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$\begingroup$

$\def\unit{{\rm unit}}\def\join{{\rm join}}$It's well known that (discrete) probability distributions form a monad. Specifically, if we let $PX$ be the set of discrete probability distributions on elements of $X$, and notate them as a set of pairs $(x,p)$ such that $\sum p=1$, then we have natural transformations

$\begin{align} \unit : X & \to PX \\ \unit : x & \mapsto \{ (x,1) \} \\ \\ \join: P(PX) & \to PX \\ \join: D & \mapsto \{(y,pq)| (x,p) \in D, (y,q)\in x \} \end{align}$

that satisfy the monad laws.

Can probability distributions be made into a comonad as well? For that, we would need to provide natural transformations

$\begin{align} {\rm counit} : PX & \to X \\ {\rm cojoin} : PX & \to P(PX) \end{align}$

that satisfy the comonad laws. It seems that the role of counit can be played by mathematical expectation (as long as $X$ is an $\mathbb{R}$-module), but in that case what is the correct definition of cojoin?


Edit:

Zhen Lin pointed out in the comments that if you want to have counit being expectation, then you need an $\mathbb{R}$-module structure on $PX$ as well as on $X$. The module operations on $PX$ are inherited from those on $X$ in the following way:

Addition

$D_1 + D_2 = \{ (x+y,pq) | (x,p)\in D_1, (y,q)\in D_2\}$

Multiplication by a scalar

$qD = \{ (qx,p) | (x,p)\in D \}$

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    I don't think `counit` is valid. `counit` would have to be a natural transformation. Take a uniform distribution over $\{-1,1\}$. If you take the expected value you get $0$, and then square that and you get $0$. Square it first and you get a guaranteed value of $1$, square that and you get $1$. Therefore, `counit` isn't a natural transformation.2016-03-03

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