Let $X$ be a random variable with F-Snedecor distribuction with parameters $(n,m)$.
Let $F_{n,m}(\alpha)$, for $\alpha \in (0,1)$, be the element such that $ F_X(F_{n,m}( \alpha )) = 1 - \alpha $. Show that $ F_{n,m}(\alpha) = \dfrac{1}{F_{m,n}(1-\alpha)}$
My first question is what does $F_{m,n}(1-\alpha)$ mean. Is it the point $y$ such that $F_{1/X}(y) = \alpha$? Because I already proved that $\frac{1}{X}$ has distribuction F-Snedecor with parameters $(m,n)$.
But even if it is that, I don't know how to approach this problem. Am I supposed to compute $F_{X}(F_{n,m}(\alpha))$ and $\dfrac{1}{F_{1/X}(F_{m,n}(1-\alpha))}$ and then compare the results?
Any help would be appreciated.
Let $X\sim F(n,m)$.
Then the notation $F_{n,m;\,\alpha}$ is often used to denote the upper $\alpha$ percent point of the $F(n,m)$ distribution. This is what you have here. In other words, this means $P(X>F_{n,m;\,\alpha})=\alpha$.
You know that $X\sim F(n,m)\implies Y=\frac{1}{X}\sim F(m,n)$.
So for any $c>0$, you have $P(X>c)=P\left(\frac{1}{X}<\frac{1}{c}\right)=P\left(Y<\frac{1}{c}\right)$.
This implies that $P(X>F_{n,m;\,\alpha})=P\left(Y<\dfrac{1}{F_{n,m;\,\alpha}}\right)=\alpha\tag{1}$
Moreover, from definition of $\alpha$ you have,
$P(Y<F_{m,n;\,1-\alpha})=\alpha\tag{2}$
From $(1)$ and $(2)$ your result follows.