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

$$\begin{equation} \rho(j) = \frac{Cov[y_t, y_{t+j}]}{\sqrt{V[y_t]}\sqrt{V[y_{t+j}]}} \end{equation}$$

$$\begin{align} V[y_t] &= V[\epsilon_t + \theta_1 \epsilon_{t-1} + \cdots + \theta_q \epsilon_{t-q}] \\ &= V[\epsilon_t] + {\theta_1}^2V[\epsilon_{t-1}] + \cdots + {\theta_q}^2V[\epsilon_{t-q}] \\ &= \sigma^2(1 + {\theta_1}^2 + \cdots + {\theta_q}^2) \end{align}$$

$$\begin{align} V[y_{t+j}] &= V[\epsilon_{t+j} + \theta_1 \epsilon_{t+j-1} + \cdots + \theta_q \epsilon_{t+j-q}] \\ &= V[\epsilon_{t+j}] + {\theta_1}^2V[\epsilon_{t+j-1}] + \cdots + {\theta_q}^2V[\epsilon_{t+j-q}] \\ &= \sigma^2(1 + {\theta_1}^2 + \cdots + {\theta_q}^2) \end{align}$$

$$\begin{equation} E[\epsilon_t] = 0 \end{equation}$$

$$\begin{equation} V[\epsilon_t] = E[{\epsilon_t}^2] = \sigma^2 \end{equation}$$

$$\begin{equation} Cov[\epsilon_t, \epsilon_s] = E[\epsilon_t\epsilon_s] = 0 \end{equation}$$

$$\begin{align} Cov[y_t，y_{t+j}] &= Cov[\epsilon_t + \theta_1 \epsilon_{t-1} + \cdots + \theta_q \epsilon_{t-q},~\epsilon_{t+j} + \theta_1 \epsilon_{t+j-1} + \cdots + \theta_q \epsilon_{t+j-q}] \\ &= E[(\epsilon_t + \theta_1 \epsilon_{t-1} + \cdots + \theta_q \epsilon_{t-q})(\epsilon_{t+j} + \theta_1 \epsilon_{t+j-1} + \cdots + \theta_q \epsilon_{t+j-q})] \end{align}$$

$$\begin{equation} Cov[y_t，y_{t+j}] = 0 \end{equation}$$

$$\begin{equation} \rho(j) = 0 \end{equation}$$

$$\begin{align} Cov[y_t，y_{t+j}] &= E[(\epsilon_t + \theta_1 \epsilon_{t-1} + \cdots + \theta_q \epsilon_{t-q})(\epsilon_{t+j} + \theta_1 \epsilon_{t+j-1} + \cdots + \theta_q \epsilon_{t+j-q})] \\ &= E[(\epsilon_t + \theta_1 \epsilon_{t-1} + \cdots + \theta_q \epsilon_{t-q})(\epsilon_{t+j} + \theta_1 \epsilon_{t+j-1} + \cdots + \theta_j \epsilon_{t} + \cdots + \theta_q \epsilon_{t+j-q})] \\ &= E[(\epsilon_t + \theta_1 \epsilon_{t-1} + \cdots + \theta_q \epsilon_{t-q})(\theta_j \epsilon_{t} + \cdots + \theta_q \epsilon_{t+j-q})] \\ &= E[(\epsilon_t + \theta_1 \epsilon_{t-1} + \cdots + \theta_{q-j} \epsilon_{t-(q-j)} + \cdots + \theta_q \epsilon_{t-q})(\theta_j \epsilon_{t} + \cdots + \theta_q \epsilon_{t+j-q})] \\ &= E[(\epsilon_t + \theta_1 \epsilon_{t-1} + \cdots + \theta_{q-j} \epsilon_{t+j-q})(\theta_j \epsilon_{t} + \theta_{j+1}\epsilon_{t-1} + \cdots + \theta_q \epsilon_{t+j-q})] \\ &= E[\theta_j{\epsilon_t}^2 + \theta_1\theta_{j+1}{\epsilon_{t-1}}^2 + \cdots + \theta_{q-j}\theta_q{\epsilon_{t+j-q}}^2] \\ &= E[\theta_j{\epsilon_t}^2] + E[\theta_1\theta_{j+1}{\epsilon_{t-1}}^2] + \cdots + E[\theta_{q-j}\theta_q{\epsilon_{t+j-q}}^2] \\ &= \theta_jE[{\epsilon_t}^2] + \theta_1\theta_{j+1}E[{\epsilon_{t-1}}^2] + \cdots + \theta_{q-j}\theta_qE[{\epsilon_{t+j-q}}^2] \\ &= \theta_j\sigma^2 + \theta_1\theta_{j+1}\sigma^2 + \cdots + \theta_{q-j}\theta_q\sigma^2 \\ &= \sigma^2(\theta_j + \theta_1\theta_{j+1} + \cdots + \theta_{q-j}\theta_q) \end{align}$$

$$\begin{align} \rho(j) &= \frac{\sigma^2(\theta_{j} + \theta_1\theta_{j+1} + \cdots + \theta_{q-j}\theta_{q})}{\sqrt{\sigma^2(1 + {\theta_1}^2 + \cdots + {\theta_q}^2)}\sqrt{\sigma^2(1 + {\theta_1}^2 + \cdots + {\theta_q}^2)}} \\ &= \frac{\theta_{j} + \theta_1\theta_{j+1} + \cdots + \theta_{q-j}\theta_{q}}{1 + {\theta_1}^2 + {\theta_2}^2 + \cdots + {\theta_q}^2} \end{align}$$