統計検定 準1級 過去問 解答/解答例と解説
2016年06月19日 (日) 試験

問題の要約

• 0 と 1 で示された位置があり，サイコロを投げて移動

• $X_t$ : 時点 $t~(t = 0, 1, \ldots)$ の位置

  $P(X_{t+1} = 0 \mid X_t = 0) = \frac{1}{6}~~~~$ ($X_t = 0$ で，3の目がでる)

  $P(X_{t+1} = 1 \mid X_t = 0) = \frac{5}{6}~~~~$ ($X_t = 0$ で，3以外の目がでる)

  $P(X_{t+1} = 0 \mid X_t = 1) = \frac{1}{2}~~~~$ ($X_t = 0$ で，偶数の目がでる)

  $P(X_{t+1} = 1 \mid X_t = 1) = \frac{5}{2}~~~~$ ($X_t = 0$ で，奇数の目がでる)

1. サイコロを 5回投げ，3, 6, 4, 3, 5 の目が出た．

   どのように移動するか．

2. 次の関係が成立する．

   $$\begin{equation} p_{t} = P(X_t = 0) \end{equation}$$
   $$\begin{equation} q_{t} = P(X_t = 1) \end{equation}$$
   $$\begin{equation} p_{t+1} = P(X_{t+1} = 0) \end{equation}$$
   $$\begin{equation} q_{t+1} = P(X_{t+1} = 1) \end{equation}$$
   $$\begin{equation} (p_{t+1}, q_{t+1}) = (p_t, q_t) \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{equation}$$
   推移行列を求めよ．

3. $t \to \infty$ のとき，$(p_{t},~q_{t})$ の値を求めよ．

解答

1. 答 : ④

   0 → 0 → 1 → 1 → 0 → 1

2. 答 : ③

   $$\begin{equation} p_{t+1} = p_t \times P(X_{t+1} = 0 \mid X_{t} = 0) + q_t \times P(X_{t+1} = 0 \mid X_{t} = 1) \end{equation}$$
   $$\begin{equation} q_{t+1} = p_t \times P(X_{t+1} = 1 \mid X_{t} = 0) + q_t \times P(X_{t+1} = 1 \mid X_{t} = 1) \end{equation}$$
   なので，
   $$\begin{align} (p_{t+1}, q_{t+1}) &= (p_t, q_t) \begin{pmatrix} P(X_{t+1} = 0 \mid X_{t} = 0) & P(X_{t+1} = 1 \mid X_{t} = 0) \\ P(X_{t+1} = 0 \mid X_{t} = 1) & P(X_{t+1} = 1 \mid X_{t} = 1) \end{pmatrix} \\ &= (p_t, q_t) \begin{pmatrix} \frac{1}{6} & \frac{5}{6} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \end{align}$$

