統計学実践ワークブックの行間埋め　第27章

コーシーシュワルツの不等式を用いる（参考）
期間tについて得られる各データを\(j\)用いて、\(y_{t,j}\)と表すとする。値を\(y_{t,j}-\mu\)として扱い、コーシーシュワルツの不等式を適用すると、 \begin{eqnarray} && \left(\displaystyle\sum_{j=1}^n (y_{t,j}-\mu)(y_{t-h,j}-\mu)\right)^2&\geq&\left(\displaystyle \sum_{j=1}^n(y_{t,j}-\mu)^2\right) \left(\displaystyle \sum_{j=1}^n(y_{t-h,j}-\mu)^2\right) \\ \\ &\Leftrightarrow&\left(\frac{1}{n}\displaystyle\sum_{j=1}^n (y_{t,j}-\mu)(y_{t-h,j}-\mu)\right)^2&geq&\left(\frac{1}{n}\displaystyle \sum_{j=1}^n(y_{t,j}-\mu)^2\right) \left(\frac{1}{n}\displaystyle \sum_{j=1}^n(y_{t-h,j}-\mu)^2\right) \\ \\ &\Leftrightarrow&\text{Cov}[Y_t,Y_{t-h}]^2&\geq&V[Y_t]V[Y_{t-h}] \\ \\ \end{eqnarray}

\begin{eqnarray} (Y_t-\mu)=\phi_1(Y_{t-1}-\mu)+U_t \end{eqnarray} の両辺に\(Y_{t-h}-\mu\)をかけて期待値をとる。 \begin{eqnarray} (Y_t-\mu)(Y_{t-h}-\mu)&=&(\phi_1(Y_{t-1}-\mu)+U_t)(Y_{t-h}-\mu) &\\ \\ \Rightarrow E[(Y_t-\mu)(Y_{t-h}-\mu)]&=&E[(\phi_1(Y_{t-1}-\mu)+U_t)(Y_{t-h}-\mu)] &\\ \\ \Rightarrow \gamma_h&=&E[\phi_1(Y_{t-1}-\mu)(Y_{t-h}-\mu)] +E[\phi_1U_t(Y_{t-h}-\mu)]& \\ \\ \Rightarrow \gamma_h&=&\phi_1\gamma_{h-1} +\phi_1E[U_tY_{t-h}]-\phi_1E[U_t\mu]& \\ \\ \Rightarrow \gamma_h&=&\phi_1\gamma_{h-1} +\phi_1(0-0) &p242中段より\\ \\ \Rightarrow \gamma_h&=&\phi_1\gamma_{h-1}\\ \\ \end{eqnarray}

\begin{eqnarray} \gamma_h&=&\phi_1\gamma_{h-1} &\\ \\ &=& \phi_1(\phi_1\gamma_{h-2}) &\\ \\ &=& \phi_1(\phi_1(\phi_1\gamma_{h-3})) &\\ \\ && \vdots &\\ \\ &=& \phi_1(\phi_1(\phi_1(\ldots(\phi_1\gamma_{0})\ldots)))& \\ \\ &=& \phi_1^h\gamma_0 &\\ \\ &=& \phi_1^h\frac{\sigma^2}{1-\phi_1^2} &...\gamma_0=\frac{\sigma^2}{1-\phi_1^2}より\\ \\ \end{eqnarray}

\begin{eqnarray} \rho_h&=&\frac{Cov[Y_t,Y_{t-h}] }{\sqrt{V[Y_t]V[Y_{t-h}]} } \\ \\ &=& \frac{\phi_1^h \gamma _0}{\sqrt{\gamma _0 \gamma _0} } \\ \\ &=& \frac{\phi_1^h\gamma _0}{\gamma _0} \\ \\ &=& \phi_1^h \end{eqnarray}

\begin{eqnarray} E[(Y_t-\mu)(Y_{t-1}-\mu)] &=& E[(U_t+\theta_1U_{t-1})(U_{t-1}+\theta_1U_{t-2})] \\ \\ &=& E[U_tU_{t-1}]+E[U_t\theta_1U_{t-2}]+E[\theta_1U_{t-1}U_{t-1}]+E[\theta_1U_{t-1}\theta_1U_{t-2}] \\ \\ &=& 0+0+\theta_1\sigma^2+0 \\ \\ &=& \theta_1\sigma^2 \\ \\ \end{eqnarray}

\begin{eqnarray} \gamma_h&=& E[(Y_t-\mu)(Y_{t-h}-\mu)] \\ \\ &=& E[(U_t+\theta_1U_{t-1}+\theta_2U_{t-2}+\ldots+\theta_qU_{t-q})(U_{t-h}+\theta_1U_{t-h-1}+\theta_2U_{t-h-2}+\ldots+\theta_qU_{t-h-q})] \\ \\ &=& E[(\displaystyle \sum_{i=0}^q\theta_iU_{t-i})(\displaystyle \sum_{j=0}^q\theta_jU_{t-h-j})] &\theta_0=1とした\\ \\ &=& E[\displaystyle \sum_{i=0}^q\sum_{j=0}^q\theta_iU_{t-i}\theta_jU_{t-h-j}]\\ \\ &=& E[\displaystyle \sum_{t-i\neq t-h-j}\theta_iU_{t-i}\theta_jU_{t-h-j}+\sum_{t-i=t-h-j}\theta_iU_{t-i}\theta_jU_{t-h-j}]\\ \\ &=& E[\displaystyle \sum_{t-i=t-h-j}\theta_iU_{t-i}\theta_jU_{t-h-j}]\\ \\ &=& E[\displaystyle \sum_{t-i=t-h-j}\theta_iU_{t-i}\theta_{i-h}U_{t-i}] &t-i=t-h-jより、j=i-hを代入\\ \\ &=& E[\displaystyle \sum_{i=h}^{q-h}\theta_i\theta_{i-h}U_{t-i}^2] &i-h\geq 0かつi\geq0より、総和はi=hからq-hまで\\ \\ &=& \theta_0\theta_h\sigma^2+\theta_{1}\theta_{h+1}\sigma^2+\theta_{2}\theta_{h+2}\sigma^2+\ldots+\theta_{q-h}\theta_{q}\sigma^2 \\ \\ &=& (\theta_h+\theta_{h+1}\theta_{1}+\theta_{h+2}\theta_{2}+\ldots+\theta_{q}\theta_{q-h})\sigma^2 &\theta_0=1より\\ \\ \end{eqnarray}

