PhyslibAlpha.ProbabilisticTheory.Classical.Channel

Given a measurable map $g : \Omega' \to \Omega$ between two measurable spaces, the function defines a classical channel (a unital positive linear map) from the space of bounded measurable functions $\text{BoundedMeasurable}(\Omega)$ to $\text{BoundedMeasurable}(\Omega')$. This channel maps an observable $f$ of the target system $\Omega$ to the observable $f \circ g$ of the source system $\Omega'$, where $(f \circ g)(x) = f(g(x))$ for all $x \in \Omega'$.

Let $\Omega$ and $\Omega'$ be measurable spaces, and let $g: \Omega' \to \Omega$ be a measurable map. For any bounded measurable function $f \in \text{BoundedMeasurable}(\Omega)$ and any point $x \in \Omega'$, the classical channel induced by $g$ (the relabeling channel $\text{comap}(g)$) maps $f$ to a function whose value at $x$ is $f(g(x))$. That is, $(\text{comap}(g)(f))(x) = f(g(x))$.

Let $\Omega$ and $\Omega'$ be measurable spaces. For any measurable map $g : \Omega' \to \Omega$, the induced classical channel $\text{comap}(g) : \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$, which maps a bounded measurable function $f$ to $f \circ g$, is normal. This means that the channel preserves the suprema of non-empty directed sets of bounded measurable functions.

Given a Markov kernel $\kappa$ from a measurable space $\Omega'$ to $\Omega$, the function `kernelAverage` maps a bounded measurable function $f$ on $\Omega$ to a bounded measurable function on $\Omega'$ defined by the integral: $$x \mapsto \int_{\Omega} f(y) \, d\kappa(x)(y)$$ for each $x \in \Omega'$. In the context of classical channels, this represents the action of the channel on an observable $f$, producing its average value at each point $x$ according to the probability distribution $\kappa(x)$.

Let $\Omega$ and $\Omega'$ be measurable spaces and let $\kappa$ be a Markov kernel from $\Omega'$ to $\Omega$. For any bounded measurable function $f$ on $\Omega$ and any $x \in \Omega'$, the value of the kernel average of $f$ at the point $x$ is given by the integral of $f$ with respect to the probability measure $\kappa(x)$: $$(\text{kernelAverage } \kappa \text{ } f)(x) = \int_{\Omega} f(y) \, d\kappa(x)(y)$$

Let $\Omega$ and $\Omega'$ be measurable spaces and let $\kappa$ be a Markov kernel from $\Omega'$ to $\Omega$. The classical channel associated with $\kappa$ is a unital positive linear map from the space of bounded measurable functions on $\Omega$ to the space of bounded measurable functions on $\Omega'$. For any bounded measurable function $f$ on $\Omega$, the channel maps it to a function on $\Omega'$ defined by the integral: $$x \mapsto \int_{\Omega} f(y) \, d\kappa(x)(y)$$ for each $x \in \Omega'$. This represents the action of the kernel on an observable by taking its average with respect to the probability distribution provided by the kernel.

Let $\Omega$ and $\Omega'$ be measurable spaces and let $\kappa$ be a Markov kernel from $\Omega'$ to $\Omega$. For any bounded measurable function $f \in \text{BoundedMeasurable}(\Omega)$ and any point $x \in \Omega'$, the value of the classical channel $\text{ofKernel}(\kappa)$ applied to $f$ and evaluated at $x$ is given by the integral of $f$ with respect to the probability measure $\kappa(x)$: $$(\text{ofKernel } \kappa)(f)(x) = \int_{\Omega} f(y) \, d\kappa(x)(y)$$

Let $\Omega$ and $\Omega'$ be measurable spaces and let $\kappa$ be a Markov kernel from $\Omega'$ to $\Omega$. The classical channel $\text{ofKernel } \kappa : \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$, which maps a bounded measurable function $f$ to the function $x \mapsto \int_{\Omega} f(y) \, d\kappa(x)(y)$, is a normal unital positive linear map. That is, it preserves the suprema of non-empty directed sets of functions.

Let $\Omega$ and $\Omega'$ be sample spaces, and let $\text{BoundedMeasurable}(\Omega)$ and $\text{BoundedMeasurable}(\Omega')$ denote the respective spaces of bounded measurable functions. For any unital positive linear map $K: \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$ and any point $x \in \Omega'$, let $\text{eval}_x$ be the evaluation map $g \mapsto g(x)$. Then the composition $\text{eval}_x \circ K$ is a normal map, meaning that for every non-empty directed subset $D \subseteq \text{BoundedMeasurable}(\Omega)$ with a least upper bound $f$, the value $(\text{eval}_x \circ K)(f)$ is the least upper bound of the set $(\text{eval}_x \circ K)(D)$ in $\mathbb{R}$.

