# Thread: showing that a quantile relation holds

1. ## showing that a quantile relation holds

Use the result of F(3,4) = 1/F(4/3) about F distribution to show that the quantile relation f'subscript'(0.95,3,4) = 1/f'subscript'(0.05,4,3).

2. Using upper percentile points....

Which means that $\displaystyle a=P(F_{3,4}>F_{3,4,a})$

we have with a=.95............

$\displaystyle .95=P(F_{3,4}>F_{3,4,.95})$

(Note that $\displaystyle F_{3,4}$ is a rv and $\displaystyle F_{3,4,a}$ is a percentile point.)

Taking the reciprocal inside the probability we have

$\displaystyle .95=P(F_{3,4}>F_{3,4,.95})$

$\displaystyle =P(F^{-1}_{3,4}<F^{-1}_{3,4,.95})$

$\displaystyle =P(F_{4,3}<F^{-1}_{3,4,.95})$

(Which follows by the definition of an F, it's a ratio of chi-squares divided by their dfs.)

And note that

$\displaystyle P(F_{4,3}<F_{4,3,.05})=.95$ also.

Thus by the continuity of the densities, these percentile points must be equal....

Hence $\displaystyle F^{-1}_{3,4,.95}$ must equal $\displaystyle F_{4,3,.05}$

and this is true for any probability, not just .05.