Say $(X_i)$ is a sequence of $n$ independent Bernoulli random variables, with parameters $p_i$, $i:1...n$. How do I prove that the random variables, $ S_{n} =\frac{1}{n} \sum _{i=1}^{n}a_iX_i $ converge in distribution to a normal as $n\to \infty$, where $0
Sum converging to a Normal
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0Thanks leonbloy. Is this for any \epsilon>0 you mean? How difficult would this be to prove it? – 2012-07-30
1 Answers
Let $e_n=E(S_n)=\frac{1}{n}\sum_{i=1}^na_i.p_i$
let $\sigma_n=\sigma(S_n)=\frac{1}{n}\sqrt{\sum_{i=1}^na_i^2p_i(1-p_i)}$
$L_n=\frac{S_n-e_n}{\sigma_n}$
(We suppose to have $0
Hence, $E(L_n)=0$ and $V(L_n)=1$. We compute the characteristic function of $L_n$ :
$\phi_{L_n}(t)=e^{i\frac{-e_n}{\sigma_n}t}\prod_{i=1}^n(1-p_i+p_ie^{i\frac{a_i}{n.\sigma_n}t}) $
We use the fact the $e^{\epsilon}\approx 1+\epsilon+\frac{\epsilon^2}{2}$ and we suppose that $\lim(n.\sigma_n)=\infty$. (we NEED it !)
$\lim(\phi_{L_n}(t))\approx e^{i\frac{-e_n}{\sigma_n}t}\prod_{i=1}^n(1+p_i.i\frac{a_i}{n.\sigma_n}t-\frac{p_i}{2}\left(\frac{a_i}{n.\sigma_n}t\right)^2)\approx_* e^{i\frac{-e_n}{\sigma_n}t}. e^{i\frac{e_n}{\sigma_n}t}.e^{-\frac{t^2}{2}}=e^{-\frac{t^2}{2}} $
Hence, this is convergent to a normal law.
for (*) you need also $\ln(1+\epsilon)\approx \epsilon-\frac{\epsilon^2}{2}$
More generally, you can see from the proof that you need $\lim (\forall i)\frac{a_i}{n\sigma_n}=0$
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0I agree, and it should be obvious that if $p_i$ are zeros, you are not convergent, so it's not surprising to see that $p_i$ and $1-p_i$ should be large enough. However, $p_i$ can have a 0 limit, if this limit is reached slowly enough to let $n.\sigma_n$ goes to $\infty$. So your conditions are enough, but not necessary... – 2012-07-31