カイ二乗\(\chi^2\)分布の導出
\(k=1\)のXの確率密度凾数
$$ \begin{eqnarray} X&=&Z_1^2\;\cdots\;Z_1:標準正規分布に従う確率変数 \\F_1(X=t)&=&\int_0^t f_1(x) \mathrm{d}x \;\cdots\;k=1のXの累積分布凾数F_1,確率密度凾数f_1 \\F_1(X=t)&=&\int_{-\sqrt{t}}^{+\sqrt{t}}g(z_1)\mathrm{d}z_1 \;\cdots\;Z_1=\pm\sqrt{X},x:0\rightarrow t,z_1:0\rightarrow\pm\sqrt{t}\;標準正規分布の確率密度凾数g \\&=&2\int_{0}^{\sqrt{t}}g(z_1)\mathrm{d}z_1 \;\cdots\;標準正規分布は偶凾数(y軸(x＝0)で左右対称),\;Z_1\geq0のみ考えればよくする \\&=&2\int_{0}^{t}g(\sqrt{x})\frac{1}{2}x^{-\frac{1}{2}}\mathrm{d}x \\&&\;\cdots\;Z_1=\sqrt{X}(Z_1\geq0)より\frac{\mathrm{d}z_1}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}=\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}},\mathrm{d}z_1=\frac{1}{2}x^{-\frac{1}{2}}\mathrm{d}x \\&&\;\cdots\;z_1:0\rightarrow\sqrt{t},\;x:0\rightarrow t \\&=&\int_{0}^{t}g(\sqrt{x})x^{-\frac{1}{2}}\mathrm{d}x \\\int_0^t f_1(x) \mathrm{d}x &=&\int_{0}^{t}g(\sqrt{x})x^{-\frac{1}{2}}\mathrm{d}x \;\cdots\;累積分布凾数同士なので等しい(0から任意のt(\gt0)までの定積分が等しい) \\f_1(x)&=&g(\sqrt{x})x^{-\frac{1}{2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/blog-post_25.html}{0から任意のt(\gt0)までの定積分が等しい\rightarrow0からtの範囲で等しい\rightarrow各tでの微分が等しい\rightarrow被積分凾数同士も等しい} \\&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{\sqrt{x}^2}{2}}\;x^{-\frac{1}{2}} \;\cdots\;標準正規分布の確率密度凾数:g(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \\&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{x}{2}}\;x^{-\frac{1}{2}} \end{eqnarray} $$\(k=2\)のXの確率密度凾数
\(k=2\) $$ \begin{eqnarray} X&=&Z_1^2+Z_2^2\;\cdots\;Z_1,Z_2:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2 \\X-Y&=&Z_2^2 \\f_2(x) &=&\int_{0}^{\infty}f_1(y)f_1(x-y)\mathrm{d}y \;\cdots\;k=2のXの確率密度凾数f_2,\;Z_1とZ_2は独立なので同時確率はf_1(Z_1)とf_1(Z_2)の積 \\&=&\int_{0}^{x} \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}\;y^{-\frac{1}{2}} \right\} \;\cdot\; \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}}e^{-\frac{(x-y)}{2}}\;y^{-\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi} \int_{0}^{x} e^{-\frac{y}{2}-\frac{(x-y)}{2}}\;y^{-\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{x}{2}} \int_{0}^{x} y^{-\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{x}{2}}\pi \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/1xa-x.html}{\int_{0}^{a}\frac{1}{\sqrt{x}\sqrt{a-x}}\mathrm{d}x=\pi} \\&=&\frac{1}{2}e^{-\frac{x}{2}} \end{eqnarray} $$\(k=3\)のXの確率密度凾数
$$ \begin{eqnarray} X&=&Z_1^2+Z_2^2+Z_3^2\;\cdots\;Z_1,Z_2,Z_3:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2+Z_2^2 \\X-Y&=&Z_3^2 \\f_3(x) &=&\int_{0}^{\infty}f_2(y)f_1(x-y)\mathrm{d}y \;\cdots\;k=3のXの確率密度凾数f_3,\;YとZ_3は独立なので同時確率はf_2(Y)とf_1(Z_3)の積 \\&=&\int_{0}^{x} \left\{ \frac{1}{2}e^{-\frac{y}{2}} \right\} \;\cdot\; \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{2}\frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\sqrt{2\pi}}e^{-\frac{x}{2}} \int_{0}^{x} (x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\sqrt{2\pi}}e^{-\frac{x}{2}} 2x^{\frac{1}{2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/1a-x.html}{\int_{0}^{a}\frac{1}{\sqrt{a-x}}\mathrm{d}x=2a^{\frac{1}{2}}} \\&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{x}{2}} x^{\frac{1}{2}} \end{eqnarray} $$\(k=4\)のXの確率密度凾数