参考
はじめにMA(\(\infty\))を考える。これは \begin{eqnarray} Y_t&=&\mu+U_t+\theta_1U_{t-1}+\theta_2U_{t-2}+\ldots \\ \\ &=& \mu+\theta(L)U_t&(\theta(L)=1+\theta_1L+\theta_2L^2+\ldots) \end{eqnarray} で表されるモデルであるとする。自己共分散\(\gamma_k\)を求める。\(\theta_0=1\)とすると、 \begin{eqnarray} \gamma_k &=& \sigma^2(\theta_k+\theta_1\theta_{k+1}+\theta_2\theta_{k+2}+\ldots) \\ \\ &=& \sigma^2(\theta_0\theta_k+\theta_1\theta_{k+1}+\theta_2\theta_{k+2}+\ldots) \\ \\ &=& \sigma^2\displaystyle\sum_{i=0}^{\infty}\theta_i\theta_{k+i} \\ \\ \end{eqnarray} が得られる。ここで、\(\theta(X)\)と\(\theta(X^{-1})\)の積において、\(X^k\)の項について考えると、 \begin{eqnarray} \theta(X)\theta(X^{-1}) &=& (1+\theta_1X+\theta_2X^2+\ldots)(1+\theta_1X^{-1}+\theta_2X^{-2}+\ldots)\\ \\ &\Rightarrow& (\theta_kX^k+\theta_{k+1}X^{k+1}+\ldots)(1+\theta_1X^{-1}+\theta_2X^{-2}+\ldots) \\ \\ &\Rightarrow& (\theta_kX^k+\theta_{k+1}X^{k+1}\theta_1X^{-1}+\theta_{k+2}X^{k+2}\theta_2X^{-2}+\ldots) \\ \\ &\Rightarrow& (\theta_kX^k+\theta_{k+1}\theta_1X^k+\theta_{k+2}\theta_2X^{k}+\ldots) \\ \\ &=& X^k(\theta_k+\theta_{k+1}\theta_1+\theta_{k+2}\theta_2+\ldots) \\ \\ &=& X^k\frac{\gamma_k}{\sigma^2} \\ \\ \end{eqnarray} が得られる。また、式(27.1)より、\(\gamma_k=\gamma_{-k}\)であることから、\(X^{-k}\)の項は \begin{eqnarray} \theta(X^{-1})\theta(X) &=& (1+\theta_1X^{-1}+\theta_2X^{-2}+\ldots)(1+\theta_1X+\theta_2X^2+\ldots)\\ \\ &\Rightarrow& (\theta_kX^{-k}+\theta_{k+1}X^{-k-1}+\ldots)(1+\theta_1X+\theta_2X^{2}+\ldots) \\ \\ &\Rightarrow& (\theta_kX^{-k}+\theta_{k+1}X^{-k-1}\theta_1X^{1}+\theta_{k+2}X^{-k-2}\theta_2X^{2}+\ldots) \\ \\ &\Rightarrow& (\theta_kX^{-k}+\theta_{k+1}\theta_1X^{-k}+\theta_{k+2}\theta_2X^{-k}+\ldots) \\ \\ &=& X^{-k}(\theta_k+\theta_{k+1}\theta_1+\theta_{k+2}\theta_2+\ldots) \\ \\ &=& X^{-k}\frac{\gamma_k}{\sigma^2} \\ \\ &=& X^{-k}\frac{\gamma_{-k}}{\sigma^2} \\ \\ \end{eqnarray} と書くことができる。これを利用して、 \begin{eqnarray} &&\theta(X)\theta(X^{-1}) &=& \displaystyle\sum_{k=-\infty}^{\infty}X^k\frac{\gamma_k}{\sigma^2} \\ \\ &\Leftrightarrow& \sigma^2\theta(X)\theta(X^{-1})&=&\displaystyle\sum_{k=-\infty}^{\infty}X^k\gamma_k \end{eqnarray} が得らえる。ここで、\(X=e^{i\lambda}\)を代入すると、 \begin{eqnarray} &&\sigma^2\theta(X)\theta(X^{-1})=\displaystyle\sum_{k=-\infty}^{\infty}X^k\gamma_k \\ \\ &\Leftrightarrow& \sigma^2\theta(e^{i\lambda})\theta(e^{-i\lambda})=\displaystyle\sum_{k=-\infty}^{\infty}e^{ik\lambda}\gamma_k \\ \\ &\Leftrightarrow& \sigma^2\theta(e^{i\lambda})\theta(e^{-i\lambda})=2\pi f(\lambda)\tag{*} \\ \\ \end{eqnarray} が得られる。(p.249の定義より)
