TITLE: Show/Prove that $F_{\alpha,n,m} =1/F_{1-\alpha,m,n}$
QUESTION [8 upvotes]: Show/Prove that $F_{\alpha,n,m} = \frac{1}{F_{1-\alpha,m,n}}$
The distributions I'm working with are the Fisher distribution and the Chi-square distribution.
I can prove that n and m switch for the F distribution as follows:
$$X = \frac{ \frac{A}{n} } { \frac{B}{m} }$$
where A and B follow $\chi^2$ distributions, are independent, and have n, m degrees of freedom respectively.  Then the following holds:
$$\frac{1}{X} = \frac{ \frac{B}{m} } { \frac{A}{n} }$$
My question is how do I show that the $F_{\alpha}$ part goes to $F_{1-\alpha}$
Is there a way to do this all at once or do I need to show the degrees of freedom switch separately from the $1-\alpha$ part of the proof?

REPLY [8 votes]: If you already established that
$$
1/F_{(m,n)} = F_{(n,m)}, 
$$
by the reciprocal of the chi square distributions,
then let $X\sim F_{(m,n)}$ hence $1/X\sim F_{(n,m)}$, so
\begin{align}
\alpha =& P(X\le F_{\alpha, (m,n)})\\
        =& P(1/X \ge 1/F_{a, (m,n)} )\\
        =& 1 - P(1/X \le 1/F_{a, (m,n)} )\\
         =&1-P( F_{(n,m)} \le 1/F_{a, (m,n)})\\
         =&1-F_{(n,m)}(1/F_{a, (m,n)})    
\end{align}
thus,
\begin{align}
1-\alpha =& F_{(n,m)}(1/F_{a, (m,n)})\\
F_{1-\alpha, (n,m)}         =& 1/F_{a, (m,n)}. 
\end{align}