$$ \begin{eqnarray} X&=&Z_1^2+Z_2^2+Z_3^2+Z_4^2\;\cdots\;Z_1,Z_2,Z_3,Z_4:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2+Z_2^2+Z_3^2 \\X-Y&=&Z_4^2 \\f_4(x) &=&\int_{0}^{\infty}f_3(y)f_1(x-y)\mathrm{d}y \;\cdots\;k=4のXの確率密度凾数f_4,\;YとZ_4は独立なので同時確率はf_3(Y)とf_1(Z_4)の積 \\&=&\int_{0}^{x} \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}y^{\frac{1}{2}} \right\} \;\cdot\; \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}}e^{-\frac{(x-y)}{2}}\;y^{\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi} \int_{0}^{x} e^{-\frac{y}{2}-\frac{(x-y)}{2}}\;y^{\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{x}{2}} \int_{0}^{x} y^{\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{x}{2}} \frac{x\pi}{2} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/xa-x.html}{\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x=\frac{a\pi}{2}} \\&=&\frac{1}{4}e^{-\frac{x}{2}} x \end{eqnarray} $$\(k=5\)のXの確率密度凾数
$$ \begin{eqnarray} X&=&Z_1^2+Z_2^2+Z_3^2+Z_4^2+Z_5^2\;\cdots\;Z_1,Z_2,Z_3,Z_4,Z_5:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2+Z_2^2+Z_3^2+Z_4^2 \\X-Y&=&Z_5^2 \\f_5(x) &=&\int_{0}^{\infty}f_4(y)f_1(x-y)\mathrm{d}y \;\cdots\;k=5のXの確率密度凾数f_5,\;YとZ_4は独立なので同時確率はf_4(Y)とf_1(Z_5)の積 \\&=&\int_{0}^{x} \left\{ \frac{1}{4}e^{-\frac{y}{2}} y \right\} \;\cdot\; \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{4}\frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}}e^{-\frac{(x-y)}{2}}\;y(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{4\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}-\frac{(x-y)}{2}}\;y(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{4\sqrt{2\pi}}e^{-\frac{x}{2}} \int_{0}^{x} y(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{4\sqrt{2\pi}}e^{-\frac{x}{2}}\frac{4}{3}x^{\frac{3}{2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/xa-x_22.html}{\int_{0}^{a}\frac{x}{\sqrt{a-x}}\mathrm{d}x=\frac{4}{3}a^{\frac{3}{2}}} \\&=&\frac{1}{3\sqrt{2\pi}}e^{-\frac{x}{2}} x^{\frac{3}{2}} \end{eqnarray} $$\(k=n\)の\(X\)の確率密度凾数の類推
$$ \begin{array}[crcccc] \\k=1&:&f_1(x)=&\frac{1}{\sqrt{2\pi}}&e^{-\frac{x}{2}}&x^{-\frac{1}{2}} \\k=2&:&f_2(x)=&\frac{1}{2}&e^{-\frac{x}{2}}&x^0 \\k=3&:&f_3(x)=&\frac{1}{\sqrt{2\pi}}&e^{-\frac{x}{2}}&x^{\frac{1}{2}} \\k=4&:&f_4(x)=&\frac{1}{4}&e^{-\frac{x}{2}}&x^1 \\k=5&:&f_5(x)=&\frac{1}{3\sqrt{2\pi}}&e^{-\frac{x}{2}}&x^{\frac{3}{2}} \\k=n&:&f_n(x)=&\alpha_n&e^{-\frac{x}{2}}&x^{\frac{n}{2}-1} \end{array} $$係数\(\alpha_n\)を求める
