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I am reading a book which talk about MCLT, in which they give one result that I don't understand.

Let A=$\sum _{s=1}^{n}x_sx'_s$, where $x'$ denote the transpose of $x$, $x_s=(x_{s(1)},...,x_{s(p)})$ is a p-variate random vector so that $x_s\sim \mathbb{N}_p(0,\Sigma)$ and $$\Sigma=\begin{pmatrix} \sigma_{11}&\sigma_{12}&... & \sigma_{1p}\\ \sigma_{11}&\sigma_{22}& ... & \sigma_{1p}\\ \sigma_{p1}&\sigma_{12}&... & \sigma_{pp}\\ \end{pmatrix}$$ If $x_1,..,x_n$ are i.i.d p-random vectors then, by the MCLT, we have $\frac{1}{\sqrt{n}}(A-n\Sigma)\sim \mathbb{N}_p(0,\Sigma_A)$ where $$ E(a_{ij})=n\sigma_{ij}\;\;(1)$$ and elements of $\Sigma_A$ are given by $$E(a_{ij}-E(a_{ij}))(a_{gh}-E(a_{gh}))=\sigma_{ih}\sigma_{jg}+\sigma_{ig}\sigma_{jh}\;\;(2)$$ (1) is easy, I understand that one, but (2) I don't know how they get it. I know that the covariance of A is given by $$\Sigma_A=E(A-E(A))(A-E(A))'.$$ I really need to understand this part. Your help is appriciated.

user229013
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  • What is the expected value of $A$? (This would be a matrix, but you should think of it as a vector). What is the covariance of $A$? This would be a tensor (a 4-dim array) but you should think of it as a matrix. – Hans Engler Apr 05 '15 at 22:08
  • Yes sir, A is a p by p matrix, so expected value of A is the expected value of each component of A which is given by (1). likewise for the covariance matrix. – user229013 Apr 05 '15 at 22:49
  • So the expected value of $A$ is given by eq. (1). Now use the same reasoning to verify (2). – Hans Engler Apr 06 '15 at 12:39
  • Thank so much Hans, finally I get it, Now I want to fill $\Sigma_A$ but I am a little bit confused because of the 4 in-dices. Assume that $$\Sigma_A=(\sigma'{ij}){1\leq i,j \leq p}$$ What is $\sigma'{ij}$? is it equal to $$\sigma'{ij}=E(a_{ij}-E(a_{ij}))(a_{gh}-E(a_{gh})),$$ I want to know if it is $\sigma'{ij}$, $\sigma'{ig}$ or something else. – user229013 Apr 06 '15 at 20:09

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