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This is a follow-up to this question.

Suppose I have a sequence of independently distributed positive random variables: $X_1\sim A_1,X_2\sim A_2, \ldots, X_n\sim A_n$, where $A_i$'s have support over $[0,\infty)$.

Furthermore, suppose that, for each $X_i$, variance $\sigma_{lb}^2<\sigma_i^2<\sigma_{ub}^2$, that is variances of $X_i$'s are bounded from above and below.

With $s_n^2=\sum_{i=1}^n\sigma_i^2$ and $s_n=\sqrt{s_n^2}$, Lindeberg's Condition is defined as:

$$\lim_{n\rightarrow\infty}\frac{1}{s_n^2}\sum_{i=1}^n\int_{\{|x-\mu_i|>\epsilon s_n\}}(x-\mu_i)^2f_i(x)dx=0$$

From the answer to my previous question, it does not hold in the case where just the variance is bounded.

Suppose we define two more possible restrictions on all $A_i$'s, in addition to the variance bound:

  1. $\mu_{lb} < \mu_i < \mu_{ub}$, that is, bound the means of all distributions from above and below;
  2. $\mathbf{P}(a\leq X_i\leq b)>0$ for all intervals $[a,b]\subseteq [0,\infty)$

Is there a subset (not necessarily proper) of the three above-defined restrictions on $A_i$'s for which Lindeberg condition holds?

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up vote 2 down vote accepted

For condition 1., an example similar to the one on the other page does it: consider $X_n$ with distribution $\frac1{2n^2}(\delta_0+\delta_n)+(1-\frac1{n^2})\delta_1$. Condition 2. is simply not relevant (regularize).

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Allright, thanks! Looks like the only way to meet the Lindeberg Condition is by requiring some conditions on higher moments and applying Lyapunov's condition. Btw, what do mean by "regularize"? I am not familiar with that term. –  M.B.M. Nov 9 '11 at 22:32
To regularize = to replace each $X_n$ by $X_n+Y_n$ with $Y_n$ independent on $X_n$, $E(Y_n^2)$ extremely small, and $Y_n$ with a density positive everywhere on $(0,+\infty)$. –  Did Nov 9 '11 at 22:35
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