式展開

バーゼル問題

準備1．$2nA_n=\left(2n-1\right)A_{n-1}$

$$\begin{eqnarray} A_n&=&\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\int_0^{\frac{\pi}{2}}\cos{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&\left[\sin{\left(x\right)\cdot\cos^{2n-1}{\left(x\right)}}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}\sin{\left(x\right)}\left\{-\left(2n-1\right)\cos^{2n-2}{\left(x\right)}\sin{\left(x\right)}\right\}\mathrm{d}x \\&&\;\dots\;(fg)^\prime=f^\prime g+fg^\prime,f^\prime g=(fg)^\prime-fg^\prime,\int f^\prime g=\int(fg)^\prime-\int fg^\prime,\int f^\prime g=\left[fg\right]-\int fg^\prime \\&&\;\dots\;f^\prime(x)=\cos{\left(x\right)} \\&&\;\dots\;f(x)=\sin{\left(x\right)} \\&&\;\dots\;g(x)=\cos^{2n-1}{\left(x\right)}=u^{2n-1}(x)\;\cdots\;u=\cos{\left(x\right)} \\&&\;\dots\;g^\prime(x)=\frac{\mathrm{d}g}{\mathrm{d}x}=\frac{\mathrm{d}g}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} =\left(2n-1\right)u^{2n-2}(x)\cdot\left(-\sin{\left(x\right)}\right) =-\left(2n-1\right)\cos^{2n-2}{\left(x\right)}\sin{\left(x\right)} \\&=&\left[0-0\right]+\left(2n-1\right)\int_0^{\frac{\pi}{2}}\color{red}{\sin^2{\left(x\right)}}\color{black}{}\cos^{2n-2}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\color{red}{\left(1-\cos^2{\left(x\right)}\right)}\color{black}{}\cos^{2n-2}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\left(\cos^{2n-2}{\left(x\right)}-\cos^{2n}{\left(x\right)}\right)\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\cos^{2(n-1)}{\left(x\right)}\mathrm{d}x-\left(2n-1\right)\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)A_{n-1}-\left(2n-1\right)A_n \\A_n+\left(2n-1\right)A_n&=&\left(2n-1\right)A_{n-1} \\2nA_n&=&\left(2n-1\right)A_{n-1} \end{eqnarray}$$

準備2．$A_n=(2n-1)nB_{n-1}-2n^2B_{n}$

$$\begin{eqnarray} A_n&=&\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\left[x\cdot\cos^{2n}{\left(x\right)}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}x\left\{2n\cos^{2n-1}{\left(x\right)\left(-\sin{\left(x\right)}\right)}\right\}\mathrm{d}x \\&&\;\dots\;(fg)^\prime=f^\prime g+fg^\prime,f^\prime g=(fg)^\prime-fg^\prime,\int f^\prime g=\int(fg)^\prime-\int fg^\prime,\int f^\prime g=\left[fg\right]-\int fg^\prime \\&&\;\dots\;f^\prime(x)=1 \\&&\;\dots\;f(x)=x \\&&\;\dots\;g(x)=\cos^{2n}{\left(x\right)}=u^{2n}\;\cdots\;u=\cos{\left(x\right)} \\&&\;\dots\;g^\prime(x)=\frac{\mathrm{d}g}{\mathrm{d}x}=\frac{\mathrm{d}g}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} =2nu^{2n-1}\cdot\left(-\sin{\left(x\right)}\right) =2n\cos^{2n-1}{\left(x\right)}\left(-\sin{\left(x\right)}\right) \\&=&\left[0-0\right]+2n\int_0^{\frac{\pi}{2}}x\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&2n\int_0^{\frac{\pi}{2}}x\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&2n\left[\left[\frac{1}{2}x^2\cdot\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}\frac{1}{2}x^2\left\{ \cos{\left(x\right)}\cdot\cos^{2n-1}{\left(x\right)} +\sin{\left(x\right)}\cdot(2n-1)\cos^{2n-2}{\left(x\right)}\left(-\sin{\left(x\right)}\right) \right\}\mathrm{d}x \right] \\&=&-2n\frac{1}{2}\int_0^{\frac{\pi}{2}}x^2\left\{ \cos^{2n}{\left(x\right)}-(2n-1)\color{red}{\sin^2{\left(x\right)}}\color{black}{}\cos^{2n-2}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\color{red}{\left\{1-\cos^2{\left(x\right)}\right\}}\color{black}{}\cos^{2n-2}{\left(x\right)} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\left\{\cos^{2n-2}{\left(x\right)}-\cos^2{\left(x\right)}\cos^{2n-2}{\left(x\right)}\right\} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\left\{\cos^{2n-2}{\left(x\right)}-\cos^{2n}{\left(x\right)}\right\} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left\{ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\cos^{2n-2}{\left(x\right)}+(2n-1)x^2\cos^{2n}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left\{ -(2n-1)x^2\cos^{2n-2}{\left(x\right)}+2nx^2\cos^{2n}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}} -(2n-1)x^2\cos^{2(n-1)}{\left(x\right)} \mathrm{d}x -n\int_0^{\frac{\pi}{2}} 2nx^2\cos^{2n}{\left(x\right)} \mathrm{d}x \\&=&(2n-1)n\int_0^{\frac{\pi}{2}} x^2\cos^{2(n-1)}{\left(x\right)} \mathrm{d}x -2n^2\int_0^{\frac{\pi}{2}} x^2\cos^{2n}{\left(x\right)} \mathrm{d}x \\&=&(2n-1)nB_{n-1}-2n^2B_{n}\;\cdots\;B_n=\int_0^{\frac{\pi}{2}}x^2\cos^{2n}{\left(x\right)}\mathrm{d}x \end{eqnarray}$$

