Quirky question about F-distribution.


We can create an F-distribution if we have two independent chi-squared random variables, and we divide them by their own degress of freedom, and divide the result by each other.

\(F = \frac{\chi^{2}_{1}/\nu_{1}}{\chi^{2}_{2}/\nu_{2}}\).

We have that:

Now intuitively I would then think that E(F)=1, but we have
E(F) = \(\frac{\nu_{2}}{\nu_{2}-2}\)

This can offcourse be proven mathematically, but is there an intuitive way of seeing that the expected value can not be 1, even though each part of the fraction has expected value 1? And is it an intuitive way to see that it has to be greater than 1, not \(\le 1\)?


Ambassador to the humans
Break it up into \((\chi_1^2 / \nu_2 ) * \frac{1}{\chi_2^2 / \nu_2}\). Since these two chi-squares are independent we see that

\(E\left[(\chi_1^2 / \nu_2 ) * \frac{1}{\chi_2^2 / \nu_2}\right] = E\left[(\chi_1^2 / \nu_2 )\right] * E\left[\frac{1}{\chi_2^2 / \nu_2}\right]\)

We know that the first expectation is 1 so your question really boils down to:

Why is \(E\left[\frac{1}{\chi_2^2 / \nu_2}\right] \neq 1\) even though \(E\left[\chi_2^2 / \nu_2\right] = 1\)

The answer is Jensen's inequality. And I'm actually going to stop here. There are some nice intuitive ways to understand Jensen's inequality but I'll let you do some research on it since that (sort of) answers your question.
Cool, from that inequality on wikipedia what it seems it seems that it is always that case that if E(X) = \(\mu\), then E(1/x) \(\ge\)1/E(X), not only for chi-squared variables, if we look at positive values, since here f(x)=1/x is convex. And since 1/x is strictly convex, we do not have equality and get strictly inequality, which then says that that E(F) is strictly greater than 1. I thought that maybe the explanation(for E(1/x)>1/E(x)) was because we have chi-squared, and since the chi-squared is skewed this could possibly provide an answer. But the answer is not in the distribution of the random variable?, it is because of the function 1/x?, and we would get the result no matter skew or symmetry?

Thanks, this was a very interesting inequality!

I guess if one think really hard, then maybe one can convince oneself intuitively that if we only have positive values, then E(1/x) must be larger than 1/E(x), but for now I do not see how to do this.
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