ここで、ARMA(p,q)モデルは定常であるとき、平均値\(\mu\)を用いて定数項cを求めると、 \begin{eqnarray} &&Y_t=c+\phi_1Y_{t-1}+\ldots+\phi_pY_{t-p}+U_t+\theta_1U_{t-1}+\ldots+\theta_qU_{t-q} \\ \\ &\Leftrightarrow& E[Y_t]=E[c+\phi_1Y_{t-1}+\ldots+\phi_pY_{t-p}+U_t+\theta_1U_{t-1}+\ldots+\theta_qU_{t-q}] \\ \\ &\Leftrightarrow& \mu=c+\phi_1\mu+\ldots+\phi_p\mu \\ \\ &\Rightarrow& c=\mu(1-\phi_1-\ldots -\phi_p) \end{eqnarray} である。cを代入することで、 \begin{eqnarray} &&Y_t=c+\phi_1Y_{t-1}+\ldots+\phi_pY_{t-p}+U_t+\theta_1U_{t-1}+\ldots+\theta_qU_{t-q} \\ \\ &\Leftrightarrow& (Y_t-\mu)=\phi_1(Y_{t-1}-\mu)+\ldots+\phi_p(Y_{t-p}-\mu)+U_t+\theta_1U_{t-1}+\ldots+\theta_qU_{t-q} \\ \\ \end{eqnarray} と書くことができるため、新しく\(Y_t-\mu\)を\(Y_t\)として扱う。
ラグ多項式を用いて式(27.7)を変形する。 \begin{eqnarray} &&Y_t=c+\phi_1Y_{t-1}+\ldots+\phi_pY_{t-p}+U_t+\theta_1U_{t-1}+\ldots+\theta_qU_{t-q} \\ \\ &\Leftrightarrow& (Y_t-\mu)=\phi_1(Y_{t-1}-\mu)+\ldots+\phi_p(Y_{t-p}-\mu)+U_t+\theta_1U_{t-1}+\ldots+\theta_qU_{t-q} \\ \\ &\Leftrightarrow& Y_t=\phi_1Y_{t-1}+\ldots+\phi_pY_{t-p}+U_t+\theta_1U_{t-1}+\ldots+\theta_qU_{t-q} &...新しいY_tに書き換えた\\ \\ &\Leftrightarrow& \phi(L)Y_t=\theta(L)U_t \end{eqnarray} と書ける。これを変形して、 \begin{eqnarray} &&\phi(L)Y_t=\theta(L)U_t \\ \\ &\Leftrightarrow& Y_t=\phi(L)^{-1}\theta(L)U_t \end{eqnarray} と書ける。これは、疑似的にMAモデルとして扱うことができる。式(*)より、\(\theta(e^{i\lambda})=\phi(e^{i\lambda})^{-1}\theta(e^{i\lambda})\)を代入して、 \begin{eqnarray} &&\sigma^2(\phi(e^{i\lambda})^{-1}\theta(e^{i\lambda}))(\phi(e^{-i\lambda})^{-1}\theta(e^{-i\lambda}))=2\pi f(\lambda) \\ \\ &\Leftrightarrow& 2\pi f(\lambda)=\sigma^2\frac{\theta(e^{i\lambda})\theta(e^{-i\lambda})}{\phi(e^{i\lambda})\phi(e^{-i\lambda})} \\ \\ &\Leftrightarrow& f(\lambda)=\frac{\sigma^2}{2\pi}\frac{\theta(e^{i\lambda})\theta(e^{-i\lambda})}{\phi(e^{i\lambda})\phi(e^{-i\lambda})} \\ \\ &\Leftrightarrow& f(\lambda)=\frac{\sigma^2}{2\pi}\frac{|\theta(e^{i\lambda})|^2}{|\phi(e^{i\lambda})|^2} \\ \\ \end{eqnarray}

\begin{eqnarray} f(\lambda)=\frac{\sigma^2}{2\pi}\frac{\theta(e^{-i\lambda})\theta(e^{i\lambda})}{\phi(e^{-i\lambda})\phi(e^{i\lambda})} \end{eqnarray} と表現される。
\(Y_t\)がホワイトノイズであるとき、\(U_t\)は平均0,分散\(\sigma^2\)であるとして、\(Y_t=U_t\)と表される。この時、ラグ多項式で表現すると、 \begin{eqnarray} &&Y_t&=&c+U_t \\ \\ &\Leftrightarrow&\phi (L)Y_t&=&c+\theta(L)U_t &\Rightarrow& \left\{ \begin{array}{l} \phi(L)=1 \\ \theta(L)=1 \\ \end{array} \right. \\ \\ \end{eqnarray} となる(\(c=\mu\)を用いて、\(Y_t-\mu\)を新しい\(Y_t\)とした)。これを式に代入して、 \begin{eqnarray} f(\lambda)&=& \frac{\sigma^2}{2\pi}\frac{\theta(e^{-i\lambda})\theta(e^{i\lambda})}{\phi(e^{-i\lambda})\phi(e^{i\lambda})} \\ \\ &=& \frac{\sigma^2}{2\pi}\frac{1\cdot 1}{1\cdot 1} \\ \\ &=& \frac{\sigma^2}{2\pi} \end{eqnarray}

\(Y_t\)がAR(1)過程であるとき、式(27.4)より、 \begin{eqnarray} Y_t=c+\phi_1Y_{t-1}+U_t \end{eqnarray} と書くことができる。ラグ多項式で記述すると、 \begin{eqnarray} &&Y_t&=&c+\phi_1Y_{t-1}+U_t \\ \\ &\Leftrightarrow&\phi (L)Y_t&=&\theta(L)U_t &\Rightarrow& \left\{ \begin{array}{l} \phi(L)=1-\phi_1L \\ \theta(L)=1 \\ \end{array} \right. \\ \\ \end{eqnarray} であるから、\(f(\lambda)\)を求めると、 \begin{eqnarray} f(\lambda)&=& \frac{\sigma^2}{2\pi}\frac{\theta(e^{-i\lambda})\theta(e^{i\lambda})}{\phi(e^{-i\lambda})\phi(e^{i\lambda})} \\ \\ &=& \frac{\sigma^2}{2\pi}\frac{1\cdot 1}{(1-\phi_1e^{-i\lambda})(1-\phi_1e^{i\lambda})} \\ \\ &=& \frac{\sigma^2}{2\pi}\frac{1}{1-\phi_1e^{-i\lambda}-\phi_1e^{i\lambda}+\phi_1e^{-i\lambda}\phi_1e^{i\lambda}} \\ \\ &=& \frac{\sigma^2}{2\pi}\frac{1}{1-\phi_1e^{-i\lambda}-\phi_1e^{i\lambda}+\phi_1^2} \\ \\ &=& \frac{\sigma^2}{2\pi}\frac{1}{1+\phi_1^2-2\phi_1((e^{-i\lambda}+e^{i\lambda})/2)} \\ \\ &=& \frac{\sigma^2}{2\pi}\frac{1}{1+\phi_1^2-2\phi_1\cos \lambda} \\ \\ \end{eqnarray}

