演習13.17の因子グラフについて \[ \newcommand\comment[1]{\color{red}{{\Tiny\mbox{#1}}}} \newcommand\l{\left} \newcommand\r{\right} \newcommand\b[1]{\pmb{\mathrm{#1}}} \newcommand\bz{\b{z}} \newcommand\bu{\b{u}} \newcommand\bx{\b{x}} \newcommand\bzn{\bz_{n}} \newcommand\bznm{\bz_{n-1}} \newcommand\fn{f_n} \newcommand\fnm{f_{n-1}} \newcommand\fnzz{\fn(\bznm,\bzn)} \newcommand\bun{\bu_n} \newcommand\bxn{\bx_n} \newcommand\ms[2]{\mu_{{#1}\to{#2}}} \newcommand\S[1]{ \sum_{#1}} \] \[ \leqalignno{ &\fnzz = p(\bun)p(\bzn | \bun, \bznm)p(\bzn | \bun, \bzn) \\ &h(\bz_1) = p(\bu_1)p(\bz_1 | \bu_1)p(\bx_1 | \bz_1, \bu_1) } \] 右向きのメッセージは (8.66),(8.69)より \[ \leqalignno{ \ms{\fn}{\bzn}(\bzn) &= \S{\bznm} \fnzz \ms{\bznm}{\fn}(\bznm)~~~\comment{※1} \\ &= \S{\bznm} \fnzz \ms{\fnm}{\bznm}(\bznm)~~~\comment{※2} } \] となる。ここで \[ \leqalignno{ &\alpha(\bzn) = \ms{\fn}{\bzn}(\bzn)\ \ (n \ge 2) } \] と定義すると \[ \leqalignno{ \alpha(\bzn) &= \S{\bznm}\fnzz\alpha(\bznm) \\ &= \S{\bznm}p(\bun)p(\bzn | \bun, \bznm)p(\bxn | \bun, \bzn)\alpha(\bznm) } \] を得る。また（8.71）より \[ \ms{h}{\bz_1}(\bz_1) = h(\bz_1) \] ここで、 \[ \alpha(\bz_1) = \ms{h}{\bz_1}(\bz_1) \] と定義すると \[ \alpha(\bz_1) = h(\bz_1) = p(\bu_1)p(\bz_1 | \bu_1)p(\bx_1 | \bz_1, \bu_1) \] を得る。