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$E(F)=\int xf_{k,m}dx$ where $f_{k,m}(t) = \Gamma(t)=\frac{\Gamma((k+m)/2)}{\Gamma (k/2)\Gamma(m/2)}k^{k/2}m^{m/2}t^{k/2 - 1}(m+kt)^{-(k+m)/2}$.

How do you find $E(F)$? Say you have to convert $x*f(k,m)$ to $C * f(k',m')$ where $f(k',m')$ is a pdf itself, which leaves $E(F)=C$ (which is$\frac{m}{m-2}$).

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One elementary way to obtain this result is to integrate the survival function. I will prepare the way with some notation and definitions, and then will integrate the survival function by changing the order of a double integral.


The cumulative distribution function of $F$ is

$$F_{k,m}(x) = I\left(\frac{x}{x + \lambda}; k/2, m/2\right)$$

where $\lambda = m/k$ and $I$ is the regularized incomplete Beta function

$$I(x; \alpha, \beta) = \frac{1}{B(\alpha, \beta)}\int_0^x t^{\alpha-1}(1-t)^{\beta-1}dt.$$

(For notational convenience, I will continue to equate $\alpha$ with $k/2$ and $\beta$ with $m/2$.) Notice that because $I$ is a CDF, its full integral must be unity:

$$1 = I(1;\alpha,\beta) = \frac{1}{B(\alpha, \beta)}\int_0^1 t^{\alpha-1}(1-t)^{\beta-1}dt.$$

If you wish, then, you may take $B(\alpha,\beta)$ to be defined by the integral on the right hand side. What will be important in the sequel is the fact that

$$\frac{B(\alpha+1,\beta-1)}{B(\alpha,\beta)} = \frac{\alpha}{\beta-1}$$

for all $\alpha \gt 0$ and $\beta \gt 1$. You can prove this in many ways (such as integration by parts).

Integration by parts demonstrates that the expectation of any positive distribution $F$ is the integral of its survival function $1-F$:

$$E[F] = \int_0^\infty x\, dF(x) = \lim_{t\to\infty}\left(t(1-F(t)) + \int_0^t (1-F(t)) dt\right) = \int_0^t (1-F(t)) dt$$

(provided the expectation is finite). Applying this to the Fisher-Snedecor distribution and changing the order of integration gives

$$\eqalign{ E[F_{k,m}]& = \int_0^\infty \left(1 - I\left(\frac{x}{x + \lambda}; \alpha,\beta\right)\right) dx \\ &= \int_0^\infty \frac{1}{B(\alpha, \beta)}\int_{x/(x+\lambda)}^1 t^{\alpha-1}(1-t)^{\beta-1}dt\,dx \\ &= \frac{1}{B(\alpha, \beta)}\int_0^\infty t^{\alpha-1}(1-t)^{\beta-1}\left(\int_0^{\lambda t/(1-t)}dx\right)dt \\ &= \frac{\lambda}{B(\alpha, \beta)}\int_0^1 t^{\alpha}(1-t)^{\beta-2}dt \\ &= \frac{\lambda}{B(\alpha, \beta)}\int_0^1 t^{\alpha}(1-t)^{\beta-2}dt \\ &= \frac{\lambda B(\alpha+1, \beta-1)}{B(\alpha,\beta)} \\ &= \lambda \frac{\alpha}{\beta-1} = \frac{m}{k}\frac{k/2}{m/2-1}=\frac{m}{m-2}. }$$

(Each step is simple and elementary or has been previously demonstrated.) The penultimate line required $\beta-1 \gt 0$; that is, $m \gt 2$. It is clear that the last integral (on the antepenultimate line) diverges when $m \le 2$.

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