3
$\begingroup$

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...

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.