$\gdef \vec#1{\boldsymbol{#1}} \\ \gdef \rank {\mathrm{rank}} \\ \gdef \det {\mathrm{det}} \\ \gdef \Bern {\mathrm{Bern}} \\ \gdef \Bin {\mathrm{Bin}} \\ \gdef \Mn {\mathrm{Mn}} \\ \gdef \Cov {\mathrm{Cov}} \\ \gdef \Po {\mathrm{Po}} \\ \gdef \HG {\mathrm{HG}} \\ \gdef \Geo {\mathrm{Geo}}\\ \gdef \N {\mathrm{N}} \\ \gdef \LN {\mathrm{LN}} \\ \gdef \U {\mathrm{U}} \\ \gdef \t {\mathrm{t}} \\ \gdef \F {\mathrm{F}} \\ \gdef \Exp {\mathrm{Exp}} \\ \gdef \Ga {\mathrm{Ga}} \\ \gdef \Be {\mathrm{Be}} \\ \gdef \NB {\mathrm{NB}}$
（再掲）
$X \sim \F(n_1, n_2)$である時、その確率密度関数$f(x)$は、
$$\begin{aligned} f(x) = \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot \dfrac{(\frac{n_1}{n_2})^{\frac{n_1}{2}} \cdot x^{\frac{n_1}{2}-1}}{(1 + \frac{n_1}{n_2} x)^{\frac{n_1+n_2}{2}}} ~~ (x \gt 0) \end{aligned}$$であり、その特性値は、
$$\begin{aligned} &{\small ①平均：}E[X] = \dfrac{n_2}{n_2 – 2} ~~ {\small (ただしn_2 \gt 2)} \\ &{\small ②分散：}V[X] = \dfrac{2 n_2^2 (n_1+n_2-2)}{n_1 (n_2-2)^2 (n_2-4)} ~~ {\small (ただしn_2 \gt 4)} \\ &③{\small 積率母関数}M^X(\theta){\small は\theta \neq 0 で存在しません} \end{aligned}$$となる。
（①平均$E[X]$の証明）
以下、$n_2 \gt 2$とする。
$$\begin{aligned} E[X] &= \int_{0}^{\infty} x f(x) dx \\[10px] &= \int_{0}^{\infty} x \cdot \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot \dfrac{(\frac{n_1}{n_2})^{\frac{n_1}{2}} \cdot x^{\frac{n_1}{2}-1}}{(1 + \frac{n_1}{n_2} x)^{\frac{n_1+n_2}{2}}} dx \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_1}{n_2})^{\frac{n_1}{2}} \int_{0}^{\infty} x^{\frac{n_1}{2}} \cdot \dfrac{1}{(1 + \frac{n_1}{n_2} x)^{\frac{n_1+n_2}{2}}} dx \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_1}{n_2})^{\frac{n_1}{2}} \int_{1}^{0} \{ \frac{n_2}{n_1} (\frac{1}{y} – 1) \}^{\frac{n_1}{2}} \cdot y^{\frac{n_1 + n_2}{2}} \cdot (\frac{n_2}{n_1})(- \frac{1}{y^2}) dy \\ &{\scriptsize (1 + \frac{n_1}{n_2} = yとおいた)} \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_2}{n_1}) \int_0^1 (\frac{1}{y} – 1)^{\frac{n_1}{2}} \cdot y^{\frac{n_1 + n_2}{2} – 2} dy \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_2}{n_1}) \int_0^1 y^{(\frac{n_2}{2} – 1) – 1} \cdot (1 – y)^{(\frac{n_1}{2} + 1) – 1} dy \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_2}{n_1}) \cdot B(\frac{n_2}{2} – 1, \frac{n_1}{2} + 1) \\ &{\scriptsize (B(\bullet,\circ) = \int_0^1 s^{\bullet – 1} (1 – s)^{\circ – 1})} \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_2}{n_1}) \cdot \dfrac{\Gamma(\frac{n_2}{2} – 1) \Gamma(\frac{n_1}{2} + 1)}{\Gamma(\frac{n_1 + n_2}{2})} \\ &{\scriptsize (B(\bullet, \circ) = \dfrac{\Gamma(\bullet) \Gamma(\circ)}{\Gamma(\bullet + \circ)})} \\[10px] &= \dfrac{1}{ (\frac{n_2}{2} – 1) \Gamma(\frac{n_2}{2} – 1) \cdot \Gamma(\frac{n_2}{2})} \cdot \dfrac{n_2}{n_1} \cdot \Gamma(\tfrac{n_2}{2} – 1) \cdot \dfrac{n_1}{2} \Gamma(\tfrac{n_1}{2}) \\ &{\scriptsize (\Gamma(\bullet+1) = \bullet \Gamma(\bullet))} \\[10px] &= \dfrac{n_2}{n_2 – 2} \end{aligned}$$
（②分散$V[X]$の証明）
以下、$n_2 \gt 4$とする。
$$\begin{aligned} E[X^2] &= \int_{0}^{\infty} x^2 f(x) dx \\[10px] &= \int_{0}^{\infty} x^2 \cdot \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot \dfrac{(\frac{n_1}{n_2})^{\frac{n_1}{2}} \cdot x^{\frac{n_1}{2}-1}}{(1 + \frac{n_1}{n_2} x)^{\frac{n_1+n_2}{2}}} dx \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_1}{n_2})^{\frac{n_1}{2}} \int_{0}^{\infty} x^{\frac{n_1}{2} + 1} \cdot \dfrac{1}{(1 + \frac{n_1}{n_2} x)^{\frac{n_1+n_2}{2}}} dx \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_1}{n_2})^{\frac{n_1}{2}} \int_{1}^{0} \{ \frac{n_2}{n_1} (\frac{1}{u} – 1) \}^{\frac{n_1}{2} + 1} \cdot u^{\frac{n_1 + n_2}{2}} \cdot (\frac{n_2}{n_1})(- \frac{1}{u^2}) du \\ &{\scriptsize (1 + \frac{n_1}{n_2} x = \frac{1}{u}とおいた)} \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_1}{n_2})^{2} \int_{0}^{1} u^{(\frac{n_2}{2} – 2) – 1} \cdot (1 – u)^{(\frac{n_1}{2} + 2) – 1} du \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_1}{n_2})^{2} \cdot B(\frac{n_2}{2} – 2, \frac{n_1}{2} + 2) \\ &{\scriptsize (B(\bullet,\circ) = \int_0^1 s^{\bullet – 1} (1 – s)^{\circ – 1})} \\[10px] &= \dfrac{\Gamma(\frac{n_1+n_2}{2})}{\Gamma(\frac{n_1}{2}) \Gamma(\frac{n_2}{2})} \cdot (\frac{n_1}{n_2})^{2} \cdot \dfrac{\Gamma(\frac{n_2}{2} – 2) \Gamma(\frac{n_1}{2} + 2)}{\Gamma(\frac{n_1 + n_2}{2})} \\ &{\scriptsize (B(\bullet, \circ) = \dfrac{\Gamma(\bullet) \Gamma(\circ)}{\Gamma(\bullet + \circ)})} \\[10px] &= \dfrac{1}{ \Gamma(\frac{n_1}{2}) \cdot (\frac{n_2}{2} – 2) (\frac{n_2}{2} – 1) \Gamma(\frac{n_2}{2} – 2)} \cdot (\frac{n_1}{n_2})^{2} \cdot \Gamma(\tfrac{n_2}{2} – 2) \cdot (\frac{n_1}{2} + 1) (\frac{n_1}{2})\Gamma(\tfrac{n_1}{2}) \\ &{\scriptsize (\Gamma(\bullet+1) = \bullet \Gamma(\bullet))} \\[10px] &= \dfrac{n_2^2 (n_1 + 2)}{n_1 (n_2 – 2)(n_2 – 4)} \end{aligned}$$
よって、
$$\begin{aligned} V[X] &= E[X^2] – (E[X])^2 \\[10px] &= \dfrac{n_2^2 (n_1 + 2)}{n_1 (n_2 – 2)(n_2 – 4)} – (\dfrac{n_2}{n_2 – 2})^2 \\[10px] &= \dfrac{2 n_2^2 (n_1+n_2-2)}{n_1 (n_2-2)^2 (n_2-4)} \end{aligned}$$