準備3．$\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0$

$$\begin{eqnarray} B_n&=&\int_0^{\frac{\pi}{2}}x^2\cos^{2n}{\left(x\right)}\mathrm{d}x \\&\leq&\int_0^{\frac{\pi}{2}}\left\{\frac{\pi}{2} \sin{\left(x\right)}\right\}^2\cos^{2n}{\left(x\right)}\mathrm{d}x \\&&\;\cdots\;\frac{2}{\pi}x\leq\sin{\left(x\right)}\;\cdots\;x\in\left[0,\frac{\pi}{2}\right] \\&&\;\cdots\;x\leq\frac{\pi}{2}\sin{\left(x\right)} \\&&\;\cdots\;(0,0)と(\frac{\pi}{2},1)を通る直線\frac{2}{\pi}xは，区間\left[0,\frac{\pi}{2}\right]では常に\sin{\left(x\right)}以下である(図を参照)． \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\sin^{2}{\left(x\right)}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\color{red}{\sin^{2}{\left(x\right)}}\color{black}{}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\color{red}{\left\{1-\cos^{2}{\left(x\right)}\right\}}\color{black}{}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\left\{1-\cos^{2}{\left(x\right)}\right\}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\left\{\cos^{2n}{\left(x\right)}-\cos^{2n+2}{\left(x\right)}\right\}\mathrm{d}x \\&=&\frac{\pi^2}{4}\left[\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x-\int_0^{\frac{\pi}{2}}\cos^{2(n+1)}{\left(x\right)}\mathrm{d}x\right] \\&=&\frac{\pi^2}{4}\left(A_n-A_{n+1}\right) \\&=&\frac{\pi^2}{4}\left(A_n-\frac{2(n+1)-1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{2(n+1)-\left\{2(n+1)-1\right\}}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{2(n+1)-2(n+1)+1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\frac{A_{n}}{2(n+1)} \\B_n&\leq&\frac{\pi^2}{4}\frac{A_n}{2(n+1)} \\B_n\frac{1}{A_n}&\leq&\frac{\pi^2}{4}\frac{A_n}{2(n+1)}\frac{1}{A_n}\;\cdots\;A_n\geq0 \\\frac{B_n}{A_n}&\leq&\frac{\pi^2}{4}\frac{1}{2(n+1)} \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&\leq&\lim_{n\rightarrow\infty}\frac{\pi^2}{4}\frac{1}{2(n+1)}=0 \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&\geq&0 \;\cdots\;A_n,B_nは共に被積分凾数\left(\cos^{2n}{\left(x\right)}, x^2\cos^{2n}{\left(x\right)}\right)が偶数乗のため，常に値は0以上． \\&&\;\cdots\;よって，その積分結果も常に0以上となる． \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&=&0\;\cdots\;前述の2つから，はさみうちの原理より極限は0． \end{eqnarray}$$

