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の章末問題を解いていきます。

# 1.1

E(x)=\sum{kPr(x=k)} ... 30講p19\\


つまり和は分解できるので

\begin{align}
\gamma_k &= E(y_t y_{t-k}) - \mu{E(y_{t-k}) + E(y_t)} + E(\mu^2) \\
&= E(y_t)E(y_{t-k}) - \mu{E(y_{t-k}) + E(y_t)} + E(\mu^2) \\
&= \mu\mu - \mu(2\mu) + E(\mu^2) \\
&= -\mu^2 + E(\mu^2) \\
\gamma_{-k} &= E(y_t y_{t+k}) - \mu{E(y_{t+k} + E(y_t)} + E(\mu^2) \\
&= E(y_t)E(y_{t+k}) - \mu{E(y_{t+k} + E(y_t)} + E(\mu^2) \\
&= \mu^2 - 2\mu^2 +E(\mu^2) \\
&= -\mu^2 + E(\mu^2) \\
ゆえに \gamma_k &= \gamma_{-k}
\end{align}


# 1.2

y_t = \mu + \epsilon_t, \epsilon \sim W.N.(\sigma^2)


\begin{align}
E(\epsilon_t) &= 0 \\
\gamma_k &= E(\epsilon_t \epsilon_{t-k}) =
\begin{cases} \sigma^2 & k = 0 \\
0 & k \neq 0
\end{cases} \\
よって \\
E(y_t) &= E(\mu + \epsilon_t) \\
&= E(\mu) + E(\epsilon_t) \\ &= \mu \\ Cov(y_t, y_{t-k}) &= E[(y_t - \mu) (y_{t-k} - \mu)] \\ &= E[(\mu + \epsilon_t - \mu) (\mu + \epsilon_{t-k} - \mu)]　\\ &= E[\epsilon_t \epsilon_{t-k}] \\ &= \gamma_k \\
\end{align}


ゆえに(1.8)のモデルは弱定常過程

# 1.4

わからず、、、

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