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『経済・ファイナンスデータの計量時系列分析』章末問題を解く-第2章ARMA過程-

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経済・ファイナンスデータの軽量時系列分析
の章末問題を解いていきます。

必要に応じ『統計学のための数学入門30講を参照しています。

Rの実施は『経済・ファイナンスデータの計量時系列分析』章末問題をRで解く-第2章ARMA過程-にあります。

2.1

 y_t = c + \phi_1 y_{t-1} + \phi_2 y_{t-2} + \epsilon_t, \epsilon_t \sim W.N.(\sigma^2) 
 \tag{2.18}

定常条件はAR特性方程式の解|z|>1だから、 (2.18)の特性方程式

1- \phi_1z-\phi_2 z^2 = 0

から

 z = \frac{-\phi_1 \pm \sqrt{\phi_1^2 + 4 \phi_2}}{2}

これから

 \phi_2 < 1 + \phi_1 \\
 \phi_2 < 1 - \phi_1

は出てくるが、

 \phi_2 > - 1

がわからない。

2.2

定常な(AR)モデルの条件は

 1 - \phi_1 z - ... - \phi_p z^p = 0
 |z| > 1

反転可能な(MA)モデルの条件は

 1 + \theta_1 z + ... + \phi_p z^p = 0
 |z| > 1

(a)MA(0) 定常かつ反転可能なモデル

(b)MA(1)

\begin{align}
 1 + z &= 0 \\
 z &= -1 
\end{align}

|z|>1ではないので反転可能ではない定常なモデル。

(c)MA(2)

\begin{align}
 1 - 0.3 z + 0.7 z^2 &= 0 \\
 z &= \frac{0.3 \pm \sqrt{0.09 - 4 \times 0.7}}{2 \times 0.7} 
\end{align}

zは虚数解なので反転可能ではない定常なモデル。

(d)AR(1)

\begin{align}
 1 - 0.5 z &= 0 \\
 z &= 2 
\end{align}

|z|>1より定常かつ反転可能なモデル。

(e)AR(2)

\begin{align}
 1 - 1.3 z + 0.4 z^2 &= 0 \\
 z &= 2.5 
\end{align}

|z|>1より定常かつ反転可能なモデル。

(f)ARMA(1,1)
AR(1)について

\begin{align}
 1 - z &= 0 \\
 z & = 1
\end{align}

よって定常ではない。MA(1)について

\begin{align}
1 + 0.5 z &= 0 \\
z &= -2 
\end{align}

よって反転可能である。

2.3

(1)

\begin{align}
 E(y_t) &= E(\mu) + E(\epsilon_t) + E(\theta_1 \epsilon_{t-1}) + E(\theta_2 \epsilon_{t-2}) \\
 &= \mu
\end{align}

(2)

\begin{align}
 \gamma_0 &= Var(y_t) \\
 &= Var(\mu + \epsilon_t + \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2}) \\
 &= Var(\epsilon_t) + \theta_1^2 Var(\epsilon_{t-1}) + \theta_2^2 Var(\epsilon_{t-2}) \\
 &= (1 + \theta_1^2 + \theta_2^2) \sigma^2
\end{align}

(3)(4)(5)わからず。

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