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Stream: theory: categorical probability

Topic: Universal property of the Giry monad


view this post on Zulip Nathaniel Virgo (Feb 04 2025 at 13:52):

This is idle wondering and I don't really expect the answer to be known, but maybe it will be an interesting conversation.

Chen's A universal characterisation of the category of standard Borel spaces characterises the category of standard Borel spaces as the universal one with certain nice properties. This is great because it means not only that we know a bunch of nice properties that this category has, but also that those are in some sense its only relevant ones when it comes to proving things about it.

However, that paper doesn't really mention Markov kernels or monads at all, so it doesn't immediately let us talk about probability.

So I'm wondering to what it extent it might be possible to construct the Giry monad, or BorelStoch\bf BorelStoch, in a similarly abstract way.

In other words, if someone hands me a category and tells me that it's the initial countably complete countably extensive category, how can I use that information to construct the notions of measure and Markov kernel?

view this post on Zulip Benedikt Peterseim (Feb 04 2025 at 14:10):

This is a very interesting question! I hadn’t seen that paper before.

view this post on Zulip Benedikt Peterseim (Feb 04 2025 at 14:15):

Of course, there are constructions of probability monads as Codensity monads, but they presuppose that you have the unit interval in your category already. On the other hand, there is also work on universal characterisations of the unit interval, so maybe using one of these, one might actually obtain a universal property for BorelStoch. This seems far from straightforward, though.

view this post on Zulip Tobias Fritz (Feb 04 2025 at 15:12):

The following is pretty speculative, but I think a characterization like the following could be true and conceptually meaningful:

Conjecture: BorelStoch\mathsf{BorelStoch} is the free Markov category with countable Kolmogorov products generated by the Markov category FinStoch\mathsf{FinStoch}.

For the morphisms, I imagine using strong symmetric monoidal functors presering copy. This would be in a bit of a different direction than Chen's result though.

view this post on Zulip Tobias Fritz (Feb 04 2025 at 15:16):

More along the lines of Chen's paper, perhaps one could consider something like the free representable Markov category that is countably distributive?

view this post on Zulip Tobias Fritz (Feb 04 2025 at 16:27):

Perhaps worth mentioning is also the idea of free affine commutative monads, which my student @Victor Bloch is working on. Similar to how the free monad generated by a binary operation is the monad of binary trees, Victor has constructed the free affine commutative monad generated by a binary operation on Set, and we seem to be able to show that it embeds into the distribution monad in a way that is topologically dense. (Very roughly, think of approximating a distribution by repeated coin flips.)

Moreover, we conjecture that the Radon monad, which is the standard probability monad on the category of compact Hausdorff spaces, literally is the free affine commutative monad generated by a binary operation. But so far we don't know how to approach this problem.

Binary operations arguably aren't special in this story, and I'd expect everything to work the same if one starts with an operation of any finite arity.

view this post on Zulip Benedikt Peterseim (Feb 04 2025 at 16:38):

Tobias Fritz said:

Perhaps worth mentioning is also the idea of free affine commutative monads, which my student Victor Bloch is working on. Similar to how the free monad generated by a binary operation is the monad of binary trees, Victor has constructed the free affine commutative monad generated by a binary operation on Set, and we seem to be able to show that it embeds into the distribution monad in a way that is topologically dense. (Very roughly, think of approximating a distribution by repeated coin flipping.)

This is interesting—so you get a construction of the dyadic rational unit interval as M(2) entirely for free? (where M is this free commutative affine monad) Also, if the conjecture about the Radon monad is true, I suppose that would give another (very natural, in my opinion) universal chracterisation of the unit interval?

view this post on Zulip Tobias Fritz (Feb 04 2025 at 16:41):

It's actually not the dyadic interval -- it's all those numbers that you can get by repeatedly flipping a biased coin with transcendental bias!

view this post on Zulip Benedikt Peterseim (Feb 04 2025 at 16:41):

By the way, are you using midpoint algebras (https://ncatlab.org/nlab/show/midpoint+algebra) or variations of this in any way? I think there are characterizations of the unit interval that use this iirc, so I’m thinking it might be relevant to the conjecture regarding the Radon monad.

view this post on Zulip Benedikt Peterseim (Feb 04 2025 at 16:41):

Tobias Fritz said:

It's actually not the dyadic interval -- it's all those numbers that you can get by repeatedly flipping a biased coin with transcendental bias!

