PRMLの行間埋め 第２章
2.4 指数型分布族
- 式(2.197)の導出
- 式(2.200)の導出
- 式(2.211)の導出
\begin{align*}
\text{Bern}(x|\mu)
&=
\mu^x(1-\mu)^{1-x}&&...\text{式(2.196)} \\ \\
&=
\exp\left\{\ln\left[\mu^x(1-\mu)^{1-x}\right]\right\}\\ \\
&=
\exp\left\{\ln\left(\mu^x\right)+\ln\left((1-\mu)^{1-x}\right)\right\}\\ \\
&=
\exp\left\{x\ln\mu+(1-x)\ln(1-\mu)\right\}\\ \\
&=
\exp\left\{x(\ln\mu-\ln(1-\mu))+\ln(1-\mu)\right\}\\ \\
&=
\exp\left\{x(\ln\mu-\ln(1-\mu))\right\}\exp\left\{\ln(1-\mu)\right\}\\ \\
&=
\exp\left\{x\left(\ln\frac{\mu}{1-\mu}\right)\right\}(1-\mu)\\ \\
&=
(1-\mu)\exp\left\{\ln\left(\frac{\mu}{1-\mu}\right)x\right\}\\ \\
\end{align*}
式(2.198)より
\begin{align*}
&&\eta&=\ln\left(\frac{\mu}{1-\mu}\right) \\ \\
&\Leftrightarrow&
\exp(\eta)&=\exp\left\{\ln\left(\frac{\mu}{1-\mu}\right)\right\} \\ \\
&&
&=\frac{\mu}{1-\mu} \\ \\
&\Leftrightarrow&
(1-\mu)\exp(\eta)
&=\mu \\ \\
&\Leftrightarrow&
\exp(\eta)
&=\mu(1+\exp(\eta)) \\ \\
&\Leftrightarrow&
\mu
&=\frac{\exp(\eta)}{1+\exp(\eta)} \\ \\
\end{align*}
と変換できるので、式(2.197)より
\begin{align*}
p(x|\mu)
&=
(1-\frac{\exp(\eta)}{1+\exp(\eta)})\exp\left\{\ln\left(\frac{\mu}{1-\mu}\right)x\right\}&&...\text{式(2.197)}\\ \\
&=
\left(1-\frac{\exp(\eta)}{1+\exp(\eta)}\right)\exp(\eta x)&&...\text{式(2.198)}\\ \\
&=
\frac{1+\exp(\eta)-\exp(\eta)}{1+\exp(\eta)}\exp(\eta x)&\\ \\
&=
\frac{1}{1+\exp(\eta)}\exp(\eta x)&\\ \\
&=
\frac{1}{1+\exp(\eta)}\frac{\exp(-\eta)}{\exp(-\eta)}\exp(\eta x)&\\ \\
&=
\frac{\exp(-\eta)}{1+\exp(-\eta)}\exp(\eta x)&\\ \\
&=
\sigma(-\eta)\exp(\eta x)
\end{align*}
と導出できる。
\begin{align*}
\exp\left\{\displaystyle\sum_{k=1}^Mx_k\ln \mu_k\right\}
&=
\exp\left\{\displaystyle\sum_{k=1}^{M-1}x_k\ln \mu_k+x_M\ln\mu_M\right\} \\ \\
&=
\exp\left\{\displaystyle\sum_{k=1}^{M-1}x_k\ln \mu_k+x_M\ln\left(1-\displaystyle\sum_{j=1}^{M-1}\mu_j\right)\right\} &&...\text{式(2.209)より}\\ \\
&=
\exp\left\{\displaystyle\sum_{k=1}^{M-1}x_k\ln \mu_k+(1-\displaystyle\sum_{k=1}^{M-1} x_k)\ln\left(1-\displaystyle\sum_{j=1}^{M-1}\mu_j\right)\right\} &&...x_k\text{のうちどれか一つが}1\text{であるため、}\displaystyle\sum_{k=1}^Mx_k=1\\ \\
&=
\exp\left\{\displaystyle\sum_{k=1}^{M-1}x_k\left(\ln \mu_k-\ln\left(1-\displaystyle\sum_{j=1}^{M-1}\mu_j\right)\right)+\ln\left(1-\displaystyle\sum_{j=1}^{M-1}\mu_j\right)\right\} &\\ \\
&=
\exp\left\{\displaystyle\sum_{k=1}^{M-1}x_k\ln\left(\frac{\mu_k}{1-\displaystyle\sum_{j=1}^{M-1}\mu_j}\right)+\ln\left(1-\displaystyle\sum_{j=1}^{M-1}\mu_j\right)\right\} &\\ \\
\end{align*}