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提案のレベルを上げる - 推進力のあるエンジニアになるための思考法

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3.18(標準) カイ二乗分布

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方針

カイ二乗分布について書く.

Z_1,...,Z_n:iid\sim\mathcal{N}(0,1)
$${Z_1,...,Z_n:iid\sim\mathcal{N}(0,1) }$$

の時,

X=Z_1^2+...+Z_n^2\sim\chi_n^2=Ga(\frac{n}{2},\frac{1}{2})
$${X=Z_1^2+...+Z_n^2\sim\chi_n^2=Ga(\frac{n}{2},\frac{1}{2}) }$$

で,確率密度関数は,

f_{X}(x)=\frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}\exp(-\frac{1}{2}x)
$${f_{X}(x)=\frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}\exp(-\frac{1}{2}x) }$$

である.積率母関数は

\begin{align}
M_X(t)&= \mathbb{E}_{X\sim\chi_n^2}[e^{tX}]\\
&= \int_0^\infty\frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}\exp\{-(\frac{1}{2}-t)x\}dx\\
&= \frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}\frac{\Gamma(\frac{n}{2})}{(\frac{1}{2}-t)^{n/2}}\int_0^\infty\frac{(\frac{1}{2}-t)^{n/2}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}\exp\{-(\frac{1}{2}-t)x\}dx\\
&= \frac{1}{(1-2t)^{n/2}}
\end{align}
$${\begin{align} M_X(t)&= \mathbb{E}_{X\sim\chi_n^2}[e^{tX}]\\ &= \int_0^\infty\frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}\exp\{-(\frac{1}{2}-t)x\}dx\\ &= \frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}\frac{\Gamma(\frac{n}{2})}{(\frac{1}{2}-t)^{n/2}}\int_0^\infty\frac{(\frac{1}{2}-t)^{n/2}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}\exp\{-(\frac{1}{2}-t)x\}dx\\ &= \frac{1}{(1-2t)^{n/2}} \end{align} }$$

である.まずは

Z_1^2\sim\chi_1^2
$${Z_1^2\sim\chi_1^2 }$$

を示す.

Y_1=Z_1^2,\ Y_2=Z_2^2,\ ...,\ Y_n=Z_n^2
$${Y_1=Z_1^2,\ Y_2=Z_2^2,\ ...,\ Y_n=Z_n^2 }$$

と置くと

\begin{align}
f_{Y_1}(y)&= f_{Z_1}(\sqrt{y})\frac{1}{2\sqrt{y}}-f_{Z_1}(-\sqrt{y})(-\frac{1}{2\sqrt{y}})\\
&= \frac{1}{\sqrt{y}}\frac{1}{\sqrt{2\pi}}\exp(-\frac{y}{2})\\
&= \frac{(\frac{1}{2})^{1/2}}{\Gamma(\frac{1}{2})}y^{\frac{1}{2}-1}\exp(-\frac{1}{2}y)
\end{align}
$${\begin{align} f_{Y_1}(y)&= f_{Z_1}(\sqrt{y})\frac{1}{2\sqrt{y}}-f_{Z_1}(-\sqrt{y})(-\frac{1}{2\sqrt{y}})\\ &= \frac{1}{\sqrt{y}}\frac{1}{\sqrt{2\pi}}\exp(-\frac{y}{2})\\ &= \frac{(\frac{1}{2})^{1/2}}{\Gamma(\frac{1}{2})}y^{\frac{1}{2}-1}\exp(-\frac{1}{2}y) \end{align} }$$

ゆえ,

Y_1,...,Y_n:iid\sim\chi_1^2
$${Y_1,...,Y_n:iid\sim\chi_1^2 }$$

が示された.次に,

X=Y_1+...+Y_n
$${X=Y_1+...+Y_n }$$

と置くと

\begin{align}
M_X(t)&= \mathbb{E}_X[e^{tX}]\\
&= \mathbb{E}_{Y_1,...,Y_n:iid\sim\chi_1^2}[e^{t(Y_1+...+Y_n)}]\\
&= \mathbb{E}_{Y_1\sim\chi_1^2}[e^{tY_1}]...\mathbb{E}_{Y_n\sim\chi_1^2}[e^{tY_n}]\\
&= \frac{1}{(1-2t)^{1/2}}...\frac{1}{(1-2t)^{1/2}}\\
&= \frac{1}{(1-2t)^{n/2}}
\end{align}
$${\begin{align} M_X(t)&= \mathbb{E}_X[e^{tX}]\\ &= \mathbb{E}_{Y_1,...,Y_n:iid\sim\chi_1^2}[e^{t(Y_1+...+Y_n)}]\\ &= \mathbb{E}_{Y_1\sim\chi_1^2}[e^{tY_1}]...\mathbb{E}_{Y_n\sim\chi_1^2}[e^{tY_n}]\\ &= \frac{1}{(1-2t)^{1/2}}...\frac{1}{(1-2t)^{1/2}}\\ &= \frac{1}{(1-2t)^{n/2}} \end{align} }$$

より,

X=Y_1+...+Y_n=Z_1^2+...+Z_n^2\sim\chi_n^2
$${X=Y_1+...+Y_n=Z_1^2+...+Z_n^2\sim\chi_n^2 }$$

が示された.問題はカイ二乗分布の期待値演算なので,ガンマ分布の時と同様に計算すれば良い.

