[統計学] ガンマ分布の期待値・分散, 可視化, 積率母関数, 再生性,ベイズモデル

確率密度関数

ガンマ分布の確率密度関数は以下で表される

$$f(x∣α,β)=1βαΓ(α)xα−1e−xβif0<x,f(x∣α,β)=0otherwiseWhere,Γ(α)=∫0∞xα−1e−xdx, 0<α \begin{align*} &f(x|\alpha, \beta) = \frac{1}{\beta^\alpha \Gamma{(\alpha)}} x^{\alpha-1} e^{-\frac{x}{\beta}} \quad \text{if}\quad 0<x, \quad f(x|\alpha, \beta)=0 \quad \text{otherwise} \\ &\text{Where,}\quad \Gamma{(\alpha)}=\int_0^\infty x^{\alpha-1} e^{-x} dx,\;0<\alpha \end{align*}$$

可視化

01: $α\alpha$を変化させたときのガンマ分布

02: $β\beta$を変化させたときのガンマ分布

画像を生成するコード

# ライブラリをimport
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
from scipy.stats import gamma

#描画用の変数を用意する
x = np.arange(0, 10, 0.01)
beta = 1
ys = []
for alpha in [0.9,1,2,3,4]:
    y = [
        gamma.pdf(
            x_,
            a=alpha,
            scale=beta,
            loc=0
        )
        for x_ in x
    ]
    ys.append(y)

# 描画
sns.set()
sns.set_style("whitegrid", {'grid.linestyle': '--'})
sns.set_context("talk", 0.8, {"lines.linewidth": 2})
sns.set_palette("cool", 5, 0.9)

fig=plt.figure(figsize=(16,9))
ax1 = fig.add_subplot(1, 1, 1)
ax1.set_title('Gamma Distribution (beta=1)')
ax1.set_ylabel('probability density')

for y in ys:
    ax1.plot(x,y)

ax1.legend(['alpha=0.9','alpha=1','alpha=2','alpha=3','alpha=4'])
plt.show()

Gamma関数の公式

$Γ(⋅)\Gamma{(\cdot )}$ だが, 以下の特徴がある.

$$Γ(α)=∫0∞tα−1e−tdt \Gamma(\alpha)=\int_{0}^{\infty} t^{\alpha-1} e^{-t} d t$$

等式

ガンマ関数について以下が成り立つ．

$$Γ(α)=(α−1)Γ(α) \Gamma(\alpha)=(\alpha-1) \Gamma(\alpha)$$

証明

$$Γ(α)=∫0∞tα−1e−tdt=∫0∞tα−1{ ddt−e−t}dt=[−tα−1e−t]0∞+(α−1)∫0∞tα−2e−tdt=(0−0)+(α−1)Γ(α−1) \begin{align*} \Gamma(\alpha) & =\int_{0}^{\infty} t^{\alpha-1} e^{-t} d t \\ & \left.=\int_{0}^{\infty} t^{\alpha-1} \left\{\, \frac{d}{d t}-e^{-t}\right.\right\} d t \\ & =\left[-t^{\alpha-1} e^{-t}\right]_{0}^{\infty}+(\alpha-1) \int_{0}^{\infty} t^{\alpha-2} e^{-t} d t \\ & =(0-0)+(\alpha-1) \Gamma(\alpha-1) \end{align*}$$

期待値・分散

期待値・分散は定義からパラメータが元の式とは異なる確率密度関数の積分と定数係数に分けて導出する.

