# "Forcing" equal probabilities in the matrix exponential of a Markov intensity matrix

I have an upper-right triangular transition intensity matrix $$Q$$ for a 7-state Markov model (with states $$X_1,X_2,...,X_7$$), from which I numerically calculate the matrix exponential to derive a transition probability matrix $$P$$.

A number of the transitions into one of the states have equal rates in the intensity matrix, say: $$q_{1,6} = q_{2,6} = q_{3,6}$$ The matrix is otherwise populated such that, after taking the matrix exponential, the derived probabilities of transitioning into $$X_6$$ from states $$X_1,X_2,X_3$$ are then different, i.e.: $$p_{1,6} ≠ p_{2,6} ≠ p_{3,6}$$

This is, of course, an expected consequence of taking the matrix exponential; however, for the purposes of a simulation I am trying to run, it is undesirable as it results in different proportions in $$X_6$$ in two different "arms" of the simulation based on the same Markov structure, but with different initial distributions across $$X_1,X_2,X_3$$.

Putting concerns around competing risks aside, is there a method by which the intensity matrix can be adjusted (or indeed any other method) to ensure that the derived probabilities of transitioning into a specific state are the same where the underlying rates are the same? (i.e. to ensure that $$p_{1,6} = p_{2,6} = p_{3,6}$$ when $$q_{1,6} = q_{2,6} = q_{3,6}$$).

• The intensity matrix answers the question of "If we are in state $$1$$, with what frequency will state $$2$$ be the next state?".
• The transition matrix answers the question of "If we are in state $$1$$ now, what is the probability of being in state $$2$$ some unit of time later?".
The key difference here is that in the second case we can take any route from $$1$$ to $$2$$. (i.e. we can go $$1 \implies 3 \implies 2$$ or any other path).
So the transition probabilities $$p_{16}$$, $$p_{26}$$ and $$p_{36}$$ will be different as states $$1$$, $$2$$ and $$3$$ have different probabilities of taking the various paths.