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Given $B_1, B_2,\ldots$ are independent and bounded variables with $E(B_i) = 0$ for all $i=1,2,\ldots$. Define $S_n = B_1+ B_2+\ldots + B_n$ with variance $s_n^2\rightarrow \infty$. Prove that $\frac{S_n}{s_n}$ has a central limit.

My attempt: Due to the given condition, without i.i.d property, I try to prove that this sequence satisfies the Lindeberg condition and then applying Lindeberg-Feller theorem, we're done. So for every $\epsilon>0$ , I need to show for each positive $\varepsilon$, $$\lim_{n\to +\infty} \frac 1{\sum_{j=1}^n\sigma_j^2 }\sum_{j=1}^n\mathbb E\left[B_j^2\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n \sigma_i^2\right\}\right]=0 .$$

Since $\sigma_i^{2} = E(B_i^2)$ for all $\ i=1,2,\ldots$ and $B_is$ are bounded variables, $B_j^2\leq M=max(|B_1|,|B_2|,\ldots)$. Thus, $$\lim_{n\to +\infty} \frac 1{\sum_{j=1}^n\sigma_j^2 }\sum_{j=1}^n\mathbb E\left[B_j^2\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n \sigma_i^2\right\}\right]\leq \lim_{n\to +\infty} \frac 1{\sum_{j=1}^n\sigma_j^2 } M^2\sum_{j=1}^n \mathbb E\left[\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n \sigma_i^2\right\}\right] .$$ As $n\rightarrow \infty$, $s_n^2\rightarrow \infty$, so all the sets $\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n \sigma_i^2\right\}\rightarrow 0$. And we would be done if we could show that $\lim_{n\rightarrow \infty} \sum_{j=1}^n \mathbb E\left[\mathbf 1\left\{ \left|B_j\right|^2\gt \varepsilon\sum_{i=1}^n \sigma_i^2\right\}\right]\rightarrow 0$. But this might not be true (counter example is the harmonic series with $p=1$) unless there is something that I was missing.

My question: Could someone please help me overcome this last step? In case I was on the wrong track, please let me know as well.

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  • $\begingroup$ Didn't you forget to divide by $\sum_{j=1}^n\sigma_j^2$? $\endgroup$ – d.k.o. Nov 1 '16 at 18:55
  • $\begingroup$ @d.k.o: that sum tends to infinity, so as long as we could prove that the sum of expected values, as $n\rightarrow \infty$ bounded, we are done. $\endgroup$ – user177196 Nov 1 '16 at 19:25
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You are on the right track. Notice that (given $\varepsilon>0$) there exists $N$ such that $$ I\left( |B_j|^2>\varepsilon s_n^2\right) = 0 $$ for all $j$ and all $n\ge N$. (The reason is that $s_n$ is a deterministic sequence tending to infinity, while the $B$'s are bounded, so eventually the inequality is not satisfied.) This means the sum $$ \sum_{j=1}^n \mathbb E\left[\mathbf 1\left( \left|B_j\right|^2\gt \varepsilon s_n^2\right)\right] $$ stops at $j=N$, so it's bounded by $N$.

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  • $\begingroup$ I'm so dumb:p great observation!! Thank you. $\endgroup$ – user177196 Nov 1 '16 at 19:27
  • $\begingroup$ could you please give this more difficult problem a try? math.stackexchange.com/questions/1995678/… I'm stuck with proving the convergence in distribution to $N(0,1)$ once again, but the trick you used could not be applied there since $A_is$ are not bounded. $\endgroup$ – user177196 Nov 2 '16 at 6:41

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