Talk:Ratio distribution

I was going to try to apply it to the t-distribution case, where (as the article states) t is the ratio of a Normal variable and a Chi-distributed variable, which are independent.

But looking at the integral,

${\displaystyle p_{Z}(z)=\int _{-\infty }^{+\infty }|y|\,p_{X,Y}(zy,y)\,dy.}$

it would seem that when

${\displaystyle p_{X,Y}=p_{X}\;p_{Y}}$

the integral would simplify to

${\displaystyle p_{Z}(z)=p_{X}(x)\int _{-\infty }^{+\infty }|y|\,p_{Y}(y)\,dy}$

which seems not at all right.

Am I completely misunderstanding here? And would it be worth doing the t distribution example explicitly, as a non-pathological case? Jheald (talk) 23:09, 12 November 2012 (UTC)

It's ${\displaystyle p_{X}(zy)}$, not ${\displaystyle p_{X}(x)}$, so this factor includes the variable of integration, and cannot be brought outside the integral. FilipeS (talk)

Take the separable case of the 4-quadrant bivariate Gaussian pdf ${\displaystyle p(x,y)={\frac {1}{2\pi \sigma ^{2}}}exp(-{\frac {x^{2}}{2}})exp(-{\frac {y^{2}}{2}})}$

then -

${\displaystyle p_{Z}(z)={\frac {1}{2\pi \sigma ^{2}}}\int _{-\infty }^{\infty }\,|y|\,exp(-{\frac {y^{2}}{2}})exp(-{\frac {(zy)^{2}}{2}})dy}$

${\displaystyle ={\frac {1}{2\pi \sigma ^{2}}}\int _{-\infty }^{\infty }\,|y|\,exp(-{\frac {y^{2}(z^{2}+1)}{2}})dy}$

From the well-known integral ${\displaystyle =\int _{0}^{\infty }\,y\,exp(-{\frac {ay^{2}}{2}})dy={\frac {1}{a}}}$

we have

${\displaystyle p_{Z}(z)={\frac {1}{\pi (1+z^{2})}}}$

which is a Cauchy distribution.