The Giry monad consists of an endofunctor, $P$, on the category of measureable spaces $\mathcal{M}$, as well as two natural transformations $\mu, \eta$ known as the product and unit respectively. $P$ maps a measurable space, $X$, to the measurable space $P(X)$ of all probability measures on $X$. The unit takes a point to its point measure, and the product takes a measure on the space of measures to its integral. Thus the Giry monad captures probability measures. The question I have is whether or not this is also a comonad with suitably chosen natural transformations? If so, what are those natural transformations?


If the endofunctor does not admit a comonad structure, is there a monad that is also a comonad that captures probability measures?


It is hard to say what the exact meaning is of "capturing probability measures". One needs to take the Giry monad and abstract it away from measurable spaces and Polish spaces, and perhaps set it in a category of categories where its structures can be generalized. Abstracting probability theory might have to do with the factorizations of the Giry monad, or defining a different monad that doesn't have anything to do with measureable spaces. For instance, I believe we should be able to define a monad on the category of finite categories that could capture some aspects of probability measures.


I realize that this question I have asked, about abstracting the Giry Monad, is a much deeper question in probability theory and perhaps needs its own question.



Is the Giry Monad also a Comonad and if not, is there a probability measures (Co)monad?
The Giry monad is not a comonad because it doesn't admit a counit.




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Let $P$ be the Giry endofunctor, which assigns to a space $X$ the (suitably topologized) space of probability measures on the Borel subsets of $X$. 


For $X$ a single point $\{a\}$, the space $P(X)$ is a single point (with element the trivial measure $\delta$). For $Y$ the discrete set $\{a\_0,a\_1\}$, it is straightforward to verify that $P(Y)$ is homeomorphic to $[0,1]$, with the measure $\mu\_p\in[0,1]$ putting measure $p$ on point $a\_0$ and $(1-p)$ on point $a\_1$.


Consider the constant map $c\_i$ at point $a\_i$ from $X$ to $Y$. Then $P(c\_i)(\delta)=\delta\_i$, the probability measure on $Y$ concentrated at $a\_i$.


Now suppose there were a natural transformation $\epsilon:P\to\mathrm{id}$. Consider the two maps $\iota\_0$ and $\iota\_1$ from $X$ to $Y$. Naturality of $\epsilon$ and the previous paragraph show that $\epsilon(\delta\_{i})=a\_i$. There is a path in $P(Y)$ between $\delta\_0$ and $\delta\_1$, which are the endpoints of the interval. Then $\epsilon$ must take this path to a path in the discrete space $Y$ between $a\_0$ and $a\_1$, a contradiction.




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I don't know how to answer your "if not" question at least partially because I don't know what it means to you for a comonad to capture probability measures.