$\sum_{n=1}^{\infty}\frac{1}{n^2}$

$$\begin{eqnarray} \\\sum_{n=1}^{\infty}\frac{1}{n^2}&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{1}{n^2} \\&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\left\{\frac{(2n-1)B_{n-1}}{nA_n}-\frac{2B_{n}}{A_n}\right\} \\&&\;\cdots\;A_n=(2n-1)nB_{n-1}-2n^2B_{n}\;(準備2．) \\&&\;\cdots\;\frac{A_n}{n^2}=\frac{(2n-1)nB_{n-1}}{n^2}-\frac{2n^2B_{n}}{n^2} \\&&\;\cdots\;\frac{A_n}{n^2}=\frac{(2n-1)B_{n-1}}{n}-2B_{n} \\&&\;\cdots\;\frac{1}{n^2}=\frac{(2n-1)B_{n-1}}{nA_n}-\frac{2B_{n}}{A_n} \\&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{2B_{n-1}}{A_{n-1}}-\frac{2B_{n}}{A_n} \\&&\;\cdots\;2nA_n=\left(2n-1\right)A_{n-1}\;(準備1．) \\&&\;\cdots\;\frac{2}{A_{n-1}}nA_n=\left(2n-1\right) \\&&\;\cdots\;\frac{2}{A_{n-1}}=\frac{\left(2n-1\right)}{nA_n} \\&=&2\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{B_{n-1}}{A_{n-1}}-\frac{B_{n}}{A_n} \\&=&2\lim_{N\rightarrow\infty} \left\{ \left(\frac{B_{1-1}}{A_{1-1}}-\frac{B_{1}}{A_1}\right) +\left(\frac{B_{2-1}}{A_{2-1}}-\frac{B_{2}}{A_2}\right) +\left(\frac{B_{3-1}}{A_{3-1}}-\frac{B_{3}}{A_3}\right) +\cdots+\left(\frac{B_{N-1}}{A_{N-1}}-\frac{B_{N}}{A_N}\right) \right\} \\&=&2\lim_{N\rightarrow\infty} \left\{ \frac{B_{0}}{A_{0}}\cancel{-\frac{B_{1}}{A_1} +\frac{B_{1}}{A_{1}}}\cancel{-\frac{B_{2}}{A_2} +\frac{B_{2}}{A_{2}}}\cancel{-\frac{B_{3}}{A_3} +\cdots}+\cancel{\cdots+\frac{B_{N-1}}{A_{N-1}}}-\frac{B_{N}}{A_N} \right\} \;\cdots\;Telescoping Series \\&=&2\left(\frac{B_{0}}{A_{0}}-\lim_{N\rightarrow\infty}\frac{B_{N}}{A_N}\right) \\&=&2\left(\frac{B_{0}}{A_{0}}-0\right) \;\cdots\;\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0\;(準備3．) \\&=&2\frac{B_{0}}{A_{0}} \\&=&\left.2\frac{\int_0^{\frac{\pi}{2}}x^2\cos^{2\cdot n}{\left(x\right)}\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\cos^{2\cdot n}{\left(x\right)}\mathrm{d}x}\right|_{n=0} \\&=&2\frac{\int_0^{\frac{\pi}{2}}x^2\cos^{2\cdot0}{\left(x\right)}\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\cos^{2\cdot0}{\left(x\right)}\mathrm{d}x} \\&=&2\frac{\int_0^{\frac{\pi}{2}}x^2\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\mathrm{d}x} \\&=&2\frac{\left[\frac{1}{3}x^3\right]_0^{\frac{\pi}{2}}}{\left[x\right]_0^{\frac{\pi}{2}}} \\&=&2\frac{\left[\frac{1}{3}\left(\frac{\pi}{2}\right)^3-\frac{1}{3}0^3\right]}{\left[\frac{\pi}{2}-0\right]} \\&=&2\frac{\frac{\pi^3}{24}}{\frac{\pi}{2}} \\&=&2\frac{\pi^2}{12} \\&=&\frac{\pi^2}{6} \end{eqnarray}$$

