Consider a CTMC $(X_t)_{t\geq 0}$ with infinitesimal generator $Q=(q_{ij})$. I am looking for an expression of the $\tilde{q}_{ij}$ for the process $(X_{\sqrt{t}})_{t\geq 0}$. How does the generator change under the transformation $t\to \sqrt{t}$ for $t\geq 0$?
EDIT 1: I found the following problem, when looking for the generator $Q$. By the Book Markov Chains by Norris, a time-homogenous CTMC $(Y_t)_{t\geq 0}$ with state space $\mathbb{N}$ satisfies for all $n\in\mathbb{N}$, all times $0\leq t_1\leq ...\leq t_{n+1}$ and states $i_1,..., i_n\in \mathbb{N}$ such that $\mathbb{P}[Y_{t_n} = i_n,...,Y_{t_1}= i_1]>0$, $$\mathbb{P}[Y_{t_n} = i_n|Y_{t_{n-1}}=i_{n-1},...,Y_{t_1} = i_1] = P_{i_{n-1}\;i_n}(t_n - t_{n-1}),$$ where $P(t)$ is the associated semi group. Let $(P(t))$ be the semigroup associated to $(X_{t})_{t\geq 0}$. Then, I have for $0\leq \tilde{t}_1\leq ...\leq \tilde{t}_{n+1}$ such that $\tilde{t}_i^2 = t_i$, that $$\mathbb{P}[X_{\sqrt{t_n}} = i_n|X_{\sqrt{t_{n-1}}}=i_{n-1},...,X_{\sqrt{t_1}} = i_1] = \mathbb{P}[X_{\tilde{t}_n} = i_n|X_{\tilde{t}_{n-1}}=i_{n-1},...,X_{\tilde{t}_1} = i_1]$$ $$= P_{i_{n-1}\;i_n}(\tilde{t}_n - \tilde{t}_{n-1}) = P_{i_{n-1}\;i_n}(\sqrt{t_n} - \sqrt{t_{n-1}}).$$
Thus, to have a semigroup $(\tilde{P}(t))$ of $(X_{\sqrt{t}})$ I need $ \tilde{P}_{i_{n-1}\;i_n}(t_n - t_{n-1}) = P_{i_{n-1}\;i_n}(\sqrt{t_n} - \sqrt{t_{n-1}}1).$
EDIT 2: I use this to share my insights and maybe start a discussion. Indeed, $(X_{\sqrt{t}})_{t\geq 0}$ is not Markov. By various sources, it holds that a continous-time process with discrete state space is Markov if and only if its sojourn times are independent and exponentially distributed (with possibly different parameters). Let $(X_t)_{t\geq 0}$ be as defined above and consider its sojourn times $S_1,S_2,...$. Then, we can construct the time to the (k+1) th event $T_k$ of a copy of $(X_{\sqrt{t}})_{t\geq 0}$ via $$T_k = \sqrt{\sum_{i=1}^k S_i} .$$ In particular satisfies the first sojourn time $\tilde{S}_1$ of $(X_{\sqrt{t}})_{t\geq 0}$ that $$\tilde{S}_1 = T_1 = \sqrt{S_1}.$$ Thus for $t\geq 0$, given we start in state $i$, we have $\mathbb{P}[\tilde{S}_1 > t]=\mathrm{exp(-q_{ii}t^2)}$, which is not exponential. Therefore, $(X_{\sqrt{t}})_{t\geq 0}$ cannot be Markov.
But what I found is, that by setting $\Delta_i=\sqrt{S_i}$, the time to the (k+1)th jump corresponds to the euclidean norm of the vector $(\Delta_1,...,\Delta_k)$ of independent RVs distributed as $\mathrm{exp(-q_{ii}t^2)}$ with different $q_{ii}$ given by the state of the embedded chain $(Y_i)_i$ of $(X_t)_{t\geq 0}$. Are there results on the euclidean norm of such vectors? Note that the $\Delta_i$ belong to the exponential family. Does this maybe help to say something about $T_k$?