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Finite Factored Sets in Pictures

10kave

6Ben Pace

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Curated. I am excited about many more distillations and expositions of relevant math on the Alignment Forum. There are a lot of things I like about this post as a distillation:

- Exercises throughout. They felt like they were simple enough that they helped me internalise definitions without disrupting the flow of reading.
- Pictures! This post made me start thinking of finite factorisations as hyperrectangles, and histories as dimensions that a property does not extend fully along.
- Clear links from Finite Factored Sets to Pearl. I think these are roughly the same links made in the original, but they felt clearer and more orienting here.
- Highlighting which of Scott's results are the "main" results (even more than the "Fundamental Theorem" name already did).
- Magdalena Wache's engagement in the comments.

I do think the pictures became less helpful to me towards the end, and I thus have worse intuitions about the causal inference part. I'm also not sure about the emphasis of this post on *causal* rather than *temporal *inference. But I still love the post overall.

Finite factored setsare a new paradigm for talking about causality. You can use them to do some cool things you can’t do with Pearl’scausal graphs, for exampleinferring a causal arrow between two binary variables.Also, finite factored sets are a really neat mathematical structure: they are a way of taking a

setand expressing it as aproductof some factors.Set factorizations are analogous tointeger factorizations, in the same way thatset partitionsare analogous tointeger partitions.So, here is my current understanding of finite factored sets, in pictures.

## 1. What are Set Factorizations?

What do these “factored sets” look like? Let’s start with a set S and factor it.

The first concept we need is a

partitionof a set S. A partition is a way of chopping up S into subsets (calledparts). Here are a few examples of partitions:We usually call the partitions X,Y,Z,U,V, or W, and their parts xi,yi,… like this:

U is called the

trivial partition. It only has one part.We can think of

partitions as properties, or variablesover our set. For example, consider a set like this:and compare it to the partitions X,Y and Z from above:

Then

Exercise

Consider these two partitions X and Y on the set S. What would it look like to represent them as properties (e.g. X = shape, Y = color) instead?

Spoiler space

It could look something like this:

Here X = shape = {x1,x2,x3} = {star, circle, square}, and Y = color ={y1,y2} = {green, orange}.

I hope you can see how partitions and properties are basically the same thing. In the rest of this post, I will use “partitions” and “properties” interchangeably. Sometimes I will use the ring-style visualization of partitions, and sometimes the property style, depending on what I find more intuitive in any given example.

Now we can define

set factorizations:A

factorizationB of our set S looks like this:A factorization is a set B={b1,b2,…,bn} of partitions (called

factors). In this case B={b1,b2}={X,Y}={{x1,x2},{y1,y2}}.But it can’t just be

anyset of partitions. In the following sections, I will explain the two conditions that B needs to fulfill in order to count as a factorization:## 1. There is a unique element for all combinations of properties

Let’s look at our partitions in terms of properties again:

What we need in order for B to be a factorization, is that for all combinations(xi,yj) of properties (for example (x1,y2), which is (blue, circle)), there is a unique element with these properties.

We can see that this is the case here: We have exactly one blue square, exactly one orange square, exactly one blue circle, and exactly one orange circle.

To express it more mathematically: For B={X1,X2,…,Xn} to be a factorization, we need that for all x1∈X1,x2∈X2,…xn∈Xn, it holds that the intersection of {x1,x2,…,xn} contains exactly one element. This means the

cartesian productof our factors isbijectiveto the set S, which justifies that we say we can “express S as theproductof our factors”.## 2. No factor is trivial

Here is an example of a non-factorization:

B = {U, V} is not a factorization here, because U is trivial and factors aren’t allowed to be trivial.

This is analogous to integer factorization, where we don’t count 1 as a factor. For example, for the integer 6 we say the factorizations are {6} and {2,3}, and don’t mention {6,1} and {6,1,1} and {6,1,1,1} and so on.

## Exercise

What about this? Is B = {X, Y} a factorization here? (take a moment to think for yourself)

more spoiler space

… No. B is not a factorization.

Why not? Let’s look at it in terms of properties again:

We can see that there is not a unique element for every combination of properties. For example, there is no orange star, and also no orange circle (i.e. x1∩y2 and x2∩y2 are empty).

What about this one? Is B = {V} a factorization?

… yes it is. This one is called

trivial factorization. Each set can be factorized as B={b1} with b1 being the maximally separating partition.If you play around a bit with sets of different sizes, you will see that the possible set factorizations correspond to the integer factorizations of the set’s size

^{[1]}:In particular, sets with a prime number of elements only have the trivial factorization.