t分布の期待値と分散

t分布の期待値

\begin{eqnarray} \mathrm{E}\left[X\right]&=&\int_{-\infty}^{\infty}x\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \;\ldots\;\href{https://shikitenkai.blogspot.com/2022/10/t.html}{t分布の確率密度凾数:\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}}} \\&=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \end{eqnarray} \begin{eqnarray} 彼積分凾数(x)&=&x\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\彼積分凾数(-x)&=&(-x)\left(1+\frac{(-x)^2}{n}\right)^{-\frac{n+1}{2}} \\&=&-x\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&-彼積分凾数(x)\;\ldots\;奇凾数 \end{eqnarray} \begin{eqnarray} \mathrm{E}\left[X\right] &=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}}\cdot0 \\&&\;\ldots\;奇凾数の上端と下端の絶対値が等しい定積分は0． \\&&\;\ldots\;\int_{-a}^{a}f(x)\mathrm{d}x=0,f(x)が奇凾数の場合 \\&=&0 \end{eqnarray}

t分布の分散

\begin{eqnarray} \mathrm{E}\left[X^2\right] &=&\int_{-\infty}^{\infty}x^2\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&=&\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x^2 \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \end{eqnarray} \begin{eqnarray} 彼積分凾数(x)&=&x^2\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\彼積分凾数(-x)&=&(-x)^2\left(1+\frac{(-x)^2}{n}\right)^{-\frac{n+1}{2}} \\&=&x^2\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&彼積分凾数(x)\;\ldots\;偶凾数 \end{eqnarray} \begin{eqnarray} \mathrm{E}\left[X^2\right]&=&\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x^2 \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&=&2\int_{0}^{\infty}x^2\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&&\;\ldots\;偶凾数の上端と下端の絶対値が等しい定積分は2\int_{0}^{a}f(x)\mathrm{d}x． \\&&\;\ldots\;\int_{-a}^{a}f(x)\mathrm{d}x=2\int_{0}^{a}f(x)\mathrm{d}x,f(x)が偶凾数の場合 \\&=&2\int_{1}^{0}\left(n\left(t^{-1}-1\right)\right)\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{n\left(t^{-1}-1\right)}{n}\right)^{-\frac{n+1}{2}} n^{\frac{1}{2}}\frac{1}{2}\left(t^{-1}-1\right)^{-\frac{1}{2}}\left(-t^{-2}\right) \mathrm{d}t \\&&\;\ldots\;x^2=n\left(t^{-1}-1\right) \\&&\;\ldots\;x=n^{\frac{1}{2}}\left(t^{-1}-1\right)^{\frac{1}{2}}=n^{\frac{1}{2}}s^{\frac{1}{2}}\;\ldots\;s=t^{-1}-1 \\&&\;\ldots\;x:0\rightarrow\infty, t:1\rightarrow0 \\&&\;\ldots\;\frac{\mathrm{d}{x}}{\mathrm{d}{t}}=\frac{\mathrm{d}{x}}{\mathrm{d}{s}}\frac{\mathrm{d}{s}}{\mathrm{d}{t}} =n^{\frac{1}{2}}\frac{1}{2}s^{\frac{1}{2}-1}\frac{\mathrm{d}{s}}{\mathrm{d}{t}}=n^{\frac{1}{2}}\frac{1}{2}\left(t^{-1}-1\right)^{-\frac{1}{2}}\left(-t^{-2}\right) \\&=&\cancel{2}\frac{1}{\cancel{\sqrt{n}}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}}\cancel{\frac{1}{2}} n\cancel{n^{\frac{1}{2}}}(-1) \int_{1}^{0} \left(1+\frac{\cancel{n}\left(t^{-1}-1\right)}{\cancel{n}}\right)^{-\frac{n+1}{2}} \left(t^{-1}-1