PRML演習問題 2.46(基本) www - 機械学習基礎理論独習

問題

$(2.158)$ の積分を計算し $(2.159)$ になることを確かめよ。

参照

$\begin{eqnarray} p(x|\mu,a,b)&=&\int_0^\infty{\mathcal N}(x|\mu,\tau^{-1}){\rm Gam}(\tau|a,b){\rm d}\tau\tag{2.158}\\ &=&\int_0^\infty\frac{b^ae^{(-b\tau)}\tau^{a-1}}{\Gamma(a)}\left(\frac{\tau}{2\pi}\right)^{1/2}\exp\left(-\frac{\tau}{2}(x-\mu)^2\right){\rm d}\tau\\ &=&\frac{b^a}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}\left(b+\frac{(x-\mu)^2}{2}\right)^{-a-1/2}\Gamma(a+1/2) \end{eqnarray}$

$\begin{eqnarray} {\rm St}(x|\mu,\lambda,\nu)=\frac{\Gamma(\nu/2+1/2)}{\Gamma(\nu/2)}\left(\frac{\lambda}{\pi\nu}\right)^{1/2}\left(1+\frac{\lambda(x-\mu)^2}{\nu}\right)^{-\nu/2-1/2}\tag{2.159} \end{eqnarray}$

解答

$(2.158)$ を計算します。

$\begin{eqnarray} p(x|\mu,a,b)&=&\int_0^\infty{\mathcal N}(x|\mu,\tau^{-1}){\rm Gam}(\tau|a,b){\rm d}\tau\\ &=&\int_0^\infty\frac{b^ae^{(-b\tau)}\tau^{a-1}}{\Gamma(a)}\left(\frac{\tau}{2\pi}\right)^{1/2}\exp\left(-\frac{\tau}{2}(x-\mu)^2\right){\rm d}\tau\\ &=&\frac{b^a}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}\int_0^\infty e^{(-b\tau)}\tau^{a-1}\tau^{1/2}\exp\left(-\frac{\tau}{2}(x-\mu)^2\right){\rm d}\tau\\ &=&\frac{b^a}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}\int_0^\infty \tau^{a-1/2}\exp\left(-\tau\left(b+\frac{1}{2}(x-\mu)^2\right)\right){\rm d}\tau\tag{1} \end{eqnarray}$

$z=\tau\left(b+\dfrac{1}{2}(x-\mu)^2\right)$ とおくと、以下が成り立ちます。

$\begin{eqnarray} \tau=\dfrac{1}{b+\dfrac{1}{2}(x-\mu)^2}z=\left(b+\dfrac{1}{2}(x-\mu)^2\right)^{-1}z\tag{2} \end{eqnarray}$

$\begin{eqnarray} {\rm d}\tau=\dfrac{1}{b+\dfrac{1}{2}(x-\mu)^2}{\rm d}z=\left(b+\dfrac{1}{2}(x-\mu)^2\right)^{-1}{\rm d}z\tag{3} \end{eqnarray}$

式 $(1),(2),(3)$ より、 $\tau$ から $z$ へ変数変換します。

$\begin{eqnarray} p(x|\mu,a,b)&=&\frac{b^a}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}\int_0^\infty \left(b+\dfrac{1}{2}(x-\mu)^2\right)^{-a+1/2}z^{a-1/2}\exp(-z)\left(b+\dfrac{1}{2}(x-\mu)^2\right)^{-1}{\rm d}z\\ &=&\frac{b^a}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}\left(b+\dfrac{1}{2}(x-\mu)^2\right)^{-a-1/2}\int_0^\infty z^{a-1/2}\exp(-z){\rm d}z\\ &=&\frac{b^a}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}\left(b+\dfrac{1}{2}(x-\mu)^2\right)^{-a-1/2}\int_0^\infty z^{a+1/2-1}\exp(-z){\rm d}z\\ &=&\frac{b^a}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}\left(b+\dfrac{1}{2}(x-\mu)^2\right)^{-a-1/2}\Gamma\left(a+\frac{1}{2}\right)\tag{4}\\ &=&\frac{\Gamma(a+1/2)}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}b^a\left(b\left(1+\dfrac{1}{2b}(x-\mu)^2\right)\right)^{-a-1/2}\\ &=&\frac{\Gamma(a+1/2)}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}b^ab^{-a-1/2}\left(1+\dfrac{1}{2b}(x-\mu)^2\right)^{-a-1/2}\\ &=&\frac{\Gamma(a+1/2)}{\Gamma(a)}\left(\frac{1}{2\pi}\right)^{1/2}b^{-1/2}\left(1+\dfrac{1}{2b}(x-\mu)^2\right)^{-a-1/2}\\ &=&\frac{\Gamma(a+1/2)}{\Gamma(a)}\left(\frac{1}{2\pi b}\right)^{1/2}\left(1+\dfrac{1}{2b}(x-\mu)^2\right)^{-a-1/2}\tag{5} \end{eqnarray}$

式 $(4)$ でガンマ関数の定義 $\ \Gamma(a)=\displaystyle\int_0^\infty z^{a-1}\exp(-z){\rm d}z\$ を用いました。

$\nu=2a,\ \lambda=a/b$ とおくと、以下が成り立ちます。

$\begin{eqnarray} a=\frac{\nu}{2}\tag{6} \end{eqnarray}$

$\begin{eqnarray} b=\frac{a}{\lambda}=\frac{\nu}{2\lambda}\tag{7} \end{eqnarray}$

式 $(6),(7)$ を式 $(5)$ へ代入します。

$\begin{eqnarray} p(x|\mu,a,b)&=&\frac{\Gamma(\nu/2+1/2)}{\Gamma(\nu/2)}\left(\frac{\lambda}{\pi \nu}\right)^{1/2}\left(1+\dfrac{\lambda(x-\mu)^2}{\nu}\right)^{-\nu/2-1/2}\tag{8} \end{eqnarray}$

式 $(8)$ より、式 $(2.159)$ が示せました。