二項分布の累積分布関数と部分和の導出

与式の右辺を \begin{align} I=\frac{n!}{ \left(k-1\right)! \left(n-k\right)!}\int_{0}^{p}{x^{k-1} \left(1-x\right)^{n-k}dx} \end{align} とおく。 部分積分の公式 \begin{align} \int_{a}^{b}{f \left(x\right)g^\prime \left(x\right)dx}= \left[f \left(x\right)g \left(x\right)\right]_a^b-\int_{a}^{b}{f^\prime \left(x\right)g \left(x\right)dx} \end{align} において、 \begin{gather} f \left(x\right)= \left(1-x\right)^{n-k} \quad g \left(x\right)=\frac{1}{k}x^k\\ f^\prime \left(x\right)=- \left(n-k\right) \left(1-x\right)^{n-k-1} \quad g^\prime \left(x\right)=x^{k-1} \end{gather} とすると、 \begin{align} I=&\frac{n!}{ \left(k-1\right)! \left(n-k\right)!} \left[\frac{1}{k}x^k \left(1-x\right)^{n-k}\right]_0^p\\ &-\frac{n! \left(n-k\right)}{k \left(k-1\right)! \left(n-k\right)!}\int_{0}^{p}{-x^k \left(1-x\right)^{n-k-1}dx}\\ =&{}_{n}C_kp^k \left(1-p\right)^{n-k}\\ &+\frac{n!}{k! \left(n-k-1\right)!}\int_{0}^{p}{x^k \left(1-x\right)^{n-k-1}dx} \end{align} 同様に、右辺第2項を \begin{align} i_1=\frac{n!}{k! \left(n-k-1\right)!}\int_{0}^{p}{x^k \left(1-x\right)^{n-k-1}dx} \end{align} とおき、 \begin{gather} f \left(x\right)= \left(1-x\right)^{n-k-1} \quad g \left(x\right)=\frac{1}{k+1}x^{k+1}\\ f^\prime \left(x\right)=- \left(n-k-1\right) \left(1-x\right)^{n-k-1} \quad g^\prime \left(x\right)=x^k \end{gather} として、 部分積分を行うと、 \begin{align} i_1=&\frac{n!}{k! \left(n-k-1\right)!} \left[\frac{1}{k+1}x^{k+1} \left(1-x\right)^{n-k-1}\right]_0^p\\ &-\frac{n! \left(n-k-1\right)}{ \left(k+1\right)k! \left(n-k-1\right)!}\int_{0}^{p}{x^{k+1} \left(1-x\right)^{n-k-2}dx}\\ =&{}_{n}C_{k+1}p^{k+1} \left(1-p\right)^{n-k-1}\\ &-\frac{n!}{ \left(k+1\right)! \left(n-k-2\right)!}\int_{0}^{p}{x^{k+1} \left(1-x\right)^{n-k-2}dx} \end{align} これを $n-k$ 回繰り返していくと、 \begin{align} I&={}_{n}C_kp^k \left(1-p\right)^{n-k}+{}_{n}C_{k+1}p^{k+1} \left(1-p\right)^{n-k-1}+ \cdots +{}_{n}C_np^n \left(1-p\right)^0\\ &=\sum_{x=k}^{n}{{}_{n}C_xp^x \left(1-p\right)^{n-x}}\\ &=P \left(x \le X\right) \end{align} したがって、 \begin{align} P \left(x \le X\right)=\frac{n!}{ \left(k-1\right)! \left(n-k\right)!}\int_{0}^{p}{x^{k-1} \left(1-x\right)^{n-k}dx} \end{align} $\blacksquare$

（i） \begin{align} I=\frac{n!}{k! \left(n-k-1\right)!}\int_{p}^{1}{t^k \left(1-t\right)^{n-k-1}dt} \end{align} とおく。 \begin{gather} f \left(t\right)=t^k \quad g \left(t\right)=-\frac{1}{n-k} \left(1-t\right)^{n-k}\\ f^\prime \left(t\right)=kt^{k-1} \quad g^\prime \left(x\right)= \left(1-t\right)^{n-k-1} \end{gather} として、 部分積分を行うと、 \begin{align} I=&\frac{n!}{k! \left(n-k-1\right)!} \left[-\frac{t^k \left(1-t\right)^{n-k}}{n-k}\right]_p^1\\ &-\frac{k \cdot n!}{k! \left(n-k\right) \left(n-k-1\right)!}\int_{p}^{1}{-t^{k-1} \left(1-t\right)^{n-k}dt}\\ =&\frac{n!}{k! \left(n-k\right)!}p^k \left(1-p\right)^{n-k}\\ &+\frac{n!}{ \left(k-1\right)! \left(n-k\right)!}\int_{p}^{1}{t^{k-1} \left(1-t\right)^{n-k}dt}\\ =&{}_{n}C_kp^k \left(1-p\right)^{n-k}\\ &+\frac{n!}{ \left(k-1\right)! \left(n-k\right)!}\int_{p}^{1}{t^{k-1} \left(1-t\right)^{n-k}dt} \end{align} 同様に、右辺第2項を \begin{align} i_1=\frac{n!}{ \left(k-1\right)! \left(n-k\right)!}\int_{p}^{1}{t^{k-1} \left(1-t\right)^{n-k}dt} \end{align} とおき、 \begin{gather} f \left(t\right)=t^{k-1} \quad g \left(t\right)=-\frac{1}{n-k+1} \left(1-t\right)^{n-k+1}\\ f^\prime \left(x\right)= \left(k-1\right)t^{k-2} \quad g^\prime \left(x\right)= \left(1-t\right)^{n-k} \end{gather} として、 部分積分を行うと、 \begin{align} i_1=&\frac{n!