Oh, I’m surprised and confused. :)

view this post on Zulip Tobias Fritz (Feb 04 2025 at 16:41):

The reason is that the defining operation of a midpoint algebra is commutative, so it cannot be the generating operation in a free monad, and not even in a free affine commutative monad :sweat_smile:

view this post on Zulip Tobias Fritz (Feb 04 2025 at 16:42):

This is clear once you note that not all binary operations in a commutative monad are commutative.

view this post on Zulip Benedikt Peterseim (Feb 04 2025 at 16:43):

Ah, I think I’m starting to see this, okay. :)

view this post on Zulip Tobias Fritz (Feb 04 2025 at 16:45):

The difficult logical part of this is from On convex linear forms, a 1974 paper by Fajtlowicz and Mycielski, who show that the theory of binary convex combinations with a transcendental coefficient is axiomatized by two equations:
image.png
Intuitively, these two equations say precisely that the monad is affine resp. commutative.

view this post on Zulip Tobias Fritz (Feb 04 2025 at 16:46):

A good way to understand the second equation is to note that it amounts to the operation distributing over itself, or what we would call commuting with itself. Like how in a commutative monad any two operations commute.

view this post on Zulip Tobias Fritz (Feb 04 2025 at 16:50):

So these algebraic structures are midpoint algebras without the commutativity. They seem to be called modes.

view this post on Zulip Tobias Fritz (Feb 04 2025 at 16:52):

Maybe you'd have some idea for how to approach the case of compact Hausdorff spaces @Benedikt Peterseim ? Happy to talk more in case you're interested.

view this post on Zulip Benedikt Peterseim (Feb 04 2025 at 18:05):

Not any specific idea right now, but it does seem of a similar flavour to some things I’ve thought about before. Would definitely be interested in some more details and in sharing some thoughts; feel free to DM me.

view this post on Zulip Morgan Rogers (he/him) (Feb 05 2025 at 09:14):

Benedikt Peterseim said:

Of course, there are constructions of probability monads as Codensity monads, but they presuppose that you have the unit interval in your category already. On the other hand, there is also work on universal characterisations of the unit interval, so maybe using one of these, one might actually obtain a universal property for BorelStoch.

You don't need a universal characterization of the unit interval, it's enough to have a construction of the unit interval using the limits and colimits available in the characterization of BorelStoch. The subtlety is just that while there is a unique uncountable measure space, there are non-equivalent ways of equipping it with this structure. We already have the trivial (two element) interval. Constructing the diadic interval is easy enough. Fortunately the operations involved in the standard unit interval are continuous (except division at 0, which is still okay) and hence measurable, so this should be possible.

view this post on Zulip Morgan Rogers (he/him) (Feb 05 2025 at 09:15):

A consequence would be that you can do probability theory over any countably complete Boolean countably lextensive category with the same suite of interval objects.

view this post on Zulip Nathaniel Virgo (Feb 08 2025 at 19:55):

Morgan Rogers (he/him) said:

A consequence would be that you can do probability theory over any countably complete Boolean countably lextensive category with the same suite of interval objects.

I'm kind of confused about this point actually. Assuming we can construct the Giry monad from the limits and colimits in the characterisation of StdBorel, however we do it, shouldn't that mean we can do it in any countably complete Boolean countably extensive category? Since Set is such a category (right?), this seems to imply we'd get for free a monad on Set that inherits all the nice properties of the Giry monad. That would be great if it's true, but it sounds like too much to hope for. Presumably something would prevent that - do you have an idea what it would be?

view this post on Zulip Nathaniel Virgo (Feb 08 2025 at 22:00):

With that caveat in mind, here's another conjecture about how to answer this version of my question:

Nathaniel Virgo said:

In other words, if someone hands me a category and tells me that it's the initial countably complete Boolean countably extensive category, how can I use that information to construct the notions of measure and Markov kernel?

Take an object XX and construct XNX^\mathbb{N} as a countable product. Consider the group of finite permutations of N\mathbb{N} and its action on XNX^\mathbb{N}. I think we have enough structure to construct the object of orbits of this group action. I conjecture that this is P(X)P(X), the Giry monad applied to XX.

This is because of de Finetti's theorem. Whatever the quotient of this group action is, measures over it ought to be basically the same as exchangeable measures over XNX^\mathbb{N}, which are the same as measures over P(X)P(X). So in particular, elements of it should just be elements of P(X)P(X). (Since over standard Borel spaces an element is the same as a measure concentrated on that element.)

