Let $\mathbf{A}=\sum_{\alpha=1}^nx_\alpha x_\alpha^T$ and $x_\alpha$ is distributed according to $N(0,\Sigma)$. I want to show that the first two moments of the elements of $\mathbf{A}$ are $$E\left[a_{ij}\right]=n\sigma_{ij}$$ $$E\left[\left(a_{ij}-n\sigma_{ij}\right)\left(a_{kl}-n\sigma_{kl}\right)\right]=n\left(\sigma_{il}\sigma_{jk}+\sigma_{ik}\sigma_{jl}\right)$$ I have no problem showing the first moment but I have some trouble showing the second moment.

Anderson (1963) is stating this without a proof or reference in his paper Asymptotic Theory for Principal Component Analysis http://www.jstor.org/stable/2991288?seq=1#page_scan_tab_contents. So I'm thinking that this is something quite straight forward or a standard result.

  • $\begingroup$ Your opening notation/condition is unclear, for me. $x_\alpha$ seems to be a multivariate vector. Is it to come from the specified normal distribution, or is it a sample realization (number of cases, i.e. many such vectors collected) giving exactly such distribution? Also, I didn't quite understand your definition of matrix A. Is it a vector or matrix? $\endgroup$
    – ttnphns
    Commented Apr 21, 2017 at 9:06
  • $\begingroup$ @ttnphns: $x_\alpha$ is one sample realization vector. Matrix $A$ is the scatter matrix over the sample of $n$ such vectors (note that the sum is going from $\alpha=1$ to $n$). $\endgroup$
    – amoeba
    Commented Apr 21, 2017 at 9:34
  • $\begingroup$ Note that $A/n$ is a sample covariance matrix, so one can equivalently ask about the covariance between two elements of the sample covariance matrix. I added [covariance-matrix] tag. $\endgroup$
    – amoeba
    Commented Apr 21, 2017 at 9:37
  • 1
    $\begingroup$ You need to use formulas for the fourth moments of Normal variables. It's just a matter of working with the (four) indexes in $E[a_{ij}a_{kl}]$, unpacking them in terms of the $x$'s, and paying attention to when two, three, or four of the same $x$ component appear together. $\endgroup$
    – whuber
    Commented Apr 21, 2017 at 13:51

1 Answer 1


Thank you, I managed to show this based on your comments

\begin{array}{l} E\left[ {\left( {{a_{ij}} - n{\sigma _{ij}}} \right)\left( {{a_{kl}} - n{\sigma _{kl}}} \right)} \right] = E\left[ {{a_{ij}}{a_{kl}} - {a_{ij}}n{\sigma _{kl}} - n{\sigma _{ij}}{a_{kl}} + n{\sigma _{ij}}n{\sigma _{kl}}} \right] = \\ E\left[ {\sum\limits_{\alpha = 1}^n {{x_{\alpha i}}x_{\alpha j}^{}} \sum\limits_{\alpha = 1}^n {{x_{\alpha k}}x_{\alpha l}^{}} } \right] - {n^2}{\sigma _{ij}}{\sigma _{kl}} = \\ E\left[ {\sum\limits_{\alpha = 1}^n {{x_{\alpha i}}x_{\alpha j}^{}{x_{\alpha k}}x_{\alpha l}^{}} + \sum\limits_{\alpha \ne \omega }^n {{x_{\alpha i}}x_{\alpha j}^{}{x_{\omega k}}x_{\omega l}^{}} } \right] - {n^2}{\sigma _{ij}}{\sigma _{kl}} = \\ E\left[ {\sum\limits_{\alpha = 1}^n {{x_{\alpha i}}x_{\alpha j}^{}{x_{\alpha k}}x_{\alpha l}^{}} } \right] + \sum\limits_{\alpha \ne \omega }^n {E\left[ {{x_{\alpha i}}x_{\alpha j}^{}} \right]E\left[ {{x_{\omega k}}x_{\omega l}^{}} \right]} - {n^2}{\sigma _{ij}}{\sigma _{kl}} = \\ \sum\limits_{\alpha = 1}^n {E\left[ {{x_{\alpha i}}x_{\alpha j}^{}{x_{\alpha k}}x_{\alpha l}^{}} \right]} + n\left( {n - 1} \right){\sigma _{ij}}{\sigma _{kl}} - {n^2}{\sigma _{ij}}{\sigma _{kl}} = \\ n\left( {{\sigma _{ij}}{\sigma _{kl}} + {\sigma _{il}}{\sigma _{jk}} + {\sigma _{ik}}{\sigma _{jl}}} \right) + n\left( {n - 1} \right){\sigma _{ij}}{\sigma _{kl}} - {n^2}{\sigma _{ij}}{\sigma _{kl}} = n\left( {{\sigma _{il}}{\sigma _{jk}} + {\sigma _{ik}}{\sigma _{jl}}} \right) \end{array}

The moments of $x_1,...,x_p$ with a joint normal distribution can be obtained from the characteristics function. (The first four moments of $x$ are available in Anderson (2003) An Introduction to Multivariate Statistical Analysis. 3ed.)


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