This concludes the examples for factorizations. Hopefully you now have some grasp on how factoring a set works.

If we have a tuple F = (S, B) of a set S and a factorization B of S, then we call F a

factored set. In this post I will assume that all sets are finite, and use “factored set” synonymously with “finite factored set”## 2. What does this have to do with Causality? - The Building Blocks

In this section, I will introduce three building blocks: three structures on factored sets, that will help us make the connection to causality later.

In section 3, I will then use these building blocks to model the causal structure behind a probability distribution with factored sets, similar to causal graphs.

The three building blocks are:

historyof a partition X is related to arandom variable’s set ofancestorsin a causal graphOrthogonalityof two partitions X, Y means they have no shared history. In the Pearl-paradigm we know that two variables X and Y have no common ancestors if and only if they are independent. Analogously, Scott Garrabrant proved that in the factored set paradigm, two variables areorthogonal if and only if they are independent.^{[2]}Conditional orthogonalityof two partitions: In the Pearl-paradigm we know that two variables X and Y ared-separatedif and only if they areconditionally independent(proofhereandhere). Analogously, Scott proved that in the factored set paradigm, two variables are conditionally orthogonal if and only if they are conditionally independent.^{[2]}"Time": Saying a partition A isbeforeB is related to acausal pathgoing from A to B in a causal graph.## History

Consider an 8-element set S, which is factorized into the factors “color”, “shape” and “fill”, like this:

Here, the factored set F is F = (S, B) with the factorization B = {color, shape, fill}.

Now, let’s consider a partition/property A - which does not need to be a factor! (i.e. A does not have to be color, shape or fill here):

Now assume we know some properties of an element s, and want to figure out if it is in a1 or in a2. The fill doesn’t matter for this, so the minimum required properties for finding out are {color, shape}.

If we know that the color is blue, then the color would be enough to determine that we are in a2, but in order to

reliablyfind out if we are in a2, we need both color and shape.This is what we will call the

historyof A: We say that in a factored set F, the history hF(A) of a property A is the set of properties we need in order to figure out A.Once we build up factored sets as a model for a causal structure, the history of a partition A will correspond to the set of ancestors of a random variable in a causal graph.

Note that I represent A as these red rings, and {color, shape, fill} as properties. I could just as well represent A as a property too (e.g. different sizes), but I prefer this representation because it distinguishes the factors in our factorization {color, shape, fill} from the variable A whose history we want to find.Exercise

The history hF(A) of a property A is the set of properties we need in order to figure out A. So, what is the hF(A) here?

hF(A) = {color}

You only need to know the color in order to tell if an element is in a1 or in a2.

Notice that in this case, A = color. So hF(A)= {color} is the same as hF(color) = {color}. Which is basically just saying that you just need to know the color in order to find out the color. In general, for every factor b in our factorization, it holds that hF(b)={b}.

Another exercise: What if A is the trivial partition? What is hF(A) here?

Here, the history is empty! (hF(A)=∅) We don’t need to know any properties, because we

alreadyknowthat every s is in a1.## Orthogonality

Now we have defined history, we can define

orthogonality, which is closely related to independence of random variables!We say that two partitions A, B are

orthogonal, if theirhistories don’t overlap.Exercise

Are A and B orthogonal here?

No. The history of A is hF(A) = {color, shape}. The history of B is hF(B)= {color, fill}, so the histories overlap.

In the Pearl-paradigm we know that two variables X and Y have no common ancestors if and only if they are independent. Analogously, Scott Garrabrant proved that when we use a factored set to model causal structure (I’ll explain how to do that in section 3), then two variables are

orthogonal if and only if they are independent.^{[2]}Note that in any factorization, the factors are all orthogonal to each other, because hF(b) ={b} for any factor b (we only need color to infer color, remember) so hF(b1)∩hF(b2)=∅ if b1≠b2.

