Median of the F-distribution Is the median of the F-distribution with m and n degrees of freedom decreasing in n, for any m?
From experiments it looks like it might be, but I have been unable to prove it.
 A: This officially free to download "Handbook of Statistical Distributions" shows that (using their notation) the cumulative distribution (cdf) of the F-distribution can be written as the cdf of a transformed Beta variable:
$$\int_{0}^{F_a} f\left(F;m,n\right)dF=1-a=\frac{B_x(\frac m2,\frac n2)}{B(\frac m2,\frac n2)}\equiv G_B(x;\frac m2,\frac n2),\;x=\frac{mF_a}{n+mF_a}$$
Moreover, based on this source the median of a $B(\alpha,\beta)$ distribution, for $\alpha>1, \beta>1$ is approx. equal to $ \frac {\alpha -\frac 13}{\alpha +\beta -\frac 23}$. In our case this approximate formula holds for $m\gt 2, n \gt2$. For this case then we have (setting $a=\frac 12$)
$$\frac{mF_\frac 12}{n+mF_\frac 12} \approx\frac {\frac m2 -\frac 13}{\frac m2 +\frac n2 -\frac 23}$$
Doing the algebra we arrive at 
$$F_\frac 12\approx \frac{n}{3n-2}\frac{3m-2}{m}$$
and so 
$$\text{sign} \left(\frac {\partial}{\partial n} F_\frac 12\right)=\text{sign} \left(\frac {\partial}{\partial n} \frac{n}{3n-2}\right)= \text{sign} \left(3n-2 - 3n\right) \lt 0$$
For the case $\alpha =1 \Rightarrow m=2$ we have (source again)
$$\frac {mF_\frac 12}{n+mF_\frac 12} = 1-\frac {1}{2^{(2/n)}}$$
which leads again to a negative relationship between the median of the F-distribution and the denominator degrees of freedom.
I have no proof for the remaining degrees of freedom. The $F(1,n)$ distribution is the distribution of a squared Student's-t random variable, if that helps.