3. 答 : ①

   次の連立漸化式を解く．

   $$\begin{equation} \left\{ \begin{array}{l} \ds p_{t+1} = \frac{1}{6}p_{t} + \frac{1}{2}q_{t}~\cdots\cdots~(1)\\ \ds q_{t+1} = \frac{5}{6}p_{t} + \frac{1}{2}q_{t}~\cdots\cdots~(2) \end{array} \right. \end{equation}$$
   $p_{t+2}$ を解いてみる．
   $$\begin{align} p_{t+2} &= \frac{1}{6}p_{t+1} + \frac{1}{2}q_{t+1} \\ &= \frac{1}{6}p_{t+1} + \frac{1}{2}\left(\frac{5}{6}p_{t} + \frac{1}{2}q_{t}\right) \end{align}$$
   ここで，(1)より，
   $$\begin{equation} \frac{1}{2}q_{t} = p_{t+1} - \frac{1}{6}p_{t} \end{equation}$$
   なので，
   $$\begin{align} p_{t+2} &= \frac{1}{6}p_{t+1} + \frac{1}{2}\left(\frac{5}{6}p_{t} + p_{t+1} - \frac{1}{6}p_{t}\right) \\ &= \frac{4}{6}p_{t+1} + \frac{1}{3}p_{t} \end{align}$$
   特性方程式
   $$\begin{equation} x^2 = \frac{4}{6}x + \frac{1}{3} \end{equation}$$
   の解は，
   $$\begin{equation} 3 x^2 - 2 x - 1 = 0 \end{equation}$$
   $$\begin{equation} (3x + 1)(x - 1) = 0 \end{equation}$$
   $$\begin{equation} x = 1,~-\frac{1}{3} \end{equation}$$
   よって，
   $$\begin{equation} p_{t+1} - p_{t} = -\frac{1}{3} \times (p_{t+1} - p_{t}) \end{equation}$$
   $$\begin{equation} p_{t+1} + \frac{1}{3}p_{t} = 1 \times \left(p_{t+1} + \frac{1}{3} p_{t}\right) \end{equation}$$
   これらより，
   $$\begin{equation} p_{t+1} - p_{t} = (p_{2} - p_{1})\left(-\frac{1}{3}\right)^{t-1} \end{equation}$$
   $$\begin{equation} p_{t+1} + \frac{1}{3}p_{t} = \left(p_{2} + \frac{1}{3} p_{1}\right) \times 1^{t-1} \end{equation}$$
   両辺を引くと，
   $$\begin{equation} - p_{t} - \frac{1}{3}p_{t} = (p_{2} - p_{1})\left(-\frac{1}{3}\right)^{t-1} - \left(p_{2} + \frac{1}{3} p_{1}\right) \end{equation}$$
   $$\begin{equation} p_{t} = -\frac{3}{4}(p_{2} - p_{1})\left(-\frac{1}{3}\right)^{t-1} + \frac{3}{4} \left(p_{2} + \frac{1}{3} p_{1}\right) \end{equation}$$
   $t \to \infty$ とすれば，
   $$\begin{equation} \lim_{t \to \infty}p_{t} = \frac{3}{4} \left(p_{2} + \frac{1}{3} p_{1}\right) \end{equation}$$
   ところで，
   $$\begin{align} (p_1, q_1) &= (p_0, q_0) \begin{pmatrix} \frac{1}{6} & \frac{5}{6} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \\ &= (1, 0) \begin{pmatrix} \frac{1}{6} & \frac{5}{6} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \\ &= \left(\frac{1}{6}, \frac{5}{6}\right) \end{align}$$
   $$\begin{align} (p_2, q_2) &= (p_1, q_1) \begin{pmatrix} \frac{1}{6} & \frac{5}{6} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \\ &= \left(\frac{1}{6}, \frac{5}{6}\right) \begin{pmatrix} \frac{1}{6} & \frac{5}{6} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \\ &= \left(\frac{1}{6}\times\frac{1}{6} + \frac{5}{6}\times\frac{1}{2},~\frac{1}{6}\times\frac{5}{6} + \frac{5}{6}\times\frac{1}{2}\right) \\ &= \left(\frac{1}{36} + \frac{5}{12},~\frac{5}{36} + \frac{5}{12}\right) \\ &= \left(\frac{4}{9},~\frac{5}{9}\right) \\ \end{align}$$
   なので，
   $$\begin{align} \lim_{t \to \infty}p_{t} &= \frac{3}{4} \left(\frac{4}{9} + \frac{1}{3} \times \frac{1}{6}\right) \\ &= \frac{3}{4} \left(\frac{4}{9} + \frac{1}{18}\right) \\ &= \frac{3}{4} \left(\frac{8 + 1}{18}\right) \\ &= \frac{3}{4} \times \frac{1}{2} \\ &= \frac{3}{8} \end{align}$$
   ここで，(1)より，
   $$\begin{equation} q_{t} = 2\,p_{t+1} - \frac{1}{3}\,p_{t} \end{equation}$$
   よって，
   $$\begin{align} \lim_{t \to \infty}q_{t} &= \lim_{t \to \infty}\left(2\,p_{t+1} - \frac{1}{3}\,p_{t}\right) \\ &= 2\,\lim_{t \to \infty}p_{t+1} - \frac{1}{3}\lim_{t \to \infty}p_{t} \\ &= 2 \times \frac{3}{8} - \frac{1}{3} \times \frac{3}{8} \\ &= \frac{3}{4} - \frac{1}{8} \\ &= \frac{5}{8} \end{align}$$
   以上より，
   $$\begin{equation} (p_{t},~q_{t}) \to \left(\frac{3}{8},~\frac{5}{8}\right)~~~~(t \to \infty) \end{equation}$$