Given a normal classical channel $K: \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$, the associated Markov kernel $\kappa$ from $\Omega'$ to $\Omega$ is defined such that for each $x \in \Omega'$, the probability measure $\kappa(x, \cdot)$ is the outcome distribution of the state $\omega_x(f) = (K f)(x)$. Explicitly, for any measurable set $s \subseteq \Omega$, the kernel satisfies $\kappa(x, s) = (K \mathbb{1}_s)(x)$, where $\mathbb{1}_s$ is the indicator function of $s$.

Let $\Omega$ and $\Omega'$ be sample spaces. Let $K: \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$ be a normal classical channel (a unital positive linear map that is normal), and let $\kappa$ be the Markov kernel associated with $K$ (denoted by `toKernel K`). For any measurable set $s \subseteq \Omega$ and any point $x \in \Omega'$, the value of the kernel $\kappa$ for the point $x$ and the set $s$ is given by $$\kappa(x, s) = K(\mathbb{1}_s)(x)$$ where $\mathbb{1}_s$ is the indicator function of the set $s$. (Note: the result is expressed as an extended non-negative real number).

Let $\Omega$ and $\Omega'$ be measurable spaces. Let $K: \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$ be a normal classical channel. Then the associated kernel $\kappa$ from $\Omega'$ to $\Omega$, defined such that for each measurable set $s \subseteq \Omega$ and $x \in \Omega'$, $\kappa(x, s) = (K \mathbb{1}_s)(x)$, is a Markov kernel.

Let $\Omega$ and $\Omega'$ be measurable spaces. Let $K : \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$ be a normal classical channel (a normal unital positive linear map) and let $\kappa$ be its associated Markov kernel from $\Omega'$ to $\Omega$. For any normal state $\sigma$ on $\text{BoundedMeasurable}(\Omega')$ with corresponding outcome distribution (probability measure) $\mu_\sigma$, the outcome distribution $\mu_{\sigma \circ K}$ of the composite state $\sigma \circ K$ is the probability measure on $\Omega$ obtained by the composition of the kernel $\kappa$ and the measure $\mu_\sigma$: $$\mu_{\sigma \circ K} = \kappa \circ_m \mu_\sigma$$ where the composition is defined by $(\kappa \circ_m \mu_\sigma)(s) = \int_{\Omega'} \kappa(x, s) \, d\mu_\sigma(x)$ for every measurable set $s \subseteq \Omega$.

Let $\Omega$ and $\Omega'$ be measurable spaces. Let $K : \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$ be a normal classical channel (a normal unital positive linear map). Let $\kappa$ be its associated Markov kernel from $\Omega'$ to $\Omega$, defined by $\kappa(x, s) = (K \mathbb{1}_s)(x)$ for each $x \in \Omega'$ and measurable set $s \subseteq \Omega$. Then, the classical channel reconstructed from $\kappa$, which maps a bounded measurable function $f$ to the integral $x \mapsto \int_{\Omega} f(y) \, d\kappa(x)(y)$, is equal to the original channel $K$. That is, $\text{ofKernel}(\text{toKernel } K) = K$.

Let $\Omega$ and $\Omega'$ be measurable spaces. For any Markov kernel $\kappa$ from $\Omega'$ to $\Omega$, let $\text{ofKernel } \kappa$ be the classical channel that maps a bounded measurable function $f$ to the function $x \mapsto \int_{\Omega} f(y) \, d\kappa(x)(y)$. Then, the Markov kernel recovered from this channel using the $\text{toKernel}$ construction is equal to the original kernel $\kappa$. That is, $\text{toKernel}(\text{ofKernel } \kappa) = \kappa$.

There is a bijective correspondence (equivalence) between the set of normal classical channels $K: \text{BoundedMeasurable}(\Omega) \to \text{BoundedMeasurable}(\Omega')$ and the set of Markov kernels $\kappa$ from $\Omega'$ to $\Omega$. A classical channel is a unital completely positive linear map between spaces of bounded measurable functions, and it is considered **normal** if it preserves the suprema of directed sets. The equivalence maps a normal channel $K$ to the kernel defined by $\kappa(x, s) = (K \mathbb{1}_s)(x)$ for measurable sets $s \subseteq \Omega$, and maps a Markov kernel $\kappa$ to the channel defined by $(K f)(x) = \int_{\Omega} f \, d\kappa(x)$.