[1]
\(E[X]\)
\(\displaystyle E[X] = \sum _{k=0}^\infty kP(X=k)\)
\(\displaystyle = \sum _{k=1}^\infty kP(X=k)\) (\(k=0\)の時は\(0\)のため除外)
\(\displaystyle = \sum _{k=1}^\infty \frac{\lambda^k}{(k-1)!}e^{-\lambda}\)
\(\displaystyle = \lambda \sum _{k=1}^\infty \frac{\lambda^{k-1}}{(k-1)!}e^{-\lambda}\)
\(\displaystyle = \lambda\) (和の部分は、ポアソン分布の全確率のため)
\(V[X]\)
離散型確率分布の時にあるあるの\(X(X-1)\)を掛けます。
\(\displaystyle E[X(X-1)] = \sum _{k=0}^\infty k(k-1)P(X=k)\)
\(\displaystyle = \sum _{k=2}^\infty k(k-1)P(X=k)\) (\(k=0,1\)の時は\(0\)のため除外)
\(\displaystyle = \sum _{k=2}^\infty \frac{\lambda^k}{(k-2)!}e^{-\lambda}\)
\(\displaystyle = \lambda^2 \sum _{k=2}^\infty \frac{\lambda^{k-2}}{(k-1)!}e^{-\lambda}\)
\(\displaystyle = \lambda^2\) (和の部分は、ポアソン分布の全確率のため)
ここで、
\(V[X] = E[X^2]-E[X]^2\)
\( = E[X(X-1)]+E[X]-E[X]^2\)
\( = \lambda ^2+\lambda -\lambda^2 = \lambda\)
[2]
\(E[\Lambda]\)
\(\displaystyle E[\Lambda] = \int^\infty _0 \lambda g(\lambda) d\lambda\)
\(\displaystyle = \int^\infty _0 \lambda \frac{\beta^\alpha}{\Gamma (\alpha)} \lambda ^{\alpha -1}e^{-\beta \lambda} d\lambda\)
\(\displaystyle = \int^\infty _0 \frac{\beta^\alpha}{\Gamma (\alpha)} \lambda ^{\alpha}e^{-\beta \lambda} d\lambda\)
ここで、\(\beta \lambda = t\)と置換を行い、
\(\displaystyle = \frac{1}{\beta}\int^\infty _0 \frac{\beta^\alpha}{\Gamma (\alpha)} \left(\frac{t}{\beta}\right) ^{\alpha}e^{-t} dt\)
\(\displaystyle = \frac{1}{\beta\Gamma (\alpha)}\int^\infty _0 t ^{\alpha}e^{-t} dt\)
\(\displaystyle = \frac{1}{\beta\Gamma (\alpha)}\Gamma (\alpha +1)\)
\(\displaystyle = \frac{\alpha}{\beta}\)
\(V[\Lambda]\)
まず、\(E[\Lambda^2]\)を計算していくが途中まで\(E[\Lambda]\)と同様の計算を行い、
\(\displaystyle E[\Lambda^2] = \int^\infty _0 \lambda ^2 g(\lambda) d\lambda\)
\(\displaystyle = \int^\infty _0 \frac{\beta^\alpha}{\Gamma (\alpha)} \lambda ^{\alpha+1}e^{-\beta \lambda} d\lambda\)
\(\displaystyle = \frac{1}{\beta^2 \Gamma (\alpha)}\int^\infty _0 t ^{\alpha+1}e^{-t} dt\)
\(\displaystyle = \frac{1}{\beta^2\Gamma (\alpha)}\Gamma (\alpha +2)\)
\(\displaystyle = \frac{\alpha (\alpha +1)}{\beta^2}\)
ここで、
\(V[\Lambda] = E[\Lambda^2]-E[\Lambda]^2\)
\(\displaystyle = \frac{\alpha (\alpha +1)}{\beta^2} -\frac{\alpha^2}{\beta^2} \)\(\displaystyle = \frac{\alpha}{\beta^2} \)
[3]
\(\Lambda\)が確率変数であると条件が加わったことにより、[1]の確率密度関数は\(\Lambda = \lambda\)の条件付き確率密度関数となり、
\(P(X=k|\Lambda=\lambda) = \displaystyle \frac{\lambda^k}{k!}e^{-\lambda}\)
ここで、求めたいものは\(X,\Lambda\)の同時確率分布の\(\lambda\)で積分した周辺確率で、
\(P(X=k) = \displaystyle \int^\infty_0 P(X=k,\Lambda = \lambda) d\lambda\)
\(= \displaystyle \int^\infty_0 P(X=k|\Lambda = \lambda)P(\Lambda = \lambda) d\lambda\)
