I am working on an exercise stated as follows:
Let $X_{k}$ be mutually independent random variables and suppose $S_{n}=\sum_{k=1}^{n}X_{k}$ has $\sigma_{n}^{2}=Var(S_{n})<\infty$. Given an example where there exists $C>0$, $q>2$ such that $$\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{q}\leq CVar(X_{k})^{q/2}$$ for all $k$ and $\sigma_{n}\rightarrow\infty$ as $n\rightarrow\infty$, but $(S_{n}-\mathbb{E}S_{n})/\sigma_{n}$ does not converge to $\mathcal{N}(0,1)$ in distribution.
There have been many posts concerning when CTL will fail, and I follow the example provided here: Some case when the central limit theorem fails
I don't want to use the $\alpha-$stable distribution since it is hard for me to compute all the stuffs above. However, the answer accepted in the post above cannot work since in that post, $$\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{q}\leq CVar(X_{k})^{q/2}$$ cannot be satisfied (you get RHS as $2^{4n}$ if $q=4$).
Thus, I changed a little bit, it seems working but I was stuck in the end.
Consider $X_{k}$ mutually independent such that $$\mathbb{P}(X_{k}=2^{-k})=\mathbb{P}(X_{k}=-2^{-k})=2^{2k-1}\ \text{and}\ \mathbb{P}(X_{k}=0)=1-2^{2k}.$$
Then, it is clear that $\mathbb{E}X_{k}=0$ and $Var(X_{k})=1$, so that $$Var(S_{n})=\sum_{k=1}^{n}Var(X_{k})=n,$$ since they are mutually independent, and therefore $\sigma_{n}=\sqrt{n}\longrightarrow\infty$ as $n\longrightarrow\infty$.
Now, for $q=4$, $$\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{4}=\mathbb{E}|X_{k}|^{4}=2^{-4k}\cdot 2^{2k-1}\cdot 2=2^{-2k},$$ so that to make $$\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{q}\leq CVar(X_{k})^{q/2},$$ we just need to find a $C<\infty$ such that $$2^{-2k}\leq CVar(X_{k})^{q/2}=C,$$ so if $C=1$, this inequality holds for all $k\in\mathbb{Z}_{\geq 0}$.
Okay, I've showed all the required things, now I want to show it does not converge to standard normal in distribution, from which the problem came.
I wanted to follow the computation in the link above to show that $\mathbb{P}((S_{n}-\mathbb{E}S_{n})/\sigma_{n}=0)\neq 0,$ so that it will never converge to standard normal which requires $\mathbb{P}(Z=0)=0$.
Now, since $\mathbb{E}X_{k}=0$, we know that $\mathbb{E}S_{n}=0$, and thus we compute $$\mathbb{P}(S_{n}/\sigma_{n}=0)=\mathbb{P}(S_{n}=0)=\mathbb{P}\Big(\sum_{k=1}^{n}X_{k}=0\Big)\geq\mathbb{P}(X_{k}=0\ \text{for all}\ k),$$ since they are mutual independent, we can further have \begin{align*} \mathbb{P}(X_{k}=0\ \text{for all}\ k)&=\mathbb{P}(X_{1}=0,X_{2}=0,\cdots,X_{n}=0)\\ &=\prod_{k=1}^{n}\mathbb{P}(X_{k}=0)\\ &=\prod_{k=1}^{n}(1-2^{2k}), \end{align*} but I don't how to compute further.
Can someone please tell me:
(1) Am I correct so far?
(2) How to compute further?
Thank you!
Edit 1:
By the way, you are really welcome to provide other possible examples, because I am not even sure if my example can work.
Thank you :)
Edit 2:
Following zhoraster's suggestion, I came up with this example, but I don't know how to show it is not convergence in distribution to $\mathcal{N}(0,1)$.
Consider $Z_{k}$ mutually independent random variables such that $$\mathbb{E}Z_{k}=0,\ \mathbb{E}Z_{k}^{2}=1$$ and for some $q>2$ and $C<\infty$, we have $$\mathbb{E}|Z_{k}|^{q}\leq C.$$
Now, set $X_{k}:=2^{k}Z_{k}$ and $S_{n}=\sum_{k=1}^{N}X_{k}$.
Note that $X_{k}$ is well-defined for each $k$, i.e. the probability will not blow up, since $$\mathbb{P}(X_{k}\leq x)=\mathbb{P}(2^{k}Z_{k}\leq x)=\mathbb{P}(Z_{k}\leq 2^{-k}x).$$
Then, it is immediate that $\mathbb{E}X_{k}=0$ and thus $\mathbb{E}S_{n}=0$.
Then, $$Var(X_{k})=\mathbb{E}X_{k}^{2}=\mathbb{E}2^{2k}Z_{k}^{2}=2^{2k},$$ and therefore $$Var(X_{k})^{q/2}=2^{qk}\ \text{and}\ Var(S_{n})=\sum_{k=1}^{n}4^{k}\longrightarrow\infty.$$
Thus, $\sigma_{n}$ diverges as $n\longrightarrow\infty$.
Further, $$\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{q}=\mathbb{E}|X_{k}|^{q}=2^{qk}\mathbb{E}|Z_{k}|^{q}\leq 2^{qk}C=CVar(X_{k})^{q/2},\ \text{as desired}.$$
Now, $X_{k}$ satisfies all the required properties, what should I do to show $(S_{n}-\mathbb{E}S_{n})/\sigma_{n}$ does not converge to $\mathcal{N}(0,1)$ in distribution?
I have some attempted but I don't know how to proceed. Again, I want to show $\mathbb{P}(S_{n}/\sigma_{n}=0)$ does not converge to $0$. So we compute: \begin{align*} \mathbb{P}\Big(\dfrac{S_{n}}{\sigma_{n}}=0\Big)&=\mathbb{P}(S_{n}=0)\\ &=\mathbb{P}\Big(\sum_{k=1}^{n}X_{k}=0\Big)\\ &\geq\mathbb{P}\Big(X_{k}=0\ \text{for all}\ k\Big)\\ &=\mathbb{P}(Z_{k}=0\ \text{for all}\ k\Big), \end{align*} but I don't know how to proceed.
Any idea?
Let $Z_k$ be iid Rademacher: $\mathrm{P}(Z_k=1) = \mathrm{P}(Z_k=-1) = 1/2$, and $X_k = 2^{k-1} Z_k$. Then, $$\sigma_n^2 = \mathrm{Var}(S_n) = \sum_{k=1}^n 4^{k-1} = \frac{4^{n}-1}3 \sim \frac{4^n}3, n\to\infty,$$ and $$ \left|S_n - \mathrm{E}[S_n]\right|= \left|S_n\right|\le \sum_{k=1}^n 2^{k-1}<2^n. $$ Consequently, $$ \lim_{n\to\infty}\mathrm{P}\Bigl(\frac{S_n - \mathrm{E}[S_n]}{\sigma_n} >2 \Bigr)\le \lim_{n\to\infty}\mathrm{P}\Bigl(\frac{2^n}{\sigma_n} >2 \Bigr) = \mathrm{P}(\sqrt{3} >2 )= 0, $$ so the normalized sums cannot converge to the Gaussian distribution. (Actually, they converge to the uniform distribution on $[-\sqrt{3},\sqrt{3}]$.)