If that's right then one can probably get the monad structure by considering a bijection between sequences of sequences of elements of XX and sequences of elements of XX.

(Although maybe there is some weird measure-theoretic reason why this doesn't work. Usually when I make a guess about measure theory it turns out to be wrong.)

view this post on Zulip Kevin Carlson (Feb 09 2025 at 07:32):

Doesn't the Giry monad restrict to a monad on Set along the inclusion of sets as discrete measurable spaces?

view this post on Zulip John Baez (Feb 09 2025 at 08:02):

I don't think so: the Giry monad maps a measurable space to the measurable space of probability measures on it, but the measurable space of probability measures on a discrete measurable space is not itself discrete, e.g. the probability measure space on 2 is isomorphic to [0,1] with its usual Borel sigma-algebra, which is different than the sigma-algebra of the discrete measurable space [0,1] (namely the sigma-algebra of all subsets of [0,1]).

view this post on Zulip Kevin Carlson (Feb 09 2025 at 08:52):

Ah, right.

view this post on Zulip Morgan Rogers (he/him) (Feb 09 2025 at 17:21):

Nathaniel Virgo said:

Morgan Rogers (he/him) said:

A consequence would be that you can do probability theory over any countably complete Boolean countably lextensive category with the same suite of interval objects.

I'm kind of confused about this point actually. ...shouldn't that mean we can do it in any countably complete Boolean countably extensive category? Since Set is such a category (right?), this seems to imply we'd get for free a monad on Set that inherits all the nice properties of the Giry monad.

I expect that the confusion comes from assuming that the monad thus constructed will inherit the properties of the Giry monad? We could figure this out explicitly, though: what is an example of a nice property that the Giry monad has that you expect to fail when the construction is transported to Set?

view this post on Zulip Nathaniel Virgo (Feb 10 2025 at 11:54):

If I try to be more precise I suppose that one of the following things must happen: either the monad needs initiality of StdBorel to construct it (hence you can't construct the same monad on Set), or some of its nice properties need initiality of StdBorel to prove them (hence its counterpart on Set doesn't have those properties), or the Set version does inherit all the nice properties but becomes trivial in some way, or at least isn't useful for probability in the same way that BorelStoch is. Or none of the above, which is the best possible outcome but seems unlikely. Or something else and I'm more confused than I thought, which is quite possible.

Specifically, the properties I'm thinking of are the following, which I regularly use to reason about its Kleisli category, BorelStoch:

These are basically the assumptions used in Fritz+Gonda+Perrone's abstract proof of de Finetti's theorem.

view this post on Zulip Tobias Fritz (Feb 11 2025 at 15:31):

Nathaniel Virgo said:

Take an object XX and construct XNX^\mathbb{N} as a countable product. Consider the group of finite permutations of N\mathbb{N} and its action on XNX^\mathbb{N}. I think we have enough structure to construct the object of orbits of this group action. I conjecture that this is P(X)P(X), the Giry monad applied to XX.

This is similar in spirit to what we're currently working on with the law of large numbers. This is at a sufficiently advanced stage that we could possibly share a draft if you're interested.

What I imagine that you have in mind is to consider a map XNPXX^{\mathbb{N}} \to PX which takes every sequence to the associated "empirical measure" defined by the relative frequencies of the elements in the sequence. The first problem now is that this doesn't make sense in general, because many sequences don't have well-defined relative frequencies. Therefore the transformation can at most be partially defined. Another issue is that you can have two sequences for which the relative frequencies do converge, and even be identical so that you get the same element of PXPX, even if the sequences do not differ by a finite permutation. (They don't even necessarily differ by an infinite permutation, because one may contain elements with relative frequency zero!) So although a partial transformation exists, there's no simple characterization of when two elements of its domain get mapped to the same element of the codomain.

