Markov operator

Let ${\displaystyle (E,{\mathcal {F}})}$ be a measurable space and ${\displaystyle V}$ a set of real, measurable functions ${\displaystyle f:(E,{\mathcal {F}})\to (\mathbb {R} ,{\mathcal {B}}(\mathbb {R} ))}$.

A linear operator ${\displaystyle P}$ on ${\displaystyle V}$ is a Markov operator if the following is true[1]^: 9–12

1. ${\displaystyle P}$ maps bounded, measurable function on bounded, measurable functions.
2. Let ${\displaystyle \mathbf {1} }$ be the constant function ${\displaystyle x\mapsto 1}$, then ${\displaystyle P(\mathbf {1} )=\mathbf {1} }$ holds. (conservation of mass / Markov property)
3. If ${\displaystyle f\geq 0}$ then ${\displaystyle Pf\geq 0}$. (conservation of positivity)

Some authors define the operators on the L^p spaces as ${\displaystyle P:L^{p}(X)\to L^{p}(Y)}$ and replace the first condition (bounded, measurable functions on such) with the property[2][3]

${\displaystyle \|Pf\|_{Y}=\|f\|_{X},\quad \forall f\in L^{p}(X)}$

Let ${\displaystyle {\mathcal {P}}=\{P_{t}\}_{t\geq 0}}$ be a family of Markov operators defined on the set of bounded, measurables function on ${\displaystyle (E,{\mathcal {F}})}$. Then ${\displaystyle {\mathcal {P}}}$ is a Markov semigroup when the following is true[1]^: 12

1. ${\displaystyle P_{0}=\operatorname {Id} }$.
2. ${\displaystyle P_{t+s}=P_{t}\circ P_{s}}$ for all ${\displaystyle t,s\geq 0}$.
3. There exist a σ-finite measure ${\displaystyle \mu }$ on ${\displaystyle (E,{\mathcal {F}})}$ that is invariant under ${\displaystyle {\mathcal {P}}}$, that means for all bounded, positive and measurable functions ${\displaystyle f:E\to \mathbb {R} }$ and every ${\displaystyle t\geq 0}$ the following holds

${\displaystyle \int _{E}P_{t}f\mathrm {d} \mu =\int _{E}f\mathrm {d} \mu }$.

Each Markov semigroup ${\displaystyle {\mathcal {P}}=\{P_{t}\}_{t\geq 0}}$ induces a dual semigroup ${\displaystyle (P_{t}^{*})_{t\geq 0}}$ through

${\displaystyle \int _{E}P_{t}f\mathrm {d\mu } =\int _{E}f\mathrm {d} \left(P_{t}^{*}\mu \right).}$

If ${\displaystyle \mu }$ is invariant under ${\displaystyle {\mathcal {P}}}$ then ${\displaystyle P_{t}^{*}\mu =\mu }$.

Let ${\displaystyle \{P_{t}\}_{t\geq 0}}$ be a family of bounded, linear Markov operators on the Hilbert space ${\displaystyle L^{2}(\mu )}$, where ${\displaystyle \mu }$ is an invariant measure. The infinitesimale generator ${\displaystyle L}$ of the Markov semigroup ${\displaystyle {\mathcal {P}}=\{P_{t}\}_{t\geq 0}}$ is defined as

${\displaystyle Lf=\lim \limits _{t\downarrow 0}{\frac {P_{t}f-f}{t}},}$

and the domain ${\displaystyle D(L)}$ is the ${\displaystyle L^{2}(\mu )}$-space of all such functions where this limit exists and is in ${\displaystyle L^{2}(\mu )}$ again.[1]^: 18[4]

${\displaystyle D(L)=\left\{f\in L^{2}(\mu ):\lim \limits _{t\downarrow 0}{\frac {P_{t}f-f}{t}}{\text{ exists and is in }}L^{2}(\mu )\right\}.}$

The carré du champ operator ${\displaystyle \Gamma }$ measuers how far ${\displaystyle L}$ is from being a derivation.

A Markov operator ${\displaystyle P_{t}}$ has a kernel representation

${\displaystyle (P_{t}f)(x)=\int _{E}f(y)p_{t}(x,\mathrm {d} y),\quad x\in E,}$

with respect to some probability kernel ${\displaystyle p_{t}(x,A)}$, if the underlying measurable space ${\displaystyle (E,{\mathcal {F}})}$ has the following sufficient topological properties:

1. Each probability measure ${\displaystyle \mu$ :{\mathcal {F}}\times {\mathcal {F}}\to [0,1]} can be decomposed as ${\displaystyle \mu (\mathrm {d} x,\mathrm {d} y)=k(x,\mathrm {d} y)\mu _{1}(\mathrm {d} x)}$, where ${\displaystyle \mu _{1}}$ is the projection onto the first component and ${\displaystyle k(x,\mathrm {d} y)}$ is a probability kernel.
2. There exist a countable family that generates the σ-algebra ${\displaystyle {\mathcal {F}}}$.

If one defines now a σ-finite measure on ${\displaystyle (E,{\mathcal {F}})}$ then it is possible to prove that ever Markov operator ${\displaystyle P}$ admits such a kernel representation with respect to ${\displaystyle k(x,\mathrm {d} y)}$.[1]^: 7–13