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.

  • $\begingroup$ 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? $\endgroup$
    – Glen_b
    Jul 23, 2019 at 5:25
  • $\begingroup$ I edited the question. Thank you for the remarks. I think I m searching for the best in the sense the smallest interval. $\endgroup$ Jul 23, 2019 at 13:55
  • $\begingroup$ 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. $\endgroup$
    – whuber
    Jul 23, 2019 at 15:00
  • $\begingroup$ I m not sure to understand what you mean. Can you elaborate a little please? $\endgroup$ Jul 23, 2019 at 15:04
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Jul 23, 2019 at 22:57

1 Answer 1


If you're really after shortest intervals (upper endpoint minus lower endpoint), the shortest interval won't cut off the same tail area in each tail (won't be symmetric in tail probability).

Note that if you start with a symmetric interval on an asymmetric unimodal distribution, in general the height of the density at the endpoints of the interval will be different. If you keep the total tail area constant by moving both endpoints in the same direction you will either increase or decrease the interval-length. It's not hard to see that moving both endpoints in the direction where you move a little further into the tail of the higher endpoint (while keeping the total tail area constant) the interval will be shorter. If you pursue that observation, it's relatively easy to conclude that with a unimodal distribution, shortest intervals will have the same height of density at the endpoints.


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