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Stream: theory: categorical probability

Topic: Supports/conditionals/kolmogorov products trilemma

Eigil Rischel (Jan 22 2026 at 18:10):

It's been observed for a while now that most Markov categories which have Kolmogorov products fail to have conditionals, and vice versa—essentially the only examples that have both are BorelStoch and things very similar to it. The shape of the problem seems to be that Kolmogorov products require some sort of continuity property that restricts the collection of available predicates $X^\infty \to Y$, but conditional distributions are usually not continuous.

Analogously to this, I've observed that Markov categories with all supports usually fail to have conditionals—here, the only counterexamples seem to be "discrete" Markov categories, like the Kleisli category of the finite distribution monad. But these don't have Kolmogorov products, and of course conversely BorelStoch doesn't have supports.

So I thought it might be an interesting question: does there exist any Markov category with conditionals, supports, and Kolmogorov products, which is not Cartesian?

Morgan Rogers (he/him) (Jan 23 2026 at 09:53):

Which of the counterexamples you mentioned are cartesian? (Or in other words, can you remind me what cartesian means in this context?)

Eigil Rischel (Jan 23 2026 at 10:16):

By Cartesian I just mean that the tensor product is a product in the categorical sense. For Markov categories this means every morphism is deterministic, so it's kind of a degenerate case from the point of view of abstract probability theory. None of the examples I mentioned are Cartesian, the point is that there are many Cartesian categories which have all three properties (this amounts to having infinite products and image factorizations, or something like that—conditionals are automatic in this case. Certainly the category of sets is an example), but I'm not aware of any noncartesian examples.

Paolo Perrone (Jan 23 2026 at 10:42):

That's an interesting question. I assume you mean countable Kolmogorov products, right?

Eigil Rischel (Jan 23 2026 at 10:43):

Yeah (although it would be quite surprising if there existed a counterexample with countable Kolmogorov products, but not one with higher cardinalities!)

Paolo Perrone (Jan 23 2026 at 10:55):

Yes, I'm just tying to this 'feeling' of continuity that Kolmogorov products have. And usually countable stuff translates well from continuous to measurable (e.g. the Borel $\sigma$-algebra functor preserves countable products), but uncountable stuff may not.

Eigil Rischel (Jan 23 2026 at 11:44):

I think if you further assume the existence of a coproduct 2 = 1 + 1, and the existence of a symmetric coinflip $f: I \to I + I$ of full support, then you can use conditionals to build something close enough to an indicator of the diagonal $2^\omega \otimes 2^\omega \to 2$, and use a Hewitt-Savage 0-1 argument to prove that this is almost surely equal to zero with respect to the independent product $f^\omega \otimes f^\omega$, which contradicts the claim that $\mathrm{copy} f^\omega << f^\omega \otimes f^\omega$, so $f^\omega$ can't have a support.

Ah, the problem with this is that the indicator constructed as a conditional can't be assumed deterministic, so we can't apply the 0-1 law. (It's also not totally obvious how to show it's deterministically equal to zero—I have a similar argument in my preprint on the universal property of BorelStoch, but there I use some pullbacks which we're not assuming exist here). But maybe some argument like this can be made to work.