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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.

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Is n the denominator degrees of freedom? –  Michael Chernick Sep 25 '12 at 22:19
    
@Michael yes it is –  fra Sep 25 '12 at 22:48
    
Have you looked at tables of the F distribution. I don't think the tables give percentiles close to 50 but maybe looking at the 25th and the 75th with a gross interpolation would give you an idea whether or not it is true. I am sure that there are software packages that will give the cumulative F or its inverse and you can see the result from that. –  Michael Chernick Sep 26 '12 at 0:10
    
@Michael yes from software calculation it seems like the result is true. But I am looking for a proof! –  fra Sep 27 '12 at 21:09
    
If it is not true the tables could tell you by giving you just one counterexample. If you need precision you can do numerical integration to a desired level of accuracy. –  Michael Chernick Sep 27 '12 at 21:13

1 Answer 1

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.

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