【統計】標準正規分布とカイ二乗分布からT分布を導出(証明あり)

標準正規分布とカイ二乗分布からT分布を導出します．T分布は自由度が大きくなればなるほど標準正規分布に近づいていきます．

確率変数 $X$ が標準正規分布 $N(0,1)$ に従い，確率変数 $Y$ が自由度 $n$ のカイ二乗分布に従い，互いに独立のとき，

とすると，確率変数 $T$ はT分布に従います．

証明）

以下のような $(X,Y) \longrightarrow (T,Y)$ への変換を考えます．

ヤコビアンを求めると，

となります．ここで，確率変数 $X$ の確率密度関数を $f_X(x)$ ，確率変数 $Y$ を $f_Y(y)$ とすると，それぞれ

となります．確率変数Xと確率変数Yは互いに独立なので，

ここで， $\left( \frac{t^2}{n} + 1 \right) \frac{y}{2} = z$ への変換を考えると，

となります．

以下のように， $f(t,u)$ を積分して $T$ の周辺確率密度関数を計算すると，これが確率変数 $T$ の従う確率密度関数となります。

これは，自由度nのT分布の確率密度関数です．

よって，標準正規分布とカイ二乗分布からT分布を導出できました．

\begin{eqnarray*}
\left \{ \begin{array}{l}
T=\frac{X}{\sqrt{\frac{Y}{n}}} \\
Y=Y
\end{array} \right.
\quad \Longleftrightarrow \quad
\left \{ \begin{array}{l}
X=T\sqrt{\frac{Y}{n}} \\
Y=Y
\end{array} \right.
\end{eqnarray*}

\begin{eqnarray*}
J 
&=& \mathrm{mod} \left| \begin{array}{cc}
\frac{\partial X}{\partial T} & \frac{\partial X}{\partial Y} \\
\frac{\partial Y}{\partial T} & \frac{\partial Y}{\partial Y}
\end{array} \right| \\
&=& \mathrm{mod} \left| \begin{array}{cc}
\sqrt{\frac{Y}{n}} & \frac{1}{2}\cdot \frac{T}{\sqrt{n}} \cdot \frac{1}{\sqrt{Y}} \\
0 & 1
\end{array} \right| \\
&=& \sqrt{\frac{Y}{n}}
\end{eqnarray*}

\begin{eqnarray*}
f_X(x) &=& \frac{1}{\sqrt{2\pi}}exp\left \{ -\frac{x^2}{2} \right \} \\
f_Y(y) &=& \frac{1}{2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1}e^{-\frac{y}{2}}
\end{eqnarray*}

\begin{eqnarray*}
f(t,y)
&=& f_X(x)f_Y(y) \cdot J \\
&=& \frac{1}{\sqrt{2\pi}}exp\left \{ -\frac{x^2}{2} \right \} \cdot \frac{1}{2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1}e^{-\frac{y}{2}} \cdot \sqrt{\frac{y}{n}} \\
&=& \frac{1}{\sqrt{2\pi}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1}e^{-\frac{x^2+y}{2}}\sqrt{\frac{y}{n}} \\
&& \left( \mbox{ここで，}x=t\sqrt{\frac{y}{n}} \mbox{を代入} \right) \\
&=& \frac{1}{\sqrt{2\pi}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1} exp\left \{ -\left( \frac{t^2}{n}+1 \right) \frac{y}{2} \right \} \sqrt{\frac{y}{n}} \\
&& (\mbox{ただし，} -\infty \leq x \leq \infty ,  0 \leq u \leq \infty)
\end{eqnarray*}

\begin{equation*}
y=\frac{2z}{\frac{t^2}{n}+1} \quad \Longleftrightarrow \quad dy = \frac{2}{\frac{t^2}{n}+1}dz
\end{equation*}