$$ \begin{eqnarray} F_n(t) &=&\int^{t}_0 f_n(x) \mathrm{d}x \;\cdots\;任意のnにおけるXの累積分布凾数F_n,確率密度凾数f_n \\F_n(t)&=&\int^{t}_0 \alpha_ne^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x \;\cdots\;前述の類推より \\&=&\alpha_n\int^{t}_0e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x\;\cdots\;\alpha_nはxによらない定数,\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\alpha_n\int^{t}_0e^{-t}(2u)^{\frac{n}{2}-1} 2\mathrm{d}u \;\cdots\;u=\frac{x}{2},x=2u,\frac{dx}{du}=2,x:0\rightarrow\infty,u:0\rightarrow\infty \\&=&\alpha_n\int^{t}_0e^{-t}2^{\frac{n}{2}-1}t^{\frac{n}{2}-1} 2\mathrm{d}t \\&=&\alpha_n\int^{t}_0e^{-t}2^{\frac{n}{2}}t^{\frac{n}{2}-1}\mathrm{d}t \\&=&\alpha_n2^{\frac{n}{2}}\int^{t}_0e^{-t}t^{\frac{n}{2}-1}\mathrm{d}t \\F_n(t=\infty)&=&\int^{\infty}_0 f_n(x) \mathrm{d}x \\&=&\alpha_n2^{\frac{n}{2}}\int^{\infty}_0e^{-t}t^{\frac{n}{2}-1}\mathrm{d}t \\&=&\alpha_n2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)\;\cdots\;\Gamma\left(s\right)=\int^{\infty}_0e^{-t}t^{s-1}\mathrm{d}t \\&=&\alpha_n2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right) \\&=&1\;\cdots\;全事象は1 \\\alpha_n &=&\frac{F_n(t=\infty)}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)} \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)} \end{eqnarray} $$係数\(\alpha_k\)の確認
$$ \begin{array}[crlllll] \\\alpha_1&=&\frac{1}{2^{\frac{1}{2}}\Gamma\left(\frac{1}{2}\right)}&=&\frac{1}{2^{\frac{1}{2}}\;\sqrt{\pi}} &=&\frac{1}{\sqrt{2\pi}} \\\alpha_2&=&\frac{1}{2^{\frac{2}{2}}\Gamma\left(\frac{2}{2}\right)}&=&\frac{1}{2^1 \; 1} &=&\frac{1}{2} \\\alpha_3&=&\frac{1}{2^{\frac{3}{2}}\Gamma\left(\frac{3}{2}\right)}&=&\frac{1}{2^{\frac{3}{2}}\;\frac{\sqrt{\pi}}{2}} &=&\frac{1}{\sqrt{2\pi}} \\\alpha_4&=&\frac{1}{2^{\frac{4}{2}}\Gamma\left(\frac{4}{2}\right)}&=&\frac{1}{2^2 \; 1} &=&\frac{1}{4} \\\alpha_5&=&\frac{1}{2^{\frac{5}{2}}\Gamma\left(\frac{5}{2}\right)}&=&\frac{1}{2^{\frac{5}{2}}\;\frac{3\sqrt{\pi}}{4}}&=&\frac{1}{3\sqrt{2\pi}} \end{array} $$累積分布凾数の等式より\(k=n\)の\(X\)の確率密度凾数を求める
$$ \begin{eqnarray} \\\int^{t}_0 f_n(x) \mathrm{d}x &=&\int^{t}_0 \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x \;\cdots\;累積分布凾数同士なので等しい(0から任意のt(\gt0)までの定積分が等しい) \\f_n(x) &=& \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/blog-post_25.html}{0から任意のt(\gt0)までの定積分が等しい\rightarrow0からtの範囲で等しい\rightarrow各tでの微分が等しい\rightarrow被積分凾数同士も等しい} \end{eqnarray} $$数学的帰納法による証明
\(f_1(x),f_n(x)\)を認めた上で，\(f_{n+1}(x)\)が\(f_n(x)\)の\(n\)を\(n+1\)とした式になるかを確認する． $$ \begin{eqnarray} f_1(x) &=& \frac{1}{\sqrt{2\pi}}e^{-\frac{x}{2}}\;x^{-\frac{1}{2}} \\f_{n}(x) &=& \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \\f_{n+1}(x) &=& \frac{1}{2^{\frac{\color{red}{n+1}}{2}}\Gamma\left(\frac{\color{red}{n+1}}{2}\right)}e^{-\frac{x}{2}}x^{\frac{\color{red}{n+1}}{2}-1}\;\cdots\;f_n(x)のnをn+1とした式,上2式を認めた上でこの式が導出できるか． \end{eqnarray} $$ $$ \begin{eqnarray} X&=&Z_1^2+\cdots+Z_n^2+Z_{n+1}^2\;\cdots\;Z_1,\cdots,Z_n,Z_{n+1}:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2+\cdots+Z_{n} \\X-Y&=&Z_{n+1}^2 \\f_{n+1}(x) &=& \int_0^\infty f_n(y)f_1(x-y) \mathrm{d}y \\&=& \int_0^\infty \left\{ \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{y}{2}}y^{\frac{n}{2}-1} \right\}\left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{x-y}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{2\pi}} \int_0^\infty e^{-\frac{y}{2}}e^{-\frac{x-y}{2}} y^{\frac{n}{2}-1}(x-y)^{-\frac{1}{2}} \mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{2}\sqrt{\pi}} \int_0^\infty