期待値の導出

$$E(X)=∫0∞xfX(x)dx=∫0∞xΓ(α)−1β−1(xβ)α−1e−xβdx=Γ(α)−1β−1∫0∞βxβ(xβ)α−1e−xβdx=Γ(α)−1β−1β∫0∞xβ(xβ)α−1e−xβdx=Γ(α)−1∫0∞(xβ)(α+1)−1e−xβdx=Γ(α)−1Γ(α+1)β∫0∞Γ(α+1)−1β−1(xβ)(α+1)−1e−xβdx=Γ(α)−1αΓ(α)β=αβ \begin{align*} \mathbb{E}(X) & =\int_{0}^{\infty} x f_{X}(x) d x \\ & =\int_{0}^{\infty} x \Gamma(\alpha)^{-1} \beta^{-1}\left(\frac{x}{\beta}\right)^{\alpha-1} e^{-\frac{x}{\beta}} d x \\ & =\Gamma(\alpha)^{-1} \beta^{-1} \int_{0}^{\infty} \frac{\beta x}{\beta}\left(\frac{x}{\beta}\right)^{\alpha-1} e^{-\frac{x}{\beta}} d x \\ & =\Gamma(\alpha)^{-1} \beta^{-1} \beta \int_{0}^{\infty} \frac{x}{\beta}\left(\frac{x}{\beta}\right)^{\alpha-1} e^{-\frac{x}{\beta}} d x \\ & =\Gamma(\alpha)^{-1} \int_{0}^{\infty}\left(\frac{x}{\beta}\right)^{(\alpha+1)-1} e^{-\frac{x}{\beta}} d x \\ & =\Gamma(\alpha)^{-1} \Gamma(\alpha+1) \beta \int_{0}^{\infty} \Gamma(\alpha+1)^{-1} \beta^{-1}\left(\frac{x}{\beta}\right)^{(\alpha+1)-1} e^{-\frac{x}{\beta}} d x \\ & =\Gamma(\alpha)^{-1} \alpha \Gamma(\alpha) \beta \\ & =\alpha \beta \end{align*}$$

分散の導出

$$E(X2)=∫0∞x2fX(x)dx=∫0∞x2Γ(α)−1β−1(xβ)α−1e−xβdx=Γ(α)−1β−1∫0∞β2(xβ)2(xβ)α−1e−xβdx=Γ(α)−1β−1β2∫0∞(xβ)(α+2)−1e−xβdx=Γ(α)−1βΓ(α+2)β∫0∞Γ(α+2)−1β−1(xβ)(α+2)−1e−xβdx=Γ(α)−1β2(α+1)αΓ(α)=αβ2(α+1) \begin{align*} \mathbb{E}\left(X^{2}\right) & = \int_{0}^{\infty} x^{2} f_{X}(x) d x \\ & = \int_{0}^{\infty} x^{2} \Gamma(\alpha)^{-1} \beta^{-1}\left(\frac{x}{\beta}\right)^{\alpha-1} e^{-\frac{x}{\beta}} d x \\ & = \Gamma(\alpha)^{-1} \beta^{-1} \int_{0}^{\infty} \beta^{2}\left(\frac{x}{\beta}\right)^{2}\left(\frac{x}{\beta}\right)^{\alpha-1} e^{-\frac{x}{\beta}} d x \\ & = \Gamma(\alpha)^{-1} \beta^{-1} \beta^{2} \int_{0}^{\infty}\left(\frac{x}{\beta}\right)^{(\alpha+2)-1} e^{-\frac{x}{\beta}} d x \\ & = \Gamma(\alpha)^{-1} \beta \Gamma(\alpha+2) \beta \int_{0}^{\infty} \Gamma(\alpha+2)^{-1} \beta^{-1}\left(\frac{x}{\beta}\right)^{(\alpha+2)-1} e^{-\frac{x}{\beta}} d x \\ & = \Gamma(\alpha)^{-1} \beta^{2}(\alpha+1) \alpha \Gamma(\alpha) \\ & = \alpha \beta^{2}(\alpha+1) \\ \end{align*}$$

よって分散は，

$$V(x)=E(x2)−E(x)2=αβ2(α+1)−α2β2=αβ2 \begin{align*} V(x) & =\mathbb{E}\left(x^{2}\right)-\mathbb{E}(x)^{2} \\ & =\alpha \beta^{2}(\alpha+1)-\alpha^{2} \beta^{2} \\ & =\alpha \beta^{2} \end{align*}$$