\right)^{1-\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} \left(\cancel{1}+t^{-1}\cancel{-1}\right)^{-\frac{n+1}{2}} \left(t^{-1}-1\right)^{\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n+1}{2}} \left(t^{-1}-1\right)^{\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{t}{t}\left(t^{-1}-1\right)\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{1}{t}\left(1-t\right)\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{1}{t}\right)^\frac{1}{2}\left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} t^{-\frac{1}{2}}\left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{4}{2}} \left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-1-1} \left(1-t\right)^{\frac{3}{2}-1} \mathrm{d}t \\&=&\frac{n}{ \frac{ \Gamma{\left(\frac{1}{2}\right)} \Gamma{\left(\frac{n}{2}\right)} }{ \Gamma{ \left(\frac{1}{2}+\frac{n}{2}\right) } }} \frac{ \Gamma{\left(\frac{n}{2}-1\right)} \Gamma{\left(\frac{3}{2}\right)} }{ \Gamma{ \left(\frac{n}{2}-1+\frac{3}{2}\right) } } \;\ldots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{ \beta{\left(a,b\right)}=\frac{ \Gamma{\left(a\right)} \Gamma{\left(b\right)} }{ \Gamma{ \left(a+b\right) } }=\int_0^1 x^{a-1}(1-x)^{b-1}\mathrm{d}x} \\&=&n \frac{ \cancel{\Gamma{ \left(\frac{1}{2}+\frac{n}{2}\right) }} }{ \Gamma{\left(\frac{1}{2}\right)} \color{green}{\Gamma{\left(\frac{n}{2}\right)}} } \frac{ \Gamma{\left(\frac{n}{2}-1\right)} \color{blue}{\Gamma{\left(\frac{3}{2}\right)}} }{ \cancel{\Gamma{ \left(\frac{n}{2}+\frac{1}{2}\right) }} } \\&=&n \frac{1}{ \cancel{\Gamma{\left(\frac{1}{2}\right)}} \color{green}{\left(\frac{n}{2}-1\right)\cancel{\Gamma{\left(\frac{n}{2}-1\right)}}} } \frac{ \cancel{\Gamma{\left(\frac{n}{2}-1\right)}} \color{blue}{\frac{1}{2}\cancel{\Gamma{\left(\frac{1}{2}\right)}}} }{1} \;\ldots\;\href{https://shikitenkai.blogspot.com/2020/08/s1ss.html}{\Gamma\left(z+1\right)=z\Gamma\left(z\right)} \\&=&n \frac{1}{ \left(\frac{n}{2}-1\right) } \frac{ \frac{1}{2} }{1} \\&=&\frac{n}{n-2} \end{eqnarray} \begin{eqnarray} \mathrm{V}\left[X\right]&=&\mathrm{E}\left[X^2\right]-\mathrm{E}\left[X\right]^2 \\&=&\frac{n}{n-2}-0^2 \\&=&\frac{n}{n-2} \end{eqnarray}

t分布の導出

$ X=\frac{Y}{\sqrt{\frac{Z}{n}}} $, $t$分布の確率変数

$t$分布は次の確率変数$X$を考えます． $$\begin{eqnarray} X&=&\frac{Y}{\sqrt{\frac{Z}{n}}}\;\cdots\;\frac{Y}{\mathrm{RMS}\left[Y\right]} \\Y&:&f_Y(y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \;\cdots\;z分布\left(標準正規分布\right) \\Z&:&f_Z(z)=\frac{1}{2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}}e^{-\frac{z}{2}}z^{\frac{n}{2}-1} \;\cdots\;\chi^2分布\left(Z=\sum_{i=1}^n Y_i^2\right) \\&&\end{eqnarray}$$

$ f_{YZ}(y,z) $, $Y$と$Z$の同時確率

$$\begin{eqnarray} f_{YZ}(y,z)&=&f_Y(y)\cdot f_Z(z)\;\cdots\;YとZは独立 \end{eqnarray}$$

