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This question is related to my previous question but I think it's sufficiently different to warrant a separate one.

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 below and above.

Does Lindeberg's Condition:

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

hold in this case? Here $s_n^2=\sum_{i=1}^n\sigma_i^2$ and $s_n=\sqrt{s_n^2}$.

My intuition tells me that it does, for the following reason: suppose that it doesn't hold. That means there is some random variable $X_i$ in the sequence most of whose mass isn't centered in an interval about the mean $[\mu_i-\epsilon s_n,\mu_i+\epsilon s_n]$. However, this interval gets larger as we add more random variables to the sequence, since their variances are bounded from below. Since the variance of $X_i$ is upper-bounded, at some $n$, the interval $[\mu_i-\epsilon s_n,\mu_i+\epsilon s_n]$ should "consume" most of $X_i$'s mass.

However, I am not sure how to prove (or disprove) it. One problem may arise if the interval $[\mu_i-\epsilon s_n,\mu_i+\epsilon s_n]$ does not increase fast enough... Not sure if I need additional conditions on higher moments of $A_i$ or not...

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1 Answer 1

up vote 3 down vote accepted

No, it does not. The trouble is that some families of random variables bounded in $L^1$ are not uniformly integrable.

To give a counterexample, assume that the distribution of $X_n$ is $\frac1{2n^2}(\delta_0+\delta_{2n})+(1-\frac1{n^2})\delta_n$ for every $n\geqslant1$. Then $\mu_n=\mathrm E(X_n)=n$ and $\sigma_n^2=\text{Var}(X_n)=1$, hence $s_n^2=n$. The $i$th integral in Lindeberg's condition of rank $n$ is $$ \mathrm E((X_i-\mu_i)^2;|X_i-\mu_i|\geqslant\varepsilon\sqrt{n})=i^2\cdot\mathrm P(|X_i-i|\geqslant\varepsilon\sqrt{n})=[i\geqslant\varepsilon\sqrt{n}], $$ hence the sum involved in Lindeberg's condition is $\frac1{s_n^2}\sim\frac1n$ times the number of indices $i\leqslant n$ such that $i\geqslant\varepsilon\sqrt{n}$, which is equivalent to $n$. That is, the sum converges to $1$ instead of $0$.

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Thank for answering my question: that is a very nice counterexample. Does the condition hold if I require every $A_i$ to have non-zero density over every interval $[a,b]\in [0,\infty)$? Also, bounding the mean for every $A_i$. Does that result in condition holding? I will write this up as a separate question... –  M.B.M. Nov 9 '11 at 21:36
    
A follow-up question: math.stackexchange.com/questions/80650/… –  M.B.M. Nov 9 '11 at 21:59
    
You could also try $X_n$ with distribution $(1 - \frac{1}{n^2} \delta_0 + \frac{1}{n^2} \delta_n$, which is somewhat messier but has mean $\mu_n = 1/n$ and $\sigma_n^2 = 1 - 1/n^2$, so $s_n^2 = n + O(1)$, and again the limit in Lindeberg's condition will be $1$ instead of 0. You could then "smear out" this example to have continuous densities positive on all of $[0,\infty)$. –  Robert Israel Nov 9 '11 at 22:27

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