積率母関数

積率母関数の導出

$$Mx(t)=E(etX)=∫0∞etxfX(x)dx=∫0∞etxΓ(α)−1β−1(xβ)α−1e−xβdx=Γ(α)−1β−1∫0∞(xβ)α−1e−1βxetxdx=Γ(α)−1β−1∫0∞(x1β1β−t1β−t⋅ββ)α−1exp⁡{−x(1β−t)}dx=Γ(α)−1β−1∫0∞{x(1β−t)⋅(1β−t)−1β−1}α−1exp⁡{−x(1β−t)}dx=Γ(α)−1β−1β1−α(1β−t)1−α∫0∞{x(1β−t)}α−1exp⁡{−x(1β−t)}dx=Γ(α)−1β−α(1β−t)1−α(1β−t)−1Γ(α)∫0∞(1β−t)Γ(α)−1{ x(1β−t)}α−1exp⁡{−x(1β−t)}dx=β−α(1β−t)−α={β(1β−t)}−α=(1−tβ)−α \begin{align*} M_{x}(t) & =\mathbb{E}\left(e^{t X}\right) \\ & = \int_{0}^{\infty} e^{t x} f_{X}(x) d x \\ & = \int_{0}^{\infty} e^{t x} \Gamma(\alpha)^{-1} \beta^{-1}\left(\frac{x}{\beta}\right)^{\alpha-1} e^{-\frac{x}{\beta}} d x \\ & = \Gamma(\alpha)^{-1} \beta^{-1} \int_{0}^{\infty}\left(\frac{x}{\beta}\right)^{\alpha-1} e^{-\frac{1}{\beta} x} e^{t x} d x \\ & = \Gamma(\alpha)^{-1} \beta^{-1} \int_{0}^{\infty}\left(x \frac{1}{\beta} \frac{\frac{1}{\beta}-t}{\frac{1}{\beta}-t} \cdot \frac{\beta}{\beta}\right)^{\alpha-1} \exp \left\{-x\left(\frac{1}{\beta}-t\right)\right\} d x \\ & = \Gamma(\alpha)^{-1} \beta^{-1} \int_{0}^{\infty}\left\{x\left(\frac{1}{\beta}-t\right) \cdot\left(\frac{1}{\beta}-t\right)^{-1} \beta^{-1}\right\}^{\alpha-1} \exp \left\{-x\left(\frac{1}{\beta}-t\right)\right\} d x \\ & = \Gamma(\alpha)^{-1} \beta^{-1} \beta^{1-\alpha}\left(\frac{1}{\beta}-t\right)^{1-\alpha} \int_{0}^{\infty}\left\{x\left(\frac{1}{\beta}-t\right)\right\}^{\alpha-1} \exp \left\{-x\left(\frac{1}{\beta}-t\right)\right\} d x \\ & \left.\left.=\Gamma(\alpha)^{-1} \beta^{-\alpha}\left(\frac{1}{\beta}-t\right)^{1-\alpha}\left(\frac{1}{\beta}-t\right)^{-1} \Gamma(\alpha) \int_{0}^{\infty}\left(\frac{1}{\beta}-t\right) \Gamma(\alpha)^{-1} \right\{\, x\left(\frac{1}{\beta}-t\right)\right\}^{\alpha-1} \exp \left\{-x\left(\frac{1}{\beta}-t\right)\right\} d x \\ & = \beta^{-\alpha}\left(\frac{1}{\beta}-t\right)^{-\alpha} \\ & = \left\{\beta\left(\frac{1}{\beta}-t\right)\right\}^{-\alpha} \\ & =(1-t \beta)^{-\alpha} \end{align*}$$

積率母関数を用いた期待値の導出

$$E(X)=MX(t)′ ∣t=0=ddt(1−tβ)−α ∣t=0=−α(1−tβ)−(α+1){(1−tβ)}′ ∣t=0=−α(1−tβ)−(α+1)(−β) ∣t=0=αβ(1−tβ)−(α+1) ∣t=0=αβ(1−0⋅β)−(α+1)=αβ \begin{align*} \mathbb{E}(X) &= M_X(t)' \; |_{t=0} \\ &= \frac{d}{dt}(1-t\beta)^{-\alpha} \; |_{t=0} \\ &= -\alpha (1-t\beta)^{-(\alpha+1)} \{(1-t\beta)\}' \; |_{t=0} \\ &= -\alpha (1-t\beta)^{-(\alpha+1)} (-\beta) \; |_{t=0} \\ &= \alpha\beta (1-t\beta)^{-(\alpha+1)} \; |_{t=0} \\ &= \alpha\beta (1-0\cdot\beta)^{-(\alpha+1)} \\ &= \alpha\beta \end{align*}$$

積率母関数を用いた分散の導出

$$E(X)=MX(t)′′ ∣t=0={αβ(1−tβ)−(α+1)}′ ∣t=0=αβ(α+1)(1−tβ)−(α+2){(1−tβ)}′ ∣t=0=−αβ(α+1)(1−tβ)−(α+2)(−β) ∣t=0=−αβ(α+1)(1−tβ)−(α+2)(−β) ∣t=0=−αβ(α+1)(1−0⋅β)−(α+2)(−β)=αβ2(α+1) \begin{align*} \mathbb{E}(X) &= M_X(t)'' \; |_{t=0} \\ &= \{ \alpha\beta (1-t\beta)^{-(\alpha+1)} \}' \; |_{t=0} \\ &= \alpha\beta(\alpha+1) (1-t\beta)^{-(\alpha+2)} \{ (1-t\beta) \}' \; |_{t=0} \\ &= -\alpha\beta(\alpha+1) (1-t\beta)^{-(\alpha+2)} (-\beta) \; |_{t=0} \\ &= -\alpha\beta(\alpha+1) (1-t\beta)^{-(\alpha+2)} (-\beta) \; |_{t=0} \\ &= -\alpha\beta(\alpha+1) (1-0\cdot \beta)^{-(\alpha+2)} (-\beta) \\ &= \alpha\beta^2(\alpha+1) \end{align*}$$

$$V(X)=E(X2)−E(X)2=αβ2(α+1)−(αβ)2=αβ2 \begin{align*} \mathbb{V}(X) &= \mathbb{E}(X^2)-\mathbb{E}(X)^2 \\ &= \alpha \beta^2(\alpha+1) - (\alpha \beta)^2 \\ &= \alpha\beta^2 \end{align*}$$

再生性

ガンマ分布は，パラメータ $α\alpha$ について再生性が成立する．

$${Xi}i={1,2,…,n}, i.i.d. ∼Gam⁡(αi,β) \begin{align*} \left\{X_{i}\right\}_{i=\{1,2, \ldots, n\}}, \text { i.i.d. } \sim \operatorname{Gam}\left(\alpha_i, \beta \right) \end{align*}$$

$$M∑Xi(t)=E(et∑Xi)=E(∏etXi)=∏E(etXi) ∵Xis are i.i.d. =∏(1−tβ)−αi=(1−tβ)−∑αi \begin{align*} & M_{\sum X_{i}}(t)=\mathbb{E}\left(e^{t \sum X_{i}}\right) \\ & = \mathbb{E}\left(\prod e^{t X_{i}}\right) \\ & =\prod \mathbb{E}\left(e^{t X_{i}}\right) \; \because X_{i} \text {s are i.i.d. } \\ & =\prod \left(1-t \beta \right)^{-\alpha_{i}} \\ & =(1-t \beta)^{-\sum \alpha_{i}} \\ \end{align*}$$

以上より，

$$∑Xi∼Gamma⁡(∑αi,β) \sum X_{i} \sim \operatorname{Gamma}\left(\sum \alpha_{i}, \beta\right)$$