(There are further analytical subtleties that we've had to struggle with, which is part of why it's taken us a few years to get to the point where we have a draft.)

view this post on Zulip Tobias Fritz (Feb 11 2025 at 15:31):

With that being said, I think that your construction could nevertheless give an affine commutative monad on Meas or BorelMeas, so that its Kleisli category is going to be a Markov category. Maybe this Markov category is of some interest even if it's not Stoch? For one thing, this construction clearly works on Set too.

view this post on Zulip Tobias Fritz (Feb 11 2025 at 15:36):

Nathaniel Virgo said:

This is because of de Finetti's theorem. Whatever the quotient of this group action is, measures over it ought to be basically the same as exchangeable measures over XNX^\mathbb{N}, which are the same as measures over P(X)P(X). So in particular, elements of it should just be elements of P(X)P(X). (Since over standard Borel spaces an element is the same as a measure concentrated on that element.)

This indeed gives a construction of PX, but it's different in two ways from the quotient that you've described:

view this post on Zulip Tobias Fritz (Feb 11 2025 at 15:40):

So although we can indeed characterize the Giry monad like this, doing so requires starting with the Markov category already. I still think that this is interesting, and this is what we're doing with the law of large numbers: we assume that we have a Markov category with additional structure that makes it have a law of large numbers, and then we derive representability from that.

view this post on Zulip Tobias Fritz (Feb 11 2025 at 15:41):

Conceptually, I think it's good practice to think of Markov categories as the primary structures and the monads as derived rather than the other way around.

view this post on Zulip Nathaniel Virgo (Feb 11 2025 at 18:31):

Ah, of course, that all makes complete sense. I know all these facts very well, I had just not realised I was ignoring them!

Still, I wonder if there's a way to fix it. It would be kind of neat to turn de Finetti's theorem into the definition of a probability measure, if it can be done.

I agree with the general principle of seeing Markov categories rather than monads as primary. It's just that it seems the work of characterising the deterministic subcategory of BorelStoch has already been done, so I was speculating about ways to get from there to an abstract characterisation of BorelStoch itself.

view this post on Zulip Morgan Rogers (he/him) (Feb 12 2025 at 14:15):

Reviewing the [[Giry monad]] page, I see that the definition is more complicated than I had naively misremembered it was; it's not just a matter of homming into an interval, for instance. The definition of probability measure on an object of a countably (co)complete Boolean lextensive category is a monotone function from its BA of subobjects to [0,1][0,1] satisfying countable constraints (these can either be expressed in terms of countable increasing chains or countable disjoint elements) and modularity (see [[probability valuation]] for that condition, which amounts to inclusion-exclusion). On the face of it, there is no reason for the collection of such maps to have the structure of an object of the same Boolean category, nor does the interval itself need to be a part of the Boolean category for this definition to make sense.

On the other hand, suppose that [0,1][0,1], rather than being a mere object of a Boolean category, were instead realised as the lattice of subobjects of some structure II in such category. Suppose further that this type of structure were such that every object comes with a trivial such structure (whose substructures are in correspondence with subobjects) and that the category of such structures is enriched over the Boolean category. Then the hom-object IXI \to X could recover the collection of monotone bla-bla-bla functions from subobjects of X to [0,1][0,1] by inverse images. That's the best version of what I was imagining that I can come up with, but it has too many holes to actually be usable: it's not even clear to me if the kind of structure that's needed is something resembling an order structure, something quantale-like or the action of a symmetric group, say (permutations of a countable set, say, if this is somehow directly related to di Finetti?)
If anyone wants to run with that idea or shoot it down, be my guest. :wink:

view this post on Zulip Sam Staton (Feb 12 2025 at 16:47):

Hi I just noticed this thread (late again!) @Tobias Fritz and @Victor Bloch, you were talking about the free affine commutative monad generated by a binary operation. Are you sure it embeds in the distribution monad? If I understand what you wrote correctly, it should be something like the Bernstein polynomials. We investigated this in Beta-Bernoulli processes and algebraic effects, and the characterization result is Proposition 7. If I understand what you said, you are looking at the case l=1l=1, and we notated the binary operation ?p?_p. But perhaps you had something different in mind, which would also be very interesting!

view this post on Zulip Sam Staton (Feb 12 2025 at 16:54):

(Sorry to jump back! I think Morgan's question is interesting, the reason why Borel probability kernels form a Kleisli category is a bit of a mystery, and didn't mean to deflect!)

view this post on Zulip Tobias Fritz (Feb 12 2025 at 18:43):

Sam Staton said:

Hi I just noticed this thread (late again!) Tobias Fritz and Victor Bloch, you were talking about the free affine commutative monad generated by a binary operation. Are you sure it embeds in the distribution monad? If I understand what you wrote correctly, it should be something like the Bernstein polynomials. We investigated this in Beta-Bernoulli processes and algebraic effects, and the characterization result is Proposition 7. If I understand what you said, you are looking at the case l=1l=1, and we notated the binary operation ?p?_p.

