統計学実践ワークブックの行間埋め　第29章

\(f(x)=A\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\)として、定数\(A\)を求める。 \begin{eqnarray} \int_{-\infty}^c f(x)dx &=& \int_{-\infty}^c A\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx \\ \\ &=& \int_{-\infty}^{\frac{c-\mu}{\sigma}} A\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{z^2}{2}\right)(\sigma dz)&...\frac{x-\mu}{\sigma}=zとした \\ \\ &=& A\int_{-\infty}^{\frac{c-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)dz \\ \\ &=& A\Phi(\frac{c-\mu}{\sigma}) \\ \\ &=& 1 \\ \\ \Leftrightarrow A&=&\frac{1}{\Phi(\frac{c-\mu}{\sigma})} \\ \Rightarrow f(x)&=&\frac{1}{\Phi(\frac{c-\mu}{\sigma})}\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\;(x\leq c) \end{eqnarray}

\begin{eqnarray} E[Z|Z\leq a] &=& \int_{-\infty}^{a} z\frac{\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)}{\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz}dz \\ \\ &=& \frac{\int_{-\infty}^{a} z\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)dz}{\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)dz} \\ \\ &=& \frac{\left[-\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)\right]_{-\infty}^{a}}{\Phi(a)} \\ \\ &=& \frac{-\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{a^2}{2}\right)}{\Phi(a)} \\ \\ &=& -\frac{\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{a^2}{2}\right)}{\Phi(a)} \\ \\ &=& -\frac{\varphi(a)}{\Phi(a)} \\ \\ \end{eqnarray}

\begin{eqnarray} V[Z|Z\leq a] &=& E[Z^2|Z\leq a]-(E[Z|Z\leq a])^2 \\ \\ &=& \int_{-\infty}^{a} z^2\frac{\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)}{\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz}dz-\left(-\frac{\varphi(a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\int_{-\infty}^{a} z^2\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz-\left(\frac{\varphi(a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\left(\left[\frac{1}{\sqrt{2\pi}}z\left(-\exp\left(-\frac{z^2}{2}\right)\right)\right]_{-\infty}^a+\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz\right)-\left(\frac{\varphi(a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\left(-a\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{a^2}{2}\right)+\Phi(a)\right)-\left(\frac{\varphi(a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\left(-a\varphi (a)+\Phi(a)\right)-\left(\frac{\varphi(a)}{\Phi(a)}\right)^2 \\ \\ &=& -\frac{a\varphi (a)}{\Phi(a)}+1-\left(\frac{\varphi(a)}{\Phi(a)}\right)^2 \\ \\ &=& 1-\frac{a\varphi (a)}{\Phi(a)}-\left(\frac{\varphi(a)}{\Phi(a)}\right)^2 \\ \\ \end{eqnarray}

\begin{eqnarray} E[X|X\leq c] &=& \int_{-\infty}^{c} x\frac{\frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}{\int_{-\infty}^{c} \frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx}dx \\ \\ &=& \frac{\int_{-\infty}^{a} (\sigma z+\mu)\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{z^2}{2}\right)(\sigma dz)}{\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{z^2}{2}\right)(\sigma dz)} &...\frac{x-\mu}{\sigma}=zとし、a=\frac{c-\mu}{\sigma}とした。\\ \\ &=& \frac{\int_{-\infty}^{a} (\sigma z+\mu)\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz}{\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)dz} \\ \\ &=& \frac{\int_{-\infty}^{a} (\sigma z+\mu)\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz}{\Phi(a)} \\ \\ &=& \frac{\int_{-\infty}^{a} \sigma z\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz+\int_{-\infty}^{a} \mu\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz}{\Phi(a)} \\ \\ &=& \frac{-\sigma \varphi (a)+\mu\Phi(a)}{\Phi(a)} \\ \\ &=& \mu-\sigma\frac{ \varphi (a)}{\Phi(a)} \\ \\ \end{eqnarray}