\begin{eqnarray} \int_{-\pi}^{\pi}f(\lambda)d\lambda &=& 2\int_{0}^{\pi}f(\lambda)d\lambda &...f(\lambda)は偶関数なので、f(-\lambda)=f(\lambda)が成立するため。 \\ \\ &=& 2\int_{0}^{\pi}\frac{1}{2\pi}\left(\gamma_0+2\displaystyle\sum_{h=1}^{\infty}\gamma_h\cos (\lambda h)\right)d\lambda \\ \\ &=& 2\frac{1}{2\pi}\int_{0}^{\pi}\left(\gamma_0+2\displaystyle\sum_{h=1}^{\infty}\gamma_h\cos (\lambda h)\right)d\lambda \\ \\ &=& 2\frac{1}{2\pi}\left[ \left(\gamma_0\lambda+2\displaystyle\sum_{h=1}^{\infty}\gamma_h\frac{1}{h}\sin (\lambda h)\right)\right]_0^{\pi} \\ \\ &=& \frac{1}{\pi}\left(\gamma_0(\pi-0)+2\displaystyle\sum_{h=1}^{\infty}\gamma_h\frac{1}{h}(\sin (\pi h)-\sin (0\cdot h))\right) \\ \\ &=& \frac{1}{\pi}\left(\gamma_0\pi+2\displaystyle\sum_{h=1}^{\infty}\gamma_h\frac{1}{h}(0-0)\right) &...\sin(n\pi)はnが整数の時に0になるため。 \\ \\ &=& \frac{1}{\pi}\gamma_0\pi \\ \\ &=& \gamma_0 \end{eqnarray}

\begin{eqnarray} \overline{y} &=& \frac{1}{T}\displaystyle\sum_{t=1}^T\left(\hat{\mu}+\displaystyle\sum_{h=1}^M[\hat{\alpha}_h\cos(\lambda_h(t-1))+\hat{\beta}_h\sin(\lambda_h(t-1))] \right) \\ \\ &=& \frac{1}{T}T\hat{\mu}+\frac{1}{T}\displaystyle\sum_{h=1}^M\left(\displaystyle\sum_{t=1}^T[\hat{\alpha}_h\cos(\lambda_h(t-1))+\hat{\beta}_h\sin(\lambda_h(t-1))] \right) \\ \\ &=& \hat{\mu}+\frac{1}{T}\displaystyle\sum_{h=1}^M\left(\displaystyle\sum_{t=1}^T[\hat{\alpha}_h\cos(\frac{2\pi h}{T}(t-1))+\hat{\beta}_h\sin(\frac{2\pi h}{T}(t-1))] \right) \\ \\ \end{eqnarray} ここで、 \begin{eqnarray} &&\displaystyle\sum_{t=1}^T\exp\left(i\frac{2\pi h(t-1)}{T} \right) \\ \\ &=& \frac{1-\exp\left(i\frac{2\pi h}{T}\cdot T\right)}{1-\exp\left(i\frac{2\pi h}{T}\right)} \\ \\ &=& \frac{1-\exp\left(2\pi ih\right)}{1-\exp\left(i\frac{2\pi h}{T}\right)} \\ \\ &=& \frac{1-1}{1-\exp\left(i\frac{2\pi h}{T}\right)} &...hが整数であるため\\ \\ &=& 0 \end{eqnarray} であることと、オイラーの公式より \begin{eqnarray} \exp\left(i\frac{2\pi h(t-1)}{T}\right)=\cos\left(\frac{2\pi h(t-1)}{T}\right)+i\sin\left(\frac{2\pi h(t-1)}{T}\right) \end{eqnarray} であることから、 \begin{eqnarray} &&\displaystyle\sum_{t=1}^T\exp\left(i\frac{2\pi h(t-1)}{T} \right) \\ \\ &=& \displaystyle\sum_{t=1}^T\left(\cos\left(\frac{2\pi h(t-1)}{T}\right)+i\sin\left(\frac{2\pi h(t-1)}{T}\right) \right) \\ \\ &=& 0 \end{eqnarray} であるから、 \begin{eqnarray} \overline{y} &=& \hat{\mu}+\frac{1}{T}\displaystyle\sum_{h=1}^M\left(\displaystyle\sum_{t=1}^T[\hat{\alpha}_h\cos(\frac{2\pi h}{T}(t-1))+\hat{\beta}_h\sin(\frac{2\pi h}{T}(t-1))] \right) \\ \\ &=& \hat{\mu}+\frac{1}{T}\displaystyle\sum_{h=1}^M\left(0 \right) \\ \\ &=& \hat{\mu} \end{eqnarray} が得られる。 \(y_t\)の標本分散を計算すると、 \begin{eqnarray} &&\frac{1}{T}\displaystyle\sum_{t=1}^T(y_t-\overline{y})^2 \\ \\ &=& \frac{1}{T}\displaystyle\sum_{t=1}^T(\hat{\mu}+\displaystyle\sum_{h=1}^M[\hat{\alpha}_h\cos(\lambda_h(t-1))+\hat{\beta}_h\sin(\lambda_h(t-1))]-\hat{\mu})^2 \\ \\ &=& \frac{1}{T}\displaystyle\sum_{t=1}^T(\displaystyle\sum_{h=1}^M[\hat{\alpha}_h\cos(\lambda_h(t-1))+\hat{\beta}_h\sin(\lambda_h(t-1))])^2 \\ \\ &=& \frac{1}{T}\displaystyle\sum_{t=1}^T(\displaystyle\sum_{h=1}^M[\hat{\alpha}_h^2\cos^2(\lambda_h(t-1))+\hat{\beta}_h^2\sin^2(\lambda_h(t-1))])&...(*) \\ \\ &=& \frac{1}{T}\displaystyle\sum_{t=1}^T(\displaystyle\sum_{h=1}^M[\hat{\alpha}_h^2\frac{\cos(2\lambda_h(t-1))+1}{2}+\hat{\beta}_h^2\frac{1-\cos(2\lambda_h(t-1))}{2}]) \\ \\ &=& \frac{1}{T}\displaystyle\sum_{t=1}^T(\displaystyle\sum_{h=1}^M[\hat{\alpha}_h^2\frac{0+1}{2}+\hat{\beta}_h^2\frac{1-0}{2}]) \\ \\ &=& \frac{1}{T}T(\displaystyle\sum_{h=1}^M[\hat{\alpha}_h^2\frac{1}{2}+\hat{\beta}_h^2\frac{1}{2}]) \\ \\ &=& \frac{1}{2}\displaystyle\sum_{h=1}^M(\hat{\alpha}_h^2+\hat{\beta}_h^2) \\ \\ \end{eqnarray} (*)の箇所では、 \begin{eqnarray} &&\displaystyle\sum_{t=1}^T \cos(\lambda_h(t-1))\sin(\lambda_h'(t-1)) \\ \\ &=& \displaystyle\sum_{t=1}^T\frac{1}{2}(\sin(\lambda_h(t-1)+\lambda_h'(t-1))+\sin(\lambda_h(t-1)-\lambda_h'(t-1))) \\ \\ &=& \displaystyle\sum_{t=1}^T\frac{1}{2}(\sin((\lambda_h+\lambda_h')(t-1))+\sin((\lambda_h-\lambda_h')(t-1))) \\ \\ &=& 0 \\ \\ &&\displaystyle\sum_{t=1}^T \cos(\lambda_h(t-1))\cos(\lambda_h'(t-1)) \\ \\ &=& \displaystyle\sum_{t=1}^T\frac{1}{2}(\cos(\lambda_h(t-1)+\lambda_h'(t-1))+\cos(\lambda_h(t-1)-\lambda_h'(t-1))) \\ \\ &=& \displaystyle\sum_{t=1}^T\frac{1}{2}(\cos((\lambda_h+\lambda_h')(t-1))+\cos((\lambda_h-\lambda_h')(t-1))) \\ \\ &=& 0 &\lambda_h\neq \lambda_h'の場合\\ \\ &&\displaystyle\sum_{t=1}^T \sin(\lambda_h(t-1))\sin(\lambda_h'(t-1)) \\ \\ &=& \displaystyle\sum_{t=1}^T\frac{1}{2}(\cos(\lambda_h(t-1)-\lambda_h'(t-1))-\cos(\lambda_h(t-1)+\lambda_h'(t-1))) \\ \\ &=& \displaystyle\sum_{t=1}^T\frac{1}{2}(\cos((\lambda_h-\lambda_h')(t-1))-\cos((\lambda_h+\lambda_h')(t-1))) \\ \\ &=& 0 &\lambda_h\neq \lambda_h'の場合\\ \\ \end{eqnarray} であることを利用し、\(h\neq h'\)もしくはcosとsinの組み合わせの要素をすべて0とした。

James D. Hamiltonの"Time Series Analysis"の六章に詳しい記載がある。
はじめに\(\alpha_{h}\)を求めるために\(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\cos(\lambda_h(t-1))\)を考える。 \begin{eqnarray} && \displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\cos(\lambda_h(t-1)) \\ \\ &=& \displaystyle\sum_{t =1}^T(\hat{\mu}+\sum_{k=1}^M [\hat{\alpha}_k\cos(\lambda_k(t -1))+\hat{\beta}_k\sin(\lambda_k(t -1))]-\hat{\mu})\cos(\lambda_h(t-1)) \\ \\ &=& \displaystyle\sum_{t =1}^T(\sum_{k=1}^M [\hat{\alpha}_k\cos(\lambda_k(t -1))+\hat{\beta}_k\sin(\lambda_k(t -1))])\cos(\lambda_h(t-1)) \\ \\ &=& \displaystyle\sum_{t =1}^T(\sum_{k=1}^M [\hat{\alpha}_k\cos(\lambda_k(t -1))\cos(\lambda_h(t-1))+\hat{\beta}_k\sin(\lambda_k(t -1))\cos(\lambda_h(t-1))]) \\ \\ &=& \displaystyle\sum_{t =1}^T(\hat{\alpha}_h\cos(\lambda_h(t -1))\cos(\lambda_h(t-1))) &...