}{ \left(k-1\right)! \left(n-k\right)!} \left[-\frac{t^{k-1} \left(1-t\right)^{n-k}}{n-k+1}\right]_p^1\\ &-\frac{ \left(k-1\right)n!}{ \left(k-1\right)! \left(n-k+1\right) \left(n-k\right)!}\int_{p}^{1}{-t^{k-1} \left(1-t\right)^{n-k}dt}\\ =&\frac{n!}{ \left(k-1\right)! \left(n-k+1\right)!}p^{k-1} \left(1-p\right)^{n-k+1}\\ &+\frac{n!}{ \left(k-2\right)! \left(n-k+1\right)!}\int_{p}^{1}{t^{k-2} \left(1-t\right)^{n-k+1}dt}\\ =&{}_{n}C_{k-1}p^{k-1} \left(1-p\right)^{n-k+1}\\ &+\frac{n!}{ \left(k-2\right)! \left(n-k+1\right)!}\int_{p}^{1}{x^{k-2} \left(1-x\right)^{n-k+1}dx} \end{align} 以下、同様の操作を $k$ 回繰り返していくと、 \begin{align} I&={}_{n}C_kp^k \left(1-p\right)^{n-k}+{}_{n}C_{k-1}p^{k-1} \left(1-p\right)^{n-k+1}+ \cdots +{}_{n}C_0p^0 \left(1-p\right)^n\\ &=\sum_{x=0}^{k}{{}_{n}C_xp^x \left(1-p\right)^{n-x}}\\ &=F \left(x\right) \end{align} （ii）いっぽう、 \begin{gather} t=\frac{m_1x}{m_1x+m_2}\\ k=\frac{m_1}{2}-1\\ n-k-1=\frac{m_2}{2}-1 \end{gather} とおくと、 （a） \begin{align} \frac{dt}{dx}=\frac{m_1m_2}{ \left(m_1x+m_2\right)^2}\Rightarrow dt=\frac{m_1m_2}{ \left(m_1x+m_2\right)^2} \cdot dx \end{align} （b） \begin{gather} t \left(m_1x+m_2\right)=m_1x\\ m_1tx+tm_2=m_1x\\ m_1tx-m_1x=-tm_2\\ m_1 \left(t-1\right)x=-tm_2\\ x=\frac{m_2t}{m_1 \left(1-t\right)} \end{gather} したがって、 \begin{align} t:p\rightarrow1 \quad \mathrm{when} \quad x:\frac{pm_2}{m_1 \left(1-p\right)}\rightarrow\infty \end{align} ここで、 \begin{align} F=\frac{m_2p}{m_1 \left(1-p\right)} \end{align} とおくと、 置換積分法により、 \begin{align} I=\frac{ \left(\frac{m_1}{2}+\frac{m_2}{2}-1\right)!}{ \left(\frac{m_1}{2}-1\right)! \left(\frac{m_2}{2}-1\right)!}\int_{F}^{\infty}{ \left(\frac{m_1x}{m_1x+m_2}\right)^{\frac{m_1}{2}-1} \left(\frac{m_2}{m_1x+m_2}\right)^{\frac{m_2}{2}-1} \cdot \frac{m_1m_2}{ \left(m_1x+m_2\right)^2}dx} \end{align} ガンマ関数の性質 $\Gamma \left(\alpha\right)= \left(\alpha-1\right)!$ より、 \begin{align} I=\frac{\Gamma \left(\frac{m_1}{2}+\frac{m_2}{2}\right)}{\Gamma \left(\frac{m_1}{2}\right)\Gamma \left(\frac{m_2}{2}\right)}\int_{F}^{\infty}{ \left(\frac{m_1x}{m_1x+m_2}\right)^{\frac{m_1}{2}-1} \left(\frac{m_2}{m_1x+m_2}\right)^{\frac{m_2}{2}-1} \cdot \frac{m_1m_2}{ \left(m_1x+m_2\right)^2}dx} \end{align} ガンマ関数とベータ関数の関係 $B \left(\alpha,\beta\right)=\frac{\Gamma \left(\alpha\right)\Gamma \left(\beta\right)}{\Gamma \left(\alpha+\beta\right)}$ より、 \begin{align} I&=\frac{1}{B \left(\frac{m_1}{2},\frac{m_2}{2}\right)}\int_{F}^{\infty}{ \left(\frac{m_1x}{m_1x+m_2}\right)^{\frac{m_1}{2}-1} \left(\frac{m_2}{m_1x+m_2}\right)^{\frac{m_2}{2}-1} \cdot \frac{m_1m_2}{ \left(m_1x+m_2\right)^2}dx}\\ &=\int_{F}^{\infty}{\frac{m_1^{\frac{m_1}{2}}m_2^{\frac{m_2}{2}}}{B \left(\frac{m_1}{2},\frac{m_2}{2}\right)} \cdot \frac{x^{\frac{m_1}{2}-1}}{ \left(m_1x+m_2\right)^{\frac{m_1}{2}+\frac{m_2}{2}}}dx} \end{align} 自由度 $ \left(m_1,m_2\right)$ の $\mathrm{F}$分布の確率密度関数は、 \begin{align} f_{m1,m2} \left(x\right)=\frac{m_1^{\frac{m_1}{2}}m_2^{\frac{m_2}{2}}}{B \left(\frac{m_1}{2},\frac{m_2}{2}\right)} \cdot \frac{x^{\frac{m_1}{2}-1}}{ \left(m_1x+m_2\right)^{\frac{m_1}{2}+\frac{m_2}{2}}} \end{align} したがって、 \begin{align} F \left(x\right)=\int_{F}^{\infty}{f_{m1,m2} \left(x\right)dx} \end{align} $\blacksquare$