Conditioning a Gaussian graph

Pin a node, and the conditional mean ripples through the network.

Fix a variable in a multivariate Gaussian at a known value and ask where the mean of the others moves to. That is conditioning. In precision form it holds the pinned coordinates fixed and solves the free block for the rest. Below, that solve is a spring network. Pin a bead at a value, let go, and the rest of the network settles at the conditional mean. No matrix algebra runs here. The springs do the elimination.

Six beads sit on frictionless vertical rods, joined by springs along a chain. One shortcut edge runs from node 1 to node 4. Both ends, 0 and 5, start pinned in red. So the four beads between them already show a real conditional mean, not just wherever they happened to start. Drag any bead and it pins wherever you drop it. That is you asserting a new conditioning value. Tap a bead without dragging to unpin it, or to pin it in place. Unpin all releases every constraint, and the network relaxes toward the faint dashed “flat prior” line. That line is a weak default belief every node carries. It is there so the system has a well-defined rest state when nothing else is fixed.

Keep both ends pinned and watch node 2 as you drag node 0. It moves. Correlation propagates along every path, attenuated by distance and by the shortcut. Now pin node 1 and drag it. Node 4 twitches at once, skipping past nodes 2 and 3. Conditioning on a graph is not local the way the chain smoother was. There, influence still reached distant nodes, but only by passing through every node in between.

Switch to the star topology and the same mechanics tell a collaborative story. Node 0 is a shared hub. The other five nodes are independent observers, each tied to the hub by its own spring of stiffness $\phi_i$. Two spokes start pinned at different values, and the hub already sits at their precision-weighted average. This is the fusion demo generalised to five voices instead of two. Pin a third spoke, or unpin one partway through, and the consensus shifts as that vote is added or withdrawn.

chain stiffness shortcut stiffness

pinned: none click a bead to pin/unpin; drag to move it

This is the multivariate generalisation the introduction promises in step 6. Pinning part of a spring network to the wall and solving for the rest is conditioning on a Gaussian graph. Arthur Dempster gave this structure its statistical name in 1972, in a paper called covariance selection. He showed that a zero in the precision matrix means two variables are independent once you condition on all the rest. The graph of the nonzero entries is the graph you are dragging here. The sister site schur.microprediction.org works out the block version of the same move for hierarchical portfolio allocation.

The usual explanation partition, invert, and read off the conditional mean

Split the variables into a free block $F$ and a pinned block $P$. Write the joint Gaussian in precision form, since a zero in the precision is exactly the missing edge you are dragging:

$$ -2\log p(x) = (x-\mu)^\top \Lambda\,(x-\mu) + \text{const}, \qquad \Lambda = \begin{pmatrix} \Lambda_{FF} & \Lambda_{FP} \\ \Lambda_{PF} & \Lambda_{PP} \end{pmatrix}. $$

Conditioning holds $x_P$ fixed at the pinned value. Substitute it and keep only the terms that still carry $x_F$. The quadratic form collapses to a quadratic in $x_F$ alone:

$$ -2\log p(x_F \mid x_P) = (x_F-\mu_F)^\top \Lambda_{FF}\,(x_F-\mu_F) + 2\,(x_F-\mu_F)^\top \Lambda_{FP}\,(x_P-\mu_P) + \text{const}. $$

That is a Gaussian in $x_F$ with precision $\Lambda_{FF}$. Differentiate and set the gradient to zero to find its mean:

$$ \Lambda_{FF}\,(x_F-\mu_F) + \Lambda_{FP}\,(x_P-\mu_P) = 0. $$

Solve the free block for $x_F$. The conditional mean is the answer, and the conditional precision is just the free block $\Lambda_{FF}$ left sitting there:

$$ \mu_{F\mid P} = \mu_F - \Lambda_{FF}^{-1}\,\Lambda_{FP}\,(x_P-\mu_P). $$

Note the block that gets inverted is $\Lambda_{FF}$, the free part of the precision, not a Schur complement. Conditioning solves the free block. Marginalising would take the Schur complement instead. Several lines of block algebra, and you have where the mean of the unpinned nodes moves to.

The physics proof the same mean, read off the network

The fusion page already settled the centre-of-mass rule. Take it as given. Each free bead feels one spring per neighbour and rests where those pulls cancel:

$$ \sum_j k_{ij}\,(z_j - z_i) = 0 \;\Longrightarrow\; z_i = \frac{\sum_j k_{ij}\,z_j}{\sum_j k_{ij}}. $$

One bead reading the precision-weighted average of its neighbours is exactly fusion. Every free bead wants that same balance at once. Pin the boundary, let go, and the network settles into the single configuration where all of them hold together. That joint balance is the conditional mean. The picture is the solve.