Let \(Y\) and \(Z\) be independent random variables
Let \(Y \sim \chi_{d_1}^2\), where \(\chi_{d_1}^2\) is the chi-squared distribution with \(d_1\) degrees of freedom
Let \(Z \sim \chi_{d_2}^2\), where \(\chi_{d_2}^2\) is the chi-squared distribution with \(d_2\) degrees of freedom
In the form of Snedecor’s F-distribution:
\(\frac{\frac{Y}{d_1}}{\frac{Z}{d_2}} \sim F_{d_1,d_2}\)
Let \(f_Y\) and \(f_Z\) be the probability density functions of \(Y\) and \(Z\), and \(f_{Y,Z}\) be the joint probability density function. Then: \[f_{Y,Z}(y,z) = f_Y(y)f_Z(z)\]
Then:
\[\begin{align*} E\left(\frac{\frac{Y}{d_1}}{\frac{Z}{d_2}} \right) &= \int_0^\infty \int_0^\infty \frac{\frac{y}{d_1}}{\frac{z}{d_2}} f_{Y,Z}(y,z)dydz \\ &= \frac{d_2}{d_1} \int_0^\infty \int_0^\infty\frac{y}{z}f_Y(y)f_Z(z) dydz \\ &= \frac{d_2}{d_1} \int_0^\infty \frac{1}{z}f_Z(z) dz \int_0^\infty yf_Y(y)dy \\ &= \frac{d_2}{d_1}\left(\int_0^\infty \frac{1}{2^{\frac{2}{d_2}}\Gamma(\frac{2}{d_2})}z^{\frac{d_2}{2}-2}e^{-\frac{z}{2}} dz \right)\left( \int_0^\infty \frac{1}{2^{\frac{2}{d_1}}\Gamma(\frac{2}{d_1})}y^{\frac{d_1}{2}}e^{-\frac{y}{2}} dy \right) \\ &\text{Note: } \int_0^\infty z^{\frac{d_2}{2}-2}e^{-\frac{z}{2}} dz \text{ converges if and only if } \frac{d_2}{2} - 2 > -1 \\ \Rightarrow d_2 &> 2 \\ \therefore \text{let: } z&= 2u \\ \Rightarrow dz &= 2du \\ \int_0^\infty \frac{1}{2^{\frac{2}{d_2}}\Gamma(\frac{2}{d_2})}z^{\frac{d_2}{2}-2}e^{-\frac{z}{2}} dz &= \frac{1}{2^{\frac{2}{d_2}-1}\Gamma(\frac{2}{d_2})} \int_0^\infty (2u)^{\frac{d_2}{2}-2}e^{-u} du \\ &= \frac{2^{\frac{d_2}{2}-2}}{2^{\frac{2}{d_2}-1}\Gamma(\frac{2}{d_2})} \int_0^\infty u^{\frac{d_2}{2}-2}e^{-u} du \\ &= \frac{1}{2} \frac{\Gamma(\frac{d_2}{2}-1)}{\Gamma(\frac{d_2}{2})} \\ &= \frac{\Gamma(\frac{d_2}{2}-1)}{(\frac{d_2}{2}-1)\Gamma(\frac{d_2}{2}-1)} \\ &= \frac{1}{d_2-2} \\ &\text{Note: } \int_0^\infty \frac{1}{2^{\frac{2}{d_1}}\Gamma(\frac{2}{d_1})}y^{\frac{d_1}{2}}e^{-\frac{y}{2}} dy \text{ converges if and only if } \frac{d_1}{2} > -1 \\ \Rightarrow d_1 &> -2 \\ \therefore \text{let: } z&= 2v \\ \Rightarrow dz &= 2dv \\ \int_0^\infty \frac{1}{2^{\frac{2}{d_1}}\Gamma(\frac{2}{d_1})}y^{\frac{d_1}{2}}e^{-\frac{y}{2}} dy &= \frac{1}{2^{\frac{2}{d_1}-1}\Gamma(\frac{2}{d_1})} \int_0^\infty (2v)^{\frac{d_1}{2}}e^{-v} dv \\ &= \frac{2^{\frac{d_1}{2}}}{2^{\frac{2}{d_1}-1}\Gamma(\frac{2}{d_1})} \int_0^\infty v^{\frac{d_1}{2}}e^{-v} dv \\ &= 2\frac{\Gamma(\frac{d_1}{2}+1)}{\Gamma(\frac{d_1}{2})} \\ &= 2\frac{\frac{d_1}{2}\Gamma(\frac{d_1}{2})}{\Gamma(\frac{d_1}{2})} \\ &= d_1 \\ E\left(\frac{\frac{Y}{d_1}}{\frac{Z}{d_2}} \right) &= \frac{d_1d_2}{d_1(d_2-2)} \\ &= \frac{d_2}{d_2-2}, d_2>2 \end{align*}\]
Let \(Y\) and \(Z\) be independent random variables
Let \(Y \sim \chi_{d_1}^2\), where \(\chi_{d_1}^2\) is the chi-squared distribution with \(d_1\) degrees of freedom
Let \(Z \sim \chi_{d_2}^2\), where \(\chi_{d_2}^2\) is the chi-squared distribution with \(d_2\) degrees of freedom
In the form of Snedecor’s F-distribution:
\(\frac{\frac{Y}{d_1}}{\frac{Z}{d_2}} \sim F_{d_1,d_2}\)
