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Suppose that $\{X_m\}_{m=1}^\infty$... is a sequence of independent random variables respectively distributed as $P(X_m = 1) = P(X_m = -1) = p_m$, $P(X_m = 0) = 1 - 2p_m$, $m \ge 1$, where $\sum^\infty_{m=1} p_m = \infty$. Use each of the methods of Lindeberg to give a proof that for a suitable choice of scaling constants $c_n$, the rescaled sum $c_n^{-1}(X_1 + ... + X_n)$ is asymptotically normal with mean $0$ and variance $1$ as $n \to \infty$.

My attempt at a solution: In order to use Lindeberg's CLT, I need to rescale my random variables so that $E[X_{m,j}] = 0$, $\sigma^2_{m,j} = E\left[X^2_{m,j}\right] < \infty$, and $\sum^n_{j=1} \sigma^2_{m,j} = 1$. I've tried setting $X_{m,j} = \frac{X_m}{\sigma_j}$, so that $E [X_{m,j}] = 0$, and $E[X^2_{m,j} = E\left[\left(\frac{X_m}{\sigma_j}\right)^2\right] = \frac{2}{\sigma_j^2}p_m$. But now I'm getting that $\sum^n_{j=1} \frac{2}{\sigma^2_j}p_n = n$, when I wanted it to be $1$. I have thought about maybe trying to rescale in a different, but I'm fairly certain that the scaling factor is actually ok, so I'm hoping that I did something else wrong. Thanks in advance for the help!

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What about Liapounov's CLT? $E(X_m)=0$ and $E(X_m^2)=2p_m$. Hence $s_n^2=2\sum p_m$. Now take $\delta=1$ in the CLT Theorem. Then we have $E|X_m|^3=2p_m$ as well and thus $$\frac{\sum E|X_m|^3}{s_n^3}=\frac{1}{\sqrt {2\sum p_m}} \rightarrow 0 \hspace{6mm} \text{ as } \hspace{4mm} n \rightarrow \infty$$ Hence Liapounov condition holds and $\frac{S_n}{s_n} \xrightarrow{d} N(0,1)$ i.e. $$ \frac{\sum X_m}{\sqrt{2 \sum p_m}} \xrightarrow{d} N(0,1)$$

Thus taking $c_n={\sqrt{2\sum_1^n p_m}}$ will do the job.

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  • $\begingroup$ True, and thank you, however, I have actually already verified that the Liapounov condition holds, that was part b - I'm sorry, I should have clarified my question! I am still trying to verify Lindenberg's condition specifically. $\endgroup$ – poppy3345 May 4 '15 at 2:08
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So, I've ended up figuring out the answer on my own - the constants that I was looking for are $c_n = \sqrt{2 p_m}n$, which fixes the problem.

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