ベイズ統計学

予測分布

モデル分布がベルヌーイ分布であるとする．

m (y ∣ π) = π^{y} (1 - π)^{1 - y}

$\begin{equation} m(y\mid\pi) = \pi^y (1-\pi)^{1-y} \end{equation}$

独立な

n

$n$ 回の試行によって観測値

{y_{1}, y_{2}, \dots, y_{n}}

$\{y_1,y_2,\ldots,y_n\}$ を得たものとする．このとき，

π

$\pi$ についての事前分布が

B e (a_{0}, b_{0})

$Be(a_0, b_0)$ であるとすれば，事後分布は

B e (a_{0} + n {\bar{y}}_{\cdot}, b_{0} + n - n {\bar{y}}_{\cdot})

$Be(a_0 + n\bar{y}_{\cdot}, b_0 + n - n\bar{y}_{\cdot})$ である．

そこで，既に入手した観測値 $\{y_1,y_2,\ldots,y_n\}$ の下で，次の試行によって得られるであろう観測値 $Y_{n+1}~(\in \{0,1\})$ の値を確率的に予測してみる． $Y_{n+1}$ の予測分布は，

\begin{aligned} p (y_{n + 1} ∣ y) & = \int_{0}^{1} m (y_{n + 1} ∣ π) p (π ∣ y) d π \\ = \int_{0}^{1} π^{y_{n + 1}} (1 - π)^{1 - y_{n + 1}} \frac{1}{B (a_{0} + n {\bar{y}}_{\cdot}, b_{0} + n - n {\bar{y}}_{\cdot})} π^{a_{0} + n {\bar{y}}_{\cdot} - 1} (1 - π)^{b_{0} + n - n {\bar{y}}_{\cdot} - 1} d π \\ = \frac{1}{B (a_{0} + n {\bar{y}}_{\cdot}, b_{0} + n - n {\bar{y}}_{\cdot})} \int_{0}^{1} π^{a_{0} + n {\bar{y}}_{\cdot} + y_{n + 1} - 1} (1 - π)^{b_{0} + n - n {\bar{y}}_{\cdot} + 1 - y_{n + 1} - 1} d π \\ = \frac{B (a_{0} + n {\bar{y}}_{\cdot} + y_{n + 1}, b_{0} + n - n {\bar{y}}_{\cdot} + 1 - y_{n + 1})}{B (a_{0} + n {\bar{y}}_{\cdot}, b_{0} + n - n {\bar{y}}_{\cdot})} \\ = \frac{\frac{Γ (a_{0} + n {\bar{y}}_{\cdot} + y_{n + 1}) \cdot Γ (b_{0} + n - n {\bar{y}}_{\cdot} + 1 - y_{n + 1})}{Γ (a_{0} + n {\bar{y}}_{\cdot} + y_{n + 1} + b_{0} + n - n {\bar{y}}_{\cdot} + 1 - y_{n + 1})}}{\frac{Γ (a_{0} + n {\bar{y}}_{\cdot}) \cdot Γ (b_{0} + n - n {\bar{y}}_{\cdot})}{Γ (a_{0} + n {\bar{y}}_{\cdot} + b_{0} + n - n {\bar{y}}_{\cdot})}} \\ = \frac{\frac{Γ (a_{0} + n {\bar{y}}_{\cdot} + y_{n + 1}) \cdot Γ (b_{0} + n - n {\bar{y}}_{\cdot} + 1 - y_{n + 1})}{Γ (a_{0} + b_{0} + n + 1)}}{\frac{Γ (a_{0} + n {\bar{y}}_{\cdot}) \cdot Γ (b_{0} + n - n {\bar{y}}_{\cdot})}{Γ (a_{0} + b_{0} + n)}} \\ = \frac{\frac{Γ (a_{0} + n {\bar{y}}_{\cdot} + y_{n + 1})}{Γ (a_{0} + n {\bar{y}}_{\cdot})} \cdot \frac{Γ (b_{0} + n - n {\bar{y}}_{\cdot} + 1 - y_{n + 1})}{Γ (b_{0} + n - n {\bar{y}}_{\cdot})}}{\frac{Γ (a_{0} + b_{0} + n + 1)}{Γ (a_{0} + b_{0} + n)}} \\ = \frac{(a_{0} + n {\bar{y}}_{\cdot})^{y_{n + 1}} \cdot (b_{0} + n - n {\bar{y}}_{\cdot})^{1 - y_{n + 1}}}{a_{0} + b_{0} + n} \\ = {(\frac{a_{0} + n {\bar{y}}_{\cdot}}{a_{0} + b_{0} + n})}^{y_{n + 1}} {(\frac{b_{0} + n - n {\bar{y}}_{\cdot}}{a_{0} + b_{0} + n})}^{1 - y_{n + 1}} \\ = {(\frac{a_{0} + n {\bar{y}}_{\cdot}}{a_{0} + b_{0} + n})}^{y_{n + 1}} {(\frac{a_{0} + b_{0} + n - a_{0} - n {\bar{y}}_{\cdot}}{a_{0} + b_{0} + n})}^{1 - y_{n + 1}} \\ = {(\frac{a_{0} + n {\bar{y}}_{\cdot}}{a_{0} + b_{0} + n})}^{y_{n + 1}} {(1 - \frac{a_{0} + n {\bar{y}}_{\cdot}}{a_{0} + b_{0} + n})}^{1 - y_{n + 1}} \end{aligned}

