PhyslibAlpha.Mathematics.Probability.Kernel.Factorization

A kernel $K: \alpha \to \beta$ factors through a kernel $L: \alpha \to \gamma$ if there exists a Markov kernel $\kappa: \gamma \to \beta$ such that $K$ is the composition of $\kappa$ and $L$, denoted as $K = \kappa \circ_k L$.

For any kernel $K: \alpha \to \beta$, $K$ factors through itself. That is, there exists a Markov kernel $\kappa: \beta \to \beta$ such that $K = \kappa \circ_k K$, where $\circ_k$ denotes the composition of kernels.

Let $K: \alpha \to \beta$, $L: \alpha \to \gamma$, and $M: \alpha \to \delta$ be kernels. If $K$ factors through $L$ (meaning there exists a Markov kernel $\kappa_1$ such that $K = \kappa_1 \circ_k L$) and $L$ factors through $M$ (meaning there exists a Markov kernel $\kappa_2$ such that $L = \kappa_2 \circ_k M$), then $K$ factors through $M$.

Let $K: \alpha \to \beta$ and $L: \alpha \to \gamma$ be kernels. If $K$ factors through $L$, then for any $a \in \alpha$, there exists a Markov kernel $\kappa: \gamma \to \beta$ such that the measure $K(a)$ is the bind of the measure $L(a)$ with $\kappa$, i.e., $K(a) = (L(a)).\text{bind } \kappa$.

Let $\alpha, \beta,$ and $\gamma$ be measurable spaces. Let $K : \alpha \to \gamma$ be a Markov kernel and let $\mu$ be a probability measure on $\beta$. Then the constant kernel from $\alpha$ to $\beta$ defined by $\mu$ factors through $K$. That is, there exists a Markov kernel $\kappa : \gamma \to \beta$ such that the constant kernel $C_\mu : \alpha \to \beta$ (where $C_\mu(a) = \mu$ for all $a \in \alpha$) satisfies $C_\mu = \kappa \circ_k K$.

Two kernels $K: \alpha \to \beta$ and $L: \alpha \to \gamma$ mutually factor through each other if $K$ factors through $L$ and $L$ factors through $K$. This means there exists a Markov kernel $\kappa: \gamma \to \beta$ such that $K = \kappa \circ_k L$, and there exists a Markov kernel $\kappa': \beta \to \gamma$ such that $L = \kappa' \circ_k K$.

For any kernel $K$ from a measurable space $\alpha$ to a measurable space $\beta$, $K$ mutually factors with itself.

Let $K$ be a Markov kernel from $\alpha$ to $\beta$ and $L$ be a Markov kernel from $\alpha$ to $\gamma$. If $K$ and $L$ mutually factor through each other, then $L$ and $K$ mutually factor through each other.

Let $\alpha, \beta, \gamma,$ and $\delta$ be measurable spaces. Let $K: \alpha \to \beta, L: \alpha \to \gamma,$ and $M: \alpha \to \delta$ be kernels. If $K$ and $L$ mutually factor through each other, and $L$ and $M$ mutually factor through each other, then $K$ and $M$ mutually factor through each other. (Note: Two kernels $K$ and $L$ mutually factor through each other if there exist Markov kernels $\kappa$ and $\kappa'$ such that $K = \kappa \circ_k L$ and $L = \kappa' \circ_k K$).

Let $\mu$ be a measure on a measurable space $\alpha$. For kernels $K: \alpha \to \beta$ and $L: \alpha \to \gamma$, we say $K$ **factors through** $L$ **almost everywhere** with respect to $\mu$ if there exists a Markov kernel $\kappa: \gamma \to \beta$ such that the equality $K = \kappa \circ_k L$ holds for $\mu$-almost every point in $\alpha$, where $\circ_k$ denotes the composition of kernels.

Let $\alpha, \beta,$ and $\gamma$ be measurable spaces, and let $\mu$ be a measure on $\alpha$. For kernels $K: \alpha \to \beta$ and $L: \alpha \to \gamma$, if $K$ factors through $L$ (meaning there exists a Markov kernel $\kappa: \gamma \to \beta$ such that $K = \kappa \circ_k L$), then $K$ factors through $L$ $\mu$-almost everywhere.

Let $\mu$ be a measure on a measurable space $\alpha$. For any kernel $K : \alpha \to \beta$, it holds that $K$ factors through itself $\mu$-almost everywhere.

Let $\mu$ be a measure on a measurable space $\alpha$. Let $K: \alpha \to \beta$, $L: \alpha \to \gamma$, and $M: \alpha \to \delta$ be kernels. If $K$ factors through $L$ $\mu$-almost everywhere and $L$ factors through $M$ $\mu$-almost everywhere, then $K$ factors through $M$ $\mu$-almost everywhere. (Note: A kernel $K$ factors through $L$ $\mu$-almost everywhere if there exists a Markov kernel $\kappa$ such that $K = \kappa \circ_k L$ for $\mu$-almost every point in $\alpha$, where $\circ_k$ denotes the composition of kernels.)