Thanks for the reminder about that paper, @Sam Staton ! It's a challenging read for someone with a background like mine, and I had tried and failed before, but right now I think I'm understanding a little more.

Would it be correct to say that if one phrases your formalism as an algebraic theory, then your νi,j\nu_{i,j} are like having access to the rational numbers ii+j\frac{i}{i+j} as constants, so that your algebraic theory includes the theory of rational convexity in particular?

view this post on Zulip Tobias Fritz (Feb 12 2025 at 18:49):

In Section 5, you say that your monad lives on the functor category [FinSet, Set], and indicate that this is because it's a parametrized algebraic theory. Is it possible to sketch the intuition behind this without giving the full definition of what a parametrized algebraic theory is?

view this post on Zulip Tobias Fritz (Feb 12 2025 at 18:53):

In any case, I do see some similarity with what we're doing, even if it's perhaps not exactly the same. I agree that the Bernstein polynomials are related in some way, but we haven't actually needed them explicitly: they are under the hood in the results in On convex linear forms, which we're simply using as a black box.

view this post on Zulip Tobias Fritz (Feb 12 2025 at 18:54):

Maybe it helps to look at their result:
image.png
Is this an instance of your Proposition 7?

view this post on Zulip Tobias Fritz (Feb 12 2025 at 19:05):

To my mind, the main takeaway of these results is more philosophical rather than technical, namely the idea that probability monads are very close to being free in a certain sense. For example, it seems plausible that the Radon monad on CHaus is the free affine commutative monad generated by a binary operation, or even by an operation of any arity 2\ge 2 (though we haven't been able to prove it yet). Is this kind of idea something that you've known already or seen somewhere else?

view this post on Zulip Sam Staton (Feb 13 2025 at 07:08):

Thanks @Tobias Fritz ! I don't think zulip has threads so I don't know how to discuss multiple things at once but there's lots of interesting stuff in your message! 

For our result, maybe the paper you posted could be a corollary (thanks for the reference, I didn't know it but it is nice) but I think it is quite different. Our paper is quite condensed, I'd quite like to now write a Markov category version. Here is the statement l=1l=1 extracted and without the other probabilities or ν\nu around. I hope I haven't made a mistake.

view this post on Zulip Sam Staton (Feb 13 2025 at 07:08):

As you know we can make a finite distributions monad TRT_R for any commutative semiring RR. It's the submonad of the modules monad where we only take things that sum to 11

Now consider the commutative semiring R=x,y  x+y=1,xy=yxR=\langle x,y\ |\ x+y=1,xy=yx\rangle. We should think of y=(1x)y=(1-x). This could be called the Bernstein semiring. It should be a subsemiring of the polynomial ring Z[x]\mathbb{Z}[x]

Theorem (if I'm remembering correctly): The free commutative affine monad with one binary operation is TRT_R.

view this post on Zulip Sam Staton (Feb 13 2025 at 07:09):

I can see that someone (maybe Sean Moss) has typed up a tidy proof of this very fact in our repository, we could perhaps share it. But first I wanted to check whether this is the result you are looking at with @Victor Bloch.

view this post on Zulip Tobias Fritz (Feb 13 2025 at 07:25):

Oh I see, that's a really nice and intriguing way to put it!

view this post on Zulip Tobias Fritz (Feb 13 2025 at 07:27):

No, we haven't known any explicit description like that. Perhaps it could be helpful (in combination with Stone-Weierstrass) for tackling the case of CHaus and the Radon monad.

view this post on Zulip Tobias Fritz (Feb 13 2025 at 07:30):

Maybe it would make sense to join forces on this and write it all up in one document? What we have so far is a draft of Victor's thesis with the formal construction of the free affine commutative monad as a quotient of the free monad, the fact that it's the monad of modes, and the embedding into the distribution monad.

view this post on Zulip Tobias Fritz (Feb 13 2025 at 07:39):

(I had asked Bart last year if he already knew of anything in this direction, which was not the case, but I now realize that I should have asked you too!)

view this post on Zulip Morgan Rogers (he/him) (Feb 13 2025 at 16:50):

Sam Staton said:

Thanks Tobias Fritz ! I don't think zulip has threads so I don't know how to discuss multiple things at once but there's lots of interesting stuff in your message!