(*)\\ \\ &=& \displaystyle\sum_{t =1}^T(\hat{\alpha}_h\cos^2(\lambda_h(t -1))) \\ \\ &=& \displaystyle\sum_{t =1}^T(\hat{\alpha}_h\frac{\cos 2(\lambda_h(t -1))+1}{2}) \\ \\ &=& (\hat{\alpha}_h\frac{0+T}{2}) \\ \\ &=& \hat{\alpha}_h\frac{T}{2} \\ \\ &\Rightarrow& \hat{\alpha}_h=\frac{2}{T}\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\cos(\lambda_h(t-1)) \end{eqnarray} (*)では上の項目と同様にして、総和が0になる項目を0として扱った。
同様にして\(\beta_h\)を求めるために\(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\sin(\lambda_h(t-1))\)を考える。 \begin{eqnarray} && \displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\sin(\lambda_h(t-1)) \\ \\ &=& \displaystyle\sum_{t =1}^T(\hat{\mu}+\sum_{k=1}^M [\hat{\alpha}_k\cos(\lambda_k(t -1))+\hat{\beta}_k\sin(\lambda_k(t -1))]-\hat{\mu})\sin(\lambda_h(t-1)) \\ \\ &=& \displaystyle\sum_{t =1}^T(\sum_{k=1}^M [\hat{\alpha}_k\cos(\lambda_k(t -1))+\hat{\beta}_k\sin(\lambda_k(t -1))])\sin(\lambda_h(t-1)) \\ \\ &=& \displaystyle\sum_{t =1}^T(\sum_{k=1}^M [\hat{\alpha}_k\cos(\lambda_k(t -1))\sin(\lambda_h(t-1))+\hat{\beta}_k\sin(\lambda_k(t -1))\sin(\lambda_h(t-1))]) \\ \\ &=& \displaystyle\sum_{t =1}^T(\hat{\beta}_h\sin(\lambda_h(t -1))\sin(\lambda_h(t-1))) &...(**)\\ \\ &=& \displaystyle\sum_{t =1}^T(\hat{\beta}_h\sin^2(\lambda_h(t -1))) \\ \\ &=& \displaystyle\sum_{t =1}^T(\hat{\beta}_h\frac{1-\cos 2(\lambda_h(t -1))}{2}) \\ \\ &=& (\hat{\beta}_h\frac{T-0}{2}) \\ \\ &=& \hat{\beta}_h\frac{T}{2} \\ \\ &\Rightarrow& \hat{\beta}_h=\frac{2}{T}\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\sin(\lambda_h(t-1)) \end{eqnarray} が得られる。(**)では上記の項目と同様の処置をした。
ここで、 \(\hat{\alpha}_h+i\hat{\beta}_h\)と\(\hat{\alpha}_h-i\hat{\beta}_h\)を考えると、 \begin{eqnarray} && \hat{\alpha}_h+i\hat{\beta}_h \\ \\ &=& \frac{2}{T}\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\cos(\lambda_h(t-1))+i\frac{2}{T}\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\sin(\lambda_h(t-1)) \\ \\ &=& \frac{2}{T}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\cos(\lambda_h(t-1))+i\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\sin(\lambda_h(t-1)) \right) \\ \\ &=& \frac{2}{T}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\right)\left(\cos(\lambda_h(t-1))+i\sin(\lambda_h(t-1)) \right) \\ \\ &=& \frac{2}{T}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\right)\exp\left(i\lambda_h(t-1)\right) \\ \\ && \hat{\alpha}_h-i\hat{\beta}_h \\ \\ &=& \frac{2}{T}\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\cos(\lambda_h(t-1))-i\frac{2}{T}\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\sin(\lambda_h(t-1)) \\ \\ &=& \frac{2}{T}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\cos(\lambda_h(t-1))-i\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\sin(\lambda_h(t-1)) \right) \\ \\ &=& \frac{2}{T}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\right)\left(\cos(\lambda_h(t-1))-i\sin(\lambda_h(t-1)) \right) \\ \\ &=& \frac{2}{T}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\right)\exp\left(-i\lambda_h(t-1)\right) \\ \\ \end{eqnarray} が得られる。