e^{-\frac{y}{2}-\frac{x-y}{2}} y^{\frac{n}{2}-1}(x-y)^{-\frac{1}{2}} \mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{2^{\frac{1}{2}}\sqrt{\pi}} \int_0^\infty e^{-\frac{x}{2}} y^{\frac{n}{2}-1}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2^{\frac{n}{2}}2^{\frac{1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} \int_0^\infty y^{\frac{n}{2}-1}(x-y)^{-\frac{1}{2}} \mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} \int_0^1 (xu)^{\frac{n}{2}-1}(x-xu)^{-\frac{1}{2}} x \mathrm{d}u \;\cdots\; y=xu, \frac{\mathrm{d}y}{\mathrm{d}u}=x, x:0\rightarrow\infty, u:0\rightarrow1 \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} \int_0^1 (xu)^{\frac{n}{2}-1}\left\{x(1-u)\right\}^{-\frac{1}{2}} x \mathrm{d}u \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} \int_0^1 x^{\frac{n}{2}-1}u^{\frac{n}{2}-1}x^{-\frac{1}{2}}(1-u)^{-\frac{1}{2}} x \mathrm{d}u \;\cdots\;(AB)^C=A^CB^C \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{\frac{n}{2}-1} x^{-\frac{1}{2}} x \int_0^1 u^{\frac{n}{2}-1}(1-u)^{-\frac{1}{2}} \mathrm{d}u \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n}{2}-1-\frac{1}{2}+1} \int_0^1 u^{\frac{n}{2}-1}(1-u)^{-\frac{1}{2}} \mathrm{d}u \;\cdots\;A^BA^C=A^{B+C} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \int_0^1 u^{\frac{n}{2}-1}(1-u)^{-\frac{1}{2}} \mathrm{d}u \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} B\left(\frac{n}{2}-1, -\frac{1}{2}\right) \;\cdots\;B(p,q)=\int_0^1 x^{p}(1-x)^{q} \mathrm{d}x \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\left(\frac{n}{2}-1\right)!\left(-\frac{1}{2}\right)!}{\left(\frac{n}{2}-1-\frac{1}{2}+1\right)!} \;\cdots\;B(p,q)=\frac{p!q!}{\left(p+q+1\right)!} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\left(\frac{n}{2}-1\right)!\left(-\frac{1}{2}\color{red}{+1-1}\right)!} {\left(\frac{n}{2}+\frac{1}{2}-1\right)!} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\left(\frac{n}{2}-1\right)!\left(\frac{1}{2}-1\right)!} {\left(\frac{n+1}{2}-1\right)!} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{1}{2}\right)} {\Gamma\left(\frac{n+1}{2}\right)} \;\cdots\;\Gamma(n)=\left(n-1\right)! \\&=&\frac{1}{2^{\frac{n+1}{2}}\color{red}{\Gamma\left(\frac{n}{2}\right)}} \frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\color{red}{\Gamma\left(\frac{n}{2}\right)}\color{black}{\Gamma\left(\frac{1}{2}\right)}} {\Gamma\left(\frac{n+1}{2}\right)} \\&=&\frac{1}{2^{\frac{n+1}{2}}}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\Gamma\left(\frac{1}{2}\right)} {\Gamma\left(\frac{n+1}{2}\right)} \\&=&\frac{1}{2^{\frac{n+1}{2}}} \color{red}{ \frac{1}{\sqrt{\pi}} } \color{black}{e^{-\frac{x}{2}}} x^{ \frac{n+1}{2}-1} \frac{\color{red}{\sqrt{\pi}}} {\Gamma\left(\frac{n+1}{2}\right)} \;\cdots\;\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n+1}{2}\right)} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \end{eqnarray} $$ 任意の\(n(n\geq1)\)において上記式が成り立つことを証明できた．以上により\(\chi^2\)分布の確率密度凾数が導出された．
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