統計検定 1級 過去問 解答/解答例と解説
2014年11月30日 (日) 試験

$$\begin{equation} y_t = \frac{3}{4}y_{t-1} - \frac{1}{8}y_{t-2} + \epsilon_t \end{equation}$$

$$\begin{equation} y_ty_{t-j} = \frac{3}{4}y_{t-1}y_{t-j} - \frac{1}{8}y_{t-2}y_{t-j} + \epsilon_ty_{t-j} \end{equation}$$

$$\begin{align} E[y_ty_{t-j}] &= \frac{3}{4}E[y_{t-1}y_{t-j}] - \frac{1}{8}E[y_{t-2}y_{t-j}] + E[\epsilon_ty_{t-j}] \\ Cov[y_t,~y_{t-j}] &= \frac{3}{4}Cov[y_{t-1},~y_{t-j}] - \frac{1}{8}Cov[y_{t-2},~y_{t-j}] + Cov[\epsilon_t,~y_{t-j}] \\ &= \frac{3}{4}Cov[y_{t-1},~y_{t-j}] - \frac{1}{8}Cov[y_{t-2},~y_{t-j}] + Cov\left[\epsilon_t,~\frac{3}{4}y_{t-j-1} - \frac{1}{8}y_{t-j-2} + \epsilon_{t-j}\right] \\ &= \frac{3}{4}Cov[y_{t-1},~y_{t-j}] - \frac{1}{8}Cov[y_{t-2},~y_{t-j}] \end{align}$$

$$\begin{equation} \gamma_h \equiv Cov[y_t,~y_{t + h}] \end{equation}$$

$$\begin{align} \gamma_{j} &= \frac{3}{4}\gamma_{j-1} - \frac{1}{8}\gamma_{j-2}~~~~(j > 0) \end{align}$$

$$\begin{align} \gamma_0 &= V[y_t] \\ &= Cov[y_t,~y_{t - 0}] \\ &= Cov\left[y_t,~\frac{3}{4}y_{t-1} - \frac{1}{8}y_{t-2} + \epsilon_t\right] \\ &= Cov\left[y_t,~\frac{3}{4}y_{t-1}\right] - Cov\left[y_t,~\frac{1}{8}y_{t-2}\right] + Cov\left[y_t,~\epsilon_t\right] \\ &= \frac{3}{4}Cov\left[y_t,~y_{t-1}\right] - \frac{1}{8}Cov\left[y_t,~y_{t-2}\right] + \sigma^2 \\ &= \frac{3}{4}\gamma_1 - \frac{1}{8}\gamma_2 + \sigma^2 \end{align}$$

$$\begin{align} \gamma_1 &= Cov[y_t,~y_{t + 1}] \\ &= Cov\left[y_{t},~\frac{3}{4}y_{t} - \frac{1}{8}y_{t-1} + \epsilon_{t+1}\right] \\ &= Cov\left[y_{t},~\frac{3}{4}y_{t}\right] - Cov\left[y_{t},~\frac{1}{8}y_{t-1}\right] + Cov\left[y_{t},~\epsilon_{t+1}\right] \\ &= \frac{3}{4}Cov\left[y_{t},~y_{t}\right] - \frac{1}{8}Cov\left[y_{t},~y_{t-1}\right] + 0 \\ &= \frac{3}{4}\gamma_0 - \frac{1}{8}\gamma_1 \\ \end{align}$$

$$\begin{align} \gamma_2 &= Cov[y_t,~y_{t + 2}] \\ &= Cov\left[y_{t},~\frac{3}{4}y_{t+1} - \frac{1}{8}y_{t} + \epsilon_{t+2}\right] \\ &= Cov\left[y_{t},~\frac{3}{4}y_{t+1}\right] - Cov\left[y_{t},~\frac{1}{8}y_{t}\right] + Cov\left[y_{t},~\epsilon_{t+2}\right] \\ &= \frac{3}{4}Cov\left[y_{t},~y_{t+1}\right] - \frac{1}{8}Cov\left[y_{t},~y_{t}\right] + 0 \\ &= \frac{3}{4}\gamma_1 - \frac{1}{8}\gamma_0 \end{align}$$

$$\begin{align} \gamma_1 = \frac{2}{3}\gamma_0 \end{align}$$

$$\begin{align} \gamma_2 &= \frac{3}{4}\left(\frac{2}{3}\gamma_0\right) -\frac{1}{8}\gamma_0 \\ &= \frac{1}{2}\gamma_0 -\frac{1}{8}\gamma_0 \\ &= \frac{3}{8}\gamma_0 \end{align}$$