変数変換, $Y$と$Z$から$X$と$U$へ

$$\begin{eqnarray} x&=&\frac{y}{\sqrt{\frac{z}{n}}}&より&y&=&x\sqrt{\frac{z}{n}} \end{eqnarray}$$ $$ \left\{ \begin{eqnarray} z&=&u \\y&=&x\sqrt{\frac{u}{n}} \end{eqnarray} \right.$$ \begin{eqnarray} J&=&\left|\frac{\partial(y,z)}{\partial(x,u)}\right| \\&=&\left|\begin{matrix} \frac{\partial y}{\partial x}&\frac{\partial y}{\partial u} \\\frac{\partial z}{\partial x}&\frac{\partial z}{\partial u} \end{matrix}\right| \\&=&\left|\begin{matrix} \sqrt{\frac{u}{n}}&\frac{x}{\sqrt{n}}\frac{1}{2}u^{-\frac{1}{2}} \\0&1 \end{matrix}\right| \\&=&\sqrt{\frac{u}{n}}\cdot 1 - \frac{x}{\sqrt{n}}\frac{1}{2}u^{-\frac{1}{2}} \cdot 0 \\&=&\sqrt{\frac{u}{n}} \end{eqnarray} \begin{eqnarray} &&\int_{0}^{\infty}\int_{\infty}^{-\infty}f_Y(y)f_Z(z)\mathrm{d}y\mathrm{d}z\;\cdots\;Y,Zの同時確率の全事象 \\&=&\int_{0}^{\infty}\int_{\infty}^{-\infty}f_Y\left(x\sqrt{\frac{u}{n}}\right)f_Z(u)\sqrt{\frac{u}{n}}\mathrm{d}x\mathrm{d}u \\&&\;\cdots\;変数変換:y=x\sqrt{\frac{u}{n}},\;z=u,J=\sqrt{\frac{u}{n}} \\&&\;\cdots\;x=\frac{y}{\sqrt{\frac{u}{n}}}なので,\;y:-\infty\rightarrow\infty,\;x:-\infty\rightarrow\infty \\&&\;\cdots\;u=zなので,\;z:0\rightarrow\infty,\;u:0\rightarrow\infty \\&=&\int_{0}^{\infty}\int_{\infty}^{-\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{\left(x\sqrt{\frac{u}{n}}\right)^2}{2}} \frac{1}{2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}}e^{-\frac{u}{2}}u^{\frac{n}{2}-1} \sqrt{\frac{u}{n}} \mathrm{d}x\mathrm{d}u \\&=& \frac{1}{\sqrt{2\pi}} \frac{1}{2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \frac{1}{\sqrt{n}} \int_{0}^{\infty}\int_{\infty}^{-\infty} e^{-\frac{x^2u}{2n}} e^{-\frac{u}{2}} u^{\frac{n}{2}-1} \sqrt{u} \mathrm{d}x\mathrm{d}u \\&=& \frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \int_{0}^{\infty}\int_{\infty}^{-\infty} e^{-\frac{x^2u}{2n}-\frac{u}{2}} u^{\frac{n}{2}-1+\frac{1}{2}} \mathrm{d}x\mathrm{d}u \\&=& \frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \int_{0}^{\infty}\int_{\infty}^{-\infty} e^{-\frac{u}{2}\left(\frac{x^2}{n}+1\right)} u^{\frac{n+1}{2}-1} \mathrm{d}x\mathrm{d}u \\&=&1\;\cdots\;全事象 \\&=&F_{XU}\left(x,u\right) \end{eqnarray}