\begin{eqnarray} V[X|X\leq c] &=& E[X^2|X\leq c]-(E[X|X\leq c])^2 \\ \\ &=& \int_{-\infty}^{c} x^2\frac{\frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}{\int_{-\infty}^{c} \frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx}dx-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \int_{-\infty}^{a} (\sigma z+\mu)^2\frac{\frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{z^2}{2}\right)}{\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{z^2}{2}\right)(\sigma dz)}(\sigma dz)-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 &...\frac{x-\mu}{\sigma}=zとし、a=\frac{c-\mu}{\sigma}とした。\\ \\ &=& \int_{-\infty}^{a} (\sigma z+\mu)^2\frac{\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)}{\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz}dz-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \int_{-\infty}^{a} (\sigma z+\mu)^2\frac{\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)}{\Phi(a)}dz-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\int_{-\infty}^{a} (\sigma^2 z^2+2\sigma z\mu+\mu^2)\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\int_{-\infty}^{a} (\sigma^2 z^2+2\sigma z\mu+\mu^2)\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\left(\int_{-\infty}^{a} \sigma^2 z^2\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz+\int_{-\infty}^{a} 2\sigma z\mu\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz+\int_{-\infty}^{a} \mu^2\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz\right)-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\left(\sigma^2\left(\left[-z\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)\right]_{-\infty}^a+\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz\right)+2\sigma \mu\left[-\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)\right]_{-\infty}^a+\mu^2\Phi(a)\right)-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\left(\sigma^2\left(-a\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{a^2}{2}\right)+\Phi(a)\right)-2\sigma \mu\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{a^2}{2}\right)+\mu^2\Phi(a)\right)-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \frac{1}{\Phi(a)}\left(\sigma^2\left(-a\varphi(a)+\Phi(a)\right)-2\sigma \mu\varphi(a)+\mu^2\Phi(a)\right)-\left(\mu-\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& -\frac{\sigma^2a\varphi(a)}{\Phi(a)}+\frac{\sigma^2\Phi(a)}{\Phi(a)}-\frac{2\sigma\mu\varphi(a)}{\Phi(a)}+\frac{\mu^2\Phi(a)}{\Phi(a)}-\left(\mu^2-2\mu\sigma\frac{ \varphi (a)}{\Phi(a)}+\left(\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2\right) \\ \\ &=& -\frac{\sigma^2a\varphi(a)}{\Phi(a)}+\sigma^2+\mu^2-\mu^2-\left(\sigma\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& -\frac{\sigma^2a\varphi(a)}{\Phi(a)}+\sigma^2-\sigma^2\left(\frac{ \varphi (a)}{\Phi(a)}\right)^2 \\ \\ &=& \left(1-\frac{a\varphi(a)}{\Phi(a)}-\left(\frac{ \varphi (a)}{\Phi(a)}\right)^2\right)\sigma^2 \\ \\ \end{eqnarray}

\begin{eqnarray} E[X|c\lt X] &=& \int_{c}^{\infty} x\frac{\frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}{\int_{c}^{\infty} \frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx}dx \\ \\ &=& \int_{c}^{\infty} x\frac{\frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}{\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx-\int_{-\infty}^{c} \frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)dx}dx \\ \\ &=& \int_{a}^{\infty} (\sigma z+\mu)\frac{\frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{z^2}{2}\right)}{1-\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi\sigma^2} }\exp\left(-\frac{z^2}{2}\right)(\sigma dz)}(\sigma dz) &...\frac{x-\mu}{\sigma}=zとし、a=\frac{c-\mu}{\sigma}とした。\\ \\ &=& \int_{a}^{\infty} (\sigma z+\mu)\frac{\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)}{1-\int_{-\infty}^{a} \frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz}dz \\ \\ &=& \int_{a}^{\infty} (\sigma z+\mu)\frac{\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)}{1-\Phi(a)}dz \\ \\ &=& \frac{1}{1-\Phi(a)}\left(\int_{-\infty}^{\infty} (\sigma z+\mu)\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz-\int_{-\infty}^{a} (\sigma z+\mu)\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz\right) \\ \\ &=& \frac{1}{1-\Phi(a)}\left(0+\mu-\sigma\int_{-\infty}^{a} z\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz-\mu\int_{-\infty}^{a}\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)dz\right) \\ \\ &=& \frac{1}{1-\Phi(a)}\left(\mu-\sigma\left[-\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{z^2}{2}\right)\right]_{-\infty}^{a}-\mu\Phi(a)\right) \\ \\ &=& \frac{1}{1-\Phi(a)}\left(\mu+\sigma\frac{1}{\sqrt{2\pi} }\exp\left(-\frac{a^2}{2}\right)-\mu\Phi(a)\right) \\ \\ &=& \frac{1}{1-\Phi(a)}\left(\mu(1-\Phi(a))+\sigma\varphi(a)\right) \\ \\ &=& \mu+\frac{1}{1-\Phi(a)}\left(\sigma\varphi(a)\right) \\ \\ &=& \mu+\frac{\varphi(a)}{1-\Phi(a)}\sigma\\ \\ \end{eqnarray}

