二項分布の累積分布関数と部分和の導出

与式の右辺を $$\begin{array}{r}I=\frac{n!}{(k-1)!(n-k)!}{\int }_0^p{x}^{k-1}{(1-x)}^{n-k}dx\end{array}$$ とおく。 部分積分の公式 $$\begin{array}{r}{\int }_a^{b}f\left(x\right)g^{\prime }\left(x\right)dx={\left[f\left(x\right)g\left(x\right)\right]}_a^b-{\int }_a^b{f}^{\prime }\left(x\right)g\left(x\right)dx\end{array}$$ において、 $$\begin{array}{c}f\left(x\right)={(1-x)}^{n-k}g\left(x\right)=\frac{1}{k}{x}^k\\ f^{\prime }\left(x\right)=-(n-k){(1-x)}^{n-k-1}{g}^{\prime }\left(x\right)=x^{k-1}\end{array}$$ とすると、 $$\begin{array}{rl}I=& \frac{n!}{(k-1)!(n-k)!}{\left[\frac{1}{k}{x}^k{(1-x)}^{n-k}\right]}_0^p\\ & -\frac{n!(n-k)}{k(k-1)!(n-k)!}{\int }_0^p-x^k{(1-x)}^{n-k-1}dx\\ =& {}_n{C}_k{p}^k{(1-p)}^{n-k}\\ & +\frac{n!}{k!(n-k-1)!}{\int }_0^p{x}^k{(1-x)}^{n-k-1}dx\end{array}$$ 同様に、右辺第2項を $$\begin{array}{r}{i}_1=\frac{n!}{k!(n-k-1)!}{\int }_0^p{x}^k{(1-x)}^{n-k-1}dx\end{array}$$ とおき、 $$\begin{array}{c}f\left(x\right)={(1-x)}^{n-k-1}g\left(x\right)=\frac{1}{k+1}{x}^{k+1}\\ f^{\prime }\left(x\right)=-(n-k-1){(1-x)}^{n-k-1}{g}^{\prime }\left(x\right)=x^k\end{array}$$ として、 部分積分を行うと、 $$\begin{array}{rl}{i}_1=& \frac{n!}{k!(n-k-1)!}{\left[\frac{1}{k+1}{x}^{k+1}{(1-x)}^{n-k-1}\right]}_0^p\\ & -\frac{n!(n-k-1)}{(k+1)k!(n-k-1)!}{\int }_0^p{x}^{k+1}{(1-x)}^{n-k-2}dx\\ =& {}_n{C}_{k+1}{p}^{k+1}{(1-p)}^{n-k-1}\\ & -\frac{n!}{(k+1)!(n-k-2)!}{\int }_0^p{x}^{k+1}{(1-x)}^{n-k-2}dx\end{array}$$ これを $n-k$ 回繰り返していくと、 $$\begin{array}{rl}I& ={}_n{C}_k{p}^k{(1-p)}^{n-k}+{}_n{C}_{k+1}{p}^{k+1}{(1-p)}^{n-k-1}+\cdots +{}_n{C}_n{p}^n{(1-p)}^0\\ & =\sum _{x=k}^n{}_n{C}_x{p}^x{(1-p)}^{n-x}\\ & =P(x\le X)\end{array}$$ したがって、 $$\begin{array}{r}P(x\le X)=\frac{n!}{(k-1)!(n-k)!}{\int }_0^p{x}^{k-1}{(1-x)}^{n-k}dx\end{array}$$ $◼$

（i） $$\begin{array}{r}I=\frac{n!}{k!(n-k-1)!}{\int }_p^1{t}^k{(1-t)}^{n-k-1}dt\end{array}$$ とおく。 $$\begin{array}{c}f\left(t\right)=t^{k}g\left(t\right)=-\frac{1}{n-k}{(1-t)}^{n-k}\\ f^{\prime }\left(t\right)=k{t}^{k-1}{g}^{\prime }\left(x\right)={(1-t)}^{n-k-1}\end{array}$$ として、 部分積分を行うと、 $$\begin{array}{rl}I=& \frac{n!}{k!(n-k-1)!}{[-\frac{t^k{(1-t)}^{n-k}}{n-k}]}_p^1\\ & -\frac{k\cdot n!}{k!(n-k)(n-k-1)!}{\int }_p^1-t^{k-1}{(1-t)}^{n-k}dt\\ =& \frac{n!}{k!(n-k)!}{p}^k{(1-p)}^{n-k}\\ & +\frac{n!}{(k-1)!(n-k)!}{\int }_p^1{t}^{k-1}{(1-t)}^{n-k}dt\\ =& {}_n{C}_k{p}^k{(1-p)}^{n-k}\\ & +\frac{n!}{(k-1)!(n-k)!}{\int }_p^1{t}^{k-1}{(1-t)}^{n-k}dt\end{array}$$ 同様に、右辺第2項を $$\begin{array}{r}{i}_1=\frac{n!}{(k-1)!(n-k)!}{\int }_p^1{t}^{k-1}{(1-t)}^{n-k}dt\end{array}$$ とおき、 $$\begin{array}{c}f\left(t\right)=t^{k-1}g\left(t\right)=-\frac{1}{n-k+1}{(1-t)}^{n-k+1}\\ f^{\prime }\left(x\right)=(k-1)t^{k-2}{g}^{\prime }\left(x\right)={(1-t)}^{n-k}\end{array}$$ として、 部分積分を行うと、 $$\begin{array}{rl}{i}_1=& \frac{n!}{(k-1)!(n-k)!}{[-\frac{t^{k-1}{(1-t)}^{n-k}}{n-k+1}]}_p^1\\ & -\frac{(k-1)n!