Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $R$?


share|cite|improve this question
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 $\gamma = \frac{2 \sigma_1^2 \sigma_2^2}{\sigma_1^4 + \sigma_2^4}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.