$$\begin{align} \gamma_0 &= \frac{3}{4}\left(\frac{2}{3}\gamma_0\right) - \frac{1}{8}\left(\frac{3}{8}\gamma_0\right) + \sigma^2 \\ &= \frac{1}{2}\gamma_0 - \frac{3}{16}\gamma_0 + \sigma^2 \\ &= \frac{5}{16}\gamma_0 + \sigma^2 \end{align}$$

$$\begin{equation} \therefore~\gamma_0 = \frac{16}{11}\sigma^2 \end{equation}$$

$$\begin{equation} \therefore~\gamma_1 = \frac{32}{33}\sigma^2 \end{equation}$$

$$\begin{equation} \therefore~\gamma_2 = \frac{6}{11}\sigma^2 \end{equation}$$

$$\begin{equation} \gamma_j = \frac{3}{4}\gamma_{j-1} - \frac{1}{8}\gamma_{j-2} \end{equation}$$

$$\begin{equation} t^2 - \frac{3}{4}t + \frac{1}{8} = 0 \end{equation}$$

$$\begin{equation} \left(t - \frac{1}{2}\right)\left(t - \frac{1}{4}\right) = 0 \end{equation}$$

$$\begin{equation} t = \frac{1}{2},~ \frac{1}{4} \end{equation}$$

$$\begin{equation} \gamma_j - \frac{1}{2}\gamma_{j-1} = \frac{1}{4}\left(\gamma_{j-1} - \frac{1}{2}\gamma_{j-2}\right) \end{equation}$$

$$\begin{equation} \gamma_j - \frac{1}{4}\gamma_{j-1} = \frac{1}{2}\left(\gamma_{j-1} - \frac{1}{4}\gamma_{j-2}\right) \end{equation}$$

$$\begin{equation} \gamma_j - \frac{1}{2}\gamma_{j-1} = \left(\gamma_{1} - \frac{1}{2}\gamma_{0}\right)\left(\frac{1}{4}\right)^{j-1} \end{equation}$$

$$\begin{equation} \gamma_j - \frac{1}{4}\gamma_{j-1} = \left(\gamma_{1} - \frac{1}{4}\gamma_{0}\right)\left(\frac{1}{2}\right)^{j-1} \end{equation}$$

$$\begin{equation} - \frac{1}{2}\gamma_{j-1} + \frac{1}{4}\gamma_{j-1} = \left(\gamma_{1} - \frac{1}{2}\gamma_{0}\right)\left(\frac{1}{4}\right)^{j-1} - \left(\gamma_{1} - \frac{1}{4}\gamma_{0}\right)\left(\frac{1}{2}\right)^{j-1} \end{equation}$$

$$\begin{align} \frac{1}{4}\gamma_{j-1} &= \left(\gamma_{1} - \frac{1}{4}\gamma_{0}\right)\left(\frac{1}{2}\right)^{j-1} - \left(\gamma_{1} - \frac{1}{2}\gamma_{0}\right)\left(\frac{1}{4}\right)^{j-1} \\ &= \left(\frac{32}{33}\sigma^2 - \frac{1}{4}\frac{16}{11}\sigma^2\right)\left(\frac{1}{2}\right)^{j-1} - \left(\frac{32}{33}\sigma^2 - \frac{1}{2}\frac{16}{11}\sigma^2\right)\left(\frac{1}{4}\right)^{j-1} \\ &= \left\{\frac{20}{33}\left(\frac{1}{2}\right)^{j-1} - \frac{8}{33}\left(\frac{1}{4}\right)^{j-1}\right\}\sigma^2 \\ \end{align}$$

$$\begin{equation} \gamma_{j} = \left\{\frac{80}{33}\left(\frac{1}{2}\right)^{j} - \frac{32}{33}\left(\frac{1}{4}\right)^{j}\right\}\sigma^2 \end{equation}$$

$$\begin{align} \rho(j) &= \frac{Cov[y_t, y_{t+j}]}{\sqrt{V[y_t]}\sqrt{V[y_{t+j}]}} \\ &= \frac{\gamma_{j}}{\sqrt{\gamma_0}\sqrt{\gamma_0}} \\ &= \frac{\gamma_{j}}{\gamma_{0}} \\ &= \frac{\left\{\frac{80}{33}\left(\frac{1}{2}\right)^{j} - \frac{32}{33}\left(\frac{1}{4}\right)^{j}\right\}\sigma^2}{\frac{16}{11}\sigma^2} \\ &= \frac{5}{3}\left(\frac{1}{2}\right)^{j} - \frac{2}{3}\left(\frac{1}{4}\right)^{j} \end{align}$$