## Posts

## Comments

**Ramana Kumar (ramana-kumar)**on Intelligence or Evolution? · 2021-10-17T14:49:11.703Z · LW · GW

This is a good point. Could something like Shapley value help in distributing credit for X between the humans and the impersonal mechanism? I find myself also wanting to ask about how frequent these cases are -- where it could easily be viewed both ways -- and declare that if it's mostly ambiguous then 'evolution' wins.

For "some impersonal mechanism" I'm thinking "memetic fitness of X amongst humans" (which in some cases cashes out as the first group of humans being larger?). What are other ways of thinking about it?

The story feels a little underspecified. When X happens because the first group of humans figured out how to thwart the second group, and anticipated them, etc. and furthermore if that group consistently does this for whatever they want, it seems a lot more like intelligence.

**Ramana Kumar (ramana-kumar)**on Intelligence or Evolution? · 2021-10-12T11:03:27.882Z · LW · GW

I did intend 'Evolution' here to include markets. The main contrast I'm trying to make is to a unified rational actor causing the outcomes it wants.

**Ramana Kumar (ramana-kumar)**on Intelligence or Evolution? · 2021-10-12T11:01:06.662Z · LW · GW

I agree that both are good explanations. My question is more about which will be dominant in the long run. I tried to ask this more clearly with the mutually exclusive version in the post (someone vs no one was trying to produce the outcome).

I could view the "Why not both?" response as indicating that neither is dominant and we just have to understand how both operate simultaneously (perhaps on different timescales) and interact. I think I'd view that as actually coming down mostly on the evolution side of things - i.e., this means I should understand intelligence only in the larger evolutionary context -- no intelligence will permanently outstrip and render irrelevant the selection forces. Is that right?

**Ramana Kumar (ramana-kumar)**on Finite Factored Sets: Conditional Orthogonality · 2021-09-13T13:26:35.602Z · LW · GW

That's right. A partial function can be thought of as a subset (of its domain) and a total function on that subset. And a (total) function can be thought of as a partition (of its domain): the parts are the inverse images of each point in the function's image.

**Ramana Kumar (ramana-kumar)**on The Blackwell order as a formalization of knowledge · 2021-09-12T21:56:46.003Z · LW · GW

Blackwell’s theorem says that the conditions under which can be said to be more

generallyuseful than are precisely the situations where is a post-garbling of .

Are the indices the wrong way around here?

**Ramana Kumar (ramana-kumar)**on Introduction to Cartesian Frames · 2021-05-17T08:25:51.241Z · LW · GW

A formalisation of the ideas in this sequence in higher-order logic, including machine verified proofs of all the theorems, is available here.

**Ramana Kumar (ramana-kumar)**on Time in Cartesian Frames · 2021-03-29T18:26:37.508Z · LW · GW

"subagent [] that could choose " -- do you mean or or neither of these? Since is not closed under unions, I don't think the controllables version of "could choose" is closed under coarsening the partition. (I can prove that the ensurables version is closed; but it would have been nice if the controllables version worked.)

ETA: Actually controllables do work out if I ignore the degenerate case of a singleton partition of the world. This is because, when considering partitions of the world, ensurables and controllables are almost the same thing.

**Ramana Kumar (ramana-kumar)**on Thoughts on Human Models · 2021-02-23T12:48:58.190Z · LW · GW

because whereas math-proof data-centers might result in our inadvertent death, something that refers to what humans want might result in deliberate torture.

I want to note that either case of screwing up this badly currently feels pretty implausible to me.

**Ramana Kumar (ramana-kumar)**on Thoughts on Human Models · 2021-02-23T12:45:23.961Z · LW · GW

Great summary!

**Ramana Kumar (ramana-kumar)**on Time in Cartesian Frames · 2021-02-22T11:46:55.536Z · LW · GW

I have something suggestive of a negative result in this direction:

Let be the prime-detector situation from Section 2.1 of the coarse worlds post, and let be the (non-surjective) function that "heats" the outcome (changes any "C" to an "H"). The frame is clearly in some sense equivalent to the one from the example (which deletes the temperature from the outcome) -- I am using my version just to stay within the same category when comparing frames. As a reminder, primality is not observable in but is observable in .**Claim:** No frame of the form is biextensionally equivalent to **Proof Idea**:

The kind of additional observability we get from coarsening the world seems in this case to be very different from the kind that comes from externalising part of the agent's decision.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-02-11T08:29:17.053Z · LW · GW

With the other problem resolved, I can confirm that adding an escape clause to the multiplicative definitions works out.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-02-11T08:25:29.721Z · LW · GW

Using the idea we talked about offline, I was able to fix the proof - thanks Rohin!

Summary of the fix:

When and are defined, additionally assume they are biextensional (take their biextensional collapse), which is fine since we are trying to prove a biextensional equivalence. (By the way this is why we can't take , since we might have after biextensional collapse.) Then to prove , observe that for all , which means , hence since a biextensional frame has no duplicate columns.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-02-05T10:59:31.881Z · LW · GW

I presume the fix here will be to add an explicit escape clause to the multiplicative definitions. I haven't been able to confirm this works out yet (trying to work around this), but it at least removes the counterexample.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-02-05T09:55:33.395Z · LW · GW

How is this supposed to work (focusing on the claim specifically)?

and so

Thus, .

Earlier, was defined as follows:

given by and

but there is no reason to suppose above.

**Ramana Kumar (ramana-kumar)**on Time in Cartesian Frames · 2021-02-04T08:41:56.604Z · LW · GW

It suffices to establish that

I think the and here are supposed to be and

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-30T13:50:57.192Z · LW · GW

Indeed I think the case may be the basis of a counterexample to the claim in 4.2. I can prove for any (finite) with that there is a finite partition of such that 's agent observes according to the assuming definition but does *not* observe according to the constructive multiplicative definition, if I take

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-30T09:21:39.636Z · LW · GW

Let

nit: should be here

and let be an element of .

and the second should be . I think for these and to exist you might need to deal with the case separately (as in Section 5). (Also couldn't you just use the same twice?)

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-30T00:12:52.641Z · LW · GW

UPDATE: I was able to prove in general whenever and are disjoint and both in , with help from Rohin Shah, following the "restrict attention to world " approach I hinted at earlier.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-29T19:23:42.718Z · LW · GW

this is clearly isomorphic to , where , where . Thus, 's agent can observe according to the nonconstructive additive definition of observables.

I think this is only true if partitions , or, equivalently, if is surjective. This isn't shown in the proof. Is it supposed to be obvious?

EDIT: may be able to fix this by assigning any that is not in to the frame so it is harmless in the product of s -- I will try this.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-29T10:52:33.342Z · LW · GW

And it seems we do actually need in the proof to justify:

Thus it suffices to show that .

Without it, we have to show instead.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-29T08:26:06.384Z · LW · GW

Because and are not a partition of the world here.

EDIT: but what we actually need in the proof is where the do result in a partition, so I think this will work out the same as the other comment. I'm still not convinced about biextensional equivalence between the frames without the rest of the product.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-28T12:40:18.696Z · LW · GW

I haven't yet figured out why it's true under - I'll keep trying, but let me know if there's a quick argument for why this holds. (Default next step for me would be to see if I can restrict attention to the world then do something similar to my other comment.)

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-26T18:53:34.223Z · LW · GW

and observe that

This cannot be true. I can prove in general whenever by observing that the agent cardinalities on each side differ.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-26T18:22:36.405Z · LW · GW

where (C),

Presumably two of those indices should be

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-26T14:19:37.813Z · LW · GW

Let . Thus, we also have that

I'm not seeing why this follows. I'll look for a counterexample, but in the meantime maybe there's a simple explanation for why we can combine the product of two assumes as an assume of the union? (I think the only relevant assumption in this context is that the s partition the world; but I might be missing some other important assumption.)

EDIT: I can see how maybe this will follow from the definition of observability of a partition from subsets (which we are also assuming) and the fact that is closed under union... will try to figure that out. -- Yep I think this works out. Sorry for the confusion.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-26T10:44:59.899Z · LW · GW

Let , and let

Is supposed to be here, rather than including ?

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-24T15:50:51.452Z · LW · GW

Next, assume that 's agent can observe according to the additive definition. We will show that 's agent can observe .

I might be misunderstanding this, but the proof suggests you're actually assuming the assuming definition here, not the additive definition. In which case we may be missing the proof of implication of any of the other definitions from the additive definition.

**Ramana Kumar (ramana-kumar)**on Eight Definitions of Observability · 2021-01-24T08:11:29.386Z · LW · GW

Definition:We say that 's agent can observe a finite partition of if for all functions , there exists an element such that for all , .

Claim:This definition is equivalent to the definition from subsets.

This doesn't hold in the degenerate case , since then we have an empty function but no elements of . (But the definition from subsets holds trivially.)

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-23T16:24:39.009Z · LW · GW

Claim:For any Cartesian frame over and partition of , let send each element of to its part in . If for all and we have , then .

I think this may also need to assume that is non-empty.

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-23T09:08:35.965Z · LW · GW

If

The argument is missing in several places like this from 4.2 onwards.