Scott also defines

conditional orthogonalityas an analog ofd-separation. I won’t define conditional orthogonality here in order to keep things simple, but you can find the definitionhere.In the Pearl-paradigm there is a theorem called

soundness and completeness of d-separation: Two variables X and Y are d-separated with regard to a set of variables Z if and only if X and Y areconditionally independentgiven Z (proofhereandhere)Scott’s central resultis the analog of this theorem in the factored set paradigm:Two variables X, Y are conditionally orthogonal with regard to a set of variables Z if and only if X and Y are conditionally independent given Z. (Technically it’s slightly more complicated, but this is the gist

^{[2]})## "Time"

weakly beforeB if A's history of A is asubset or equalto B's history (i.e. hF(A)⊆hF(B) ).strictly beforeB if A's history is astrict subsetof B's history (i.e. hF(A)⊊hF(B)).This notion of “time” is closely related to the concept of a causal arrow going from A to B in a causal graph.

You can imagine A's history like "everything that comes before A in time", so if everything that’s in A’s history is also in B’s history then A is before B:

Exercise

Is A or B weakly or strictly before the other here? (i.e. is one of the histories of A or B a subset of the other? Reminder: history = set of properties needed to infer our partition)

A is strictly before B! hF(A) = {color, shape} and hF(B) = {color, shape, fill}, so hF(A)⊊hF(B).

Now we have the building blocks to use factored sets for causal inference!

## 3. Causal Inference using Factored Sets

In this section I will walk through an example of inferring causality from data using factored sets.

Consider an experiment in which we collect 2 bits. The distribution P looks like this:

P(00) = 1%

P(01) = 9%

P(10) = 81%

P(11) = 9%

Let’s say X is the first bit, and Y is the second bit.

In the causal graph paradigm, we would observe that X and Y are dependent (P(X=0)=10%≠P(X=0|Y=0)=182). Thus we are not able to distinguish between these three causal graphs:

We don’t know whether X causes Y, Y causes X, or there is some common factor W that causes both.

## How do we look at this in the Factored Set Paradigm?

In the factored set paradigm, start with the sample space Ω of our distribution P:

We then say a

modelof the distribution is a factored set F = (S, B) and a function f:S→Ω:XS and YS are the “preimages” of X and Y under f. By that I mean that all their parts xSi and ySj are the

preimagesof xi and yj respectively, under f. That looks as follows:(Note that in this case f is

bijective, but in general f does not have to be bijective. If we allow f to be non-bijective, then the framework works in more generality because we can describe some processes that we couldn’t otherwise describe.^{[3]})However, we can’t just use

anyfactorization B. In order for (F, f) to count as amodelof our distribution P, it needs to be such that thedependencies and independencies of our distribution are represented in the factorization.That means:

independentin P if and only if XS and YS areorthogonalin F (i.e. their histories don’t overlap)dependentin P if and only if XS and YS arenot orthogonalin FRemember, our probability distribution P was

P(00) = 1%

P(01) = 9%

P(10) = 81%

P(11) = 9%

Is the following actually a model of P?

No, it is not: X, and Y are

dependentin P, but XS and YS areorthogonalin F! (Note that orthogonality depends on what model we are in/what factorization we use.)Here is a really tricky one: Is this a model of P?

Also no, but this is hard to see, so let’s walk through it.

Consider the variable Z = X XOR Y

We don’t

have toexpress our distribution in terms of X and Y, we can just as well define it in terms of X and Z. And then it looks like this:P(00) = P(X=0, Z=0) = 1%

P(01) = P(X=0, Z=1) = 9%

P(10) = P(X=1, Z=1) = 81%

P(11) = P(X=1, Z=0) = 9%

We can see that

Z and X are independent! (becauseP(Z=0)=1%+9%=10%, and alsoP(Z=0|X=0)=1%9%+1%=10% , andP(Z=0|X=1)=9%81%+9%=10%)What does this mean for our model? It means that ZS and XS need to be orthogonal.

Are ZS and XS orthogonal in our model? Here it is again:

No, because the history for both XS and ZS is {V}, so XS and ZS have an overlapping history.

So F = (S, {VS}) is also not a model of P.

Does our distribution P have a model at all?

Yes - here is a model that actually works for P:

Here hF(XS)={XS} and hF(ZS)= {ZS}, and hF(YS) = {XS,ZS}.

This means XS and ZS are

orthogonal, which matches our observation that X and Z areindependentin P. Also XS and YS arenot orthogonal, which matches our observation that X and Y aredependentin P.It also means that hF(XS)⊊hF(YS), so XS is strictly before YS.

