A three-variable chain

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Let X_1\sim\operatorname{N}(0,1), X_2=0.8 X_1+\epsilon_2 and X_3=0.8 X_2+\epsilon_3, with \epsilon_2,\epsilon_3\sim\operatorname{N}(0,0.36) independent of everything else. Find \Sigma and K=\Sigma^{-1}, and check k_{13}=0.

Worked answer (click to show)

\operatorname{Var} X_2=0.64\cdot 1+0.36=1, \qquad \operatorname{Var} X_3=0.64\cdot 1+0.36=1,

\operatorname{Cov}(X_1,X_2)=0.8, \operatorname{Cov}(X_2,X_3)=0.8\,\operatorname{Var} X_2=0.8, and \operatorname{Cov}(X_1,X_3)=0.8\,\operatorname{Cov}(X_1,X_2)=0.64. So

\Sigma=\begin{pmatrix} 1 & 0.8 & 0.64 \\ 0.8 & 1 & 0.8 \\ 0.64 & 0.8 & 1 \end{pmatrix},

the autoregressive form with r=0.8. Its inverse is

K=\frac{1}{1-r^2} \begin{pmatrix} 1 & -r & 0 \\ -r & 1+r^2 & -r \\ 0 & -r & 1 \end{pmatrix} =\frac{1}{0.36} \begin{pmatrix} 1 & -0.8 & 0 \\ -0.8 & 1.64 & -0.8 \\ 0 & -0.8 & 1 \end{pmatrix} =\begin{pmatrix} 2.778 & -2.222 & 0 \\ -2.222 & 4.556 & -2.222 \\ 0 & -2.222 & 2.778 \end{pmatrix}.

Indeed k_{13}=0. The first row of \Sigma times the third column of K is 0.8(-2.222)+0.64(2.778)=-1.778+1.778=0, as required for an off-diagonal entry of \Sigma K=I.