CLT for random variables with varying distributions The following is not a homework problem. Here it is:
Suppose $\{X_k\}_{k=1}^{\infty}$ is a sequence of independent random variables such that
$$
P(X_k=1)=P(X_k=-1)=\frac{1}{2}-\frac{1}{2k^2}
$$
$$
P(X_k=k)=P(X_k=-k)=\frac{1}{2k^2}.
$$
Find the limiting distribution of $\frac{1}{\sqrt{n}} \sum\limits_{k=1}^n X_k$.
Now, I am guessing that the given probabilities are for $k\geq 2$, since $k=1$ would yield an inconsistent definition. Furthermore, I am assuming the case $k=1$ corresponds to only the second line.
The first sign of trouble comes from the fact that these random variables are not identically distributed (otherwise, we could just use CLT). One thing I noticed is that $\mathbb{E}(X_k)=0$ for each $k$. Also, it appears as though the limit of the sequence is a random variable $X$ with $P(X=1)=P(X=-1)=\frac{1}{2} \hspace{2mm}$  (not that this result is necessarily of much help).
I'd very much appreciate any suggestions on how to approach this problem. Thanks!
 A: Let $Y_k=\mathrm{sgn}(X_k)$, then $(Y_k)$ is a centered i.i.d. Bernoulli sequence hence, by the most usual CLT, $T_n=\frac1{\sqrt{n}}\sum\limits_{k=1}^nY_k$ converges in distribution to a standard normal random variable $T$. By Borel-Cantelli lemma, the random set $\{n\geqslant1\mid X_n\ne Y_n\}$ is almost surely finite hence there exists some almost surely finite random variable $Z$ such that $\left|\sum\limits_{k=1}^nX_k-\sum\limits_{k=1}^nY_k\right|\leqslant Z$ for every $n$. In particular $R_n-T_n\to0$ almost surely, where $R_n=\frac1{\sqrt{n}}\sum\limits_{k=1}^nX_k$.
Now it is a general fact that if $T_n\to T$ in distribution and if $R_n-T_n\to0$ almost surely then $R_n\to T$ in distribution. Thus, CLT holds for $(X_n)$ as well.
A: The limiting distribution is normal.
You're correct that you can't use the classical CLT here. But you can use the Lindeberg CLT. That version of the theorem does not require identically distributed random variables, provided that certain conditions are met.
To use this theorem, you have to set up your problem using a triangular array. Basically, in the first row of the array you will have just $X_1$. In the second row, you'll have $X_1, X_2$. In the third row, $X_1, X_2, X_3$, and so on. Variables are repeated across rows, so any two rows contain non-independent variables. However, within each row, all the variables are independent, which is what the theorem requires.
The only other thing you need to check is the Lindeberg condition. This might seem tricky at first, but if you calculate the variances of a handful of $X_k$'s, you'll find that as $k \to \infty$, Var$(X_k) \to 2$, and it never exceeds 2. A little more thinking will give you the Lindeberg condition easily.
A: Suppose $\{X_k\}_{k=1}^{\infty}$ is a sequence of independent random
variables such that $$ P(X_k=1)=P(X_k=-1)=\frac{1}{2}-\frac{1}{2k^2}
$$
$$ P(X_k=k)=P(X_k=-k)=\frac{1}{2k^2}. $$ Find the limiting distribution of $\frac{1}{\sqrt{n}} \sum\limits_{k=1}^n X_k$.
The characteristic function of $X_k$ is
$$\phi_k(\xi)=(\frac{1}{2}-\frac{1}{2k^2})e^{i\xi}+(\frac{1}{2}-\frac{1}{2k^2})e^{-i\xi}+\frac{1}{2k^2}e^{ik\xi}+\frac{1}{2k^2}e^{-ik\xi}$$
$$=(1-\frac{1}{k^2})\cos \xi +\frac{1}{k^2}\cos k\xi = \cos \xi +\frac{1}{k^2}(\cos k\xi-\cos\xi)$$
Then the characteristic function of $X_k/\sqrt{n}$ is
$$Ee^{i\xi X_k/\sqrt{n}}=\cos \frac{\xi}{\sqrt{n}} +\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})$$
and
$$Ee^{i\xi(\sum_{k=1}^n X_k)/\sqrt{n}}=\prod_{k=1}^n\left(\cos \frac{\xi}{\sqrt{n}} +\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})\right)$$
$$=\exp\left[\log \sum_{k=1}^n\left(1-(1-\cos \frac{\xi}{\sqrt{n}} -\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))\right)\right]$$
Then
$$\log \sum_{k=1}^n\left(1-(1-\cos \frac{\xi}{\sqrt{n}} -\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))\right)$$
$$\sim -\sum_{k=1}^n(1-\cos \frac{\xi}{\sqrt{n}}-\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}}))$$
$$\sim -\sum_{k=1}^n(1-\cos \frac{\xi}{\sqrt{n}})+\sum_{k=1}^n\frac{1}{k^2}(\cos \frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})=S_1+S_2$$
For $S_1$ we have
$$S_1=-n(1-\cos\frac{\xi}{\sqrt{n}})=-2n
\sin^2\frac{\xi}{\sqrt{n}}=-2n\frac{\sin^2\frac{\xi}{\sqrt{n}}}{\frac{\xi^2}{n}}\times\frac{\xi^2}{n}
\to -2\xi^2 , \ \ n \to \infty.$$
$$S_2=\sum_{k=1}^n\frac{1}{k^2}(\cos
\frac{k\xi}{\sqrt{n}}-\cos\frac{\xi}{\sqrt{n}})=-2\sum_{k=1}^n\frac{1}{k^2}\sin\frac{(k+1)\xi}{2\sqrt{n}}\sin\frac{(k-1)\xi}{2\sqrt{n}}$$
$$=-2\sum_{k=1}^n\frac{1}{k^2}\frac{(k^2-1)\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }$$
$$=-2\sum_{k=1}^n\frac{\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }+
2\sum_{k=1}^n\frac{1}{k^2}\frac{\xi^2}{n}\frac{\sin\frac{(k+1)\xi}{2\sqrt{n}}}{\frac{(k+1)\xi}{2\sqrt{n}}}\frac{\sin\frac{(k-1)\xi}{2\sqrt{n}}}{\frac{(k-1)\xi}{2\sqrt{n}} }$$
$$=S_{21}+S_{22}.$$
Now it is clear that $S_{21} \to -2\xi^2$ and $S_{22} \to 0$ as $n \to \infty$.
Therefore
$$Ee^{i\xi(\sum_{k=1}^n X_k)/\sqrt{n}} \to e^{-4\xi^2}, \ \ n \to \infty.$$
This is the characteristic function of r.v. $X=2Z \sim N(0,4)$.