ただし，パラメータ $β\beta$ について再生性が成立しないので注意．

$${Xi}i={1,2,…,n}, i.i.d. ∼Gam⁡(α,βi) \begin{align*} \left\{X_{i}\right\}_{i=\{1,2, \ldots, n\}}, \text { i.i.d. } \sim \operatorname{Gam}\left(\alpha, \beta_{i}\right) \end{align*}$$

$$M∑Xi(t)=E(et∑Xi)=E(∏etXi)=∏E(etXi) ∵xi s are i.i.d. =∏(1−tβi)−α≠(1−t∑βi)−α \begin{align*} M_{\sum X_{i}}(t) & =\mathbb{E}\left(e^{t \sum X_{i}}\right) \\ & = \mathbb{E}\left(\prod e^{t X_{i}}\right) \\ & = \prod \mathbb{E}\left(e^{t X_{i}}\right) \; \because x_{i} \text { s are i.i.d. } \\ & = \prod\left(1-t \beta_{i}\right)^{-\alpha} \\ & \neq\left(1-t \sum \beta_{i}\right)^{-\alpha} \end{align*}$$

ガンマ分布を用いたベイズモデル (Poasson-Gamma Model)

以下の問題設定を考える．

$$x∣λ∼Po⁡(λ),fx∣λ(x)=(x!)−1λxe−λλ∼Gam⁡(α,β),fλ(λ)=Γ(α)−1β−1(λβ)α−1e−λβλ∣X∼Fλ∣X(λ) \begin{align*} & x \mid \lambda \sim \operatorname{Po}(\lambda), \quad f_{x \mid \lambda}(x)=(x!)^{-1} \lambda^{x} e^{-\lambda} \\ & \lambda \sim \operatorname{Gam}(\alpha, \beta), f_{\lambda}(\lambda)=\Gamma(\alpha)^{-1} \beta^{-1}\left(\frac{\lambda}{\beta}\right)^{\alpha-1} e^{-\frac{\lambda}{\beta}} \\ & \lambda \mid X \sim F_{\lambda | X}(\lambda) \end{align*}$$

事後分布の導出

$$log⁡fλ∣X=log⁡fX∣λ(X)fλ(λ)=log⁡fX∣λ+log⁡fλ(λ)=log⁡∏i=1Nfx(xi)+log⁡fλ(λ)∝∑{−log⁡(xi!)+xilog⁡λ−λ}+(α−1)log⁡λβ−λβ∝{∑xi}log⁡λ−Nλ+(α−1)log⁡λ−1βλ={α+(∑xi)−1}log⁡λ−Nβ+1βλ \begin{align*} \log f_{\lambda|\mathbf{X}} & = \log f_{\mathbf{X} | \lambda}(\mathbf{X}) f_\lambda(\lambda) \\ & =\log f_{X \mid \lambda}+\log f_{\lambda}(\lambda) \\ & =\log \prod_{i=1}^{N} f_{x}\left(x_{i}\right)+\log f_{\lambda}(\lambda) \\ & \propto \sum\left\{-\log \left(x_{i}!\right)+x_{i} \log \lambda-\lambda\right\}+(\alpha-1) \log \frac{\lambda}{\beta}-\frac{\lambda}{\beta} \\ & \propto\left\{\sum x_{i}\right\} \log \lambda-N \lambda+(\alpha-1) \log \lambda-\frac{1}{\beta} \lambda \\ & =\left\{\alpha+\left(\sum x_{i}\right)-1\right\} \log \lambda-\frac{N \beta+1}{\beta} \lambda \end{align*}$$

以上より，事後分布がいかに定まる．

$$λ∣X∼Gam⁡(α+∑xi,β(Nβ+1)−1) \lambda \mid X \sim \operatorname{Gam}\left(\alpha+\sum x_{i}, \quad \beta(N \beta+1)^{-1}\right)$$

