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If $\mathbf{X} \sim \mathcal{N}_N(\mathbf{m}, \mathbf{C})$ is an $N$-dimensional gaussian vector, where $\mathbf{m} \in \mathbb{R}^{N}$ and $\mathbf{C} \in \mathbb{R}^{N \times N}$, what is the distribution of $$ Y = \| \mathbf{X} \|^2 $$ where $\| \cdot \|$ denotes the $L_2$-norm (Euclidean norm) ?

It may be useful to know that the mean can be easily calculated via $$ \mathbb{E}[ \| \mathbf{X} \|^2 ] = \mathbb{E}\left[\sum_{i=1}^N X_i^2 \right] = \sum_{i=1}^N \mathbb{E}[X_i^2] = \sum_{i=1}^N (\sigma^2_i + m_i^2) = \sum_{i=1}^N\sigma_i^2 + \sum_{i=1}^N m_i^2 = \mathrm{tr}(\mathbf{C}) + \| \mathbf{m} \|^2 $$ where $\mathrm{tr}(\cdot)$ denotes the trace of a matrix.

EDIT: Related question: link.

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If $\mathbf{m}=0$ and $\mathbf{C}$ is the identity matrix, then $Y$ is (by definition) distributed according to a chi-squared distribution.

We can relax the assumption that $\mathbf{m}=0$ and obtain the non-central chi-squared distribution.

On the other hand, if we maintain the assumption that $\mathbf{m}=0$ but allow for general $\mathbf{C}$, we have the Wishart distribution.

Finally, for general $(\mathbf{m},\mathbf{C})$, Y has a generalised chi-squared distribution.

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  • $\begingroup$ Thank you! I needed the general $(\mathbf{m}, \mathbf{C})$ case. Can you please edit your answer by writing the CDF of the generalized chi-squared distribution? Or some other useful representation. The Wikipedia page doesn't have much information about it. $\endgroup$ – PseudoRandom Apr 5 '18 at 12:17
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    $\begingroup$ @PseudoRandom Unfortunately that's the extent of my knowledge on it! I'd suggest you start looking at the references on the Wikipedia page; that might get you somewhere. I'll try to have a look and update my answer if I find something helpful. Otherwise, you might also want to look at the latest edit to my answer, which might be of some help. $\endgroup$ – Theoretical Economist Apr 5 '18 at 12:43
  • $\begingroup$ Thanks! In my specific case, $\mathbf{m} \neq \mathbf{0}$, but others may find your edit very useful. I will have a look at the references, thanks for that. $\endgroup$ – PseudoRandom Apr 5 '18 at 17:50

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