対数尤度関数\(l(\mu)\)を\(\mu\)で微分して0と置くことで求められる。 \begin{eqnarray} l(\mu) &=& \log L(\mu) \\ \\ &=& \log \displaystyle\prod_{i=1}^m (\frac{1}{\Phi((c-\mu)/\sigma)}\frac{1}{\sqrt{2\pi}\sigma }\exp \left(-\frac{(x_i-\mu)^2}{2\sigma^2}\right)) \\ \\ &=& \displaystyle\sum_{i=1}^m\log (\frac{1}{\Phi((c-\mu)/\sigma)}\frac{1}{\sqrt{2\pi}\sigma }\exp \left(-\frac{(x_i-\mu)^2}{2\sigma^2}\right)) \\ \\ &=& \displaystyle\sum_{i=1}^m\left( -\log \Phi((c-\mu)/\sigma)-\log(\sqrt{2\pi}\sigma)-\frac{(x_i-\mu)^2}{2\sigma^2}\right) \\ \\ \frac{\partial}{\partial \mu}l(\mu) &=& \displaystyle\sum_{i=1}^m\left( \frac{1}{\sigma}\frac{\varphi((c-\mu)/\sigma)}{\Phi((c-\mu)/\sigma)} -0+\frac{x_i-\mu}{\sigma^2}\right) \\ \\ &=& \frac{1}{\sigma^2}\displaystyle\sum_{i=1}^m\left( \sigma\frac{\varphi((c-\mu)/\sigma)}{\Phi((c-\mu)/\sigma)} +x_i-\mu\right) \\ \\ &=& \frac{1}{\sigma^2}\left( m\sigma\frac{\varphi((c-\mu)/\sigma)}{\Phi((c-\mu)/\sigma)} +\displaystyle\sum_{i=1}^mx_i-m\mu\right) \\ \\ &=& \frac{m}{\sigma^2}\left( \sigma\frac{\varphi((c-\mu)/\sigma)}{\Phi((c-\mu)/\sigma)} +\frac{1}{m}\displaystyle\sum_{i=1}^mx_i-\mu\right) \\ \\ &=& 0 \\ \\ \Leftrightarrow \mu&=&\frac{1}{m}\displaystyle\sum_{i=1}^mx_i+\sigma\frac{\varphi((c-\mu)/\sigma)}{\Phi((c-\mu)/\sigma)} \\ \\ &=&\overline{x}+\sigma\frac{\varphi(a)}{\Phi(a)} \end{eqnarray}