Splitting off new topics is encouraged. Linking is easy enough to connect different topics when needed :innocent:

view this post on Zulip Victor Bloch (Feb 15 2025 at 06:11):

Sam Staton schrieb:

As you know we can make a finite distributions monad TRT_R for any commutative semiring RR. It's the submonad of the modules monad where we only take things that sum to 11

Now consider the commutative semiring R=x,y  x+y=1,xy=yxR=\langle x,y\ |\ x+y=1,xy=yx\rangle. We should think of y=(1x)y=(1-x). This could be called the Bernstein semiring. It should be a subsemiring of the polynomial ring Z[x]\mathbb{Z}[x]

Theorem (if I'm remembering correctly): The free commutative affine monad with one binary operation is TRT_R.

This construction of the free affine commutative monad looks a lot like our construction as the free mode monad. The constructions seem to be isomorphic: We can think of the elements of TRXT_RX as formal convex combinations in XX with parameters in RR. This viewpoint then gives (it at least looks like it does) the free mode on XX.

view this post on Zulip Sam Staton (Feb 15 2025 at 11:17):

Thanks very much! I see the connection with the other paper now, the variable xx can be thought of as a trancendental.

view this post on Zulip Sam Staton (Feb 15 2025 at 11:17):

I just remembered that the version of the Theorem above is Example 5 in our LAFI abstract semirings, monads, and tensors, which we ought to get back to. Happy to collaborate! The idea there was to combine monads by combining semirings, partly as a way of building Markov categories, but also with other applications.

view this post on Zulip Sam Staton (Feb 15 2025 at 11:19):

I was interested in the polynomials-as-functions view because I wanted to think about this operation as tossing an unknown coin: what can we know about it? And then these equations are all that we can say. I was a bit inspired by the "Bernoulli factory" work. 

I think perhaps this is philosophically different from the transcendental interpretation, do you agree? Even though the theory is the same. This might come out when we add limits and recursion. I was expecting to get continuous functions, or something like it, coming from the idea of the Bernstein polynomial basis or the Bernoulli factory results. Whereas I think you were hoping to just get all real probabilities? One route we were thinking of was a possible connection between *-semirings and iteration theories, which we were looking at but which I think is still hanging.

view this post on Zulip Sam Staton (Feb 15 2025 at 11:20):

(The Beta-Bernoulli paper is also in this unknown-coin direction, basically we have a collection of these unknown coins, ie a tensor of the free affine monad with one binary generator with itself ll times, and this indexing ll is why it is presheaf-enriched; plus ν\nu has a complete interpretation as integrating-out one of the polynomials, even though the axiomatization is finite and simple. Happy to discuss more!)

view this post on Zulip Tobias Fritz (Feb 16 2025 at 17:18):

Sam Staton said:

I just remembered that the version of the Theorem above is Example 5 in our LAFI abstract semirings, monads, and tensors, which we ought to get back to.

Cool! So the proof of that isomorphism isn't yet available anywhere?

view this post on Zulip Tobias Fritz (Feb 16 2025 at 17:20):

Sam Staton said:

I was expecting to get continuous functions, or something like it, coming from the idea of the Bernstein polynomial basis or the Bernoulli factory results. Whereas I think you were hoping to just get all real probabilities?

I see. Yes, we've been hoping to get all real probabilities. Concerning my conjecture that the free affine commutative monad generated by a binary operation on CHaus would be the Radon monad, @Benedikt Peterseim has had some nice insights that go in the direction of a disproof.

view this post on Zulip Tobias Fritz (Feb 16 2025 at 17:21):

But even if that was true, it would be less convincing than obtaining such a universal property for the Giry monad on standard Borel spaces directly. Here's one way in which this might work:

view this post on Zulip Tobias Fritz (Feb 16 2025 at 17:22):

Conjecture: The Giry monad on BorelMeas is freely generated by a binary operation, among all affine commutative monads which moreover preserve countable cofiltered limits.

view this post on Zulip Tobias Fritz (Feb 16 2025 at 17:27):

