# Robust estimation of tree structured Gaussian Graphical Model

arXiv: https://arxiv.org/abs/1901.08770 · HTML: https://mlsanity.com/paper/1901.08770/

## Key facts

- Cited by 7 papers in the map (data through 14 Oct 2026).
- First posted on arXiv 25 Jan 2019.

## Built on by (most cited first, 7 of 7)

- [SGA: A Robust Algorithm for Partial Recovery of Tree-Structured Graphical Models with Noisy Samples](https://mlsanity.com/paper/2101.08917/) (2021)
- [Robust estimation of tree structured models](https://mlsanity.com/paper/2102.05472/) (2021)
- [Robustifying Algorithms of Learning Latent Trees with Vector Variables](https://mlsanity.com/paper/2106.00885/) (2021)
- [Learning Linear Polytree Structural Equation Models](https://mlsanity.com/paper/2107.10955/) (2021)
- [Recoverability Landscape of Tree Structured Markov Random Fields under Symmetric Noise](https://mlsanity.com/paper/2102.08554/) (2021)
- [Decentralized Learning of Tree-Structured Gaussian Graphical Models from Noisy Data](https://mlsanity.com/paper/2109.10642/) (2021)
- [Robust Model Selection of Gaussian Graphical Models](https://mlsanity.com/paper/2211.05690/) (2022)

## Abstract (arXiv)

Consider jointly Gaussian random variables whose conditional independence structure is specified by a graphical model. If we observe realizations of the variables, we can compute the covariance matrix, and it is well known that the support of the inverse covariance matrix corresponds to the edges of the graphical model. Instead, suppose we only have noisy observations. If the noise at each node is independent, we can compute the sum of the covariance matrix and an unknown diagonal. The inverse of this sum is (in general) dense. We ask: can the original independence structure be recovered? We address this question for tree structured graphical models. We prove that this problem is unidentifiable, but show that this unidentifiability is limited to a small class of candidate trees. We further present additional constraints under which the problem is identifiable. Finally, we provide an O(n^3) algorithm to find this equivalence class of trees.