\(\hat{\alpha}_h^2+\hat{\beta}_h^2=(\hat{\alpha}_h+i\hat{\beta}_h)(\hat{\alpha}_h-i\hat{\beta}_h)\)を計算すると、 \begin{eqnarray} && \hat{\alpha}_h^2+\hat{\beta}_h^2 \\ \\ &=& (\hat{\alpha}_h+i\hat{\beta}_h)(\hat{\alpha}_h-i\hat{\beta}_h) \\ \\ &=& (\frac{2}{T}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})\right)\exp\left(i\lambda_h(t-1)\right))(\frac{2}{T}\left(\displaystyle\sum_{\tau=1}^T(y_{\tau}-\overline{y})\right)\exp\left(-i\lambda_h(\tau-1)\right)) \\ \\ &=& \frac{2^2}{T^2}\displaystyle\sum_{t=1}^T\sum_{\tau=1}^T(y_{t}-\overline{y})(y_{\tau}-\overline{y})\exp\left(i\lambda_h((t-1)(\tau-1))\right) \\ \\ &=& \frac{2^2}{T^2}\displaystyle\sum_{t=1}^T\sum_{\tau=1}^T(y_{t}-\overline{y})(y_{\tau}-\overline{y})\exp\left(i\lambda_h(t-\tau)\right) \\ \\ \end{eqnarray} ここで、\(\tau=t, \tau=t\pm 1, \tau=t\pm 2,\ldots\)の項に分ける。 \begin{eqnarray} && \hat{\alpha}_h^2+\hat{\beta}_h^2 \\ \\ &=& \frac{2^2}{T^2}\displaystyle\sum_{t=1}^T\sum_{\tau=1}^T(y_{t}-\overline{y})(y_{\tau}-\overline{y})\exp\left(i\lambda_h(t-\tau)\right) \\ \\ &=& \frac{2^2}{T^2}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})(y_{t}-\overline{y})\exp\left(i\lambda_h(t-t)\right)\right.\\ \\ &&+\displaystyle\sum_{t=1}^{T-1}(y_{t}-\overline{y})(y_{t+1}-\overline{y})\exp\left(i\lambda_h(t-(t+1))\right) \\ &&+\displaystyle\sum_{t=2}^T(y_{t}-\overline{y})(y_{t-1}-\overline{y})\exp\left(i\lambda_h(t-(t-1))\right) \\ \\ &&+\displaystyle\sum_{t=1}^{T-2}(y_{t}-\overline{y})(y_{t+2}-\overline{y})\exp\left(i\lambda_h(t-(t+2))\right) \\ &&+\displaystyle\sum_{t=3}^T(y_{t}-\overline{y})(y_{t-2}-\overline{y})\exp\left(i\lambda_h(t-(t-2))\right) \\ \\ &&\vdots \\ \\ &&+\displaystyle\sum_{t=1}^{1}(y_{t}-\overline{y})(y_{t+(T-1)}-\overline{y})\exp\left(i\lambda_h(t-(t+T-1))\right) \\ &&+\displaystyle\sum_{t=T}^T(y_{t}-\overline{y})(y_{t-(T-1)}-\overline{y})\exp\left(i\lambda_h(t-(t-(T-1)))\right) \left.\right)\\ \\ &=& \frac{2^2}{T^2}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})(y_{t}-\overline{y})\right.\\ \\ &&+\displaystyle\sum_{t=1}^{T-1}(y_{t}-\overline{y})(y_{t+1}-\overline{y})\exp\left(-i\lambda_h\right) \\ &&+\displaystyle\sum_{t=2}^T(y_{t}-\overline{y})(y_{t-1}-\overline{y})\exp\left(i\lambda_h\right) \\ \\ &&+\displaystyle\sum_{t=1}^{T-2}(y_{t}-\overline{y})(y_{t+2}-\overline{y})\exp\left(-2i\lambda_h\right) \\ &&+\displaystyle\sum_{t=3}^T(y_{t}-\overline{y})(y_{t-2}-\overline{y})\exp\left(2i\lambda_h\right) \\ \\ &&\vdots \\ \\ &&+\displaystyle\sum_{t=1}^{T-(T-1)}(y_{t}-\overline{y})(y_{t+(T-1)}-\overline{y})\exp\left(-(T-1)i\lambda_h\right) \\ &&+\displaystyle\sum_{t=T}^T(y_{t}-\overline{y})(y_{t-(T-1)}-\overline{y})\exp\left((T-1)i\lambda_h\right) \left.\right)\\ \\ &=& \frac{2^2}{T^2}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})(y_{t}-\overline{y})\right.