# Ratio of perfectly correlated gaussian distributions

Let $M$ be a positive definite matrix and let $w \in S^{d-1}$ be a unit vector uniformly distributed over the sphere. I want to understand the distribution of the quadratic form $\frac{w^T M^3 w}{w^T M^2 w}$. Since the $w$ is uniformly distributed over the unit $(d-1)$-sphere, the coordinate values are Gaussian distributed. It seems that both $w^T M^3 w$ and $w^T M^2 w$ are Gaussian distributed random variables. The ratio distribution of two Gaussian distributions is a Cauchy distribution, however in this case the distributions have correlation coefficient $\rho = 1$. Every formula for the ratio distribution I've seen has the expression $\frac{1}{\sqrt{1-\rho^2}}$.

Experimentally the distribution of $\frac{w^T M^3 w}{w^T M^2 w}$ behaves very similarly to a Gaussian. Is there a simple expression for the distribution of this random variable or for its mean and variance?

## 1 Answer

$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\la}{\lambda} \newcommand{\La}{\Lambda} \newcommand{\Si}{\Sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

Let \begin{equation} R:=\frac{w^T M^3 w}{w^T M^2 w}. \end{equation} The distribution of none of the random variables (r.v.'s) $w^T M^3 w$, $w^T M^2 w$, $R$ is Gaussian -- because each of them is positive.

However, $R$ is the ratio of two quadratic forms in iid standard normal r.v.'s $Z_1,\dots,Z_d$. Indeed, by the spherical symmetry of the distribution of the standard normal random vector $Z:=(Z_1,\dots,Z_d)$, the random vector $w$ equals $Z/\|Z\|$ in distribution, where $\|Z\|$ is the Euclidean norm of $Z$. So, we may write $R$ as the ratio of two quadratic forms in $Z_1,\dots,Z_d$: \begin{equation} R=\frac{Q_3}{Q_2},\quad Q_j:=Z^T M^j Z. \end{equation}

Consider now the spectral decomposition $M=O^T\La O$ of the positive definite matrix $M$, where $O$ is an orthogonal matrix and $\La$ is the diagonal matrix with the eigenvalues $\la_1\ge\cdots\ge\la_d>0$ of $M$ on the diagonal. Then \begin{equation} Q_j=Z^T O^T \La^j O Z=V^T \La^j V=\sum_1^d\la_i^j V_i^2, \end{equation} where $V=(V_1,\dots,V_d):=OZ$, so that $V$ is also a standard normal random vector, and hence $V_1,\dots,V_d$ are iid standard normal r.v.'s. Note also that $Q_3=\sum_1^d\la_i^3 V_i^2\le\la_1\sum_1^d\la_i^2 V_i^2=\la_1Q_2$ and, similarly, $Q_3\ge\la_dQ_2$, so that we have the specific bounds on the values of $R$, namely $\la_d\le R\le\la_1$.

In particular, if $\la_1=\la_d=:\la$, then $R=\la$, a constant. To avoid this triviality, let us assume that from now on we have $\la_d<\la_1$.

Then, using transformations of distributions of random vectors (see e.g. formula (10)), one can find integral expressions for the density (pdf) $f_R$ of $R$.

For $d=2$, this pdf can be found explicitly. Indeed, then we can write \begin{equation} R=\frac{\la_1^3 V_1^2+\la_2^3 V_2^2}{\la_1^2 V_1^2+\la_2^2 V_2^2} =\frac{\la_1^3 F+\la_2^3}{\la_1^2 F+\la_2^2}=:\rho(F),\quad F:=\frac{V_1^2}{V_2^2}. \end{equation} Let $u:=\rho^{-1}$, so that \begin{equation} u(r)=\frac{\la_2^2 (r-\la_2)}{\la_1^2 \left(\la_1-r\right)} \end{equation} for $r\in(\la_2,\la_1)$. Note also that $F$ has the $F_{1,1}$ distribution, with pdf $f_F$ given by \begin{equation} f_F(x)=\frac1{\pi(1+x)\sqrt x} \end{equation} for $x>0$. So, by formula (5), the pdf $f_R$ of $R$ is given by \begin{equation} f_R(r)=f_V(u(r))|u'(r)|= \frac{\la_1 \la_2}{\pi \sqrt{\left(\la_1-r\right) \left(r-\la_2\right)} \,[\la_1^2+(\la_1+\la_2) \left(\la_2-r\right)]} \end{equation} for $r\in(\la_2,\la_1)$. Here is the graph of $f_R$ for $\la_1=2$ and $\la_2=1$; the shape is certainly not Gaussian. 