\begin{eqnarray*}
f(t)
&=& \int _0 ^{\infty} f(t,y)dy \\
&=& \int _0 ^{\infty} \frac{1}{\sqrt{2\pi}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1} exp\left \{ -\left( \frac{t^2}{n}+1 \right) \frac{y}{2} \right \} \sqrt{\frac{y}{n}} dy \\
&=& \int _0 ^{\infty} \frac{1}{\sqrt{2\pi n}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)}  \left( \frac{2z}{\frac{t^2}{n}+1} \right) ^{\frac{n}{2}-1} e^{-z} \left( \frac{2z}{\frac{t^2}{n}+1} \right) ^{\frac{1}{2}} \frac{2}{\frac{t^2}{n}+1}dz \\
&=& \int _0 ^{\infty} \frac{1}{\sqrt{2\pi n}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)}  \left( \frac{2z}{\frac{t^2}{n}+1} \right) ^{\frac{n-1}{2}} e^{-z} \cdot \frac{2}{\frac{t^2}{n}+1}dz \\
&=& \int _0 ^{\infty} \frac{1}{\sqrt{2\pi n}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} \left( \frac{t^2}{n}+1 \right) ^{-\frac{n+1}{2}} z ^{\frac{n-1}{2}}\cdot 2^{\frac{n+1}{2}} \cdot e^{-z}dz \\
&=& \frac{2^{\frac{n+1}{2}}}{\sqrt{2\pi n}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right) \left( \frac{t^2}{n}+1 \right) ^{\frac{n+1}{2}}} \int _0 ^{\infty} z^{\frac{n-1}{2}}\cdot e^{-z}dz \\
&=& \frac{1}{\sqrt{\pi n}\Gamma \left( \frac{n}{2} \right) \left( \frac{t^2}{n}+1 \right) ^{\frac{n+1}{2}}} \Gamma \left( \frac{n-1}{2}+1 \right) \qquad \left( \because \Gamma (x) = \int _0 ^{\infty} t^{x-1} e^{-t} dt \right) \\
&=& \frac{1}{\sqrt{\pi n}\Gamma \left( \frac{n}{2} \right) \left( \frac{t^2}{n}+1 \right) ^{\frac{n+1}{2}}} \Gamma \left( \frac{n+1}{2} \right) \\
&=& \frac{\Gamma \left( \frac{n+1}{2} \right) }{\sqrt{n} \Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{n}{2} \right)} \left( \frac{t^2}{n}+1 \right) ^{-\frac{n+1}{2}} \qquad \left(\because \Gamma \left( \frac{1}{2} \right) = \sqrt{\pi} \right) \\
&=& \frac{1}{\sqrt{n} B \left( \frac{n}{2},\frac{1}{2} \right)} \left( \frac{t^2}{n}+1 \right) ^{-\frac{n+1}{2}} \qquad \left( \because B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma (p+q)} \right) \\
&=& n^{-\frac{1}{2}} \left\{  B \left( \frac{n}{2},\frac{1}{2} \right) \right\} ^{-1} \left( \frac{t^2}{n}+1 \right) ^{-\frac{n+1}{2}}
\end{eqnarray*}

\begin{eqnarray*}

\left \{ \begin{array}{l}

T=\frac{X}{\sqrt{\frac{Y}{n}}} \\

Y=Y

\end{array} \right.

\quad \Longleftrightarrow \quad

\left \{ \begin{array}{l}

X=T\sqrt{\frac{Y}{n}} \\

Y=Y

\end{array} \right.

\end{eqnarray*}

\begin{eqnarray*}

&=& \mathrm{mod} \left| \begin{array}{cc}

\frac{\partial X}{\partial T} & \frac{\partial X}{\partial Y} \\

\frac{\partial Y}{\partial T} & \frac{\partial Y}{\partial Y}

\end{array} \right| \\

&=& \mathrm{mod} \left| \begin{array}{cc}

\sqrt{\frac{Y}{n}} & \frac{1}{2}\cdot \frac{T}{\sqrt{n}} \cdot \frac{1}{\sqrt{Y}} \\

0 & 1

\end{array} \right| \\

&=& \sqrt{\frac{Y}{n}}

\end{eqnarray*}

\begin{eqnarray*}

f_X(x) &=& \frac{1}{\sqrt{2\pi}}exp\left \{ -\frac{x^2}{2} \right \} \\

f_Y(y) &=& \frac{1}{2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1}e^{-\frac{y}{2}}

\end{eqnarray*}

\begin{eqnarray*}

f(t,y)

