"General" non centered Chi distribution (having correlated random variables)? Let $\mathbf{X} = [X_0, X_1]^t \sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$ with $\boldsymbol{\mu} = [\mu_0, \mu_1]^t \in \mathbb{R}^2$ and $
 \boldsymbol{\Sigma} = \begin{bmatrix}
       \sigma_0^2 & \rho\sigma_0\sigma_1 \\[0.3em]
       \rho\sigma_0\sigma_1 & \sigma_1^2           \\[0.3em]
     \end{bmatrix}$. Can we derive a probability density function for the random variable $Y = \|\mathbf{X}\| = \sqrt{X_0^2+X_1^2}$, having $\rho \neq 0$ ? I would expect the solution -that is, if it exists- to be valid for higher dimensions.
 A: Yes, you can!  Make set of transformations to effectively uncorrelate each term and obtain a statistic with a noncentral $\chi^{2}$ distribution. I am going to give you something more general.I will call your random vector $\mathbf{x}$. I will also call your mean vector, $\boldsymbol{\mu}$ the expected value of $\mathbf{x}$, $E\left\{\mathbf{x} \right\}$, to emphasize the role each distribution parameter plays. Now consider $s$ $=$ $\mathbf{x}^{\text{T}} \mathbf{A} \mathbf{x}$, where $A$ has rank $r$.
\begin{equation}
\begin{split}
\mathbf{y} &\triangleq \mathbf{\Sigma}^{- \frac{1}{2} } \cdot \mathbf{x}
\\
\mathbf{z} &\triangleq  \mathbf{y} -  \mathbf{\Sigma}^{- \frac{1}{2} } \cdot E\left\{  \mathbf{x}\right\}
\\
\mathbf{\Sigma}^{\frac{1}{2} } \,  \mathbf{P}_{\mathbf{1}}^{\perp} \, \mathbf{\Sigma}^{ \frac{1}{2} } &\triangleq \mathbf{U}^{\text{T}} \boldsymbol{\Lambda} \mathbf{U}
\quad \text{(Spectral Theorem)}
\\
\mathbf{w} &\triangleq \mathbf{U}^{\text{T}} \mathbf{z}
\\
\mathbf{b} &\triangleq \mathbf{U}^{\text{T}} \mathbf{\Sigma}^{- \frac{1}{2} } \cdot E\left\{  \mathbf{x} \right\}
\end{split}
\end{equation}
Using these definitions, write:
\begin{equation}
s \triangleq \mathbf{x}^{\text{T}} \mathbf{A} \mathbf{x} = \left( \mathbf{w} + \mathbf{b} \right)^{\text{T}} \boldsymbol{\Lambda}  \left( \mathbf{w} + \mathbf{b} \right)
\end{equation}
Because $\text{cov}(\mathbf{w})$ $=$ $\mathbf{I}$ and $\mathbf{b}$ is a deterministic vector, $s$ is now a $\chi_{r}^{2}$ distribution with noncentrality parameter $\vert \vert \, \mathbf{b}\, \vert \vert^{2}$.
Your objective was to find the $\chi$-distribution. In that case, you do a simple transformation:
\begin{equation}
t = \sqrt{s} \implies \quad p_{T}(t) = 2 \,\vert s\vert \,p_{S}\left( t^{2} \right)
\end{equation}
When you set $\mathbf{A}$ $=$ $\mathbf{I}$, you are are done.
