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I've just calculated an exercise with confidence intervals. My solution is correct if $$\frac{1}{F_{n,m;1-\alpha}} = F_{m,n;\alpha}$$

Is that the case?

If not: Is there any relationship between $F_{n,m}$ and $F_{m,n}$?

Notation

$F_{m,n;\alpha}$ denotes the $\alpha$ quantile of the Fisher distribution with parameters $m$ and $n$. The Fisher distribution can be created by two independent random variables $X \sim \chi^2_m, Y \sim \chi^2_n$: $$\frac{\frac{1}{m} X}{\frac{1}{n} Y} \sim F_{m,n}$$

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    Personally I am not familiar enough with things as "quantiles" to say yes or no this. But yes, there is a relationship between $F_{n,m}$ and $F_{m,n}$. Note that $F_{n,m}$ and $1/F_{m,n}$ must have equal distributions, which follows directly from the description you gave yourself from the Fisher distribution (I added that $X$ and $Y$ must be independent here). – 2017-02-25
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    Yes, it's correct. If $U\sim F_{n,m}$, then $1/U\sim F_{m,n}$. And since the map $U\mapsto 1/U$ is strictly decreasing, the $\alpha$ quantile $F_{n,m}$ maps to the $1-\alpha$ quantile of $1/F_{n,m}=F_{m,n}$: this is just the same as saying thay if $\alpha=\Pr(U – 2017-02-25
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    @EinarRødland I just found a proof. I guess it is the same what you wrote, but a bit more detailed. – 2017-02-26

1 Answers 1

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The quantile function $F_{n,m;\alpha} = F_{n,m}(\alpha) : [0, 1] \rightarrow \mathbb{R}$ is defined by

$$P(X \leq F_{n,m;\alpha}) = \alpha$$

Let $X \sim F_{n,m}$, $Y \sim F_{m,n}$. By definition of the F distribution: $Y^{-1} \sim F_{n,m}$. It follows:

\begin{align} P(X \leq F_{n,m;\alpha}) &= \alpha \\ &= 1 - (1-\alpha)\\ &=1 - P(Y \leq F_{m,n;1-\alpha})\\ &= P(Y \geq F_{m, n; 1-\alpha})\\ &= P \left (\frac{1}{Y} \leq \frac{1}{F_{m,n;1-\alpha}} \right )\\ &= P \left ( X \leq \frac{1}{F_{m,n;1-\alpha}} \right ) \end{align}

And hence

$$F_{n,m;\alpha} = \frac{1}{F_{m,n;1-\alpha}}$$