[1]
\(\displaystyle E[Y_t]=E\left[ \sum_{j=0}^\infty g_j \epsilon_{t-j}\right]\)
\(\displaystyle =\sum_{j=0}^\infty g_j E\left[ \epsilon_{t-j}\right]\)\(=0\)
[2]
\(Cov[Y_t,Y_{t-s}]\)\(=E[Y_tY_{t-s}]-E[Y_t]E[Y_{t-s}]\)
\(=E[Y_tY_{t-s}]\)
\(\displaystyle =E\left[\sum_{j=0}^\infty g_j \epsilon_{t-j}\sum_{k=0}^\infty g_k \epsilon_{t-s-k}\right]\)
\(\displaystyle =\sum_{j=0}^\infty \sum_{k=0}^\infty g_jg_k E\left[ \epsilon_{t-j} \epsilon_{t-s-k}\right]\)
\(t-j=t-s-k\)すなわち、\(j=k+s\)の時のみ\(E[ \epsilon_{t-j} \epsilon_{t-s-k}]=\sigma^2\)で、それ以外の時は、\(E[ \epsilon_{t-j} \epsilon_{t-s-k}]=0\)なので、
\(\displaystyle = \sigma^2 \sum_{k=0}^\infty g_k g_{k+s} \)
[3]
(1)の両辺に\(Y_{t-1}\)を掛けて期待値を取ると、
\(E[Y_tY_{t-1}]=a_1E[Y_{t-1}^2]+a_2E[Y_{t-1}Y_{t-2}]\)\(+E[\epsilon_tY_{t-1}]\)
\(C(1)=a_1C(0)+a_2C(1)\)
[4]
(1)の両辺に\(Y_{t-2}\)を掛けて期待値を取ると、
\(E[Y_tY_{t-2}]=a_1E[Y_{t-1}Y_{t-2}]+a_2E[Y_{t-2}^2]\)\(+E[\epsilon_tY_{t-2}]\)
\(C(2)=a_1C(1)+a_2C(0)\)
[5]
[3], [4]で得られた式を未定係数\(a_1,a_2\)を求めるための行列方程式の形にすると、
\(\begin{bmatrix}C(0) & C(1) \\ C(1) & C(0) \end{bmatrix}\begin{bmatrix}a_1 \\ a_2 \end{bmatrix}\)\(=\begin{bmatrix}C(1) \\ C(2) \end{bmatrix}\)
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