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 $F$-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:  :$\displaystyle \int_0^\infty z^{m / 2 - 2} e^{-z / 2} \rd z$ {{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:  :$\displaystyle \int_0^\infty y^{n / 2} e^{-y / 2} \rd y$ 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}} Category:F-Distribution Category:Expectation 8vt3k3grqsum36goho997r6cu334gor