予測分布の導出

データ $X\mathbf{X}$ を観測した後の $xx$の分布を考える．

$$X∗∣X,α′β′∼F X_{*} \mid X, \alpha^{\prime} \beta^{\prime} \sim F \\$$

このとき確率密度関数は，

$$fXk(x∗)=∫0∞fX∣λ(x)fλ(λ)dλ=∫0∞(x!)−1λxe−λΓ(α′)−1β′−1(λβ′)α−1e−λβdλ=(x!)−1Γ(α′)−1β′−1∫0∞λxe−λ(λβ′)α−1e−λβdλ=(x!)−1Γ(α′)−1β′−1∫0∞exp⁡{−λ(1+1β′)}β′x(λβ′)x(λβ′)α−1dλ=(x!)−1Γ(α′)−1β′x−1∫0∞exp⁡{−λ(1+1β′)}(λ1β′)α+x−1dλ=(x!)−1Γ(α′)−1β′x−1∫0∞exp⁡{−λ(1+1β′)}(λ1β′1+1β′1+1β′⋅β′β′)α+x−1dλ=(x!)−1Γ(α′)−1β′x−1∫0∞exp⁡{−λ(1+1β′)}λ(1+1β′)(β′+1)−1}α+x−1dλ=(x!)−1Γ(α′)−1β′x−1(β′+1)1−α−x∫0∞exp⁡{−λ(1+1β)}{λ(1+1β′)}α+x−1dλ=(x!)−1Γ(α′)−1β′x−1(β′+1)1−α−xΓ(α′+x)(1+1β′)−1∫0∞Γ(α′+x)−1(1+1β′)exp⁡{−λ(1+1β′)}{ λ(1+1β′)}α+x−1dλ=(x!)−1Γ(α′)−1Γ(α′+x)β′x−1(β′+1)1−α−x(β′β′+1)=(x!)−1Γ(α′)−1Γ(α′+x)β′(β′+1)−(α+x) \begin{align*} f_{X_{k}}\left(x_{*}\right) & = \int_{0}^{\infty} f_{X \mid \lambda}(x) f_{\lambda}(\lambda) d \lambda \\ & = \int_{0}^{\infty}(x!)^{-1} \lambda^{x} e^{-\lambda} \Gamma\left(\alpha^{\prime}\right)^{-1} \beta^{\prime-1}\left(\frac{\lambda}{\beta^{\prime}}\right)^{\alpha-1} e^{-\frac{\lambda}{\beta}} d \lambda \\ & = (x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \beta^{\prime-1} \int_{0}^{\infty} \lambda^{x} e^{-\lambda}\left(\frac{\lambda}{\beta^{\prime}}\right)^{\alpha-1} e^{-\frac{\lambda}{\beta}} d \lambda \\ & = (x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \beta^{\prime-1} \int_{0}^{\infty} \exp \left\{-\lambda \left(1+\frac{1}{\beta^{\prime}}\right) \right\} \beta^{\prime x}\left(\frac{\lambda}{\beta^{\prime}}\right)^{x}\left(\frac{\lambda}{\beta^{\prime}}\right)^{\alpha-1} d \lambda \\ & = (x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \beta^{\prime x-1} \int_{0}^{\infty} \exp \left\{-\lambda\left(1+\frac{1}{\beta^{\prime}}\right)\right\}\left(\lambda \frac{1}{\beta^{\prime}}\right)^{\alpha+x-1} d \lambda \\ = & (x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \beta^{\prime x-1} \int_{0}^{\infty} \exp \left\{-\lambda\left(1+\frac{1}{\beta^{\prime}}\right)\right\}\left(\lambda \frac{1}{\beta^{\prime}} \frac{1+\frac{1}{\beta^{\prime}}}{1+\frac{1}{\beta^{\prime}}} \cdot \frac{\beta^{\prime}}{\beta^{\prime}}\right)^{\alpha+x-1} d \lambda \\ = & \left.(x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \beta^{\prime x-1} \int_{0}^{\infty} \exp \left\{-\lambda\left(1+\frac{1}{\beta^{\prime}}\right)\right\} \lambda\left(1+\frac{1}{\beta^{\prime}}\right)\left(\beta^{\prime}+1\right)^{-1}\right\}^{\alpha+x-1} d \lambda \\ = & (x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \beta^{\prime x-1}\left(\beta^{\prime}+1\right)^{1-\alpha-x} \int_{0}^{\infty} \exp \left\{-\lambda\left(1+\frac{1}{\beta}\right)\right\}\left\{\lambda\left(1+\frac{1}{\beta^{\prime}}\right)\right\}^{\alpha+x-1} d \lambda \\ = & (x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \beta^{\prime x-1}\left(\beta^{\prime}+1\right)^{1-\alpha-x} \Gamma\left(\alpha^{\prime}+x\right)\left(1+\frac{1}{\beta^{\prime}}\right)^{-1} \\ & \left.\left.\int_{0}^{\infty} \Gamma\left(\alpha^{\prime}+x\right)^{-1}\left(1+\frac{1}{\beta^{\prime}}\right) \exp \left\{-\lambda\left(1+\frac{1}{\beta^{\prime}}\right)\right\} \right\{\, \lambda\left(1+\frac{1}{\beta^{\prime}}\right)\right\}^{\alpha+x-1} d \lambda \\ = & (x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \Gamma\left(\alpha^{\prime}+x\right) \beta^{\prime} x-1\left(\beta^{\prime}+1\right)^{1-\alpha-x}\left(\frac{\beta^{\prime}}{\beta^{\prime}+1}\right) \\ = & (x!)^{-1} \Gamma\left(\alpha^{\prime}\right)^{-1} \Gamma\left(\alpha^{\prime}+x\right) \beta^{\prime}\left(\beta^{\prime}+1\right)^{-(\alpha+x)} \end{align*}$$