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-23T06:03:37.212Z · LW · GW

is given by

There's a prime missing on . I'd also have expected instead of as the variable (doesn't affect correctness).

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-22T18:00:58.207Z · LW · GW

while is given by

should be applied to above, I think

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-22T13:48:25.786Z · LW · GW

Let , send each element of to its part in , so .

Presuming the here should be a

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-22T08:50:25.359Z · LW · GW

Oh I see this has also been fixed. Thanks!

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-22T08:45:24.911Z · LW · GW

This is also suspicious in section 2.2 about Assuming. I think it should be the other way around and about Assume rather than Commit, and I don't think that's equivalent to what's written here. (But I'm not confident about this.)

Claim:For all , and .

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-21T12:05:16.405Z · LW · GW

Claim:For all , and .

Are these the wrong way around?

I believe is indeed trivial, but the opposite is less obvious.

**Ramana Kumar (ramana-kumar)**on Committing, Assuming, Externalizing, and Internalizing · 2021-01-21T11:54:36.580Z · LW · GW

Similarly, for all , and .

We don't need to be a proper subset here (i.e., I think is a typo for ). Also in my view all the isomorphisms in this section are actually equalities (but it's also reasonable to never consider equality of frames).

**Ramana Kumar (ramana-kumar)**on Additive and Multiplicative Subagents · 2021-01-20T17:34:24.929Z · LW · GW

We let , where .

minor typo: I think should be above.

**Ramana Kumar (ramana-kumar)**on Additive and Multiplicative Subagents · 2021-01-19T19:02:22.102Z · LW · GW

I'm not sure why you started using the equivalence symbol on morphisms, e.g.,

in the categorical definition of multiplicative subagent.

I think equality () is the correct concept to be using here instead, as in the categorical definition of (original) subagent.

**Ramana Kumar (ramana-kumar)**on Additive and Multiplicative Subagents · 2021-01-19T15:01:27.522Z · LW · GW

and let and compose to something homotopic to the identity in both orders.

Pretty sure the s should be s here (though they're notation for the same function anyway).

**Ramana Kumar (ramana-kumar)**on Sub-Sums and Sub-Tensors · 2021-01-14T06:30:34.058Z · LW · GW

In the first claim of 3.4, the last bit of the claim is but the proof actually shows .

**Ramana Kumar (ramana-kumar)**on Multiplicative Operations on Cartesian Frames · 2020-12-20T07:15:56.214Z · LW · GW

a few more typos. (do say if these nits aren't wanted.)

is the natural bijection

(should be: )

while the right hand side is

(should be: in )

**Ramana Kumar (ramana-kumar)**on Introduction to Cartesian Frames · 2020-12-17T15:26:39.953Z · LW · GW

To see that *some* restriction is required here (not imposed by HOL), consider that if may contain arbitrary Cartesian frames over then we would have an injection that, for example, encodes a set as the Cartesian frame with (the environment and evaluation function are unimportant), which runs afoul of Cantor's theorem regarding the cardinality of .

I wouldn't be surprised if a similar encoding/injection could be made using just the operations used to construct Cartesian frames that appear in this sequence - though I have not found one explicitly myself yet.

**Ramana Kumar (ramana-kumar)**on Multiplicative Operations on Cartesian Frames · 2020-12-17T14:49:18.012Z · LW · GW

Since we have already established commutativity, it suffices to show that .

For the confused reader, the argument in more detail here is:

where is commutativity of tensor, is the fact claimed to suffice above, and is the implicitly assumed lemma that and implies (this is proved later but only for ).

**Ramana Kumar (ramana-kumar)**on Multiplicative Operations on Cartesian Frames · 2020-12-16T08:46:05.321Z · LW · GW

minor typo in the indices here:

We will show that , and since the definition of is symmetric in swapping and , it will follow that

**Ramana Kumar (ramana-kumar)**on Functors and Coarse Worlds · 2020-12-10T00:18:41.746Z · LW · GW

Nice! I'm glad I asked, since I hadn't realised those sufficient conditions for being homotopic to the identity. That's useful in several proofs I suspect. (For anyone following along, I believe this only holds for a morphism from a frame to itself.)

**Ramana Kumar (ramana-kumar)**on Functors and Coarse Worlds · 2020-12-09T18:22:14.889Z · LW · GW

minor typo:

and take to be the set of all such that

should have

Also I think later in that proof some of the 's (like in ) should be 's instead.

**Ramana Kumar (ramana-kumar)**on Introduction to Cartesian Frames · 2020-12-05T08:34:02.460Z · LW · GW

Do we lose much by restricting attention to finite Cartesian frames (i.e., with finite agent and environment)? I ask because I'm formalising these results in higher-order logic (HOL), and the category