Schur damping

How far do you trust a floating joint's recoil?

Eliminating the free joint from the chain in the reduced mass demo is a genuine Schur complement. The energy's Hessian in the particle $z$ and the joint $w$ is the precision matrix. Minimising out $w$ takes the Schur complement of the $w$ block. Issai Schur wrote the determinant identity behind this in 1917. Emilie Haynsworth attached his name to the operation in 1968.

$$ \Lambda = \begin{pmatrix} p+\rho & -\rho \\ -\rho & \rho+\phi \end{pmatrix}, \qquad (p+\rho) \;-\; \frac{\rho^2}{\rho+\phi} \;=\; p \;+\; \underbrace{\frac{\phi\rho}{\phi+\rho}}_{S(1)} . $$

The prior's $p$ rides through untouched. So the particle's effective stiffness toward the observation is $S(1) = \rho - \rho^2/(\rho+\phi)$. That is the channel stiffness minus a correction fed back by the joint's recoil. Schur damping keeps only a fraction $\gamma$ of that correction,

$$ S(\gamma) \;=\; \rho \;-\; \gamma\,\frac{\rho^2}{\rho+\phi}, $$

and the dial has a mechanical reading. At $\gamma = 1$ the joint floats free and the particle feels the true marginal, which is the reduced mass. At $\gamma = 0$ the joint is clamped to the data. The particle inherits the channel's full stiffness $\rho$ and is pulled hard toward $y$. It is overconfident, because the joint's own uncertainty has been ignored. An interior $\gamma$ trusts the recoil partway. That is the right posture when the coupling $\rho$ is estimated rather than known. The same dial damps the off-diagonal feedback of a partitioned precision matrix. At the sister site schur.microprediction.org it interpolates hierarchical and optimisation-based portfolio construction.

The top rail is the true chain. The bottom rail is the $\gamma$-damped consolidation. Slide $\gamma$ to 1 and the two particles meet on the dashed line. Slide it toward 0 and the bottom body swells to mass $\rho$ and drags its particle toward the observation. Drag any body to move the anchors.

damping $\gamma$ obs. mass $\phi$ channel $\rho$ prior mass $P^{-1}$

damped mass $S(\gamma)$ exact (reduced) mass $S(1)$ chain particle damped particle

The damped rail is still a real spring system. Its stiffness toward the data is $S(\gamma)$ instead of $S(1)$. One honest caveat. The algebra fixes the damped stiffness, but where the damped spring anchors is a modelling choice. Here the undamped remainder stays anchored at the data. That is the plug-in reading of $\gamma = 0$, with the joint clamped. Damping does not break the physics. It changes which physics you assert. See the Schur damping papers through the schur site's bibliography.

The usual explanation integrate the joint out, term by term

Write the chain as one joint Gaussian in the particle $z$ and the floating joint $w$. The energy is $\tfrac12 p(z-m)^2 + \tfrac12\rho(z-w)^2 + \tfrac12\phi(w-y)^2$, and its Hessian is the precision matrix:

$$ \Lambda = \begin{pmatrix} p+\rho & -\rho \\ -\rho & \rho+\phi \end{pmatrix}. $$

Marginalising the joint means integrating $w$ out of the density. Only the quadratic form sets the leftover precision, so work in deviations from the mean and complete the square in $w$:

$$ (p+\rho)\,z^2 - 2\rho\,z w + (\rho+\phi)\,w^2 = (\rho+\phi)\Big(w - \tfrac{\rho}{\rho+\phi}\,z\Big)^2 + \Big[(p+\rho) - \tfrac{\rho^2}{\rho+\phi}\Big]z^2. $$

The Gaussian integral over $w$ eats the first bracket and leaves the second. That leftover coefficient is the precision the particle keeps. It is the Schur complement of the $w$ block:

$$ (p+\rho) - \frac{\rho^2}{\rho+\phi} \;=\; p + \frac{\rho(\rho+\phi) - \rho^2}{\rho+\phi} \;=\; p + \frac{\phi\rho}{\phi+\rho}. $$

The prior's $p$ survives untouched and the coupling collapses to $\phi\rho/(\phi+\rho)$. That is the marginal stiffness $S(1)$ the demo damps. A little integration, and you have the reduced coupling.

The physics proof the same stiffness, two springs in series

The reduced mass page already did the elimination. Two springs in series, $\rho$ then $\phi$, combine into one whose compliances add. Take that as given:

$$ \frac{1}{S(1)} = \frac{1}{\rho} + \frac{1}{\phi} \;\Longrightarrow\; S(1) = \frac{\phi\rho}{\phi+\rho}. $$

That series combination is the joint recoiling instead of holding firm. The prior sits in parallel and simply adds its $p$. The $\gamma$ dial keeps a fraction of the recoil, sliding the joint from free at $\gamma=1$ to clamped at $\gamma=0$. No Schur algebra required. You are turning a knob on how much the joint is allowed to move.