Theorem: Variance 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 variance of $X$ is given by: 
:$\var X = \dfrac {2 m^2 \paren {m + n - 2} } {n \paren {m - 4} \paren {m - 2}^2}$
for $m > 4$, and does not exist otherwise.

Proof:
Since $m > 4 > 2$, we have by Expectation of Snedecor's F-Distribution:
:$\expect X = \dfrac m {m - 2}$
We now aim to compute $\expect {X^2}$ with a view to apply Variance as Expectation of Square minus Square of Expectation.
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:
:$\dfrac {Y / n} {Z / m} \sim F_{n, m}$
Therefore:
:$\expect {X^2} = \expect {\paren {\dfrac {Y / n} {Z / m} }^2}$
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 {\paren {\dfrac {Y / n} {Z / m} }^2}
      | r = \int_0^\infty \int_0^\infty \frac {y^2 / n^2} {z^2 / m^2} \map {f_{Y, Z} } {y, z} \rd y \rd z
}}
{{eqn | r = \frac {m^2} {n^2} \int_0^\infty \int_0^\infty \frac {y^2} {z^2} \map {f_Y} y \map {f_Z} z \rd y \rd z
}}
{{eqn | r = \frac {m^2} {n^2} \paren {\int_0^\infty \frac {\map {f_Z} z} {z^2} \rd z} \paren {\int_0^\infty y^2 \map {f_Y} y \rd y}
      | c = rewriting
}}
{{eqn | r = \frac {m^2} {n^2} \paren {\frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 3} e^{-z / 2} \rd z} \paren {\frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2 + 1} e^{-y / 2} \rd z}
      | c = {{Defof|Chi-Squared Distribution}}
}}
{{end-eqn}}
Note that the integral: 
:$\ds \int_0^\infty z^{m / 2 - 3} e^{-z / 2} \rd z$
converges {{iff}}:
:$\dfrac m 2 - 3 > -1$
That is: 
:$m > 4$
With that, we have for $m > 4$: 
{{begin-eqn}}
{{eqn | l = \frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 3} e^{-z / 2} \rd z
      | r = \frac 2 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty \paren {2 u}^{m / 2 - 3} e^{-u} \rd u
      | c = substituting $z = 2 u$
}}
{{eqn | r = \frac {2^{m / 2 - 3} } {2^{m / 2 - 1} \map \Gamma {\frac m 2} } \int_0^\infty u^{m / 2 - 3} e^{-u} \rd u
}}
{{eqn | r = \frac 1 4 \times \frac {\map \Gamma {\frac m 2 - 2} } {\map \Gamma {\frac m 2} }
      | c = {{Defof|Gamma Function}}
}}
{{eqn | r = \frac 1 4 \times \frac {\map \Gamma {\frac m 2 - 2} } {\paren {\frac m 2 - 1} \paren {\frac m 2 - 2} \map \Gamma {\frac m 2 - 2} }
      | c = Gamma Difference Equation
}}
{{eqn | r = \frac 1 {\paren {m - 2} \paren {m - 4} }
}}
{{end-eqn}}
Note that the integral: 
:$\ds \int_0^\infty y^{n / 2 + 1} e^{-y / 2} \rd z$
converges {{iff}}:
:$\dfrac n 2 + 1 > -1$
That is: 
:$n > -4$
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 + 1} e^{-y / 2} \rd z
      | r = \frac 2 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty \paren {2 v}^{n / 2 + 1} e^{-v} \rd v
      | c = substituting $y = 2 v$
}}
{{eqn | r = \frac {2^{n / 2 + 1} } {2^{n / 2 - 1} \map \Gamma {\frac n 2} } \int_0^\infty v^{n / 2 + 1} e^{-v} \rd v
}}
{{eqn | r = 4 \times \frac {\map \Gamma {\frac n 2 + 2} } {\map \Gamma {\frac n 2} }
      | c = {{Defof|Gamma Function}}
}}
{{eqn | r = 4 \times \frac n 2 \paren {\frac n 2 + 1} \frac {\map \Gamma {\frac n 2} } {\map \Gamma {\frac n 2} }
      | c = Gamma Difference Equation
}}
{{eqn | r = n \paren {n + 2}
}}
{{end-eqn}}
We therefore have: 
{{begin-eqn}}
{{eqn | l = \expect {X^2}
      | r = \frac {m^2 n \paren {n + 2} } {n^2 \paren {m - 2} \paren {m - 4} }
}}
{{end-eqn}}
We therefore have: 
{{begin-eqn}}
{{eqn | l = \var X
      | r = \expect {X^2} - \paren {\expect X}^2
      | c = Variance as Expectation of Square minus Square of Expectation
}}
{{eqn | r = \frac {m^2 \paren {n + 2} } {n \paren {m - 2} \paren {m - 4} } - \frac {m^2} {\paren {m - 2}^2}
}}
{{eqn | r = \frac {m^2 \paren {n + 2} \paren {m - 2} } {n \paren {m - 2}^2 \paren {m - 4} } - \frac {m^2 n \paren {m - 4} } {n \paren {m - 2}^2 \paren {m - 4} }
      | c = aiming to write as a single fraction
}}
{{eqn | r = \frac {m^2 \paren {\paren {n + 2} \paren {m - 2} - n \paren {m - 4} } } {n \paren {m - 2}^2 \paren {m - 4} }
      | c = factoring $m^2$
}}
{{eqn | r = \frac {m^2 \paren {n m + 2 m - 2 n - 4 - n m + 4 n} } {n \paren {m - 2}^2 \paren {m - 4} }
      | c = expanding brackets
}}
{{eqn | r = \frac {m^2 \paren {2 m + 2 n - 4} } {n \paren {m - 2}^2 \paren {m - 4} }
      | c = simplifying
}}
{{eqn | r = \frac {2 m^2 \paren {m + n - 2} } {n \paren {m - 2}^2 \paren {m - 4} }
      | c = factoring $2$
}}
{{end-eqn}}
{{qed}}