p.8と第六章の該当箇所 を参考 \begin{eqnarray} f_{1}(x) &=& \int_{-\infty}^{\infty} \frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp \left( -\frac{1}{2(1-\rho^2)}\left( \left(\frac{x-\mu_X}{\sigma_X} \right)^2-2\rho\left(\frac{x-\mu_X}{\sigma_X}\right)\left(\frac{y-\mu_Y}{\sigma_Y}\right) +\left(\frac{Y-\mu_Y}{\sigma_Y} \right)^2\right) \right)dy \\ \\ &=& \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\int_{-\infty}^{\infty} \exp \left( -\frac{1}{2(1-\rho^2)}\left( \left(\frac{x-\mu_X}{\sigma_X} \right)^2-2\rho\left(\frac{x-\mu_X}{\sigma_X}\right)X +Y^2\right) \right)(\sigma _Y dY)&...\left(Y=\frac{y-\mu_Y}{\sigma_Y},\;dy=\sigma_Y dYを利用\right) \\ \\ &=& \frac{1}{2\pi\sigma_1\sqrt{1-\rho^2}}\int_{-\infty}^{\infty} \exp \left( -\frac{1}{2(1-\rho^2)}\left( \left( Y-\rho\left(\frac{x_1-\mu_1}{\sigma_1}\right) \right)^2+\left(\frac{x-\mu_X}{\sigma_X} \right)^2-\left(\rho\left(\frac{x-\mu_X}{\sigma_X}\right)\right)^2\right) \right)dY \\ \\ &=& \frac{1}{2\pi\sigma_1\sqrt{1-\rho^2}}\int_{-\infty}^{\infty} \exp \left( -\frac{1}{2(1-\rho^2)}\left( \left( Y-\rho\left(\frac{x-\mu_X}{\sigma_X}\right) \right)^2\right)\exp\left(-\frac{(1-\rho^2)}{2(1-\rho^2)}\left(\frac{x-\mu_X}{\sigma_X} \right)^2\right) \right)dY \\ \\ &=& \frac{1}{2\pi\sigma_1\sqrt{1-\rho^2}}\cdot \sqrt{2\pi(1-\rho^2)}\exp\left(-\frac{1}{2}\left(\frac{x-\mu_X}{\sigma_X} \right)^2\right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\sigma_X}\exp\left(-\frac{(x-\mu_X)^2}{2\sigma_X^2} \right) \end{eqnarray}

\begin{eqnarray} f_2(y|x) &=& \frac{f(x,y)}{f(x)} \\ \\ &=& \frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp \left( -\frac{1}{2(1-\rho^2)}\left( \left(\frac{x-\mu_X}{\sigma_X} \right)^2-2\rho\left(\frac{x-\mu_X}{\sigma_X}\right)\left(\frac{y-\mu_Y}{\sigma_Y}\right) +\left(\frac{y-\mu_Y}{\sigma_Y} \right)^2\right) \right)}{\frac{1}{\sqrt{2\pi\sigma_X^2}}\exp\left(-\frac{(x-\mu_X)^2}{2\sigma_X^2} \right)} \\ \\ &=& \frac{\sqrt{2\pi}\sigma_X}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp \left( -\frac{1}{2(1-\rho^2)}\left( \left(\frac{x-\mu_X}{\sigma_X} \right)^2-2\rho\left(\frac{x-\mu_X}{\sigma_X}\right)\left(\frac{y-\mu_Y}{\sigma_Y}\right) +\left(\frac{y-\mu_Y}{\sigma_Y}\right)^2-(1-\rho^2)\left(\frac{x-\mu_X}{\sigma_X^2}\right)^2\right) \right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\sigma_Y\sqrt{1-\rho^2}}\exp \left( -\frac{1}{2(1-\rho^2)}\left( \rho^2\left(\frac{x-\mu_X}{\sigma_X} \right)^2-2\rho\left(\frac{x-\mu_X}{\sigma_X}\right)\left(\frac{y-\mu_Y}{\sigma_Y}\right) +\left(\frac{y-\mu_Y}{\sigma_Y} \right)^2\right) \right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\sigma_Y\sqrt{1-\rho^2}}\exp \left( -\frac{1}{2(1-\rho^2)}\left( \left( \frac{y-\mu_Y}{\sigma_Y}\right)-\rho\left(\frac{x-\mu_X}{\sigma_X}\right) \right)^2\right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\sigma_Y\sqrt{1-\rho^2}}\exp \left( -\frac{1}{2(1-\rho^2)\sigma_Y^2}\left( y-\mu_Y-\rho\frac{\sigma_Y}{\sigma_X}\left(x-\mu_X\right) \right)^2\right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\sqrt{\sigma_Y^2(1-\rho^2)}}\exp \left( -\frac{1}{2(1-\rho^2)\sigma_Y^2}\left( y-\mu_Y-\frac{\sigma_{XY}}{\sigma_X\sigma_Y}\frac{\sigma_Y}{\sigma_X}\left(x-\mu_X\right) \right)^2\right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\sqrt{\tau^2}}\exp \left( -\frac{1}{2\tau^2}\left( y-\mu_Y-\beta\left(x-\mu_X\right) \right)^2\right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\tau^2}\exp \left( -\frac{1}{2\tau^2}\left( y-(\mu_Y-\beta\mu_X)-\beta x \right)^2\right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\tau^2}\exp \left( -\frac{1}{2\tau^2}\left( y-\alpha-\beta x \right)^2\right) \\ \\ &=& \frac{1}{\sqrt{2\pi}\tau^2}\exp \left( -\frac{\left( y-(\alpha+\beta x) \right)^2}{2\tau^2}\right) \\ \\ \end{eqnarray}