$\begin{align} p(y_{n+1} \mid \bm{y}) &= \int_{0}^{1}m(y_{n+1} \mid \pi)\,p(\pi \mid \bm{y})\,d\pi \\ &= \int_{0}^{1}\pi^{y_{n+1}}(1-\pi)^{1-y_{n+1}}\frac{1}{B(a_0 + n\bar{y}_{\cdot}, b_0 + n - n\bar{y}_{\cdot})} \pi^{a_0 + n\bar{y}_{\cdot}-1}(1-\pi)^{b_0 + n - n\bar{y}_{\cdot}-1} d\pi \\ &= \frac{1}{B(a_0 + n\bar{y}_{\cdot}, b_0 + n - n\bar{y}_{\cdot})}\int_{0}^{1}\pi^{a_0 + n\bar{y}_{\cdot}+y_{n+1}-1}(1-\pi)^{b_0 + n - n\bar{y}_{\cdot}+1-y_{n+1}-1} d\pi \\ &= \frac{B(a_0 + n\bar{y}_{\cdot}+y_{n+1},~b_0 + n - n\bar{y}_{\cdot}+1-y_{n+1})}{B(a_0 + n\bar{y}_{\cdot},~b_0 + n - n\bar{y}_{\cdot})} \\ &= \frac{\ds\frac{\varGamma(a_0 + n\bar{y}_{\cdot}+y_{n+1})\cdot\varGamma(b_0 + n - n\bar{y}_{\cdot}+1-y_{n+1})}{\varGamma(a_0 + n\bar{y}_{\cdot}+y_{n+1} + b_0 + n - n\bar{y}_{\cdot}+1-y_{n+1})}}{\ds\frac{\varGamma(a_0 + n\bar{y}_{\cdot})\cdot\varGamma(b_0 + n - n\bar{y}_{\cdot})}{\varGamma(a_0 + n\bar{y}_{\cdot}+b_0 + n - n\bar{y}_{\cdot})}} \\ &= \frac{\ds\frac{\varGamma(a_0 + n\bar{y}_{\cdot}+y_{n+1})\cdot\varGamma(b_0 + n - n\bar{y}_{\cdot}+1-y_{n+1})}{\varGamma(a_0 + b_0 + n + 1)}}{\ds\frac{\varGamma(a_0 + n\bar{y}_{\cdot})\cdot\varGamma(b_0 + n - n\bar{y}_{\cdot})}{\varGamma(a_0 + b_0 + n)}} \\ &= \frac{\ds\frac{\varGamma(a_0 + n\bar{y}_{\cdot}+y_{n+1})}{\varGamma(a_0 + n\bar{y}_{\cdot})}\cdot\frac{\varGamma(b_0 + n - n\bar{y}_{\cdot}+1-y_{n+1})}{\varGamma(b_0 + n - n\bar{y}_{\cdot})}}{\ds\frac{\varGamma(a_0 + b_0 + n + 1)}{\varGamma(a_0 + b_0 + n)}} \\ &= \frac{(a_0 + n\bar{y}_{\cdot})^{y_{n+1}}\cdot(b_0 + n - n\bar{y}_{\cdot})^{1-y_{n+1}}}{a_0 + b_0 + n} \\ &= \left(\frac{a_0 + n\bar{y}_{\cdot}}{a_0 + b_0 + n}\right)^{y_{n+1}}\left(\frac{b_0 + n - n\bar{y}_{\cdot}}{a_0 + b_0 + n}\right)^{1-y_{n+1}} \\ &= \left(\frac{a_0 + n\bar{y}_{\cdot}}{a_0 + b_0 + n}\right)^{y_{n+1}}\left(\frac{a_0 + b_0 + n - a_0 - n\bar{y}_{\cdot}}{a_0 + b_0 + n}\right)^{1-y_{n+1}} \\ &= \left(\frac{a_0 + n\bar{y}_{\cdot}}{a_0 + b_0 + n}\right)^{y_{n+1}}\left(1-\frac{a_0 + n\bar{y}_{\cdot}}{a_0 + b_0 + n}\right)^{1-y_{n+1}} \end{align}$

ここで，

\hat{π} = \frac{a_{0} + n {\bar{y}}_{\cdot}}{a_{0} + b_{0} + n}

$\begin{equation} \hat{\pi} = \frac{a_0 + n\bar{y}_{\cdot}}{a_0 + b_0 + n} \end{equation}$

とおくと，

\hat{π}

$\hat{\pi}$ は

π

$\pi$ の事後平均であり，

p (y_{n + 1} ∣ y) = {\hat{π}}^{y_{n + 1}} (1 - \hat{π})^{1 - y_{n + 1}}

$\begin{equation} p(y_{n+1}\mid\bm{y}) = \hat{\pi}^{y_{n+1}}(1-\hat{\pi})^{1-y_{n+1}} \end{equation}$

となる．新たな試行で成功する予測確率と期待値は，

P (Y_{n + 1} = 1 ∣ y) = \hat{π}

$\begin{equation} P(Y_{n+1} = 1\mid\bm{y}) = \hat{\pi} \end{equation}$

E [Y_{n + 1} ∣ y] = \hat{π}

$\begin{equation} E[Y_{n+1}\mid\bm{y}] = \hat{\pi} \end{equation}$