$f_X\left(x\right)$, $X$の確率密度凾数

\begin{eqnarray} f_{XU}\left(x,u\right) &=&\frac{\partial^2}{\partial x \partial u}F_{XU}\left(x,u\right) \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} e^{-\frac{u}{2}\left(1+\frac{x^2}{n}\right)} u^{\frac{n+1}{2}-1} \end{eqnarray} $X$の確率密度凾数$f_X\left(x\right)$がすなわち$t$分布なので，uについて積分することで周辺確率を求めます． \begin{eqnarray} t(x)=f_X(x)&=&\int_{0}^{\infty}f_{XU}\left(x,u\right)\mathrm{d}u \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \int_{0}^{\infty} e^{-\frac{u}{2}\left(\frac{x^2}{n}+1\right)} u^{\frac{n+1}{2}-1} \mathrm{d}u \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \int_{0}^{\infty} e^{-t} \cdot\left(\frac{2t}{\frac{x^2}{n}+1}\right)^{\frac{n+1}{2}-1} \cdot\frac{2}{\frac{x^2}{n}+1}\mathrm{d}t \\&&\;\cdots\;t=\frac{u}{2}\left(\frac{x^2}{n}+1\right) \\&&\;\cdots\;u:0\rightarrow\infty,t:0\rightarrow\infty \\&&\;\cdots\;u=\frac{2t}{\frac{x^2}{n}+1} \\&&\;\cdots\;\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{2}{\frac{x^2}{n}+1} \\&&\;\cdots\;\mathrm{d}u=\frac{2}{\frac{x^2}{n}+1}\mathrm{d}t \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \frac{2}{\frac{x^2}{n}+1} \left(\frac{2}{\frac{x^2}{n}+1}\right)^{\frac{n+1}{2}-1} \int_{0}^{\infty} e^{-t} t^{\frac{n+1}{2}-1} \mathrm{d}t \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \left(\frac{2}{\frac{x^2}{n}+1}\right)^{\frac{n+1}{2}} \Gamma{\left(\frac{n+1}{2}\right)} \\&=&\frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} 2^{\frac{n+1}{2}} \left(\frac{x^2}{n}+1\right)^{-\frac{n+1}{2}} \\&=&\frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\sqrt{\cancel{2}\pi n}\;\cancel{2^{\frac{n}{2}}}\Gamma{\left(\frac{n}{2}\right)}} \cancel{2^{\frac{n}{2}}}\cancel{2^{\frac{1}{2}}}\left(\frac{x^2}{n}+1\right)^{-\frac{n+1}{2}} \\&=&\frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\sqrt{\pi n}\Gamma{\left(\frac{n}{2}\right)}} \left(\frac{x^2}{n}+1\right)^{-\frac{n+1}{2}} \\&=&\frac{1}{\sqrt{n}}\frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\sqrt{\pi}\;\Gamma{\left(\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&\frac{1}{\sqrt{n}} \frac{ \Gamma{ \left(\frac{n+1}{2}\right) } }{ \Gamma{ \left(\frac{1}{2}\right) } \Gamma{ \left(\frac{n}{2}\right) } } \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \;\cdots\;\Gamma{ \left(\frac{1}{2}\right) }=\sqrt{\pi} \\&=&\frac{1}{\sqrt{n}} \frac{ \Gamma{ \left(\frac{1}{2}+\frac{n}{2}\right) } }{ \Gamma{ \left(\frac{1}{2}\right) } \Gamma{ \left(\frac{n}{2}\right) } } \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&\frac{1}{\sqrt{n}} \frac{1} { \beta{ \left(\frac{1}{2},\frac{n}{2}\right) } } \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{ \beta{\left(a,b\right)}=\frac{ \Gamma{\left(a\right)} \Gamma{\left(b\right)} }{ \Gamma{ \left(a+b\right) } }=\int_0^1 x^{a-1}(1-x)^{b-1}\mathrm{d}x} \end{eqnarray}

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式展開

バーゼル問題

準備1．\(2nA_n=\left(2n-1\right)A_{n-1}\)

準備2．\(A_n=(2n-1)nB_{n-1}-2n^2B_{n}\)

準備3．\(\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0\)

\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)

t分布の期待値と分散

t分布の期待値

t分布の分散

t分布の導出

\( X=\frac{Y}{\sqrt{\frac{Z}{n}}} \), \(t\)分布の確率変数

\( f_{YZ}(y,z) \), \(Y\)と\(Z\)の同時確率

変数変換, \(Y\)と\(Z\)から\(X\)と\(U\)へ

\(f_X\left(x\right)\), \(X\)の確率密度凾数