# Show/Prove that $F_{\alpha,n,m} =1/F_{1-\alpha,m,n}$

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?

## 2 Answers

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}

This relationship is used to get lower tail probabilties of F-distributions. Most tables of F-distributions give only upper tail probabilities.

The succinct answer to your main question is as follows: When you take reciprocals, the sense of the inequality is changed: $P(X \le c) = P(1/X \ge 1/c).$

Statistical software usually gives the complete PDF of a distribution. Suppose $X \sim \mathsf{F}(5, 10)$ and you want to cut off the lower 5% of the probability in this distribution. That is, you seek $c$ such that $P(X \le c) = 0.05.$ In R statistical software quantile 0.05 of $X$ is $c = 0.2112$ to four places.

qf(.05, 5, 10)       # in R 'qf' is the inverse CDF or quantile function
## 0.2111904
pf(.2111904, 5,10)   # in R 'pf' is the CDF
## 0.04999999


Your result shows that this is the same as cutting 5% of the probability of the probability in the distribution $\mathsf{F}(10, 5)$ in the upper tail and taking the reciprocal.

That is, $Y = 1/X \sim \mathsf{F}(10, 5),$ so that $P(Y \ge d) = .95$ and using $c=1/d$ as the lower 5% cutoff of $X.$ Here $d = 4.735,$ which you should verify from the F-table in your textbook. Also, $c = 1/d = 0.2112$ cuts 5% from the lower tail of $\mathsf{F}(5, 10),$ as previously found directly from R software.

 d = qf(.95, 10, 5);  d;  1/d
## 4.735063   # d = 4.735 is the 95th quantile of F(10, 5)
## 0.2111904  # c = 1/d = 0.2112 is the 5th quantile of F(5, 10)


Note: In the notation $F_{\alpha, n, m},$ the probability $\alpha$ is not a quantile; it is an upper-tail probability.

The figure below shows the density function of $\mathsf{F}(5,10).$ The areas beneath the density to the left of 0.2112 and to the right of 3.326 are both 5%. You can find the right-hand cutoff from printed tables of the F-distribution.

• I like the discussion but it misses the point of the question. How do I prove/show that $F_{\alpha,n,m} = \frac{1}{F_{1-\alpha,m,n}}$ mathematically? In the question I show how to prove part of it but I am curious as to how to mathematically prove the $1-\alpha$ piece. Commented Jul 26, 2017 at 0:25
• Please refer the comment by Vancak... He/she has already proved that... Commented Sep 8, 2018 at 5:11