I am working on a ratings transition matrix and I wondered how people scale it down to shorter time periods (although one should more or less stick to the estimation period i know).

It is clear that one can scale it to a longer time period by repeatedly multiplying the matrix with itself.

How would you scale the matrix down to a, say monthly, transition matrix?

I found a result on reversible Markov Chains and thought the following:

Let $M$ be the transition matrix of a reversible markov chain, then one can factor it into $$ M = S D$$

with a symmetric matrix $S$ and a diagonal matrix $D$. It appears that also the converse holds true.

Now, if we had a reversible markov chain, we could raise the matrix to a fraction by taking the eigenvalue decomposition of $S = E^{T}\Lambda E$ and raising $S$ as well as $D$ to the appropriate power.

So we define

$$ \bar{M}:= E^T \Lambda^{\frac{1}{n}}E D^{\frac{1}{n}}$$ and say that $\bar{M}$ is now $M^{\frac{1}{n}}$.

The question is: Is this way appropriate? Does it make any sense (from a mathematical viewpoint - I know that downscaling a transition matrix is debatable)? Are ratings transition matrices from a reversible markov chain?

The main question is: What are the commonly used method to raise ratings transition matrices to a fraction?


1 Answer 1


You are right, the rules to time-scale a T-years transition matrix $M_T$ are:

  • $M_{k·T} = M_T^k$
  • $M_{T/k} = \sqrt[k]{M_T}$

The root of a matrix M can be obtained using the spectral decomposition:

$M = P·D·P^{-1} \Longrightarrow M^k = P·D^k·P^{-1}$

where $P$ and $D$ are the eigenvectors and eigenvalue matrices of $M_T$.

Note: The Perron-Frobenius tells that transition matrix eigenvalues satisfies |λ|≤1. This allows negative and complex eigenvalues. The following R code shows a transition matrix with a negative eigenvalue:

M = matrix(ncol=3, nrow=3, 0)
M[1,1]=0.1; M[1,2]=0.8; M[1,3]=0.1
M[2,1]=0.8; M[2,2]=0.1; M[2,3]=0.1
M[3,1]=0.0; M[3,2]=0.0; M[3,3]=1.0

Note: As indicated in Regularization Algorithms for Transition Matrices, the root of a transition matrix can be an invalid transition matrix or may not be unique. In this case, we must transform this matrix to the relevant Markov matrix. This process is called regularization. In the referenced document there exist some algorithms to regularize a transition matrix.

Note: I suspect that the transition matrices with a single absorbent state and strictly diagonal dominant has ​​real and positive eigenvalues, but I have no proof.

You can obtain more info at CCruncher technical document.

  • $\begingroup$ Algorithms for computing these roots stably differ a bit from this approach. Nick Higham does good work in this field. Try this paper $\endgroup$
    – Brian B
    Mar 20, 2014 at 19:15
  • $\begingroup$ @Gerard Torrent OK I think I am missing something: Why should the eigenvalues of $D$ be nonnegative or even real? Why should all fractional powers exist (like 1/12 for example)? $\endgroup$
    – vanguard2k
    Mar 21, 2014 at 10:29
  • $\begingroup$ @BrianB Nice!! Now that I read it, the Schur decomposition looks like the natural approach for this... $\endgroup$
    – vanguard2k
    Mar 21, 2014 at 10:42
  • $\begingroup$ @vanguard2k: I edited the entry trying to respond your question. $\endgroup$ Mar 22, 2014 at 7:41
  • $\begingroup$ @Brian B: As I understand, the Schur-Pade algorithm obtains the same root matrix but with a superior accuracy. It can be usefull when the transition matrix is ill-conditioned. $\endgroup$ Mar 22, 2014 at 7:42

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