Let \(f_Y\) and \(f_Z\) be the probability density functions of \(Y\) and \(Z\), and \(f_{Y,Z}\) be the joint probability density function. Then: \[f_{Y,Z}(y,z) = f_Y(y)f_Z(z)\]
Then:
\[\begin{align*} E(X^2) &= E\left(\left(\frac{\frac{Y}{d_1}}{\frac{Z}{d_2}} \right)^2\right) \\ &= \int_0^\infty \int_0^\infty \frac{\frac{y^2}{d_1^2}}{\frac{z^2}{d_2^2}} f_{Y,Z}(y,z) dydz \\ &= \frac{d_2^2}{d_1^2}\int_0^\infty \int_0^\infty \frac{y^2}{z^2} f_{Y,Z}(y,z) dydz \\ &= \frac{d_2^2}{d_1^2}\int_0^\infty \frac{1}{z^2} f_{Z}(z) dz \int_0^\infty y^2 f_{Y}(y) dy\\ &= \frac{d_2^2}{d_1^2}\int_0^\infty \frac{1}{2^{\frac{d_2}{2}}\Gamma(\frac{d_2}{2})} z^{\frac{d_2}{2}-3}e^{-\frac{z}{2}} dz \int_0^\infty \frac{1}{2^{\frac{d_1}{2}}\Gamma(\frac{d_1}{2})} z^{\frac{d_1}{2}-3}e^{-\frac{y}{2}} dy\\ &= \frac{d_2^2}{d_1^2} \left[\left(\frac{1}{2^{\frac{d_2}{2}}\Gamma(\frac{d_2}{2})}\int_0^\infty z^{\frac{d_2}{2}-3}e^{-\frac{z}{2}} dz \right) \left(\frac{1}{2^{\frac{d_1}{2}}\Gamma(\frac{d_1}{2})} \int_0^\infty y^{\frac{d_1}{2}+1}e^{-\frac{y}{2}} dy \right)\right]\\ &\text{Note: } \int_0^\infty z^{\frac{d_2}{2}-2}e^{-\frac{z}{2}} dz \text{ converges if and only if } \frac{d_2}{2} - 3 > -1 \\ \Rightarrow d_2 &> 4 \\ \therefore \text{let: } z&= 2u \\ \Rightarrow dz &= 2du \\ \frac{1}{2^{\frac{d_2}{2}}\Gamma(\frac{d_2}{2})}\int_0^\infty z^{\frac{d_2}{2}-3}e^{-\frac{z}{2}} dz &= \frac{2}{2^{\frac{d_2}{2}}\Gamma(\frac{d_2}{2})}\int_0^\infty (2u)^{\frac{d_2}{2}-3}e^{-u} dz \\ &= \frac{2^{\frac{d_2}{2}-3}}{2^{\frac{d_2}{2}-1}\Gamma(\frac{d_2}{2})}\int_0^\infty u^{\frac{d_2}{2}-3}e^{-u} \\ &= \frac{1}{4} \frac{\Gamma(\frac{d_2}{2}-2)}{\Gamma(\frac{d_2}{2})} \\ &= \frac{1}{4} \frac{\Gamma(\frac{d_2}{2}-2)}{(\frac{d_2}{2}-1)(\frac{d_2}{2}-2)\Gamma(\frac{d_2}{2}-2)} \\ &= \frac{1}{(d_2-2)(d_2-4)}\\ &\text{Note: } \int_0^\infty y^{\frac{d_1}{2}+1}e^{-\frac{y}{2}} dy \text{ converges if and only if } \frac{d_1}{2} + 1 > -1 \\ \Rightarrow d_1 & > -4 \\ \therefore \text{let: } y&= 2v \\ \Rightarrow dy &= 2dv \\ \frac{1}{2^{\frac{d_1}{2}}\Gamma(\frac{d_1}{2})} \int_0^\infty y^{\frac{d_1}{2}+1}e^{-\frac{y}{2}} dy &= \frac{1}{2^{\frac{d_1}{2}-1}\Gamma(\frac{d_1}{2})} \int_0^\infty (2v)^{\frac{d_1}{2}+1}e^{-v} dv \\ &= \frac{2^{\frac{d_1}{2}+1}}{2^{\frac{d_1}{2}}\Gamma(\frac{d_1}{2})} \int_0^\infty v^{\frac{d_1}{2}+1}e^{-v} dv \\ &= 4 \cdot \frac{\Gamma(\frac{d_1}{2}+2)}{\Gamma(\frac{d_1}{2})} \\ &= 4 \cdot \frac{d_1}{2}\left(\frac{d_1}{2} + 1 \right) \frac{\Gamma(\frac{d_1}{2})}{\Gamma(\frac{d_1}{2})} \\ &= d_1(d_1+2) \\ E(X^2) &= \frac{d_2^2}{d_1^2} \cdot \frac{d_1(d_1+2)}{(d_2-2)(d_2-4)} \\ &= \frac{d_2^2}{d_1} \cdot \frac{(d_1+2)}{(d_2-2)(d_2-4)} \\ \mathrm{Var}(X) &= \frac{d_2^2}{d_1} \cdot \frac{(d_1+2)}{(d_2-2)(d_2-4)} - \frac{d_2^2}{(d_2-2)^2} \\ &= \frac{d_2^2(d_1+2)(d_2-2) - d_2^2 d_1(d_2-4)}{d_1(d_2-2)^2(d_2-4)} \\ &= \frac{d_2^2(d_1d_2 +2d_2-2d_1 - 4 - d_1d_2+4d_1)}{d_1(d_2-2)^2(d_2-4)} \\ &= \frac{d_2^2(2d_1 + 2d_2 - 4)}{d_1(d_2-2)^2(d_2-4)} \\ &= \frac{2d_2^2(d_1 + d_2 - 2)}{d_1(d_2-2)^2(d_2-4)}, d_2>4 \end{align*}\]