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Stream: theory: categorical probability

Topic: characterization of disintegrations

Matteo Capucci (he/him) (Jan 22 2025 at 14:09):

Consider a probability space $(B, db)$ and a family of probability spaces $(E_b, de_b)$ indexed by the points of $b$. I think you can equip $E = \sum_{b \in B} E_b$ with a measurable structure, generated by measurables of the form $\sum_{b \in U} E_b$ for $U \subseteq B$ measurable and of the form $\sum_{b \in B} V_b$ for $V_b \subseteq E_b$ measurable.
I think you can also equip $E$ with a measure $de$, defined on generators as

$$de\left(\sum_{b \in U} E_b\right) = db(U)\\ de\left(\sum_{b \in B} V_b\right) = \int_b de_b(V_b)\,db$$

In this way there is an obvious measure-preserving projection map $p:E \to B$.

Matteo Capucci (he/him) (Jan 22 2025 at 14:12):

Two questions:

1. Does this construction make sense in generality? Is there a more general one in which I don't need to work pointwise eith B?
2. Does anyone know a characterization of the maps $p:E \to B$ arising in this way? I know these are called 'disintegrations'. Do they have a lifting property of some kind?

Paolo Perrone (Jan 22 2025 at 14:14):

Is the space B countable or uncountable?

Matteo Capucci (he/him) (Jan 22 2025 at 14:15):

Does it make a difference?

Paolo Perrone (Jan 22 2025 at 14:16):

In general, yes. (I know it shouldn't, but that's how measure theory works...)

Matteo Capucci (he/him) (Jan 22 2025 at 14:16):

Uh, why?

Paolo Perrone (Jan 22 2025 at 14:21):

Basically, in the second equation, for the right-hand side to type-check and exist, we want the integrand to be measurable (= integrable, in this case) as a function of $b$. This is automatic if $B$ is discrete, but not in general.

Matteo Capucci (he/him) (Jan 22 2025 at 14:25):

I see, it is not a cardinality problem though

Matteo Capucci (he/him) (Jan 22 2025 at 14:26):

I found this definition which seems to be more careful about that point but I don't get it. It seems to be defining disintegrations only for 'trivial' bundles $X \times Y \to Y$.
image.png

Paolo Perrone (Jan 22 2025 at 14:27):

Consider the (easier) case where all the $E_b$ are equal, so that the total space is just a product. (But we allow the measures $de_b$ to vary, so that we still have a "dependent product" feeling.)
In this case we want $de_b$ to be measurable in $b$ in the sense that for every measurable $A\subseteq E_b$, the function $b\mapsto de_b(A)$ is measurable. This is precisely saying that we have a Markov kernel, (equivalently, a regular conditional distribution, or a disintegration).

Paolo Perrone (Jan 22 2025 at 14:27):

It's not a cardinality issue, you are right. It's just an issue that becomes visible once we leave the discrete case.

Paolo Perrone (Jan 22 2025 at 14:28):

Measurability in $b$ is crucial.

Matteo Capucci (he/him) (Jan 22 2025 at 14:28):

Yeah I see what you mean

Paolo Perrone (Jan 22 2025 at 14:28):

This is one of the main reasons why in categorical probability we are so fixated with Markov kernels. The good news is that once you do have that measurability, a lot of things work very well.

Matteo Capucci (he/him) (Jan 22 2025 at 14:29):

Okok so I'll add that assumption

Paolo Perrone (Jan 22 2025 at 14:29):

In terms of Markov categories, this is precisely the idea of having conditionals.

Paolo Perrone (Jan 22 2025 at 14:30):

Now if the $E_b$ are not all equal, it becomes tricky to state the measurability condition: who is $A$?

Matteo Capucci (he/him) (Jan 22 2025 at 14:32):

I guess you have to consider $B$-indexed families of $V_b \subseteq E_b$ and ask measurability for the measures $b \mapsto de_b(V_b)$

Matteo Capucci (he/him) (Jan 22 2025 at 14:33):

Though yeah it's easier said than done :thinking:

Paolo Perrone (Jan 22 2025 at 14:35):

Well, the good news is that measurable spaces are very "coarse", and a lot of nice measurable spaces turn out to be isomorphic. (For example, $\mathbb{R}$, $\mathbb{R}^2$, $S^1$ and every separable Banach space of dimension >0 are isomorphic as measurable spaces!)
In particular, the only standard Borel spaces (which always admit disintegrations) are, up to iso, $\mathbb{R}$ and its retracts (which are $\mathbb{N}$ and the finite sets).
So, in very very many situations, it is not a real loss of generality to assume that the $E_b$ are all the same (take them to be $\mathbb{R}$).

Matteo Capucci (he/him) (Jan 22 2025 at 14:36):

Bold words to write on a CT chatroom XD

Paolo Perrone (Jan 22 2025 at 14:36):

Even if you want some fibers to be discrete, you can take $\mathbb{R}$, and as measure you take one supported, say, on $\mathbb{N}\subseteq\mathbb{R}$

Paolo Perrone (Jan 22 2025 at 14:36):

(That's at least what probabilists usually do.)

Matteo Capucci (he/him) (Jan 22 2025 at 14:38):

I'm more interested in characterizing when such a situation is available than constructing one, i.e. my question (2). That is, suppose you have a measure-preserving map $p:E \to B$ already, when can you disintegrate the measure on $E$ such that $\int_{(b,e)} \varphi(b,e) d(b,e) = \int_b \int_e \varphi(b,e)d b de_b$?

