I'm reading a proof of Multivariate CLT using Lindeberg Theorem.

Let $X_n = (X_{ni},... ,X_{nk})$ be independent random vectors all having the same distribution. Suppose that $E[X_{nu}]<\infty$; let the vector of means be $c=(c_1,..., c_k)$, where $c_u=E[X_{nu}],$ and let the covariance matrix be $\Sigma = [\sigma_{uv}],$ where $\sigma_{uv}=E[(X_{nu} — c_u)(X_{nv} — c_v)].$ Put $Sn=X_{1}+\cdots X_{n}.$ Under these assumptions, the distribution of the random vector $(S_n — nc)/\sqrt{n}$ converges weakly to the centered normal distribution with covariance matrix $\Sigma$.

The proof is as follows:

Let $Y =(Y_1,...,Y_{n})$ be a normally distributed random with $0$ means and covariance matrix $\Sigma.$ For given $t=(t_1,...,t_k)$ let $Z_n=\displaystyle\sum_{u=1}^{k}t_u(X_{nu}-c_{u})$ and $Z=\displaystyle\sum_{u=1}^{k}t_uY_u.$ Then it suffices prove that $n^{1/2}\displaystyle\sum_{j=1}^{n}Z_j$ converges in distribution to $Z$ (for arbitrary $t$). But this is an instant consequence of the Lindeberg-Levy theorem.

I'm stuck following this proof. I'm not sure if Lindeberg condition is satisfied, i.e. $$\displaystyle\lim_{n\rightarrow\infty}\displaystyle\sum_{k=1}^{n}\frac{1}{s_n}\int_{\{|Z_k/\sqrt{n}|>\epsilon s_{n}\}}\frac{|Z_k|^2}{n} dP=0.$$

My idea is that $\{|Z_k/\sqrt{n}|>\epsilon s_{n}\}$ decreases to $\emptyset$; that's the reason of the integral converges to $0,$ but what about of the convergence or divergence of $s_{n}$ and the sum that tends to infinity?

Any kind of help is thanked in advanced.

  • $\begingroup$ I think there's a mistake in your formula for the Lindberg condition. $\endgroup$ – Ryan Warnick Feb 9 '18 at 18:33
  • $\begingroup$ ${\displaystyle \lim _{n\to \infty }{\frac {1}{s_{n}^{2}}}\sum _{k=1}^{n}\mathbb {E} \left[(X_{k}-\mu _{k})^{2}\cdot \mathbf {1} _{\{|X_{k}-\mu _{k}|>\varepsilon s_{n}\}}\right]=0}$ is how I've seen it represented. I don't know if the two are equivalent, but the summation in your expression doesn't have any indexed terms inside of it. $\endgroup$ – Ryan Warnick Feb 9 '18 at 18:49
  • $\begingroup$ You're right @RyanWarnick. Thanks to note it. $\endgroup$ – Squird37 Feb 11 '18 at 19:10

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.