Probability density function of a quotient of two normal random variables I have the following expression:
$$R=\frac{\sigma_1^2\nu_1(t)-\sigma_2^2\nu_2(t)}{\sigma_1^2\nu_1(t)+\sigma_2^2\nu_2(t)}$$
where:
$$[\nu_1(t),\nu_2(t)]$$ are two independent normally distributed random variables.
My question is: how can I find an expression for the probability density function $(pdf)$ of $R$? 
Thanks.
 A: Let $\nu_1$ and $\nu_2$ be independent standard normal random variables. Then $U=\frac{\nu_1}{\nu_2}$ is well known to follow Cauchy distribution with pdf:
$$
    f_U(u) = \frac{1}{\pi} \frac{1}{1+u^2}
$$
Let $X = \frac{\sigma_1^2 - U  \cdot  \sigma_2^2}{\sigma_1^2 + U  \cdot  \sigma_2^2}$. Assuming $\sigma_1>0$ and $\sigma_2 > 0$, it is evident that the mapping $u \mapsto \frac{\sigma_1^2 - u \cdot \sigma_2^2}{\sigma_1^2 + u  \cdot  \sigma_2^2}$ maps $\mathbb{R}\backslash \{ -\frac{\sigma_1^2}{\sigma_2^2} \}$ to $\mathbb{R}\backslash \{-1\}$. Indeed, for $x \not= -1$,
$$
   \frac{\sigma_1^2 - u \cdot \sigma_2^2}{\sigma_1^2 + u  \cdot  \sigma_2^2} = x \qquad  \implies \qquad u(x) = \frac{1}{1+x} \left( \frac{\sigma_1^2}{\sigma_2^2} - x \right)
$$
Thus we readily read off $f_X(x)$ from the measure:
$$ \begin{eqnarray}
    \mathrm{d} F_U(u) &=& f_U(u) \mathrm{d} u = \frac{1}{\pi} \frac{|u^\prime(x)|}{1+u^2(x)} \mathrm{d}x = \frac{2}{\pi}  \frac{\sigma_1^2 \sigma_2^2}{(1+x)^2 \sigma_2^4 + \sigma_1^4(1-x)^2} \mathrm{d}x \\
   &=& \frac{2}{\pi}  \frac{\sigma_1^2 \sigma_2^2}{\left(\sigma_1^4 + \sigma_2^4\right)\left(x - \frac{\sigma_1^4-\sigma_2^4}{\sigma_1^4+\sigma_2^4} \right)^2 + \frac{4 \sigma_1^4 \sigma_2^4}{\sigma_1^4+\sigma_2^4}} \mathrm{d}x = \mathrm{d}F_X(x)
\end{eqnarray}
$$
We therefore see that $X$ follows a Cauchy distribution with location parameters $\mu = \frac{\sigma_1^4-\sigma_2^4}{\sigma_1^4+\sigma_2^4}$ and scale parameter $\gamma = \frac{2 \sigma_1^2 \sigma_2^2}{\sigma_1^4 + \sigma_2^4}$.
