# Proving one side of multidimensional CLT

I am having some trouble in proving one part of multi-dimensional Central limit Theorem: If $X_n$ follow CLT then $X_n$ satisfy Lindeberg Condition.

We have $k$ dimensional random vectors $X_{n,j}$, $j=1,2,...,r_n$ with $r_n\to\infty$ as $n\to\infty$. Assume $E(X_{n,j})=0$ for all $j$ and for all $n$. Suppose $V_{n,j}=Cov(X_{n,j})$ and then define $Y_{n,j}=W_n^{-1/2}X_{n,j}$ where $W_n=\sum_{j=1}^{r_n}V_{n,j}$. Then it is obvious that $\sum_{j=1}^{r_n}Cov(Y_{n,j})=I_k$.

Now suppose $Y_{n,j}$ satisfy CLT i.e. $S_n\to N(0,I_k)$ (where $S_n=\sum_{j=1}^{r_n}Y_{n,j}$) and $\max_{j\leq r_n}P(||Y_{n,j}||\geq\epsilon)\to0$ as $n\to\infty$. That is, basically I am trying to prove the Lindeberg Condition with these two assumptions.

I have shown that if for any $t\in\mathbb R^k$, we define $U_{n,j}=\dfrac{(W_n^{1/2}t)'Y_{n,j}}{\sqrt{t'W_nt}}$ then $\sum_{j=1}^{r_n}U_{n,j}\to N(0,1)$and $\max_{j\leq r_n}P(|U_{n,j}|\geq\epsilon)\to0$ as $n\to\infty$.

Hence by univariate Central Limit Theorem, Lindeberg Condition holds for $U_{n,j}$ i.e. $\sum_{j=1}^{r_n}EU_{n,j}^21_{\{|U_{n,j}|\geq\epsilon\}}\to0$ as $n\to\infty$.

I now want to show that this proves that $\sum_{j=1}^{r_n}E||Y_{n,j}||^21_{\{||Y_{n,j}||\geq\epsilon\}}\to0$ as $n\to\infty$.

I will be done if I can show that there exists constant $M$ (depending on $k$ possibly) such that $(\sum_{i=1}^ka_i^2)(\sum_{i=1}^kb_i^2)\leq M(\sum_{i=1}^ka_ib_i)^2$ where $a_i,b_i\in\mathbb R$.

If anyone can help me prove this final inequality, I will be grateful. If it doesn't hold, then can you help me prove the Lindeberg Condition in another way? As you can see, I am interested in using Cramer Wold device to prove Lindeberg condition. Thank you.

• How are the first paragraph and all the rest even related? – Did Sep 27 '15 at 19:55
• You are right. Sorry for the confusion: I meant to say in the first paragraph that if $X_n$ follow CLT then $X_n$ satisfy Lindeberg Condition. Editing. – Landon Carter Sep 28 '15 at 3:32
• BTW I realised the inequality I am looking for does not exist: consider $(a_1,...,a_k)$ and $(b_1,...,b_k)$ orthogonal. So I will have to prove Lindeberg Condition in some other way. – Landon Carter Sep 28 '15 at 3:36