Recovering a One-Year Transition Matrix from a Three-Year Matrix

Summary

The document asks how to infer an average one-year transition matrix when multiplying three annual matrices produces a three-year transition matrix. The response proposes taking a matrix cube root: if the relevant matrix can be diagonalized with distinct eigenvalues, transform it to diagonal form, take cube roots of the diagonal entries, and transform back. The result is a candidate one-year matrix whose third power equals the given three-year matrix.

The response gives a linear-algebra procedure rather than an application to a particular transition matrix. It notes an eigenvalue condition but does not explain how to select among possible roots, whether the recovered matrix remains a valid transition matrix with nonnegative entries and rows summing to one, or how to handle non-diagonalizable cases. Consequently, the procedure does not by itself establish that a unique or probabilistically meaningful annual matrix exists.

Key ideas

• A three-year transition matrix can be approached by finding a matrix cube root.
• Diagonalization reduces the calculation to taking cube roots of eigenvalues.
• Distinct eigenvalues are given as a condition for the proposed procedure.
• A matrix root may not be unique or satisfy the constraints of a transition matrix.

Tags

# How do I get the average transition matrix for three consecutive years?


# How do I get the average transition matrix for three consecutive years?












I have a one year transition matrix for three consecutive years. Multiplying these three matrices together yields the three year transition matrix. I want to obtain the average transition matrix for the three years (average^3 = 3yrtransition)

What is the procedure to be used? Is this possible at all? (I kind of realize that there might be multiple solutions to this problem due to possible multiple paths to achieve the end state).

## Answer by Johann Hibschman (score 5)

https://quant.stackexchange.com/a/1183

If the transition matrix has distinct eigenvalues, you can diagonalize it and then take the cube root of the diagonal. E.g., you can compute the SVD, verify that the eigenvalues are distinct, take the cube root of the diagonal matrix, then re-multiply it together.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.