Given $A_1, A_2,\ldots $ are independent and $A_k \sim U[-a_k, a_k]$ where $0<a_k\leq 1$ for each $k$. Prove that $(A_k)$ satisfies Lindeberg's condition IFF $\sum_{k=1}^{\infty} a_k^2=\infty$.

My attempt: $(\Leftarrow)$ I tried to use Lindeberg-Feller theorem. So we need to show $A_1, A_2,...$ is uniformly asymptotically negligible (u.a.n), which is: $\lim_{n\rightarrow \infty} \max_{k\leq n} \frac{\sigma_k^2}{\sum_{i=1}^n \sigma_{i}^2} = 0$. First, $A_k \sim U[-a_k, a_k]$, $\sigma_{k}^2 = \frac{a_k^2}{3}$. Thus by dividing both the numerator and denominator by $\max_{k\leq n} (\sigma_k^2)$, and since $\frac{1}{\sum_{i=1}^{n} \sigma_{i}^2}\geq \frac{1}{\frac{\sum_{i=1}^{n} \sigma_{i}^2}{\max_{k\leq n} (\sigma_k^2)}}\geq \frac{1}{n}$, by letting $n\rightarrow \infty$ and using the hypothesis, we get: $\lim_{n\rightarrow \infty} \max_{k\leq n} \frac{\sigma_k^2}{\sum_{i=1}^n \sigma_{i}^2} = 0$.

Now, I would need to show $\frac{\sigma_{n}}{\sum_{i=1}^{n} \sigma_{i}^2}$ converges in distribution to $N(0,1)$. This is equivalent to show $$\phi_{\frac{\sigma_{n}}{\sum_{i=1}^{n} \sigma_{i}^2}}(t) = \prod_{i=1}^{n} \phi_ {\frac{a_k}{\sum_{i=1}^{n}}}(t) = \prod_{i=1}^{n} \frac{\sin\left(\frac{ta_k}{\sum_{i=1}^{n} \sigma_{i}^2}\right)}{\frac{ta_k}{\sum_{i=1}^{n} \sigma_{i}^2}}$$ (as $\phi_{a_k}(t) = \frac{\sin(ta_k)}{ta_{k}}$). But I could not show the last identity equal to $e^{\frac{-t^2}{2}}$, which is the characteristic function of $N(0,1)$

$(\Rightarrow)$ Assume $(A_k)$ satisfies the Lindeberg condition. By contradiction, assume $\sum_{k=1}^{\infty} a_k^2 < \infty$. This implies that $\lim_{n\rightarrow \infty} a_n^2 = 0$ (by the negation of nth term test for convergence). This means $\lim_{n\rightarrow \infty} \sigma_{n}^{2} = 0$, so $\max_{k\leq n} \sigma_{n}^2 = \sigma_{k_i}^2$ where $k_i$ is some finite number not depending on $n$. So by testing the uniformly asymptotically negligible, $\lim_{n\rightarrow \infty} \max_{k\leq n} \frac{\sigma_{k}^2}{\sum_{i=1}^{n} \sigma_{i}^2} = \lim_{n\rightarrow \infty} m$ where m is some positive number (since $\sum_{k=1}^{\infty} a_k^2 < \infty$, and $\max_{k\leq n} \sigma_{k}^2 = \frac{a_{k_i}^2}{3} > 0$. So the sequence $A_1, A_2,\ldots$ fails to be u.a.n, so it cannot satisfy Lindeberg condition (this is the contradiction).

My question: Could anyone please help me complete the proof of convergence in distrubtion to $N(0,1)$ for the backward direction? Also, any thoughts about my proof above would be greatly appreciated. I don't find the proof for the forward direction quite nice though...


In this context, Lindeberg's condition reads as follows: for each positive $\varepsilon$, $$\lim_{n\to +\infty} \frac 1{\sum_{j=1}^na_j^2 }\sum_{j=1}^n\mathbb E\left[A_j^2\mathbf 1\left\{ \left|A_j\right|^2\gt \varepsilon\sum_{i=1}^n a_i^2\right\}\right]=0 .$$ Note that the distribution of $A_j$ is equal to that of $a_j U$, where $U$ is uniformly distributed on $[-1,1]$. Therefore, $$\mathbb E\left[A_j^2\mathbf 1\left\{ \left|A_j\right|^2\gt \varepsilon\sum_{i=1}^n a_i^2\right\}\right]=a_j^2\mathbb E\left[U^2\mathbf 1\left\{ a_j^2\left|U\right|\gt \varepsilon \sum_{i=1}^n a_i^2\right\}\right]\\ \leqslant a_j^2\mathbb E\left[U^2\mathbf 1\left\{ \left|U\right|\gt \varepsilon \sum_{i=1}^n a_i^2\right\}\right], $$ since $a_j^2\leqslant 1$. Now, it suffices to sum and notice that $\mathbb E\left[U^2\mathbf 1\left\{ \left|U\right|\gt \varepsilon \sum_{i=1}^n a_i^2\right\}\right]=0$ for $n$ large enough.

  • $\begingroup$ What a slick proof to cancel the sum! Thank you so much. $\endgroup$ – user177196 Nov 1 '16 at 18:01
  • $\begingroup$ could you help confirm if my proof above for the forward direction was correct? $\endgroup$ – user177196 Nov 5 '16 at 5:57

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