Theorem: Expectation of Snedecor's F-Distribution

Let $n, m$ be strictly positive integers. 
Let $X \sim F_{n, m}$ where $F_{n, m}$ is Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac m {m - 2}$
for $m > 2$, and does not exist otherwise.

Proof:
Let $Y$ and $Z$ be independent random variables.
Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. 
Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom.
Then, from Quotient of Independent Random Variables with $\chi^2$ Distribution Divided by Degrees of Freedom has Snedecor's $F$-Distribution:
:$\dfrac {Y / n} {Z / m} \sim F_{n, m}$
So we aim to compute:
:$\expect {\dfrac {Y / n} {Z / m} }$
Let $f_Y$ and $f_Z$ be the probability density functions of $Y$ and $Z$ respectively.
Let $f_{Y, Z}$ be the joint probability density function of $Y$ and $Z$. 
From Condition for Independence from Joint Probability Density Function, we have for each $y, z \in \R_{\ge 0}$: 
:$\map {f_{Y, Z} } {y, z} = \map {f_Y} y \map {f_Z} z$
We therefore have: 
{{begin-eqn}}
{{eqn | l = \expect {\dfrac {Y / n} {Z / m} }
      | r = \int_0^\infty \int_0^\infty \frac {y / n} {z / m} \map {f_{Y, Z} } {y, z} \rd y \rd z
}}
{{eqn | r = \frac m n \int_0^\infty \int_0^\infty \frac y z \map {f_Y} y \map {f_Z} z \rd y \rd z
}}
{{eqn | r = \frac m n \paren {\int_0^\infty \frac {\map {f_Z} z} z \rd z} \paren {\int_0^\infty y \map {f_Y} y \rd y}
      | c = rewriting
}}
{{eqn | r = \frac m n \paren {\frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z} \paren {\frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2} e^{-y / 2} \rd y}
      | c = {{Defof|Chi-Squared Distribution}}
}}
{{end-eqn}}
Note that the integral converges: 
:$\ds \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z$
{{iff}}: 
:$\dfrac m 2 - 2 > -1$
That is: 
:$m > 2$
With that, we have: 
{{begin-eqn}}
{{eqn | l = \frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z
      | r = \frac 1 {2^{m / 2 - 1} \map \Gamma {\frac m 2} } \int_0^\infty \paren {2 u}^{m / 2 - 2} e^{-u} \rd u
      | c = substituting $z = 2 u$
}}
{{eqn | r = \frac {2^{m / 2 - 2} } {2^{m / 2 - 1} \map \Gamma {\frac m 2} } \int_0^\infty u^{m / 2 - 2} e^{-u} \rd u
}}
{{eqn | r = \frac 1 2 \frac {\map \Gamma {\frac m 2 - 1} } {\map \Gamma {\frac m 2} }
      | c = {{Defof|Gamma Function}}
}}
{{eqn | r = \frac {\map \Gamma {\frac m 2 - 1} } {\paren {\frac m 2 - 1} \map \Gamma {\frac m 2 - 1} }
      | c = Gamma Difference Equation
}}
{{eqn | r = \frac 1 {m - 2}
}}
{{end-eqn}}
The integral: 
:$\ds \int_0^\infty y^{n / 2} e^{-y / 2} \rd y$
converges {{iff}} $n > -2$. 
This is ensured by the fact that $n \in \N$. 
With that, we have: 
{{begin-eqn}}
{{eqn | l = \frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2} e^{-y / 2} \rd y
      | r = \frac 2 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty \paren {2 v}^{n / 2} e^{-v} \rd v
      | c = substituting $y = 2 v$
}}
{{eqn | r = \frac {2^{n / 2} } {2^{n / 2 - 1} \map \Gamma {\frac n 2} } \int_0^\infty v^{n / 2} e^{-v} \rd v
}}
{{eqn | r = 2 \times \frac {\map \Gamma {\frac n 2 + 1} } {\map \Gamma {\frac n 2} }
}}
{{eqn | r = 2 \times \frac {\frac n 2 \map \Gamma {\frac n 2} } {\map \Gamma {\frac n 2} }
      | c = Gamma Difference Equation
}}
{{eqn | r = n
}}
{{end-eqn}}
So: 
{{begin-eqn}}
{{eqn | l = \expect {\dfrac {Y / n} {Z / m} }
      | r = \frac {n m} {n \paren {m - 2} }
}}
{{eqn | r = \frac m {m - 2}
}}
{{end-eqn}}
{{qed}}