この分布をPoisson-Gamma分布という．負の二項分布に対応していることに注意する．最後にこの分布の期待値と分散を求める．

$$E(X∗)=Eλ[EX∗∣λ(X∗∣λ)]=Eλ[λ]=αβE(X∗2)=Eλ[EX∗∣λ(X∗2∣λ)]=Eλ[λ2+λ]=Ex[λ2]+Eλ[λ]=αβ2(1+α)+αβV(X∗2)=αβ2(1+α)+αβ−α2β2=αβ2+αβ=αβ2(1+β) \begin{align*} \mathbb{E}\left(X_{*}\right) & =\mathbb{E}_{\lambda}\left[\mathbb{E}_{X_{*} \mid \lambda}\left(X_{*} \mid \lambda\right)\right] \\ & =\mathbb{E}_{\lambda}[\lambda] \\ & =\alpha \beta \\ \mathbb{E}\left(X_{*}{ }^{2}\right) & =\mathbb{E}_{\lambda}\left[\mathbb{E}_{X_{*} \mid \lambda}\left(X_{*}{ }^{2} \mid \lambda\right)\right] \\ & =\mathbb{E}_{\lambda}\left[\lambda^{2}+\lambda\right] \\ & =\mathbb{E}_{x}\left[\lambda^{2}\right]+\mathbb{E}_{\lambda}[\lambda] \\ & =\alpha \beta^{2}(1+\alpha)+\alpha \beta \\ V\left(X_{*}{ }^{2}\right) & =\alpha \beta^{2}(1+\alpha)+\alpha \beta-\alpha^{2} \beta^{2} \\ & =\alpha \beta^{2}+\alpha \beta \\ & =\alpha \beta^{2}(1+\beta) \end{align*}$$

参考文献

(1) 竹村彰通.”新装改訂版 現代数理統計学”.2020.学術図書出版社

(2) 久保川."現代数理統計学の基礎".2017.共立出版

(3) C.M.ビショップ.”パターン認識と機械学習 上 ベイズ理論による統計的予測”.2019.丸善出版

(4) 須山敦志.”機械学習スタートアップシリーズ ベイズ推論による機械学習入門”.2018.講談社サイエンティフィク

更新

2024.08.14: 構成を一部変更，再生性，ベイズモデルの章を追加