# Do probability distributions form a comonad?

$\def\unit{{\rm unit}}\def\join{{\rm join}}$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}

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:

$$D_1 + D_2 = \{ (x+y,pq) | (x,p)\in D_1, (y,q)\in D_2\}$$
$$qD = \{ (qx,p) | (x,p)\in D \}$$
Your description of the unit of $P$ monad seems to be incorrect. Do you mean to have $1 / |{X}|$ instead of $1$? In that case, I suppose you must also assume that $X$ is finite. –  Zhen Lin Jun 28 '12 at 7:42
No - here $X$ is a set (the sample space), and ${\rm unit}(x)$ is the trivial distribution over $x\in X$, i.e. the distribution which always selects $x$. –  Chris Taylor Jun 28 '12 at 7:46
Ah, yes. That makes sense. I don't think can make expectation into the comonad unit, because $P X$ is not a vector space even when $X$ is. –  Zhen Lin Jun 28 '12 at 7:49
It's possible to define a vector space structure on $PX$. I'll edit the question to include that. –  Chris Taylor Jun 28 '12 at 7:59
Your formula for $D_1 + D_2$ will misbehave when there are $x_1, x_2, y_1, y_2$ such that $x_1 + y_1 = x_2 + y_2$. –  Zhen Lin Jun 28 '12 at 9:13