1.答案

\begin{align}
\mathbb{E}_{X\sim\chi_n^2}[X^\nu]&= \int_0^\infty\frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}x^{\nu+\frac{n}{2}-1}\exp(-\frac{1}{2}x)dx\\
&= \frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}\frac{\Gamma(\nu+\frac{n}{2})}{(\frac{1}{2})^{\nu+\frac{n}{2}}}\int_0^\infty\frac{(\frac{1}{2})^{\nu+\frac{n}{2}}}{\Gamma(\nu+\frac{n}{2})}x^{\nu+\frac{n}{2}-1}\exp(-\frac{1}{2}x)dx\\
&= \frac{2^\nu\Gamma(\nu+\frac{n}{2})}{\Gamma(\frac{n}{2})}
\end{align}
$${\begin{align} \mathbb{E}_{X\sim\chi_n^2}[X^\nu]&= \int_0^\infty\frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}x^{\nu+\frac{n}{2}-1}\exp(-\frac{1}{2}x)dx\\ &= \frac{(\frac{1}{2})^{n/2}}{\Gamma(\frac{n}{2})}\frac{\Gamma(\nu+\frac{n}{2})}{(\frac{1}{2})^{\nu+\frac{n}{2}}}\int_0^\infty\frac{(\frac{1}{2})^{\nu+\frac{n}{2}}}{\Gamma(\nu+\frac{n}{2})}x^{\nu+\frac{n}{2}-1}\exp(-\frac{1}{2}x)dx\\ &= \frac{2^\nu\Gamma(\nu+\frac{n}{2})}{\Gamma(\frac{n}{2})} \end{align} }$$

2.答案

自由度nのカイ二乗分布に従う確率変数をX_nと置くことにする.

X_n\sim\chi_n^2
$${X_n\sim\chi_n^2 }$$

この時,X_{n+2}の確率密度関数は

\begin{align}
f_{X_{n+2}}(x)&= \frac{(\frac{1}{2})^{\frac{n}{2}+1}}{\Gamma(\frac{n}{2}+1)}x^{n/2}\exp(-\frac{1}{2}x)\\
&= \frac{\frac{1}{2}x}{\frac{n}{2}}\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}\exp(-\frac{1}{2}x)\\
&= \frac{x}{n}f_{X_n}(x)
\end{align}
$${\begin{align} f_{X_{n+2}}(x)&= \frac{(\frac{1}{2})^{\frac{n}{2}+1}}{\Gamma(\frac{n}{2}+1)}x^{n/2}\exp(-\frac{1}{2}x)\\ &= \frac{\frac{1}{2}x}{\frac{n}{2}}\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}\exp(-\frac{1}{2}x)\\ &= \frac{x}{n}f_{X_n}(x) \end{align} }$$

と表せ,

\begin{align}
f_{X_n}(x)&= \frac{n}{x}f_{X_{n+2}}(x)\\
f_{X_n}(x)&= \frac{x}{n-2}f_{X_{n-2}}(x)
\end{align}
$${\begin{align} f_{X_n}(x)&= \frac{n}{x}f_{X_{n+2}}(x)\\ f_{X_n}(x)&= \frac{x}{n-2}f_{X_{n-2}}(x) \end{align} }$$

が成り立つ.ゆえ,

\begin{align}
\mathbb{E}_{X_n\sim\chi_n^2}[h(X_n)]&= \int_0^\infty h(x)f_{X_n}(x)dx\\
&= \int_0^\infty h(x)\frac{n}{x}f_{X_{n+2}}(x)dx\\
&= n\mathbb{E}_{X_{n+2}\sim\chi_{n+2}^2}[\frac{h(X_n)}{X_n}]\\
\mathbb{E}_{X_n\sim\chi_n^2}[h(X_n)]&= \int_0^\infty h(x)f_{X_n}(x)dx\\
&= \int_0^\infty h(x)\frac{x}{n-2}f_{X_{n-2}}(x)dx\\
&= \frac{1}{n-2}\mathbb{E}_{X_{n-2}\sim\chi_{n-2}^2}[X_{n-2}h(X_{n-2})]
\end{align}
$${\begin{align} \mathbb{E}_{X_n\sim\chi_n^2}[h(X_n)]&= \int_0^\infty h(x)f_{X_n}(x)dx\\ &= \int_0^\infty h(x)\frac{n}{x}f_{X_{n+2}}(x)dx\\ &= n\mathbb{E}_{X_{n+2}\sim\chi_{n+2}^2}[\frac{h(X_n)}{X_n}]\\ \mathbb{E}_{X_n\sim\chi_n^2}[h(X_n)]&= \int_0^\infty h(x)f_{X_n}(x)dx\\ &= \int_0^\infty h(x)\frac{x}{n-2}f_{X_{n-2}}(x)dx\\ &= \frac{1}{n-2}\mathbb{E}_{X_{n-2}\sim\chi_{n-2}^2}[X_{n-2}h(X_{n-2})] \end{align} }$$

参考文献

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