TITLE: Convergence of independent random variables with different distributions
QUESTION [1 upvotes]: 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.

REPLY [2 votes]: 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 $N(0,\frac13)$.
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