Let $\mu$ be a measure on a measurable space $\alpha$. Two kernels $K: \alpha \to \beta$ and $L: \alpha \to \gamma$ **mutually factor almost everywhere** with respect to $\mu$ if $K$ factors through $L$ $\mu$-almost everywhere and $L$ factors through $K$ $\mu$-almost everywhere. This means there exist Markov kernels $\kappa_1: \gamma \to \beta$ and $\kappa_2: \beta \to \gamma$ such that the equalities $K = \kappa_1 \circ_k L$ and $L = \kappa_2 \circ_k K$ hold for $\mu$-almost every point in $\alpha$, where $\circ_k$ denotes the composition of kernels.

Let $\mu$ be a measure on a measurable space $\alpha$. For any kernel $K: \alpha \to \beta$, the kernel $K$ mutually factors with itself $\mu$-almost everywhere.

Let $\mu$ be a measure on a measurable space $\alpha$. For any two kernels $K: \alpha \to \beta$ and $L: \alpha \to \gamma$, if $K$ and $L$ mutually factor $\mu$-almost everywhere, then $L$ and $K$ mutually factor $\mu$-almost everywhere.

Let $\mu$ be a measure on a measurable space $\alpha$. Let $K: \alpha \to \beta$, $L: \alpha \to \gamma$, and $M: \alpha \to \delta$ be kernels. If $K$ and $L$ mutually factor $\mu$-almost everywhere, and $L$ and $M$ mutually factor $\mu$-almost everywhere, then $K$ and $M$ mutually factor $\mu$-almost everywhere. (Two kernels $K$ and $L$ mutually factor $\mu$-almost everywhere if there exist Markov kernels $\kappa_1$ and $\kappa_2$ such that $K = \kappa_1 \circ_k L$ and $L = \kappa_2 \circ_k K$ hold for $\mu$-almost every point in $\alpha$, where $\circ_k$ denotes the composition of kernels.)

Let $\mu$ be a measure on a measurable space $\alpha$, and let $K, L : \alpha \to \beta$ be two kernels. If $K$ and $L$ are equal $\mu$-almost everywhere ($K = L$ $\mu$-a.e.), then they mutually factor $\mu$-almost everywhere. That is, there exist Markov kernels $\kappa_1, \kappa_2 : \beta \to \beta$ such that the compositions $K = \kappa_1 \circ_k L$ and $L = \kappa_2 \circ_k K$ hold for $\mu$-almost every $x \in \alpha$.

Let $\alpha$ be a measurable space and $\beta$ be a countably generated measurable space. Let $\mu$ be a measure on $\alpha$, and let $K, L : \alpha \to \beta$ be kernels where $K$ is a finite kernel. If for every measurable set $s \subseteq \beta$, the measures assigned to $s$ satisfy $K(x, s) = L(x, s)$ for $\mu$-almost every $x \in \alpha$, then $K$ and $L$ are equal $\mu$-almost everywhere (i.e., $K(x, \cdot) = L(x, \cdot)$ as measures for $\mu$-almost every $x$).

Two kernels $K: \alpha \to \beta$ and $L: \alpha \to \gamma$ have a **common refinement** if there exists a measurable space $\Omega$ and a kernel $G: \alpha \to \Omega$ such that both $K$ and $L$ factor through $G$. This means there exist Markov kernels $\kappa_1: \Omega \to \beta$ and $\kappa_2: \Omega \to \gamma$ such that the compositions satisfy $K = \kappa_1 \circ_k G$ and $L = \kappa_2 \circ_k G$.

Let $\alpha$ and $\beta$ be measurable spaces. For any kernel $K : \alpha \to \beta$, $K$ and $K$ have a common refinement. That is, there exists a measurable space $\Omega$, a kernel $G : \alpha \to \Omega$, and Markov kernels $\kappa_1, \kappa_2 : \Omega \to \beta$ such that $K = \kappa_1 \circ_k G$ and $K = \kappa_2 \circ_k G$.

Let $K: \alpha \to \beta$ and $L: \alpha \to \gamma$ be kernels. If $K$ and $L$ have a common refinement, then $L$ and $K$ also have a common refinement. Two kernels have a common refinement if there exists a kernel $G: \alpha \to \Omega$ such that both $K$ and $L$ factor through $G$.

Let $K: \alpha \to \beta$, $L: \alpha \to \gamma$, $K': \alpha \to \delta$, and $L': \alpha \to \epsilon$ be kernels. If $K$ and $L$ have a common refinement, and $K'$ factors through $K$ and $L'$ factors through $L$, then $K'$ and $L'$ also have a common refinement.

Let $K: \alpha \to \beta$ and $L: \alpha \to \gamma$ be kernels. If $K$ factors through $L$, meaning there exists a Markov kernel $\kappa: \gamma \to \beta$ such that $K = \kappa \circ_k L$, then $K$ and $L$ have a common refinement (i.e., there exists a kernel through which both $K$ and $L$ factor).

Let $K: \alpha \to \beta$ and $L: \alpha \to \gamma$ be kernels. If $L$ factors through $K$, then $K$ and $L$ have a common refinement.