p.60参考

対数尤度関数\(l(\alpha,\beta,\tau)\)を求め、最尤推定量を求める。 \begin{eqnarray} l(\alpha,\beta,\tau) &=& \log \displaystyle\prod_{i=1}^m f(y_i|x_i) \\ \\ &=& \log \displaystyle\prod_{i=1}^m \frac{1}{\sqrt{2\pi}\tau}\exp \left( -\frac{\left( y_i-(\alpha+\beta x_i) \right)^2}{2\tau^2}\right) \\ \\ &=& \displaystyle\sum_{i=1}^m \log \frac{1}{\sqrt{2\pi\tau^2}}\exp \left( -\frac{\left( y_i-(\alpha+\beta x_i) \right)^2}{2\tau^2}\right) \\ \\ &=& \displaystyle\sum_{i=1}^m \left( -\log\sqrt{2\pi\tau^2} -\frac{\left( y_i-(\alpha+\beta x_i) \right)^2}{2\tau^2}\right) \\ \\ &=& \displaystyle\sum_{i=1}^m \left( -\log\sqrt{2\pi}-\log\sqrt{\tau^2} -\frac{\left( y_i-(\alpha+\beta x_i) \right)^2}{2\tau^2}\right) \\ \\ &=& \displaystyle\sum_{i=1}^m \left( -\log\sqrt{2\pi}-\frac{1}{2}\log\tau^2 -\frac{\left( y_i-(\alpha+\beta x_i) \right)^2}{2\tau^2}\right) \\ \\ &\Rightarrow& \left\{ \begin{array}{l} \frac{\partial}{\partial \alpha}l(\alpha,\beta,\tau)&=&0 \\ \\ \frac{\partial}{\partial \beta}l(\alpha,\beta,\tau)&=&0 \\ \\ \frac{\partial}{\partial (\tau^2)}l(\alpha,\beta,\tau)&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \displaystyle\sum_{i=1}^m \frac{(y_i-(\alpha+\beta x_i))}{\tau^2}&=&0 \\ \\ \displaystyle\sum_{i=1}^m \frac{x_i(y_i-(\alpha+\beta x_i))}{\tau^2}&=&0 \\ \\ \displaystyle\sum_{i=1}^m -\frac{1}{2\tau^2}+\frac{(y_i-(\alpha+\beta x_i))^2}{2(\tau^2)^2}&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \displaystyle\sum_{i=1}^m (y_i-(\alpha+\beta x_i))&=&0 \\ \\ \displaystyle\sum_{i=1}^m x_i(y_i-(\alpha+\beta x_i))&=&0 \\ \\ \displaystyle\sum_{i=1}^m -\tau^2+(y_i-(\alpha+\beta x_i))^2&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \displaystyle\sum_{i=1}^m (y_i-\beta x_i)-m\alpha&=&0 \\ \\ \displaystyle\sum_{i=1}^m x_i(y_i-(\alpha+\beta x_i))&=&0 \\ \\ \displaystyle\sum_{i=1}^m -\tau^2+(y_i-(\alpha+\beta x_i))^2&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\beta \overline{x}-\alpha&=&0 \\ \\ \displaystyle\sum_{i=1}^m x_i(y_i-(\alpha+\beta x_i))&=&0 \\ \\ \displaystyle\sum_{i=1}^m -\tau^2+(y_i-(\alpha+\beta x_i))^2&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\beta \overline{x}-\alpha&=&0 \\ \\ \displaystyle\sum_{i=1}^m x_i(y_i-(\overline{y}-\beta \overline{x}+\beta x_i))&=&0 \\ \\ \displaystyle\sum_{i=1}^m -\tau^2+(y_i-(\overline{y}-\beta \overline{x}+\beta x_i))^2&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\beta \overline{x}-\alpha&=&0 \\ \\ \displaystyle\sum_{i=1}^m x_i((y_i-\overline{y})-\beta (x_i-\overline{x}))&=&0 \\ \\ \displaystyle\sum_{i=1}^m -\tau^2+((y_i-\overline{y})-\beta (x_i-\overline{x}))^2&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\beta \overline{x}-\alpha&=&0 \\ \\ \displaystyle\sum_{i=1}^m (x_i-\overline{x})((y_i-\overline{y})-\beta (x_i-\overline{x}))&=&\displaystyle\sum_{i=1}^m -\overline{x}((y_i-\overline{y})-\beta (x_i-\overline{x})) \\ \\ -\tau^2m+\displaystyle\sum_{i=1}^m ((y_i-\overline{y})-\beta (x_i-\overline{x}))^2&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\beta \overline{x}-\alpha&=&0 \\ \\ \frac{1}{m}\displaystyle\sum_{i=1}^m (x_i-\overline{x})(y_i-\overline{y})-\beta (x_i-\overline{x})^2&=& -\overline{x}((\overline{y}-\overline{y})-\beta (\overline{x}-\overline{x})) \\ \\ -\tau^2+\frac{1}{m}\displaystyle\sum_{i=1}^m (y_i-\overline{y})^2-2\beta (x_i-\overline{x})(y_i-\overline{y})+\beta^2 (x_i-\overline{x})^2&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\beta \overline{x}-\alpha&=&0 \\ \\ \frac{1}{m}\displaystyle\sum_{i=1}^m (x_i-\overline{x})(y_i-\overline{y})-\beta (x_i-\overline{x})^2&=& -\overline{x}((\overline{y}-\overline{y})-\beta (\overline{x}-\overline{x})) \\ \\ -\tau^2+ s_Y^2-2\beta s_{XY}+\beta^2 s_X^2&=&0 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\beta \overline{x}-\alpha&=&0 \\ \\ s_{XY}-\beta s_{X}^2&=& 0 \\ \\ \tau^2&=& s_Y^2-2\beta s_{XY}+\beta^2 s_X^2 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\beta \overline{x}-\alpha&=&0 \\ \\ \beta&=& \frac{s_{XY}}{s_X^2} \\ \\ \tau^2&=& s_Y^2-2\beta s_{XY}+\beta^2 s_X^2 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \overline{y}-\frac{s_{XY}}{s_X^2} \overline{x}-\alpha&=&0 \\ \\ \beta&=& \frac{s_{XY}}{s_X^2} \\ \\ \tau^2&=& s_Y^2-2\frac{s_{XY}}{s_X^2} s_{XY}+(\frac{s_{XY}}{s_X^2})^2 s_X^2 \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \alpha&=&\overline{y}-\frac{s_{XY}}{s_X^2} \overline{x} \\ \\ \beta&=& \frac{s_{XY}}{s_X^2} \\ \\ \tau^2&=& s_Y^2-2\frac{s_{XY}}{s_X^2} s_{XY}+\frac{s_{XY}^2}{s_X^2} \\ \\ \end{array} \right. \\ \\ &\Leftrightarrow& \left\{ \begin{array}{l} \alpha&=&\overline{y}-\frac{s_{XY}}{s_X^2} \overline{x} \\ \\ \beta&=& \frac{s_{XY}}{s_X^2} \\ \\ \tau^2&=& s_Y^2-\frac{s_{XY}^2}{s_X^2} \\ \\ \end{array} \right. \\ \\ \end{eqnarray}