\\ \\ &&+\displaystyle\sum_{t-1=2}^{T}(y_{t-1}-\overline{y})(y_{(t-1)+1}-\overline{y})\exp\left(-i\lambda_h\right) \\ &&+\displaystyle\sum_{t=2}^T(y_{t}-\overline{y})(y_{t-1}-\overline{y})\exp\left(i\lambda_h\right) \\ \\ &&+\displaystyle\sum_{t-2=3}^{T}(y_{t-2}-\overline{y})(y_{(t-2)+2}-\overline{y})\exp\left(-2i\lambda_h\right) \\ &&+\displaystyle\sum_{t=3}^T(y_{t}-\overline{y})(y_{t-2}-\overline{y})\exp\left(2i\lambda_h\right) \\ \\ &&\vdots \\ \\ &&+\displaystyle\sum_{t-(T-1)=T}^{T}(y_{t-(T-1)}-\overline{y})(y_{(t-(T-1))+(T-1)}-\overline{y})\exp\left(-(T-1)i\lambda_h\right) \\ &&+\displaystyle\sum_{t=T}^T(y_{t}-\overline{y})(y_{t-(T-1)}-\overline{y})\exp\left((T-1)i\lambda_h\right) \left.\right)\\ \\ &=& \frac{2^2}{T^2}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})(y_{t}-\overline{y})\right.\\ \\ &&+\displaystyle\sum_{t-1=2}^{T}(y_{(t-1)+1}-\overline{y})(y_{t-1}-\overline{y})\exp\left(-i\lambda_h\right) \\ &&+\displaystyle\sum_{t=2}^T(y_{t}-\overline{y})(y_{t-1}-\overline{y})\exp\left(i\lambda_h\right) \\ \\ &&+\displaystyle\sum_{t-2=3}^{T}(y_{(t-2)+2}-\overline{y})(y_{t-2}-\overline{y})\exp\left(-2i\lambda_h\right) \\ &&+\displaystyle\sum_{t=3}^T(y_{t}-\overline{y})(y_{t-2}-\overline{y})\exp\left(2i\lambda_h\right) \\ \\ &&\vdots \\ \\ &&+\displaystyle\sum_{t-(T-1)=T}^{T}(y_{(t-(T-1))+(T-1)}-\overline{y})(y_{t-(T-1)}-\overline{y})\exp\left(-(T-1)i\lambda_h\right) \\ &&+\displaystyle\sum_{t=T}^T(y_{t}-\overline{y})(y_{t-(T-1)}-\overline{y})\exp\left((T-1)i\lambda_h\right) \left.\right)\\ \\ &=& \frac{2^2}{T^2}\left(\displaystyle\sum_{t=1}^T(y_{t}-\overline{y})(y_{t}-\overline{y})\right.\\ \\ &&+\displaystyle\sum_{t-1=2}^{T}(y_{t}-\overline{y})(y_{t-1}-\overline{y})\exp\left(-i\lambda_h\right) \\ &&+\displaystyle\sum_{t=2}^T(y_{t}-\overline{y})(y_{t-1}-\overline{y})\exp\left(i\lambda_h\right) \\ \\ &&+\displaystyle\sum_{t-2=3}^{T}(y_{t}-\overline{y})(y_{t-2}-\overline{y})\exp\left(-2i\lambda_h\right) \\ &&+\displaystyle\sum_{t=3}^T(y_{t}-\overline{y})(y_{t-2}-\overline{y})\exp\left(2i\lambda_h\right) \\ \\ &&\vdots \\ \\ &&+\displaystyle\sum_{t-(T-1)=T}^{T}(y_{t}-\overline{y})(y_{t-(T-1)}-\overline{y})\exp\left(-(T-1)i\lambda_h\right) \\ &&+\displaystyle\sum_{t=T}^T(y_{t}-\overline{y})(y_{t-(T-1)}-\overline{y})\exp\left((T-1)i\lambda_h\right) \left.\right)\\ \\ &=& \frac{2^2}{T^2}\left(T\gamma_0\right.\\ \\ &&+T\gamma_{1}\exp\left(-i\lambda_h\right) \\ &&+T\gamma_{1}\exp\left(i\lambda_h\right) \\ \\ &&+T\gamma_{2}\exp\left(-2i\lambda_h\right) \\ &&+T\gamma_{2}\exp\left(2i\lambda_h\right) \\ \\ &&\vdots \\ \\ &&+T\gamma_{T+1}\exp\left(-(T-1)i\lambda_h\right) \\ &&+T\gamma_{T+1}\exp\left((T-1)i\lambda_h\right) \left.\right)&\gamma_h=\gamma_{-h}より\\ \\ &=& \frac{2^2}{T}\left( \displaystyle\sum_{h=-T+1}^{T-1}\hat{\gamma}_he^{-i\lambda_h h}\right)\\ \\ &=& 2\pi\frac{4}{T}\hat{f}(\lambda_h) \\ \\ &\Leftrightarrow& \frac{1}{2}(\hat{\alpha}_h^2+\hat{\beta}_h^2)=\frac{4\pi}{T}\hat{f}(\lambda_h) \end{eqnarray}