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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 $R$?


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Are $\nu_1,\nu_2$ independent ? Try to look at $P(R\leq r)$ by interpreting this number as a double integral on a certain domain. Then differentiate w.r.t. $r$. – vanna Jul 30 '12 at 14:40
@vanna: thank you for the suggestion. Yes the $\nu_1$ and $\nu_2$ are independent. – Riccardo.Alestra Jul 30 '12 at 14:43
1 In the section "Gaussian Ratio distribution" you can find an expression for the pdf of the distribution - for the correlated case (your case) where it says that the form becomes "even more complicated"... Hope this helps! – vanguard2k Jul 30 '12 at 14:46
it is cauchy if they are mean o – mike Jul 30 '12 at 15:21
up vote 1 down vote accepted

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 $\gamma = \frac{2 \sigma_1^2 \sigma_2^2}{\sigma_1^4 + \sigma_2^4}$.

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