Talk:Ratio distribution

Noncentral normal (correlated or uncorrelated) RV ratio

It appears that Springer 1979 contains the exact PDF for a ratio of two RVs with normally distributed, non-central, non-identical means and arbitrary correlation coefficient. It's in Problem 4.28 on page 166 and it appears that this is the result from Hinkley 1969, but it is not stated in the "Exact correlated noncentral normal ratio". Should this be clarified in this section? — Preceding unsigned comment added by Kowfrancis (talk • contribs) 23:50, 8 December 2021 (UTC)

Distribution ratio

help me for my exam! Does the value of the distribution ratio depend on 1.the relative volumes of the solvents in equilibrium 2.the mass of the solute dissolved 3.temperature?

The "distribution ratio" is an concept in chemistry (as far as I read), see, e.g., liquid-liquid extraction. The "ratio distribution" is something different, and the present article does not explain this difference. — fnielsen 08:18, 16 May 2007 (UTC)

Proper definition of variables

nowhere are lowercase x,y,z, or t defined, and they should be. —Preceding unsigned comment added by 134.107.13.58 (talk) 12:14, 13 January 2011 (UTC)

It's standard in probability to use lowercase letters to denote a particular values of the corresponding random variable. FilipeS (talk) 16:32, 6 October 2012 (UTC)

Ratio vs. product distributions

Someone wrote in the article:

Interestingly, this distribution does not depend on $a$ and it should be noted that the result stated by Springer (p158 Question 4.6) is not correct. The ratio distribution is similar to but not the same as the product distribution of the random variable $W=XY$

The statement is nevertheless correct in the case where the numerator and the denominator are independent standard Cauchy random variables, which seems to be what the article is saying earlier. FilipeS (talk) 16:31, 6 October 2012 (UTC)

Is the formula correct?

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,

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

it would seem that when

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

the integral would simplify to

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

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 $p_{X}(zy)$, not $p_{X}(x)$, 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 $p(x,y)={\frac {1}{2\pi \sigma ^{2}}}exp(-{\frac {x^{2}}{2}})exp(-{\frac {y^{2}}{2}})$

then -

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

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

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

we have

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

which is a Cauchy distribution.