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Stream: theory: categorical probability

Topic: strengths of the Giry monad

Matteo Capucci (he/him) (Aug 29 2022 at 09:01):

Hi there,
could you help me check these facts?

1. On the category of measurable spaces and measurable maps there is a monad $D$ sending $(X, \Sigma_X)$ to the measurable space of probability measures on it, whose unit is given by Dirac measures, and whose multiplication is

$$DDX \to DX\\ \mathrm{d} \mathbb P \mapsto \lambda A \,.\, \int_{DX} \mathbb P(A)\,\mathrm{d} \mathbb P$$

2. This monad has a right 'tensorial strength'

$$DX \times Y \to D(X \times Y)\\ (\mathrm{d} \mathbb P, y) \mapsto \mathbb P \otimes \delta_y$$

where $\otimes$ denotes 'pointwise' product of measures: $\mathbb P \otimes \mathbb Q(A) = \mathbb P(A)\mathbb Q(A)$

3. From the right strength we can get a 'closure strength' (enrichment), as explained at [[strong monad]]

$$\sigma : D(X \to Y) \to (X \to DY) \mathrm{d} f \mapsto \lambda x \,.\, ev^{X,Y}_* (\mathrm{d} f \otimes \delta_x) = \lambda x \,.\, \lambda A \,.\,\int_{X \to Y} {\mathbf 1}_A(f(x)) \,\mathrm{d} f$$

EDIT: now I wonder, what do I mean by '$\to$' in the category of measurable spaces? Could I get away with $X \to Y = \prod_{x \in X} Y$? But then I suspect $ev$ might not even exist...

From now on I'm less sure of the correctness (to address the point above: pretend I'm working in a sufficiently nice category of measurable spaces now, such as qB spaces or Radon spaces or even finite spaces):

4. There is another map given by taking cartesian products of probability spaces

$$K : (X \to DY) \to D(X \to Y)\\ (\mathbb P_x)_{x \in X} \mapsto \bigotimes_{x \in X} \mathbb P_x$$

If all the $\mathbb P_x$ have density (call it $p_x$), then such a product measure has density $\lambda (y_x)_{x \in X} \,.\, \prod_{x \in X} p_x(y_x)$ (notice its resemblance to an $\mathsf S$ combinator!). Otherwise you define by extension of the product measures defined on finite cylinders, see here.

5. The composite $K ; \sigma$ is the identity. I have a proof of this when $X$ and $Y$ are finite (and '$Y$ is sober'), and I think it should extend in the same way the product of prob measures extends from finite to infinite products:

$$\sigma(K((\mathbb P_x)_{x \in X})) = \sigma(\bigotimes_{x \in X} \mathbb P_x) = \lambda \bar x \,.\, \lambda A\,.\, (\bigotimes_{x \in X} (\mathbb P_x)_{x \in X} \otimes \delta_{\bar x})(ev^{-1} A)\\= \lambda \bar x \,.\, \lambda A \,.\, \prod_{x \in X} \mathbb P_x(\underbrace{\pi_{\bar x}\{f\mid f(\bar x) \in A\}}_{\text{$Y$ if $x \neq \bar x$, $A$ otherwise}}) = \lambda \bar x \,.\, \lambda A\,.\, \mathbb P_{\bar x}(A) = (\mathbb P_x)_{x \in X}$$

Hence $K$ is a section of $\sigma$ and I don't think it admits a left-inverse too: given an $X$-indexed family of prob measures on $Y$, this induces a prob measure on the function space $X \to Y$ but the same prob measure could be induced by many different prob measures on $X \to Y$.

Sam Staton (Sep 01 2022 at 09:46):

Hi Matteo, your $K$ looks like what we have been calling stochastic memoization, is that right? e.g. in this short abstract.

Matteo Capucci (he/him) (Sep 02 2022 at 08:06):

Ha, that's interesting!

Matteo Capucci (he/him) (Sep 02 2022 at 08:06):

It does look exactly the same, yeah