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.


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.


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