FL): $\lim_{n\to\infty} \frac{\max_{j=1,\dots,k_n}\sigma_{nj}^2}{s_n^2} = 0.$

Now there are two versions of Lindeberg-Feller Central Limit Theorem:

1. Theorem 7.2.1 from Kai Lai Chung's A Course in Probability Theory, $$(L) \Leftrightarrow (CL) + (H)'';$$

2. From Wikipedia, $$\text{when (FL) is true: } (L) \Leftrightarrow (CL).$$ (Note: the Wikipedia article deals with a sequence of random variables, but here I take the liberty to generalize it for a double array of random variables. Please correct me if I am wrong.)

Questions:

1. If I understand correctly, the version of Lindebrg-Feller CLT in Chung's implies that $$\text{when (H) is ture: }(L) \Leftrightarrow (CL),$$ doesn't it?

2. I wonder what relations are between these two versions of Lindeberg-Feller CLT?

Does $$(FL) \Rightarrow (H),$$ under the assumptions of independence within each row and finite mean and finite variance, given at the begining of this post? So is the version in Wikipedia a special case of that in Chung's?

3. Is it also right that $$(L) \Leftrightarrow (CL) + (FL)''?$$ (I saw this was written in a note in an obscure way to me, so I am not sure If I understand it correctly.)

Thanks and regards!

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Aren't you autoduplicating yourself? –  Did Jul 5 '12 at 14:02
@did: Am I?    –  Tim Jul 5 '12 at 14:03
Tell me (you should know whether you are or not, shouldn't you?). –  Did Jul 5 '12 at 15:04
@did: Is this what you meant by autoduplicating: if answer to the second question in 2 is yes, then <= in 3 is correct? –  Tim Jul 5 '12 at 15:09