The idea here is that the Giry monad first of all does preserve countable cofiltered limits; that's one version of the Kolmogorov extension theorem. For the intuition behind the conjectured universality, think of building the free affine commutative monad inductively by repeatedly applying the generating binary operation and imposing the relevant equations. Then the preservation of cofiltered limits gives us the additional power to create generalized elements of {0,1}N\{0,1\}^{\mathbb{N}} by constructing a compatible family of marginals. In terms of a coin with transcendental bias, this amounts to allowing protocols with infinitely many coin flips! The point now is that if we allow infinite protocols, then we should be able to use that coin to simulate any other coin with arbitrary bias, and in particular get all real probabilities. Formally, this is by combining the preservation of cofiltered limits with functoriality with respect to measurable maps {0,1}N{0,1}\{0,1\}^{\mathbb{N}} \to \{0,1\}, which will create new generalized elements of {0,1}\{0,1\}. This will hopefully make the resulting set over {0,1}\{0,1\} become exactly [0,1][0,1].

view this post on Zulip Tobias Fritz (Feb 16 2025 at 17:29):

Sam Staton said:

I think perhaps this is philosophically different from the transcendental interpretation, do you agree?

I admittedly don't understand the Bernoulli factory stuff :see_no_evil: I do think of the free affine commutative monad, or rather of its Kleisli category, as a Markov category formalizing the processes that one can build by having access to a coin of unknown bias.

view this post on Zulip Tobias Fritz (Feb 16 2025 at 17:37):

Perhaps we should discuss how to proceed with regards to writing, authorship etc in private? Victor and I are preparing an ACT submission on this (with deadline on Feb 24) and should be able to share it by tomorrow. It could be nice if you wanted to contribute the Bernstein semiring observations to that. On the other hand, if I understand correctly, you're working on including this in an extended version of the LAFI paper? Of course that would be fine for us too.

It should also be okay if we don't have any definite plan by Feb 24, since our ACT submission is only for a talk and we'll try to publish elsewhere. Perhaps JPAA could be a good venue for the sort of thing that we have in mind, but a CS journal could be a good option as well especially if there is more content in that direction. (I generally try to publish in journals rather than conferences due to the higher quality of peer review.)

view this post on Zulip Nathaniel Virgo (Feb 16 2025 at 19:03):

Tobias Fritz said:

The idea here is that the Giry monad first of all does preserve countable cofiltered limits; that's one version of the Kolmogorov extension theorem.

Ooh, is there a reference/proof for this fact? I've been wondering about it recently.

view this post on Zulip Sam Staton (Feb 17 2025 at 17:09):

One reference is Dirichlet is natural, where it is called Bochner extension, there may be earlier ones too. There it is restricted to surjective diagrams, which may be important?

view this post on Zulip Sam Staton (Feb 17 2025 at 17:25):

Tobias Fritz said:

Conjecture: The Giry monad on BorelMeas is freely generated by a binary operation, among all affine commutative monads which moreover preserve countable cofiltered limits.

Thanks, this is an interesting conjecture. I'll think about it!

We had previously thought of freely adding some cofiltered limits to FinStoch\mathbf{FinStoch}, and then I think we arrived at Stone spaces and Radon probability kernels. Maybe that is analogous to what @Nathaniel Virgo was asking in the first place.

But this conjecture about BorelMeas\mathbf{BorelMeas} is an interesting idea, to start from BorelMeas\mathbf{BorelMeas} rather than freely adding the limits themselves.

I know there is @Tobias Fritz's paper with @Paolo Perrone about probability as a colimit, which seems similar in spirit to me? Also in domain theory something like this happens, the probabilisitic powerdomain includes all measures but can be generated by one binary operation, although it is a bit complicated.

view this post on Zulip Nathaniel Virgo (Feb 17 2025 at 19:06):

I might as well throw another conjecture/question into this thread, as someone here will probably know the answer: does BorelStoch have all countable cofiltered limits? I think this is a bit stronger than claiming the Giry monad preserves countable cofiltered limits, in that so far I haven't been able to prove abstractly from the things I know about BorelStoch (e.g. that it's representable, that BorelStochdet\textbf{BorelStoch}_{det} has countable limits, BorelStoch has Kolmogorov products etc.) and it does't seem like assuming the Giry monad preserves all cofiltered limits immediately helps. (It gives you some of them but not all of them, at least not straightforwardly.) But at the same time it doesn't seem much stronger so I'm still wondering if it might be true.

view this post on Zulip Paolo Perrone (Feb 17 2025 at 22:03):

This proposition might help, if I understand correctly.