&=& f_X(x)f_Y(y) \cdot J \\

&=& \frac{1}{\sqrt{2\pi}}exp\left \{ -\frac{x^2}{2} \right \} \cdot \frac{1}{2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1}e^{-\frac{y}{2}} \cdot \sqrt{\frac{y}{n}} \\

&=& \frac{1}{\sqrt{2\pi}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1}e^{-\frac{x^2+y}{2}}\sqrt{\frac{y}{n}} \\

&& \left( \mbox{ここで，}x=t\sqrt{\frac{y}{n}} \mbox{を代入} \right) \\

&=& \frac{1}{\sqrt{2\pi}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1} exp\left \{ -\left( \frac{t^2}{n}+1 \right) \frac{y}{2} \right \} \sqrt{\frac{y}{n}} \\

&& (\mbox{ただし，} -\infty \leq x \leq \infty , 0 \leq u \leq \infty)

\end{eqnarray*}

\begin{equation*}

y=\frac{2z}{\frac{t^2}{n}+1} \quad \Longleftrightarrow \quad dy = \frac{2}{\frac{t^2}{n}+1}dz

\end{equation*}

\begin{eqnarray*}

f(t)

&=& \int _0 ^{\infty} f(t,y)dy \\

&=& \int _0 ^{\infty} \frac{1}{\sqrt{2\pi}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} y^{\frac{n}{2}-1} exp\left \{ -\left( \frac{t^2}{n}+1 \right) \frac{y}{2} \right \} \sqrt{\frac{y}{n}} dy \\

&=& \int _0 ^{\infty} \frac{1}{\sqrt{2\pi n}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} \left( \frac{2z}{\frac{t^2}{n}+1} \right) ^{\frac{n}{2}-1} e^{-z} \left( \frac{2z}{\frac{t^2}{n}+1} \right) ^{\frac{1}{2}} \frac{2}{\frac{t^2}{n}+1}dz \\

&=& \int _0 ^{\infty} \frac{1}{\sqrt{2\pi n}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} \left( \frac{2z}{\frac{t^2}{n}+1} \right) ^{\frac{n-1}{2}} e^{-z} \cdot \frac{2}{\frac{t^2}{n}+1}dz \\

&=& \int _0 ^{\infty} \frac{1}{\sqrt{2\pi n}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right)} \left( \frac{t^2}{n}+1 \right) ^{-\frac{n+1}{2}} z ^{\frac{n-1}{2}}\cdot 2^{\frac{n+1}{2}} \cdot e^{-z}dz \\

&=& \frac{2^{\frac{n+1}{2}}}{\sqrt{2\pi n}2^{\frac{n}{2}}\Gamma \left( \frac{n}{2} \right) \left( \frac{t^2}{n}+1 \right) ^{\frac{n+1}{2}}} \int _0 ^{\infty} z^{\frac{n-1}{2}}\cdot e^{-z}dz \\

&=& \frac{1}{\sqrt{\pi n}\Gamma \left( \frac{n}{2} \right) \left( \frac{t^2}{n}+1 \right) ^{\frac{n+1}{2}}} \Gamma \left( \frac{n-1}{2}+1 \right) \qquad \left( \because \Gamma (x) = \int _0 ^{\infty} t^{x-1} e^{-t} dt \right) \\

&=& \frac{1}{\sqrt{\pi n}\Gamma \left( \frac{n}{2} \right) \left( \frac{t^2}{n}+1 \right) ^{\frac{n+1}{2}}} \Gamma \left( \frac{n+1}{2} \right) \\

&=& \frac{\Gamma \left( \frac{n+1}{2} \right) }{\sqrt{n} \Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{n}{2} \right)} \left( \frac{t^2}{n}+1 \right) ^{-\frac{n+1}{2}} \qquad \left(\because \Gamma \left( \frac{1}{2} \right) = \sqrt{\pi} \right) \\

&=& \frac{1}{\sqrt{n} B \left( \frac{n}{2},\frac{1}{2} \right)} \left( \frac{t^2}{n}+1 \right) ^{-\frac{n+1}{2}} \qquad \left( \because B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma (p+q)} \right) \\