F-distribution likelihood interval, does it have to be symetric?

F-distribution : https://en.wikipedia.org/wiki/F-distribution

I was told that the confidence interval for the ratio of variances (the $$F$$-function) is a symmetric interval for instance :

$$[ F_{1 - \alpha / 2, n-1 , m-1} \times \frac{S_Y^2 }{S_X^2} ; F_{\alpha / 2, n-1 , m-1} \times \frac{S_Y^2 }{S_X^2} ]$$

I know how to prove that a symmetric interval is the best one for a symmetric cumulative distribution, is it still true for $$F$$ ?

i think the only notion of "best" I saw is according to the cramer rao bound.

• 1. You don't have a confidence interval for a function there. That's a CI for a particular parameter, please correct it. 2, while editing, fix the spelling of 'symmetric'. 3, There's no compulsion to use $\alpha/2$ in each tail, though it's a common choice; the symmetric interval is an interval, not "the" interval. 4. In what sense do you intend "best" in your question? Best at what? – Glen_b Jul 23 at 5:25
• I edited the question. Thank you for the remarks. I think I m searching for the best in the sense the smallest interval. – Marine Galantin Jul 23 at 13:55
• You need to consider what you mean by "smallest," because the most appropriate meaning in this context is not the difference between the endpoints: it's the log ratio of the endpoints. – whuber Jul 23 at 15:00
• I m not sure to understand what you mean. Can you elaborate a little please? – Marine Galantin Jul 23 at 15:04
• Certainly the shortest interval (upper endpoint minus lower endpoint) won't cut off the same tail area; with a unimodal distribution, shortest intervals will instead have the same height of density at the endpoints. – Glen_b Jul 23 at 22:57