Distribution of the $L^{2}$ norm of a vector of components drawn from Gaussian distributions I recently asked this question involving uniform distributions. I am wondering what would be the equivalent for Gaussian distributions. The problem states as follows.
We consider a random vector $\vec{v} = \left(x_{1}, x_{2}, \dots, x_{n}\right)$ built from $n$ real random variables drawn from a Gaussian distribution $\mathcal{N}\left(\mu, \sigma\right)$, $\mu$ and $\sigma$ being the same for all $x_{i}$.
What is the distribution $D$ of the $L^{2}$-norm of such random vectors $\vec{v}$: $\left\lVert\vec{v}\right\rVert_{2} = \sqrt{x_{1}^{2}+x_{2}^{2}+\dots+x_{n}^{2}}$?
In other words, what is the analytical expression of the distribution obtained through this numerical experiment:
# Packages
import numpy as np
import random as rd
import matplotlib.pyplot as plt

# Parameters
mu = 5
sigma = 2
n = 10
count = 100000

# Compute a random norm
def random_norm(mu, sigma, n):
    v = [rd.gauss(mu, sigma) for i in range(0, n)]
    return sum([x ** 2 for x in v]) ** (1./2.)

# Generate random vectors and compute their norm
norms = [random_norm(mu, sigma, n) for i in range(0, count)]

# Plot the resulting distribution
plt.hist(norms, 100)
plt.show()

 A: The non-central Chi distribution with $k$ degrees of freedom gives the distribution of $\sqrt{\sum_{i = 1}^{k}\left(X_{i}/\sigma_{i}\right)^{2}}$ for independent Gaussian random variables with variances $\sigma_{1}^{2}, \dots, \sigma_{k}^{2}$ and means $\mu_{1}, \dots, \mu_{k}$. It has the density function
$$f\left(x;k,\lambda\right)=\dfrac{e^{-\left(x^2+\lambda^2\right)/2}x^k\lambda}
{\left(\lambda x\right)^{k/2}} I_{k/2-1}\left(\lambda x\right)$$
where $\lambda = \sqrt{\sum_{i = 1}^{k}\mu_{i}^{2}}$ and $I_{k/2-1}\left(\lambda x\right)$ is a modified Bessel function of the first kind.
So it follows that in our case with standard deviation $\sigma$ and mean $\mu$, the density function of the $L^{2}$ norm will be that of a scaled non-central Chi distribution with $n$ degrees of freedom $f\left(x/\sigma;n,\lambda\right)$, where $\lambda = \sqrt{n}\mu/\sigma$.
To see this, we can view our desired norm as being formed by
$$\sqrt{X_{1}^{2} + \dots + X_{n}^{2}} = \sigma\sqrt{Y_{1}^{2} + \dots + Y_{n}^{2}}$$
where $Y_{i} \sim  \mathcal{N}\left(\mu/\sigma, 1\right)$.
