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For the first $i$ observations, writeBy Gaussian conjugacy for a normal linear model, the prefix posterior isCompute a Cholesky decomposition of $A_0={\Sigma }^{-1}$ once. If $x_i^T$ is row $i$ of the design matrix, thenThe Rank-one Cholesky update obtains a triangular factor $A_i=L_i{L}_i^T$ from $L_{i-1}$ in $O(p^2)$ operations. Two triangular solves give $m_i=A_i^{-1}{b}_i$, and, for $z_i\sim N(0,I_p)$, another solve givesThe initial factorization costs $O(p^3)$ and all $n$ updates and samples cost $O(n{p}^2)$. This is within the requested $O(p^3+n{p}^2+n^{2}p)$ bound.

Put $z_i=\log Y_i$. The Weighted graph Laplacian $L_J$ of the tree is defined byMultiplication of the Gaussian likelihood by the prior shows that, conditionally on the precision parameter $\tau$,Completing the square therefore givesAs a function of $\tau$, the posterior density isso, in shape-rate notation,The precision matrix $M$ has the sparsity pattern of a tree. A sparse Cholesky decomposition and its triangular solves have $O(n)$ cost and storage on this graph, while the gamma update also costs $O(n)$. Hence each systematic-scan Gibbs sampler iteration costs $O(n)$.

Given a Cholesky decomposition $A=L{L}^T$, a rank-one Cholesky update computes a triangular factor of $A+x{x}^T$ in $O(n^2)$ operations without refactorizing the matrix from scratch.