1
$\begingroup$

Prove that if $ (X_n),\: n \ge 1$ are independent r.v.s with $ E X_n = 0, E X_n^2 = \sigma_n^2$ which obey central limit theorem and $$\lim_{n\rightarrow\infty}E\bigg(\frac{S_n }{\sqrt{s_n}}\bigg)^{2k} = \frac{(2k)!}{2^kk!},$$

then $ (X_n) $ obeys a Lindeberg condition of order $2k$, i.e. $$\sum_{j = 1}^n\int_{|x| > \varepsilon s_n}|x|^{2k}dF_j(x) = o(s_{n}^{2k}),\: \quad \forall \varepsilon > 0, $$ where $ s_n^2 = E S_n^2 = \sum_{i=1}^n\sigma_i^2. $

Here is what I've got so far.

  1. Lindeberg condition of order $r > 2$ is equivalent to Lyapunov condition of order $r \;(\sum_{j = 1}^n E|X_j|^r = o(s_n^r))$.

Proof:

Clearly, Lyapunov condition implies Lindeberg condition. Suppose now that $$\sum_{j = 1}^n\int_{|x| > \varepsilon s_n}|x|^{2k}dF_j(x) = o(s_{n}^{2k}).\: $$ Then \begin{multline*} \sum_{j = 1}^n E|X_j|^r \le \sum_{j = 1}^n E((\varepsilon s_n)^{r-2}X_j^2I\{|X_j|\le \varepsilon s_n\} + |X_j|^rI\{|X_j|>\varepsilon s_n\})\le \\ \le \varepsilon ^{r-2}s_n^r + o(s_n^r) \:\; \forall \varepsilon > 0 .\: \end{multline*}

  1. $$\frac{(2k)!}{2^kk!} = (2k - 1)!! = E N^{2k},\: \quad N \sim N(0,\,1).\: $$

    It would be logical to assume that $$ \lim_{n\rightarrow\infty}E\bigg(\frac{S_n}{s_n}\bigg)^{2k} =E N^{2k} = \lim_{n\rightarrow\infty}E\bigg(\frac{S_n}{\sqrt{s_n}}\bigg)^{2k} , $$ but I don't know whether it's true and how it would help.

    Could anyone hive me a hint for solving this? Thank you!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.