# Distribution of Squared Euclidean Norm of Gaussian Vector

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.

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.
• 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. – PseudoRandom Apr 5 '18 at 12:17
• 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. – PseudoRandom Apr 5 '18 at 17:50