Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a markov kernel K. From this I find the invariant probability $\pi$. The question is to design a "dream" matrix K*, that converges in one step. Such that $\lambda_{SLEM}=0$ (second largest eigen-value modulus). I am not sure how to go about designing the dream matrix. Any pointers to literature/pointers would be welcome

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

Consider three states $A,B,C$. We have that $A$ almost surely goes to itself and no other states. Let $B$ almost surely go to $A$ and to no other states. Let $C$ go to $A$ with probability $0.5$ and to $B$ with probability $0.5$.

The transition matrix may be seen as $M = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 0.5 & 0.5 & 0 \end{array} \right) $

The eigenvalues of the associated transition matrix $M$ are $1,0,0$. The powers of the transition matrix $M$ are $M^2, M^3, \dots$, these just give us the matrix $A$ defined by

$A= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right) $

This converges in just one step.

EDIT:

Suppose we didn't know how to get there. We seek a transition matrix with an existing limiting distribution, then the Perron-Frobenius theorem tells us the spectral radius of this matrix must be $1$ with unique multiplicity one and all other eigenvalues smaller. So simply pick a matrix of order three by three with nine coefficients to be found and calculate what you need to satisfy those conditions.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.