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A stochastic matrix, with elements $\in[0,1]$ and rows summing to 1 are known to have one eigenvalue 1 (stationary distribution) and the rest of lower magnitude. However I don't know about many results regarding their inverses.

  1. In which cases is it possible to find another stochastic matrix being the inverse to the first?
  2. If we can not find an inverse fulfilling the requirements, can we find some "pseudo" inverse?
  3. Can we "expand" the probability space to find a larger space where it is possible to find a matrix ?

EDIT: The closest I have come so far is to expand the space to double number of states and allowing elements > 1. Let us call the $N\times N$ probability transition matrix $\bf P$, then assuming $\bf P$ nonsingular we can calculate ${\bf P}^{-1}$, build the new matrices :

$${\bf P_e = P \otimes I}_2\hspace{1cm}{\bf P_{ie}} = |{\bf P}^{-1}|\otimes ({{\bf 1}_2{\bf 1}_2}^T-{\bf I}_2)^{(1-\text{sgn}({\bf P}))\otimes {\bf I}_2}$$

Where the $|\cdot|$ and power are scalar wise operations.

If we do this then of course $\bf P_{ie}P_e \neq I\hspace{1cm}\bf P_{e}P_{ie} \neq I$, but that is because we are not finished yet!

If we now turn each $2\times2$ block $\bf A$ into a scalar by the following calculation: $[1,0] {\bf A} [1,-1]^T$

And systematically apply to all blocks:

$$\cases{({\bf I}_N \otimes [1,0])({\bf P_{ie}P_e})({\bf I}_N \otimes [1,-1]^T)\\ ({\bf I}_N \otimes [1,0])({\bf P_{e}P_{ie}})({\bf I}_N \otimes [1,-1]^T)}$$

  1. Now we do indeed get identities!
  2. Row sums of $\bf P_{ie}$ equals 1 if we treat every second bin as $-1$ (as the construction above does).
  3. But how to interpret the elements of $\bf {P_{ie}}$ which can be >1 ?

Furthermore $\sum_j |({\bf P_{ie}})_{ij}|$ seems to be a measure of the confusion, it goes to $\infty$ as ${\bf P} \to \frac{1}{N^2}{{\bf 1}_N} {{\bf 1}_N}^T$ which is indeed the maximally confusing distribution as we lose all information except the mean values ( all other eigenvalues except stationary distribution will be 0 ).

Rodrigo de Azevedo
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mathreadler
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1 Answers1

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The only stochastic matrices with a stochastic inverse are the permutation matrices. If $A$ and $B$ are stochastic, then all entries of $AB$ are non-negative. The only way $(AB)_{ij}$ can be zero is if the "support" of row $i$ of $A$ and column $j$ of $B$ are disjoint. So if $AB=I$ we need row $1$ of $A$ to be orthogonal to columns $2,\ldots,n$ of $B$. As these columns must be linearly independent, then row $1$ of $A$ can only have one non-zero entry. Continuing, all rows of $A$ have only one nonzero entry etc.

Angina Seng
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  • This is as I suspected. Kind of sad. I went the other direction determined to make ${\bf P}^{-1}$ interpretable in a probability context, but it is not very intuitive how to do it. – mathreadler Aug 14 '17 at 08:44
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    @mathreadler It's actually counterintuitive (from the probabilistic perspective) that you would think it could happen at all. If the chain is deterministic, then you have either a permutation matrix (which is invertible) or there are transient states (so you cannot go backward in time because you can't tell whether you got to your current absorbing state from a transient state or because you started there). If the chain isn't deterministic, then asking for an inverse is like asking for a recipe for "un-diffusion", which is badly defined even in the discrete situation. – Ian Aug 17 '18 at 15:53
  • @Ian it's ok I found another fruitful approach. We need to blow up the probability space a factor of 2 to avoid negative numbers. You can think of it in a non-probabilistic context. We have non-negative quantities which that are being "moved around" i.e. "transported" by percentages. – mathreadler Aug 17 '18 at 20:30