Paolo Perrone (Jan 22 2025 at 14:39):

That's an excellent question. That's precisely (one of the many version of) the disintegration theorem.

Paolo Perrone (Jan 22 2025 at 14:40):

One positive answer is: if $E$ is standard Borel, then you always can.

Paolo Perrone (Jan 22 2025 at 14:40):

(Abstractly, the Markov category of standard Borel spaces has all conditional distributions.)

Matteo Capucci (he/him) (Jan 22 2025 at 14:41):

Paolo Perrone said:

(Abstractly, the Markov category of standard Borel spaces has all conditional distributions.)

Can you really phrase non-trivial disintegration as a Markov-categorical conditional?

Paolo Perrone (Jan 22 2025 at 14:41):

Yes!

Matteo Capucci (he/him) (Jan 22 2025 at 14:41):

Uhm maybe you are saying that in that case, every such bundle is trivial so it doesn't matter

Matteo Capucci (he/him) (Jan 22 2025 at 14:41):

Else can you show me how?

Paolo Perrone (Jan 22 2025 at 14:44):

Matteo Capucci (he/him) said:

Can you really phrase non-trivial disintegration as a Markov-categorical conditional?

Yes, that's one of the reasons which "sold me" to Markov categories.
(The other one being: these measurability issues are clearly necessary, but I find them very tricky to work with. Using Markov categories everything becomes string diagrams, and I find it much more suggestive and intuitive that way.)

Tobias Fritz (Jan 22 2025 at 14:45):

I'm just catching up on this, so I hope you don't mind me adding 2 cents. The fact that conditionals give you disintegrations is essentially Prop 11.7 in my 'synthetic' paper:
image.png
(Take $A = I$ to get your situation, with your $p$ in place of $f$, which happens to be deterministic and surjective, but neither of these is actually relevant.)

Nowadays I guess we'd call this 'parametric Bayesian inversion' instead of disintegration.

Paolo Perrone (Jan 22 2025 at 14:45):

Matteo Capucci (he/him) said:

Uhm maybe you are saying that in that case, every such bundle is trivial so it doesn't matter

I don't think it's a matter of "triviality of the bundle" here. If we see the fibers as measure spaces, they are allowed to be very different, so our bundle is not even locally trivial. Still disintegrations exist.

Paolo Perrone (Jan 22 2025 at 14:47):

Tobias Fritz said:

The fact that conditionals give you disintegrations is essentially Prop 11.7 in my 'synthetic' paper

Exactly. The non-parametric case is drawn here on the nlab (which does not allow triangles in string diagrams):
image.png

Matteo Capucci (he/him) (Jan 22 2025 at 14:49):

Ah, interesting!!

Matteo Capucci (he/him) (Jan 22 2025 at 14:50):

The trick is to give the family of spaces $(E_b, de_b)$ as a Markov kernel $B \to \sum_{b \in B} E_b$, I see.

Matteo Capucci (he/him) (Jan 22 2025 at 14:51):

In Tobias notation, my $E$ s are $Y$ and $B$ is $X$, with the assignment given by $f:X \to Y$ and the base measure being $s:I \to X$

Paolo Perrone (Jan 22 2025 at 14:51):

Yes! So a Markov kernel, "morally", is a "dependent probability measure", with all the technicalities worked out.

Paolo Perrone (Jan 22 2025 at 14:51):

Matteo Capucci (he/him) said:

In Tobias notation, my $$E$$s are $Y$ and $B$ is $X$, with the assignment given by $f:X \to Y$ and the base measure being $s:I \to X$

No, wait. Your total space is X and your B is Y.

Matteo Capucci (he/him) (Jan 22 2025 at 14:52):

The confusing thing is that you never say $f:B \to \sum_b E_b$ has to split the projection map. And I guess I understand why: you are going to condition on $E_b$ anyway.

Matteo Capucci (he/him) (Jan 22 2025 at 14:52):

That is, $de_b$ is actually $f(b)\vert_{E_b}$

Matteo Capucci (he/him) (Jan 22 2025 at 14:53):

Paolo Perrone said:

Matteo Capucci (he/him) said:

In Tobias notation, my $$E$$s are $Y$ and $B$ is $X$, with the assignment given by $f:X \to Y$ and the base measure being $s:I \to X$

No, wait. Your total space is X and your B is Y.

Uh-oh did I get it backwards?

Matteo Capucci (he/him) (Jan 22 2025 at 14:54):

Right... you specify a measure on $E$ and then show it disintegrates, what I describe is integration. Guess I am constructively minded :P

Paolo Perrone (Jan 22 2025 at 14:54):

Oh, you can prove using string diagrams (try!) that the following conditions are equivalent:

• f is almost surely deterministic
• its Bayesian inverse is almost surely a section.

Matteo Capucci (he/him) (Jan 22 2025 at 14:55):

Are you talking now about the disintegrating $f$ or the integrating $f$? :sweat_smile:

Paolo Perrone (Jan 22 2025 at 14:55):

If you need a reference, see Proposition 2.5 here (but it's not where it was first proven)

Matteo Capucci (he/him) (Jan 22 2025 at 14:55):

I guess the disintegrating one, i.e. the 'bundle projection', right?

Paolo Perrone (Jan 22 2025 at 14:56):

f is the projection. Its Bayesian inverse/disintegration is the conditional