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$Z_1, Z_2,...$ are iid uniformly distributed on $[-1;1]$,

$\lim_{n \to \infty} a_n=0$ and $\lim_{n \to \infty} na_n=\infty$ also $a_n>0$ $\forall n$,

$X_{n,j}= \frac{1}{a_n}I(|Z_j| \le a_n)$ $\forall j = 1,...,n$ where $I$ is the indicator function,

$S_n=\sum_{j=1}^{n}X_{n,j}$.

One needs to show that $(S_n-n)\sqrt{\frac{a_n}{n}} \to N(0,1)$ in distribution.

Easy to show that:

$E[S_n]=n$ if $n$ large enough,

also $Var[S_n]= \frac{(1-a_n)n}{a_n}$ if $n$ large enough,

and the Lyapunov condition is satisfied for $\delta=2$, i.e. $$\lim_{n \to \infty} \sum_{j=1}^{n} \frac{1}{Var(S_n)^2}E[|X_{n,j}-E(X_{n,j})|^4]=0.$$

But then it follows that $\frac{S_n-E[S_n]}{\sqrt{Var(S_n)}}=(S_n-n)\sqrt{\frac{a_n}{n(1-a_n)}} \to N(0,1)$ in distribution. What am I doing wrong?

Reaction to Did's comment:

$Z_n=(S_n-n)\sqrt{\frac{a_n}{n(1-a_n)}}$ and $Y_n=Z_n*\sqrt{(1-a_n)}$, Clearly if $F_{Z_n}(x)$ is a cdf for $Z_n$, then $F_{Z_n}(\frac{x}{\sqrt{1-a_n}})$ is the cdf for $Y_n$. Now we need to prove that $\lim_{n \to \infty}F_{Z_n}(\frac{x}{\sqrt{1-a_n}})=\Phi(x)$ $\forall x$. And here I am stuck.

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  • $\begingroup$ "What am I doing wrong?" Nothing, since $\sqrt{1/(1-a_n)}\to1$. $\endgroup$ – Did Jan 24 '15 at 16:40
  • $\begingroup$ @Did I intuitively see that, but I am lacking a formal proof. $\endgroup$ – Sergey Zykov Jan 24 '15 at 18:09
  • $\begingroup$ @Did Please see my edit above for an attempt of the formal proof. $\endgroup$ – Sergey Zykov Jan 24 '15 at 18:19
  • $\begingroup$ @SergeyZykov Can't you apply Slutsky's Theorem where $Y_n=\sqrt{1-a_n}$ with prob $1$? $\endgroup$ – Saty Jul 27 '15 at 3:03

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