Exercise 10.4 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 10.4 解答例

\(\bm{\phi}\)は\(d\)次元の確率ベクトルとする。

\(\bm{\phi}_a\), \(\bm{\phi}_b\)をそれぞれ\(\mathbb{R}^d\)の任意の確率ベクトルとし、 \(\bm{\phi}_a = (a_1, \dots, a_d)\), \(\bm{\phi}_b = (b_1, \dots, b_d)\)とおくと、

\begin{equation} \label{eq:10-4-dbc} \begin{aligned} & p(\bm{\phi}_a) \mathrm{Pr}( \bm{\phi}^{(t)} = \bm{\phi}_a, \bm{\phi}^{(t+1)} = \bm{\phi}_b ) \\ = & p(a_1, \dots, a_d) p(b_1 | a_2, \dots, a_d) p(b_2 | b_1, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & p(a_1|a_2, \dots, a_d) p(a_2, \dots, a_d) p(b_1 | a_2, \dots, a_d) p(b_2 | b_1, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & p(a_1|a_2, \dots, a_d) p(b_1, a_2, \dots, a_d) p(b_2 | b_1, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & p(a_1|a_2, \dots, a_d) p(a_2 | b_1, a_3, \dots, a_d) p(b_1, a_3, \dots, a_d) p(b_2 | b_1, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & p(a_1|a_2, \dots, a_d) p(a_2 | b_1, a_3, \dots, a_d) p(b_1, b_2, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & \ldots \\ = &p(a_1 | a_2, \dots, a_d) p(a_2 | b_1, a_3, \dots, a_d) \dots p(a_d | a_1, \dots, a_{d-1}) p(b_1, \dots, b_d) \\ = & p(\bm{\phi}_b) \mathrm{Pr}( \bm{\phi}^{(t)} = \bm{\phi}_b, \bm{\phi}^{(t+1)} = \bm{\phi}_a ) \\ \end{aligned} \end{equation}

が成り立つ。

よって、

\begin{equation*} \begin{aligned} \text{Pr}(\bm{\phi}^{(s+1)} \in A) &= \int_A \left[\int_{\mathbb{R}^d} p(\bm{\phi}^{(s)}) \mathrm{Pr}(\bm{\phi}^{(t)} = \bm{\phi}^{(s)}, \bm{\phi}^{(t+1)} = \bm{\phi}^{(s+1)}) d\bm{\phi}^{(s)}\right] d\bm{\phi}^{(s+1)} & (\text{by assumption}) \\ &= \int_A \left[\int_{\mathbb{R}^d} p(\bm{\phi}^{(s+1)}) \mathrm{Pr}(\bm{\phi}^{(t)} = \bm{\phi}^{(s+1)}, \bm{\phi}^{(t+1)} = \bm{\phi}^{(s)}) d\bm{\phi}^{(s)}\right] d\bm{\phi}^{(s+1)} &(\because \eqref{eq:10-4-dbc}) \\ &= \int_A p(\bm{\phi}^{(s+1)}) \left[\int_{\mathbb{R}^d} \mathrm{Pr}(\bm{\phi}^{(t)} = \bm{\phi}^{(s+1)}, \bm{\phi}^{(t+1)} = \bm{\phi}^{(s)}) d\bm{\phi}^{(s)}\right] d\bm{\phi}^{(s+1)} \\ &= \int_A p(\bm{\phi}^{(s+1)}) d\bm{\phi}^{(s+1)} \\ &= \int_A p(\bm{\phi}) d\bm{\phi} \end{aligned} \end{equation*}

\(\bm{\phi}\) is a \(d\)-dimensional vector of parameters.

Let \(\bm{\phi}_a\), \(\bm{\phi}_b\) be arbitrary probability vectors in \(\mathbb{R}^d\), and \(\bm{\phi}_a = (a_1, \dots, a_d)\), \(\bm{\phi}_b = (b_1, \dots, b_d)\). Then, we have

\begin{equation} \label{eq:10-4-dbc-eng} \begin{aligned} & p(\bm{\phi}_a) \mathrm{Pr}( \bm{\phi}^{(t)} = \bm{\phi}_a, \bm{\phi}^{(t+1)} = \bm{\phi}_b ) \\ = & p(a_1, \dots, a_d) p(b_1 | a_2, \dots, a_d) p(b_2 | b_1, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & p(a_1|a_2, \dots, a_d) p(a_2, \dots, a_d) p(b_1 | a_2, \dots, a_d) p(b_2 | b_1, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & p(a_1|a_2, \dots, a_d) p(b_1, a_2, \dots, a_d) p(b_2 | b_1, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & p(a_1|a_2, \dots, a_d) p(a_2 | b_1, a_3, \dots, a_d) p(b_1, a_3, \dots, a_d) p(b_2 | b_1, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & p(a_1|a_2, \dots, a_d) p(a_2 | b_1, a_3, \dots, a_d) p(b_1, b_2, a_3, \dots, a_d) \\ & \times p(b_3 | b_1, b_2, a_4, \dots, a_d) \dots p(b_d | b_1, \dots, b_{d-1}) \\ = & \ldots \\ = &p(a_1 | a_2, \dots, a_d) p(a_2 | b_1, a_3, \dots, a_d) \dots p(a_d | a_1, \dots, a_{d-1}) p(b_1, \dots, b_d) \\ = & p(\bm{\phi}_b) \mathrm{Pr}( \bm{\phi}^{(t)} = \bm{\phi}_b, \bm{\phi}^{(t+1)} = \bm{\phi}_a ) \\ \end{aligned} \end{equation}

Therefore, the following holds:

\begin{equation*} \begin{aligned} \text{Pr}(\bm{\phi}^{(s+1)} \in A) &= \int_A \left[\int_{\mathbb{R}^d} p(\bm{\phi}^{(s)}) \mathrm{Pr}(\bm{\phi}^{(t)} = \bm{\phi}^{(s)}, \bm{\phi}^{(t+1)} = \bm{\phi}^{(s+1)}) d\bm{\phi}^{(s)}\right] d\bm{\phi}^{(s+1)} & (\text{by assumption}) \\ &= \int_A \left[\int_{\mathbb{R}^d} p(\bm{\phi}^{(s+1)}) \mathrm{Pr}(\bm{\phi}^{(t)} = \bm{\phi}^{(s+1)}, \bm{\phi}^{(t+1)} = \bm{\phi}^{(s)}) d\bm{\phi}^{(s)}\right] d\bm{\phi}^{(s+1)} &(\because \eqref{eq:10-4-dbc-eng}) \\ &= \int_A p(\bm{\phi}^{(s+1)}) \left[\int_{\mathbb{R}^d} \mathrm{Pr}(\bm{\phi}^{(t)} = \bm{\phi}^{(s+1)}, \bm{\phi}^{(t+1)} = \bm{\phi}^{(s)}) d\bm{\phi}^{(s)}\right] d\bm{\phi}^{(s+1)} \\ &= \int_A p(\bm{\phi}^{(s+1)}) d\bm{\phi}^{(s+1)} \\ &= \int_A p(\bm{\phi}) d\bm{\phi}. \end{aligned} \end{equation*}