Let
- $(E,\mathcal E)$ be a measurable space;
- $\kappa$ be a Markov kernrel on $(E,\mathcal E)$;
- $\mu$ be a probability measure on $(E,\mathcal E)$ invariant with respect to $\kappa$
- $A:=\kappa-\operatorname{id}_{L^2(\mu)}$;
- $\kappa_t:=e^{tA}$ for $t\ge0$;
- $(\Omega,\mathcal A)$ be a measurable space;
- $\operatorname P_x$ be a probability measure on $(\Omega,\mathcal A)$ for $x\in E$;
- $(M_n)_{n\in\mathbb N_0}$ be a Markov chain started at $x$ on $(\Omega,\mathcal A,\operatorname P_x)$ with transition kernel $\kappa$ and $$S_nf:=\frac1n\sum_{i=1}^nf(M_i)\;\;\;\text{for }n\in\mathbb N$$ for $f\in L^2(\mu)$;
- $(X_t)_{t\ge0}$ be a Markov process started at $x$ on $(\Omega,\mathcal A,\operatorname P_x)$ with transition semigroup $(\kappa_t)_{t\ge0}$ and $$T_tf:=\int_0^tf(X_s)\:{\rm d}s\;\;\;\text{for }t>0$$ for $f\in L^2(\mu)$.
$(\kappa_t)_{t\ge0}$ is a contraction semigroup on $L^2(\mu)$ with generator $A$. Let $f\in L^2(\mu)$ and $f_0:=f-\mu f$. We know that $$\operatorname{Var}_\mu[S_nf]=\frac1n\|f_0\|_{L^2(\mu)}^2+\frac2n\sum_{m=1}^n\left(1-\frac mn\right)\langle\kappa^mf_0,f_0\rangle_{L^2(\mu)}\tag1$$ for all $n\in\mathbb N$ and $$\operatorname{Var}[T_tf]=\frac2t\int_0^t\left(1-\frac st\right)\langle\kappa_sf_0,f_0\rangle_{L^2(\mu)}\:{\rm d}s\tag2$$ for all $t>0$.
Since $(\kappa_t)_{t\ge0}$ is the natural embedding of $\kappa$ into continuous time, I wonder how $(1)$ and $(2)$ are related. Does $(2)$ somehow boil down to $(1)$?
I've tried to compute $(2)$ using $$\kappa_t=\sum_{k\in\mathbb N_0}\frac{t^kA^k}{k!}\tag3.$$ Using this, I obtain $$\operatorname{Var}[T_tf]=\left\langle\sum_{k\in\mathbb N_0}\frac{t^{k+1}A^k}{(k+2)!}f_0,f_0\right\rangle_{L^2(\mu)}\tag4.$$ But how can we simplify the sum? If I formally treat $A$ as a real number, I would obtain $$\sum_{k\in\mathbb N_0}\frac{t^{k+1}A^k}{(k+2)!}=\frac{-tA+\kappa_t-1}{tA^2}\tag5.$$ This obviously does not generalize to the operator case (maybe $\frac1{A^2}$ would be the inverse of $A^2$; but that would not be really helpful in deriving a relation to $(1)$).