}{(k-1)!(n-k+1)(n-k)!}{\int }_p^1-t^{k-1}{(1-t)}^{n-k}dt\\ =& \frac{n!}{(k-1)!(n-k+1)!}{p}^{k-1}{(1-p)}^{n-k+1}\\ & +\frac{n!}{(k-2)!(n-k+1)!}{\int }_p^1{t}^{k-2}{(1-t)}^{n-k+1}dt\\ =& {}_n{C}_{k-1}{p}^{k-1}{(1-p)}^{n-k+1}\\ & +\frac{n!}{(k-2)!(n-k+1)!}{\int }_p^1{x}^{k-2}{(1-x)}^{n-k+1}dx\end{array}$$ 以下、同様の操作を $k$ 回繰り返していくと、 $$\begin{array}{rl}I& ={}_n{C}_k{p}^k{(1-p)}^{n-k}+{}_n{C}_{k-1}{p}^{k-1}{(1-p)}^{n-k+1}+\cdots +{}_n{C}_0{p}^0{(1-p)}^n\\ & =\sum _{x=0}^k{}_n{C}_x{p}^x{(1-p)}^{n-x}\\ & =F\left(x\right)\end{array}$$ （ii）いっぽう、 $$\begin{array}{c}t=\frac{m_{1}x}{m_{1}x+m_2}\\ k=\frac{m_1}{2}-1\\ n-k-1=\frac{m_2}{2}-1\end{array}$$ とおくと、 （a） $$\begin{array}{r}\frac{dt}{dx}=\frac{m_1{m}_2}{{(m_{1}x+m_2)}^2}\Rightarrow dt=\frac{m_1{m}_2}{{(m_{1}x+m_2)}^2}\cdot dx\end{array}$$ （b） $$\begin{array}{c}t(m_{1}x+m_2)=m_{1}x\\ m_{1}tx+t{m}_2=m_{1}x\\ m_{1}tx-m_{1}x=-t{m}_2\\ m_1(t-1)x=-t{m}_2\\ x=\frac{m_{2}t}{m_1(1-t)}\end{array}$$ したがって、 $$\begin{array}{r}t:p\to 1\mathrm{when}x:\frac{p{m}_2}{m_1(1-p)}\to \mathrm{\infty }\end{array}$$ ここで、 $$\begin{array}{r}F=\frac{m_{2}p}{m_1(1-p)}\end{array}$$ とおくと、 置換積分法により、 $$\begin{array}{r}I=\frac{(\frac{m_1}{2}+\frac{m_2}{2}-1)!}{(\frac{m_1}{2}-1)!(\frac{m_2}{2}-1)!}{\int }_F^{\mathrm{\infty }}{\left(\frac{m_{1}x}{m_{1}x+m_2}\right)}^{\frac{m_1}{2}-1}{\left(\frac{m_2}{m_{1}x+m_2}\right)}^{\frac{m_2}{2}-1}\cdot \frac{m_1{m}_2}{{(m_{1}x+m_2)}^2}dx\end{array}$$ ガンマ関数の性質 $\mathrm{\Gamma }\left(\alpha \right)=(\alpha -1)!$ より、 $$\begin{array}{r}I=\frac{\mathrm{\Gamma }(\frac{m_1}{2}+\frac{m_2}{2})}{\mathrm{\Gamma }\left(\frac{m_1}{2}\right)\mathrm{\Gamma }\left(\frac{m_2}{2}\right)}{\int }_F^{\mathrm{\infty }}{\left(\frac{m_{1}x}{m_{1}x+m_2}\right)}^{\frac{m_1}{2}-1}{\left(\frac{m_2}{m_{1}x+m_2}\right)}^{\frac{m_2}{2}-1}\cdot \frac{m_1{m}_2}{{(m_{1}x+m_2)}^2}dx\end{array}$$ ガンマ関数とベータ関数の関係 $B(\alpha ,\beta )=\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)}{\mathrm{\Gamma }(\alpha +\beta )}$ より、 $$\begin{array}{rl}I& =\frac{1}{B(\frac{m_1}{2},\frac{m_2}{2})}{\int }_F^{\mathrm{\infty }}{\left(\frac{m_{1}x}{m_{1}x+m_2}\right)}^{\frac{m_1}{2}-1}{\left(\frac{m_2}{m_{1}x+m_2}\right)}^{\frac{m_2}{2}-1}\cdot \frac{m_1{m}_2}{{(m_{1}x+m_2)}^2}dx\\ & ={\int }_F^{\mathrm{\infty }}\frac{m_1^{\frac{m_1}{2}}{m}_2^{\frac{m_2}{2}}}{B(\frac{m_1}{2},\frac{m_2}{2})}\cdot \frac{x^{\frac{m_1}{2}-1}}{{(m_{1}x+m_2)}^{\frac{m_1}{2}+\frac{m_2}{2}}}dx\end{array}$$ 自由度 $(m_1,m_2)$ の $\mathrm{F}$分布の確率密度関数は、 $$\begin{array}{r}{f}_{m1,m2}\left(x\right)=\frac{m_1^{\frac{m_1}{2}}{m}_2^{\frac{m_2}{2}}}{B(\frac{m_1}{2},\frac{m_2}{2})}\cdot \frac{x^{\frac{m_1}{2}-1}}{{(m_{1}x+m_2)}^{\frac{m_1}{2}+\frac{m_2}{2}}}\end{array}$$ したがって、 $$\begin{array}{r}F\left(x\right)={\int }_F^{\mathrm{\infty }}{f}_{m1,m2}\left(x\right)dx\end{array}$$ $◼$