Convergence of independent random variables with different distributions Let $Y_1,Y_2,...,$ be independent r.v. with distribution $P(Y_k=k)=P(Y_k=−k)=1/2$ for all k. Here, $S_n=Y_1+Y_2+...+Y_n \ \ ∀k>0$.
Does ${S_n/n^{3/2}}$ converge in distribution? If yes, please write the limit distribution.
Note: each Yi has a different distribution
I know that the Law of Large Numbers requires i.i.d. random variables but I don't know how to start given this condition.
 A: A simple calculation shows $E\Big(\frac{S_n}{n\sqrt{n}}\Big)=0$ while $$V\bigg(\frac{S_n}{n\sqrt{n}}\bigg)=\frac{n(n+1)(2n+1)}{6n^3}\longrightarrow\frac{1}{3}$$ as $n \longrightarrow \infty$. The latter calculation suggests $\frac{S_n}{n\sqrt{n}}$ won't converge in distribution to the constant random variable $0$ since its limiting variance is non$-$vanishing. Moreover, since $\frac{S_n}{n\sqrt{n}}$ resembles an average of some sort, we may hypothesize that $\frac{S_n}{n\sqrt{n}}$ converges in distribution to a $N(0,1/3)$ random variable. Let's use moment generating function to see this. The moment generating function of $\frac{S_n}{n\sqrt{n}}$ is $$E\bigg(e^{\frac{tS_n}{n\sqrt{n}}}\bigg)=E\bigg(e^{\frac{tY_1}{n\sqrt{n}}}\bigg)\times \ldots \times E\bigg(e^{\frac{tY_n}{n\sqrt{n}}}\bigg)=\prod_{k=1}^n\cosh\bigg(\frac{kt}{n\sqrt{n}}\bigg)$$ You can see here that the right hand side in the above expression  converges to the function $y=e^{x^2/6}$ which is precisely the moment generating function of a $N(0,1/3)$ random variable.
A: You can apply the Lindeberg central limit theorem to the sequence $(Y_n)$. The $i$th $Y_i$ has mean $0$ and variance $i^2$ so the variance of the sum $S_n:=Y_1+\cdots+Y_n$ is
$$
s_n^2 := \operatorname{Var}(S_n)=1^2+\cdots+n^2=\frac 16n(n+1)(2n+1)\sim \frac{n^3}3.\tag1$$
Check that the Lindeberg condition holds: For every $\epsilon>0$,
$$
\lim_{n\to\infty} \frac1{s_n^2}\sum_{i=1}^n E\left[Y_i^2 I\left(|Y_i|>\epsilon s_n\right)\right]=0.\tag2
$$
Indeed, observe for every $i=1,\ldots,n$:
$$
Y_i^2 = i^2 \le n^2=n^3\frac1n\le 3s_n^2\frac1n=\frac 3ns_n^2.\tag3$$
In view of (3) we see that given $\epsilon>0$ the condition $|Y_i|>\epsilon s_n$ is never true when $n$ is sufficiently large (namely, when $\frac3n<\epsilon^2$), hence for $n$ sufficiently large each term in the sum in (2) is zero, and the Lindeberg condition holds.
The conclusion of the Lindeberg CLT is that $S_n/s_n$ converges in distribution to standard normal, hence the limit distribution for $S_n/n^{3/2}$ is $N(0,\frac13)$.
