Continuity of Markov chain' trasnition matrix functions

Question

I'm studying Continuous-time Markov chains.

Given the definition of continous-time transition functions matrix:

$P(t) = (P\small{ij}(t)) = \mathbb{P}(X\small{t} = j | X\small{0} = i)$

So I have two questions related to continuity of this functions:

1. Can there be transitional functions which are non-continous?
2. Can there be transitional functions which are continuous, but non-differentiable?

Answer 1

1. No. For the matrix $P(t)$ to generate a Markov process, $\{P(t)\}_{t\geq 0}$ must be what is called a semigroup. This property is powerful as it essentially relates the Markov matrix at different times in a very particular way that enforces a lot of regularity. It is true that any semigroup associated with a Markov process is what is called strongly continuous, which translates to each component of $P(t)$ being continuous if $P(t)$ is a matrix.

2. I believe the answer to this is also no, at least if the Markov process is time-homogeneous. It is also true that Markov semigroups are what are called contraction semigroups, which implies that they have a densly defined infinitesimal generator, which is basically $P'(0)$. Because of the semigroup property, $P'(0) = P'(t)$ for any $t\geq 0$. If this operator is densly defined, then this implies that each component is differentiable. I'm not sure about the non-time homogeneous case.

This stuff is usually discussed in much more general infinite-dimensional frameworks but I believe these results translate to the answers you are looking for.