Example. Exact covariance on the alternating four-cycle [fdip-0037]

Take the four-cycle on \(V=(0,1,2,3)\), with forward edges and weights \((1,t,1,t)\), \(t>0\). Use counting pairing and \(Q=L+m^2I\), \(m>0\). Put \(a=m^2\) and \(s=1+t+a\). The following functions on vertices form an orthonormal eigenbasis; the middle column lists the unnormalized rows \(2u_i^T\): \[\begin {array}{c|c|c} i&2u_i^T&\lambda _i\\\hline 0&(1,1,1,1)&0\\ 1&(1,-1,-1,1)&2\\ 2&(1,1,-1,-1)&2t\\ 3&(1,-1,1,-1)&2+2t \end {array}\]

Set \[\begin {aligned} q_0&=a,\\ q_1&=a+2,\\ q_2&=a+2t,\\ q_3&=a+2+2t,\\ \Delta &=a(a+2)(a+2t)(a+2+2t). \end {aligned}\] All 4 \(q_i\) are positive. With \(P_{u_i}=u_i u_i^T\), normalization and inverse covariance give \[\begin {aligned} \det Q&=\Delta ,& Z_Q&=4\pi ^2/\sqrt \Delta ,\\ G&=\sum _{i=0}^3q_i^{-1}P_{u_i}. \end {aligned}\] With \(O\) the array whose columns are \(u_i\), this is \(G=O\operatorname {diag}(q_i^{-1})O^T\).

Write \(d_i=q_i^{-1}\). Multiplying the signs of the relevant coordinates in each projector gives \[\begin {aligned} 4G_{00}&=d_0+d_1+d_2+d_3,\\ 4G_{01}&=d_0-d_1+d_2-d_3,\\ 4G_{02}&=d_0-d_1-d_2+d_3,\\ 4G_{03}&=d_0+d_1-d_2-d_3. \end {aligned}\] Thus the complete covariance, with \((A,B,C,D)=(G_{00},G_{01},G_{02},G_{03})\), is \[G=\begin {pmatrix} A&B&C&D\\B&A&D&C\\C&D&A&B\\D&C&B&A \end {pmatrix}.\] In particular the variance and adjacent two-point expectation are the first two formulas, with their indicated signs.

For the opposite entry, the common denominator \(4\Delta \) has numerator \[\begin {aligned} N&=(a+2)(a+2t)(a+2+2t)\\ &\quad -a(a+2t)(a+2+2t)\\ &\quad -a(a+2)(a+2+2t)\\ &\quad +a(a+2)(a+2t)\\ &=2[(a+2t)(a+2+2t)-a(a+2)]\\ &=8t(a+1+t). \end {aligned}\] Consequently \(G_{02}=2ts/\Delta \). At \((t,m)=(1,1)\), \(q=(1,3,3,5)\), \(\Delta =45\), \(Z_Q=4\pi ^2/(3\sqrt 5)\), and \[\begin {aligned} &(G_{00},G_{01},G_{02},G_{03})\\ &\qquad =(7/15,1/5,2/15,1/5). \end {aligned}\] At \((t,m)=(2,1)\), \(q=(1,3,5,7)\), \(\Delta =105\), and \[(G_{00},G_{01},G_{02},G_{03}) =(44,19,16,26)/105.\] These numbers are obtained by direct substitution in the 4 signed sums above. They give, in particular, the opposite covariance used in block elimination and the two distinct adjacent covariances used in relabeling.

More generally, weights \((b,c,b,c)\) with \(b,c>0\) keep the same 4 eigenvectors and give eigenvalues \(0,2b,2c,2b+2c\), by the same coordinate products. This is a real spectral calculation. The arithmetic columns are \(2u_i\), whose Gram matrix is \(4I\); they are not an orthonormal basis over an arbitrary coefficient ring.

The eigenvectors and signed sums are computed here from the weighted cycle.

[boyd2004convex, Section A.5.2, pp. 646-647] gives general spectral decomposition.

[friedli2017statistical, Proposition B.55, p. 505] gives the Gaussian density. Covariance equals inverse precision by Theorem [fdip-000B].