Let $f(m,n,w)$ be the probability density function of F variable with m numerator df and n denominator df, i.e.

$$f(m,n,w)=\frac{\Gamma\left(\frac{m+n}{2}\right)(m/n)^{m/2}}{\Gamma(m/2)\Gamma(n/2)}w^{(m/2)-1}\left(1+\frac{mw}{n}\right)^{-(m+n)/2}$$

I am interested in the infimum of $\int_1^\infty f(m,n,w) dw$ over all $m,n$. From my exploration, this seems to be $$1-erf(1/\sqrt{2})\approx 0.3173$$ where $erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt$.

Can anyone prove this or point me to a reference?

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What is the significance of this result? – PEV Mar 11 '11 at 14:35
This is related to testing hypothesis about equality of variances. – TCL Mar 11 '11 at 14:40
Do you require $m$ and $n$ to be positive integers? – Mike Spivey Mar 11 '11 at 21:35
Yes. They are positive integers. It appears that the inf (min) attains at m=1, n=infinity, i.e. when it becomes $\chi^2$ with one degree of freedom. – TCL Mar 11 '11 at 21:39
I think your conjecture is correct, but unfortunately I don't see how to prove it, either. – Mike Spivey Mar 11 '11 at 23:58

(I will write first the variable and then the degrees of freedom, and use $F$ as the symbol for the cdf of the F-distribution).
I believe your conjecture can be proven in the following (not fully rigorous) way:

We are examining the function

$$\int_1^\infty f(w;m,n) dw = F(w\ge1;m;n) = 1- F(1;m,n)$$

The cdf of the F-distribution is written in terms of the regularized incomplete beta (RIB) function: $$F(w;m,n) = I_{\frac {mw}{mw+n}}(\frac m2,\frac n2)$$

In our case $w=1$ so we have

$$1- F(1;m,n) = 1- I_{\frac {m}{m+n}}(\frac m2,\frac n2)$$

By the symmetry properties of the RIB function, we have

$$1- I_{\frac {m}{m+n}}(\frac m2,\frac n2) = I_{\frac {n}{m+n}}(\frac n2,\frac m2) = F(1;n,m)$$ Linking the beginning and the end of these equalities, we search for the infimum of

$$\inf_{n,m} \int_1^\infty f(w;m,n) dw = \inf_{n,m} F(1;n,m)$$

Now $n$ has become the nominator df, and $m$ has become the denominator df. I have shown in this post that the median of the F-distribution is decreasing in the denominator df, (and it is immediate to show that it is increasing in the nominator df). When the median is increasing, it means that probability mass is moved to the right of the real line - so $F(1;n,m)$ should decrease. So we want the median to increase, hence $n$ to increase "as much as possible", meaning $n\rightarrow \infty$. Analogously, when the median is decreasing, probability mass moves to the left of the real line, and so $F(1;n,m)$ should increase, something that we don't want. Therefore, we would want the negatively related denominator df (which here is $m$) to be "as small as possible" so as for the median to be as large as possible. Since there is a lower bound of unity for df's, we are lead to $m=1$. These combinations of df's lead indeed to a chi-square with one df, and the actual value is the one you indicated involving the error function.

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