How Can Markov Models Be Used to Compare Transition State Matrices?

  • Context: Graduate 
  • Thread starter Thread starter fsteveb
  • Start date Start date
  • Tags Tags
    Compare Matrices
Click For Summary
SUMMARY

This discussion focuses on comparing transition state matrices using Markov models. The user seeks methods to assess similarity between 3x4 and 4x4 matrices, noting that averaging the diagonal elements is inadequate due to the zero value in the 4,4 position. Steve Brailsford suggests utilizing eigenvalues, emphasizing that if the processes lack absorbing states, the eigenvectors associated with eigenvalue 1 represent the stationary state, providing a more reliable comparison method.

PREREQUISITES
  • Understanding of Markov models and their applications
  • Familiarity with matrix operations and properties
  • Knowledge of eigenvalues and eigenvectors
  • Concept of stationary states in Markov processes
NEXT STEPS
  • Research methods for comparing Markov transition matrices
  • Learn about eigenvalue decomposition and its applications
  • Explore the concept of stationary distributions in Markov chains
  • Investigate alternative similarity measures for matrices
USEFUL FOR

Researchers in applied mathematics, data scientists, and anyone working with Markov models and transition state analysis.

fsteveb
Messages
3
Reaction score
0
I have a problem where I get 3 or 4x4 matrices and I'd like to compare them. The matrices are transition states so markov models are applicable, but I can't find anything about how to compare the matrices for similarity. One solution that has been done is to agv the diagonal, but since the 4,4 element is always zero, your only using 3 numbers of the 16 and throwing the rest away. The determinant has no correlation between the system so can't be used since it is too affected by single value changes. Does anyone know of another method I might be able to use?
Steve Brailsford
 
Physics news on Phys.org
I would use eigenvalues. If the processes don't have any absorbing states then the eigenvectors corresponding to eigenvector 1 is the stationary state.
 

Similar threads

Replies
24
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 65 ·
3
Replies
65
Views
20K
Replies
0
Views
891
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K