CLT for $P\{X_k = \pm 2^k\} = \frac{1}{2}$

Question

I want to show that the following family of random variables either satisfy or don't satisfy the central limit theorem $P\{X_k = \pm 2^k\} = \frac{1}{2}$.

For the CLT I know that I want to use the Lindeberg condition. However, I'm having trouble understanding the definition given in equation 5.6 of Feller's book volume 1.

Answer 1

The moment generating function for $\bar{X}_n$ is $$M_{\bar{X}_n}(t)=\prod_{k=1}^n\cosh\Big(\frac{t2^n}{n}\Big)$$ Note for any $t\neq 0$ we have $$M_{\bar{X}_n}(t)\geq\cosh\Big(\frac{t2^n}{n}\Big)\longrightarrow \infty$$ as $n \longrightarrow \infty$ so the sequence of random variables $\{\bar{X}_n\}_{n=1}^{\infty}$ doesn't converge to any normal distribution.

Answer 2

This is nice to illustrate Lindeberg-Feller but you can also solve it with your bare hands.

Consider a the following operation in probability distributions on R:

Starting with a random variable with mean zero and unit variance, add to it an independent random variable equal to ( \pm \sqrt{3} ) with probability 1/2 each, and divide the result by two to get again unit variance.

Check that the Gaussian is not a fixed point of this operation.

This will imply that your sequence does not satisfy the CLT.

Answer 3

Let $S_n=\sum_{k=1}^n X_k$. Then $$\require{cancel} \frac{S_n}{\sqrt{\text{Var }S_n}} =\frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n2^{2k}}} =\frac{2^{n+1}\sum_{k=1}^nX_k2^{-(n+1)}}{\sqrt{(2^{2(n+1)}-1)/3}} = \cancelto{\to 1}{\color{gray}{\frac{2^{n+1}}{\sqrt{2^{2(n+1)}-1}}}}\sqrt{3}\cdot\sum_{k=1}^n X_{n+1-k}2^{-k} $$ Now, notice that $X_{n+1-k}2^{-(n+1)}\sim \text{Unif}\{+2^{-k},-2^{-k}\}$. Therefore, $S_n/\sqrt{\text{Var }S_n}$ converges in distribution to $$ \sqrt{3}\sum_{i=1}^\infty \xi_i\left(\frac12\right)^i, \qquad \xi_i\sim \text{Unif} \{-1,+1\}\;\text{i.i.d} $$ One can verify this summation is not normally distributed, so CLT does not apply. In fact, it is distributed like $2C-1$, where $C$ has the Cantor distribution.