Since $m > 4 > 2$, we have by Expectation of 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:  :$\displaystyle \int_0^\infty z^{m / 2 - 3} e^{-z / 2} \rd z$ 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:  :$\displaystyle \int_0^\infty y^{n / 2 + 1} e^{-y / 2} \rd z$ 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}} Category:Variance Category:F-Distribution nh8427ywsvbpmohhel3n2kc6kdzgm1g