If XS is strictly before YS, then the causal arrow goes from X to Y, so we have found a causal direction! It’s this one:

From what we know so far, this causal direction X→Y only holds for this particular model (F, f), but in fact,

you can also provethat X→Y holds foranymodel of this distribution P.So, we just inferred causality from

observational(as opposed tointerventional) data, in a way that Pearl’s causal models wouldn’t have inferred!## Sanity-checking the result in Pearl’s paradigm

I have encountered a lot of skepticism that we can infer the causality X→Y here. So I’m going to switch back to the Pearl paradigm, and explain why X causes Y if our distribution is P, and we can actually infer that from only observational data without needing interventional data.

This section will assume you know how to determine (in-)dependence in probability distributions and in causal graphs. (If you don’t, you can either just believe me, or learn about ithereandhere).Again, say X is the first bit, Y is the second bit, and Z = X XOR Y. Here is our distribution again, in table-form:

XYZP(X, Y, Z)The (in-)dependencies we can read from this are:

dependentindependentdependentThe possible causal graphs which fulfill these dependencies and independencies are these four:

So, we know that Y does not cause X! This matches our finite factored set finding that X causes Y, but it's weaker. Can we also infer that X causes Y?

Let’s concretize the above graphs by adding the conditional probabilities. Graph 1 then looks like this:

Graph 2 is somewhat trickier, because W is not uniquely determined. But one possibility is like this:

Note that W is just the negation of Z here (W=¬Z). Thus, W and Z are information equivalent, and that means graph 2 is actually just graph 1.

Can we find a different variable W such that graph 2 does

notreduce to graph 1? I.e. can we find a variable W such that Z is not deterministic given W?No, we can’t. To see that, consider the distribution P(Z|X,Y,W). By definition of Z, we know that

P(Z|X,Y,W)={1if Z=X XOR Y,0otherwise..

In other words, P(Z|X,Y,W) is deterministic.

We also know that W d-separates Z from X,Y in graph 2:

This d-separation implies that Z is independent from X, Z given W:

P(Z|W)=P(Z|W,X,Y)

As P(Z|X,Y,W) is deterministic, P(Z|W) also has to be deterministic.

So, graph 2 always reduces to graph 1, no matter how we choose W. Analogously, graph 3 and graph 4 also reduce to graph 1, and we know that our causal structure is graph 1:

Which means we also know that X causes Y.

The reason why we usually wouldn’t have found this causal direction using causal graphs is that we

wouldn’t even have consideredZ as potentially interesting. This is what factored sets give us: They make us consider every possible way of defining variables, so wedon’t miss out on any informationthat may be hidden if we just look at a predetermined set of variables.## Summary

Set factorizations are a way of expressing sets as a

product of some factors, similar to how integer factorization is about expressing integers as a product of some factors.We can define a“before”other properties. We say that two variables are

historyon them, that tells us which properties cameorthogonalif they have no shared history. Using these notions of history and orthogonality, we can define a mathematical structure calledmodelof a probability distribution. With this model, we can do causal inference (inferring causal structure from data).Factored sets let us infer causal relations that we usually wouldn’t have found using causal graphs. For example, if we have two binary variables X and Y, and X is independent from X XOR Y, then we can infer the causal direction X→Y.

## Further Reading

I hope you got a bit of a grasp on finite factored sets, and see why they are really neat. If you want to read more, the best entry point is probably

this edited transcriptfrom a talk by Scott Garrabrant.For a non-mathematical intuition how Scott relates the concepts of time, causality, abstraction, and agency, see his

Saving Timepost.I haven’t looked closely into AI alignment specific applications of factored sets, but it looks like they can be used to better talk about

embedded agency,decision theory, andELK._____

This post is a result of adistillationworkshop led by John Wentworth atSERI MATS. I’d like to thank Leon Lang, Scott Garrabrant, Matt MacDermott, Jesse Hoogland, and Marius Hobbhahn for feedback and discussions on this post.^{^}Note that the number of set factorizations of an n-element set is not the same as the number of integer factorizations of n, because elements are distinguishable, so for example these two factorizations do not count as the same factorization:

^{^}Actually, it’s somewhat more complicated than “X and Y are (conditionally) orthogonal if and only if they are independent”. The full version is more like “If XS and YS are partitions on a set S, which has a mapping f:S→Ω to the sample space, then XS and YS are (conditionally) orthogonal if and only if the images X and Y of XS and YS are (conditionally) independent”. But for the sake of this explanation, if you just remember that “orthogonality ⇔ independence”, that's enough.

^{^}Even if f is not bijective, the “preimages” XS and YS of X and Y are always well-defined partitions. Here are two examples in which f is not bijective: