Suppose that $x\sim N(0, V)$ is $p$ dimensional with $V$ diagonal having elements $v\_i^2$. Then


\begin{align}
f\_x(x) & \propto \left(\prod\_p v\_i\right)^{-1} \exp\left(-\frac{1}{2}\sum\_p \frac{x^2\_i}{v\_i^2}\right) \\[10pt]
& \propto \left(\prod\_p v\_i\right)^{-1} \exp\left(-\frac{1}{2}\sum\_p \frac{\|x\|\_2^2x^2\_i}{\|x\|\_2^2v\_i^2}\right)
\end{align}


Now let $y\_i = x\_i/\|x\|\_2$ and $u=\|x\|\_2$. Making the transformation gives


$$f\_{u,y}(u,y) \propto \left(\prod\_p v\_i\right)^{-1} u^{p-1}\exp\left(-\frac{u^2}{2} \sum\_p \frac{y^2\_i}{v\_i^2}\right)$$


where $u\in (0, \infty)$ and $y'y=1$ (with $u^{p-1}$ coming in through the Jacobian). The density doesn't factor (unless $V\propto I$), so $u$ and $y$ are dependent. This is perfectly sensible to me; informally, in the 2-dimensional case if $V=\operatorname{diag}(10000, 1)$ then clearly if the direction is near $(1,0)$ the magnitude will be larger than if it were near $(0,1)$. Similarly, it's intuitive that the dependence disappears if $V \propto I$ (in which case $y$ falls out of the density entirely).


My question is as follows: First, is my reasoning (and math!) correct? Second, in the first case where $V\not \propto I$ is it possible to reparameterize in terms of independent quantities analogous to the direction and magnitude ( maybe something like, for example, requiring $y$ to lie on an ellipsoid determined by $V$)? It seems like there should be but it's eluding me.



Dependence between direction and magnitude of multivariate normal random vector
Your reasoning looks right, although I'm not that familiar with the exact notation you're using, except that the $v\_i$ should be in the denominator, not the numerator.


In the second case the answer is yes. In general, say you have any norm $\| \cdot \|$ on $\mathbb{R}^p$. There is a measure $\mu$ on the boundary of the unit ball $B$ of $\| \cdot \|$, called the *cone measure*, with the property that there is the following version of integration in spherical coordinates:
$$
\int\_{\mathbb{R}^p} f(x) \ dx = \int\_0^\infty u^{p-1} \int\_{\partial B} f(uy) \ d\mu(y) \ du
$$
for any integrable function $f$.


Now in your case your density can be written in the form $f(x) = F(\| x \|)$, where $\| x \| = \sqrt{\sum (x\_i/v\_i)^2}$. This means that a random vector $X \sim N(0,V)$ has the property that $\| X \|$ and $X/\|X\|$ are independent, and the latter is distributed according to the cone measure on the surface of the ellipsoid $\{ x : \| x \|\_V \le 1\}$.


