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Let $x_i\in\mathbb{R}^N$ be multivariate normal distributed with mean vector $\mu\in\mathbb{R}^N$ and no correlation: $x_i\sim N(\mu,\sigma^2 I_N)$. Given $T$ iid samples, define the matrix $$X:=\left( \begin{array}{c} x_1' \\ \vdots \\ x_T'\end{array} \right)\in\mathbb{R}^{T \times N}$$ I am interested how the resulting distribution of the elements of the matrix $X'X$ can be characterized. Clearly, the diagonal element $X'X_{i,i}$ contain the sum of independent squared normal variables with mean vector $\mu_i$ and variance $\sigma^2$. Therefore $\sigma^2X'X_{i,i}$ corresponds to a non-central chi squared distribution with $T$ degrees of freedom and non-centrality parameter $\lambda=T\frac{\mu_i^2}{\sigma^2}$. But what can we say about the distribution of the off-diagonal elements $X'X_{i,j}$?

Update: For a better understanding I appreciate every comment that helps to solve the simplified problem with $\mu=0$.

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    $\begingroup$ (+1) A great deal can be said, and already has been (although not to the extent of giving the full multivariate distribution): see stats.stackexchange.com/questions/5399, stats.stackexchange.com/questions/51699, and stats.stackexchange.com/questions/85916. $\endgroup$
    – whuber ♦
    Commented May 29, 2015 at 17:29
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    $\begingroup$ One natural way of approaching this problem would be to express $X_i$ in spherical coordinates as the product of a unit vector and a length. This works out nicely when all means $\mu$ are zero, for then the unit vectors are uniformly distributed and the lengths have a $\chi(n)$ distribution. It's going to get trickier with nonzero means... . $\endgroup$
    – whuber ♦
    Commented Jun 1, 2015 at 16:22
  • $\begingroup$ Okay, I understand that point. The problem with non-zero $\mu$ would be that we do not have a uniform distribution on the unit sphere any more...is there any way to handle this? $\endgroup$
    – Stefan Voigt
    Commented Jun 1, 2015 at 16:55
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    $\begingroup$ Not only is the distribution of unit vectors non-nonuniform, the distribution of the radius becomes a non-central $\chi$ distribution. There are ways to simplify the situation and make it more tractable. For instance, by adopting a suitable coordinate system you may assume $\sigma=1$ and $\mu$ has only one non-zero component. That effectively reduces the heart of the problem to a one-dimensional one. $\endgroup$
    – whuber ♦
    Commented Jun 1, 2015 at 16:58
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    $\begingroup$ What whuber is suggesting it to apply a rotation which sends $\mu$ to $(||\mu||, 0, \dots, 0)'$ without modifying the variance-covariance. You can combine that with a multiplication by $1/\sigma$ to get rid (momentarily) of the variance. $\endgroup$
    – Elvis
    Commented Jun 2, 2015 at 9:13

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