\(= \displaystyle \int^\infty_0 \frac{\lambda^k}{k!}e^{-\lambda} \frac{\beta^\alpha}{\Gamma (\alpha)} \lambda ^{\alpha -1}e^{-\beta \lambda}d\lambda\)
\(= \displaystyle\frac{\beta^\alpha}{k! \Gamma (\alpha)} \int^\infty_0 \lambda ^{k+\alpha -1}e^{-(\beta+1) \lambda}d\lambda\)
\(= \displaystyle\frac{\Gamma(k+\alpha)}{k! \Gamma (\alpha)} \frac{\beta^\alpha}{(\beta+1)^{k+\alpha}}\)
[4]
\(E[X]\)
\(\displaystyle E[X] = \sum _{k=0}^\infty kP(X=k)\)
\(\displaystyle = \sum _{k=1}^\infty kP(X=k)\) (\(k=0\)の時は\(0\)のため除外)
\(\displaystyle = \sum _{k=1}^\infty \frac{\Gamma(k+\alpha)}{(k-1)! \Gamma (\alpha)} \frac{\beta^\alpha}{(\beta+1)^{k+\alpha}}\)
\(k-1 = l\)で置換して、
\(\displaystyle = \sum _{l=0}^\infty \frac{\Gamma(l+1+\alpha)}{l! \Gamma (\alpha)} \frac{\beta^\alpha}{(\beta+1)^{l+1+\alpha}}\)
全確率の形を作るために、\(\Gamma (\alpha) \rightarrow \Gamma (\alpha+1)\)\(,\beta^\alpha \rightarrow \beta^{\alpha+1}\)となるように調整して、
\(\displaystyle = \frac{\alpha}{\beta}\sum _{l=0}^\infty \frac{\Gamma(l+1+\alpha)}{l! \Gamma (\alpha+1)} \frac{\beta^{\alpha+1}}{(\beta+1)^{l+1+\alpha}}\)
\(\displaystyle = \frac{\alpha}{\beta}\)
\(V[X]\)
まず、\(E[X(X-1)]\)を計算していくが途中まで\(E[\Lambda]\)と同様の計算を行い、
\(\displaystyle E[X(X-1)] = \sum _{k=0}^\infty k(k-1)P(X=k)\)
\(\displaystyle = \sum _{k=2}^\infty k(k-1)P(X=k)\) (\(k=0,1\)の時は\(0\)のため除外)
\(\displaystyle = \sum _{k=2}^\infty \frac{\Gamma(k+\alpha)}{(k-2)! \Gamma (\alpha)} \frac{\beta^\alpha}{(\beta+1)^{k+\alpha}}\)
\(k-2 = m\)で置換して、
\(\displaystyle = \sum _{m=0}^\infty \frac{\Gamma(m+2+\alpha)}{m! \Gamma (\alpha)} \frac{\beta^\alpha}{(\beta+1)^{m+2+\alpha}}\)
全確率の形を作るために、\(\Gamma (\alpha) \rightarrow \Gamma (\alpha+2)\)\(,\beta^\alpha \rightarrow \beta^{\alpha+2}\)となるように調整して、
\(\displaystyle = \frac{\alpha(\alpha +1)}{\beta^2}\sum _{m=0}^\infty \frac{\Gamma(m+2+\alpha)}{m! \Gamma (\alpha+2)} \frac{\beta^{\alpha+2}}{(\beta+1)^{m+2+\alpha}}\)
\(\displaystyle = \frac{\alpha(\alpha +1)}{\beta^2}\)
ここで、
\(V[X] = E[X^2]-E[X]^2\)
\( = E[X(X-1)]+E[X]-E[X]^2\)
\( \displaystyle = \frac{\alpha(\alpha +1)}{\beta^2} +\left(\frac{\alpha}{\beta}\right) -\left(\frac{\alpha}{\beta}\right)^2 \)\(\displaystyle = \frac{\alpha}{\beta}\left(1+\frac{1}{\beta}\right)\)
[5]
\(n\)個のデータから得られた平均、分散を\(\bar{X}\)\(,S^2\)とすると、
\( \left\{ \begin{align} &\bar{X} =\frac{\alpha}{\beta} \\ & S^2 = \frac{\alpha}{\beta}\left(1+\frac{1}{\beta}\right) \end{align} \right. \)
これを解いて、
\( \left\{ \begin{align} \hat{\alpha} = \frac{\bar{X}^2}{S^2-\bar{X}} \\ \hat{\beta} = \frac{\bar{X}}{S^2-\bar{X}} \end{align} \right. \)
また、これらが正となる条件は、
\(\bar{X} < S^2\)
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